NCERT Exemplar Class 12 Maths Solutions Chapter 5 Continuity and Differentiability

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NCERT Exemplar Class 12 Maths Solutions Chapter 5 Continuity and Differentiability

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The chapter has only 1 exercise with 107 questions in total that are all problem based with variable weightage.

Edited By Ravindra Pindel | Updated on Sep 15, 2022 04:55 PM IST | #CBSE Class 12th

NCERT solutions for Class 12 Maths chapter 5 continuity and differentiability-Solving the book of NCERT for Class 12 students is a must. NCERT Class 12 Mathematics book covers everything that will come in the board exams and helps in creating a base of the topics. Therefore, if one wants to earn the topic for exams and also for higher education, solving the NCERT questions is a must. We are here to provide you with all the NCERT Exemplar Class 12 Math solutions chapter 5. This is one of the most crucial chapters that need to be covered by the student to score well. That is why we are trying to make sure that the students have the answers with them for practice and reference.

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Question:1

Examine the continuity of the function

$f(x) = x^3 + 2x - 1 $at x = 1$

Answer:

A function f(x) is said to be continuous at x = c if,
Left hand limit (LHL at x = c) = Right hand limit (RHL at x = c) = f(c).
Mathematically we can present it as-
$\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)$
Where h is a very small number very close to 0 (h→0)
Now according to above theory-
$f(x) = x^3 + 2x - 1 $at x = 1$ is continuous if -
$\lim _{h \rightarrow 0} \mathrm{f}(1-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(1+\mathrm{h})=\mathrm{f}(1)$
But we can see that,
$\begin{array}{l} {\mathrm{LHL}}=\lim _{h \rightarrow 0} \mathrm{f}(1-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}\left\{(1-\mathrm{h})^{3}+2(1-\mathrm{h})-1\right\} \\ \therefore \mathrm{LHL}=(1-0)^{3}+2(1-0)-1=2 \ldots(1) \end{array}$
Similarly, we proceed for RHL-
${\mathrm{RHL}}=\lim _{h \rightarrow 0}\left\{(1+h)=\lim _{h \rightarrow 0}\left\{(1+h)^{3}+2(1+h)-1\right\}\right.$
$\therefore \mathrm{RHL}=(1+0)^{3}+2(1+0)-1=2 \ldots(2)$$
$\\And \\$f(1)=(1+0)^{3}+2(1+0)-1=2 \ldots(3)$ \\ $Now from equation 1,2 and 3 we can conclude that \\$$ \lim _{h \rightarrow 0} \mathrm{f}(1-\mathrm{h})=\lim _{h \rightarrow 0} \mathrm{f}(1+\mathrm{h})=\mathrm{f}(1)=2 $$ \\$\therefore f(x)$ is continuous at $x=1$$

Question:2

Find which of the functions is continuous or discontinuous at the indicated points:
$f(x)=\left\{\begin{array}{cl} 3 x+5, & \text { if } x \geq 2 \\ x^{2}, & \text { if } x<2 \end{array}\right. \text { at } x=2$

Answer:

Given,
$f(x)=\left\{\begin{array}{cl} 3 x+5, & \text { if } x \geq 2 \\ x^{2}, & \text { if } x<2 \end{array}\right.$
We need to check its continuity at x = 2
A function f(x) is said to be continuous at x = c if,
Left hand limit (LHL at x = c) = Right hand limit (RHL at x = c) = f(c).
Mathematically we can represent it as-
$\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)$
Where h is a very small number very close to 0 (h→0)
Now according to above theory-
f(x) is continuous at x = 2 if -
$\begin{aligned} &\lim _{h \rightarrow 0} \mathrm{f}(2-\mathrm{h})=\lim _{h \rightarrow 0} \mathrm{f}(2+\mathrm{h})=\mathrm{f}(2)\\ & \text { Now we can see that, }\\ &LH L=\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0}(2-h)^{2} \text {Using equation } \\ & \therefore L H L=(2-0)^{2}=4 \ldots (2) \end{aligned}$
Similarly, we proceed for RHL-
$\begin{aligned} &\lim _{\mathrm{RHL}}=\underset{\mathrm{h} \rightarrow 0}{\mathrm{f}(2+\mathrm{h})}=\lim _{\mathrm{h} \rightarrow 0}\{3(2+\mathrm{h})+5\}\\ &\therefore \mathrm{RHL}=3(2+0)+5=11 \ldots(3)\\ &\text { then, }\\ &f(2)=3(2)+5=11 \ldots(4) \end{aligned}$
Now from equation 2, 3 and 4 we can conclude that
$\begin{aligned} &\lim _{h \rightarrow 0} f(2-h) \neq \lim _{h \rightarrow 0} f(2+h)\\ &\therefore f(x) \text { is discontinuous at } x=2 \end{aligned}$

Question:3

Find which of the functions is continuous or discontinuous at the indicated points:

$f(x)=\left\{\begin{array}{cl} \frac{1-\cos 2 x}{x^{2}}, & \text { if } x \neq 0 \\ 5, & \text { if } x=0 \end{array}\right. \text { at } x=0$

Answer:

Given,
$f(x)=\left\{\begin{array}{cl} \frac{1-\cos 2 x}{x^{2}}, & \text { if } x \neq 0 \\ 5, & \text { if } x=0 \end{array}\right.$
We need to check its continuity at x = 0
A function f(x) is said to be continuous at x = c if,
Left hand limit (LHL at x = c) = Right hand limit (RHL at x = c) = f(c).
Mathematically we can present it as
$\lim _{h \rightarrow 0} \mathrm{f}(\mathrm{c}-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(\mathrm{c}+\mathrm{h})=\mathrm{f}(\mathrm{c}) \\Where $h$ is a very small number very close to $0(\mathrm{~h} \rightarrow 0)$ \\ Now according to above theory- \\$f(x)$ is continuous at $x=0$ if $-$ \\$$ \lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(0+h)=f(0) $$
$\begin{aligned} &\text { then, }\\ &\lim _{\mathrm{h} \mathrm{HL}=\mathrm{h} \rightarrow 0} \mathrm{f}(-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \frac{1-\cos 2(-\mathrm{h})}{(-\mathrm{h})^{2}} \underbrace{\{\text { using equation } 1\}}\\ &\text { As we know } \cos (-\theta)=\cos \theta\\ &\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{1-\cos 2 \mathrm{~h}}{\mathrm{~h}^{2}}\\ &\because 1-\cos 2 x=2 \sin ^{2} x\\ &\therefore \mathrm{lim}_{\mathrm{h} \rightarrow 0} \frac{2 \sin ^{2} \mathrm{~h}}{\mathrm{~h}^{2}}=2 \lim _{\mathrm{h} \rightarrow 0}\left(\frac{\sin \mathrm{h}}{\mathrm{h}}\right)^{2} \end{aligned}$
As this limit can be evaluated directly by putting value of h because it is taking indeterminate form (0/0)
As we know,
$\begin{aligned} &\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\\ &\therefore \mathrm{LHL}=2 \times 1^{2}=2 \ldots(2)\\ &\text { Similarly, we proceed for RHL- }\\ &\lim _{\mathrm{RHL}} \mathrm{h} \rightarrow 0^+{\mathrm{f}}(\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \frac{1-\cos 2(\mathrm{~h})}{(\mathrm{h})^{2}}\\ &\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{1-\cos 2 \mathrm{~h}}{\mathrm{~h}^{2}}\\ &\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{2 \sin ^{2} \mathrm{~h}}{\mathrm{~h}^{2}}=2 \lim _{\mathrm{h} \rightarrow 0}\left(\frac{\sin \mathrm{h}}{\mathrm{h}}\right)^{2} \end{aligned}$
Again, using sandwich theorem, we get -
$RHL = 2 \times 1^2 = 2...(3)$
And,
f (0) = 5 …(4)
From equation 2, 3 and 4 we can conclude that
$\lim _{h \rightarrow 0} \mathrm{f}(0-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(0+\mathrm{h}) \neq \mathrm{f}(0)$
∴ f(x) is discontinuous at x = 0

Question:4

Find which of the functions is continuous or discontinuous at the indicated points:
$f(x)=\left\{\begin{aligned} \frac{2 x^{2}-3 x-2}{x-2}, \text { if } & x \neq 2 \\ 5,\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: & \text { if } x=2 \end{aligned}\right.$
at x = 2

Answer:

Given,
$f(x)=\left\{\begin{aligned} \frac{2 x^{2}-3 x-2}{x-2}, \text { if } & x \neq 2 \\ 5,\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: & \text { if } x=2 \end{aligned}\right.$
We need to check its continuity at x = 2
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
$\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)$
Where h is a very small number very close to 0 (h→0)
Now according to above theory-
f(x) is continuous at x = 2 if -
$\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} f(2+h)=f(2)$
Then,
$\begin{aligned} &\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} \frac{2(2-h)^{2}-3(2-h)-2}{(2-h)-2} \quad\{\text { using equation } 1\}\\ &\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{2\left(4+\mathrm{h}^{2}-4 \mathrm{~h}\right)-6+3 \mathrm{~h}-2}{-\mathrm{h}}\\ &\Rightarrow \mathrm{lim} \frac{8+2 \mathrm{~h}^{2}-8 \mathrm{~h}-6+3 \mathrm{~h}-2}{-\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \frac{2 \mathrm{~h}^{2}-5 \mathrm{~h}}{-\mathrm{h}}\\ &\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{-\mathrm{h}(5-2 \mathrm{~h})}{-\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0}(5-2 \mathrm{~h})\\ &\therefore \mathrm{LHL}=5-2(0)=5 \ldots(2) \end{aligned}$
Similarly, we proceed for RHL-
$\lim _{\mathrm{RHL}=\mathrm{h} \rightarrow 0} \mathrm{f}(2+\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \frac{2(2+\mathrm{h})^{2}-3(2+\mathrm{h})-2}{(2+\mathrm{h})-2} \text { [using equation } \left.1\right\}$
$\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{2\left(4+\mathrm{h}^{2}+4 \mathrm{~h}\right)-6-3 \mathrm{~h}-2}{\mathrm{~h}}$ \\$\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{8+2+8 \mathrm{~h}-6-3 \mathrm{~h}-2}{-\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \frac{2 \mathrm{~h}^{2}+5 \mathrm{~h}}{\mathrm{~h}}$ \\$\Rightarrow \lim _{h \rightarrow 0} \frac{h(5+2 h)}{h}=\lim _{h \rightarrow 0}(5+2 h)$ \\$\therefore \mathrm{RHL}=5+2(0)=5 \ldots(3)$ \\And, $f(2)=5\{$ using egn 1$\} \ldots(4)$$
From the above equation 2 , 3 and 4 we can say that
$\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} f(2+h)=f(2) =5$
∴ f(x) is continuous at x = 2

Question:5

Find which of the functions is continuous or discontinuous at the indicated points:
$f(x)=\left\{\begin{array}{cl} \frac{|x-4|}{2(x-4)}, \text { if } & x \neq 4 \\ 0, & \text { if } x=4 \end{array}\right.$
at x = 4

Answer:

Given,
$f(x)=\left\{\begin{array}{cl} \frac{|x-4|}{2(x-4)}, \text { if } & x \neq 4 \\ 0, & \text { if } x=4 \end{array}\right.$ ...(1)
We need to check its continuity at x = 4
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
$\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)$
Where h is a very small number very close to 0 (h→0)
Now according to above theory-
f(x) is continuous at x = 4 if -
$\lim _{h \rightarrow 0} f(4-h)=\lim _{h \rightarrow 0} f(4+h)=f(4)$
Clearly,
$\begin{aligned} &\lim _{\mathrm{h} \mathrm{HL}} \mathrm{f}(4-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \frac{|4-\mathrm{h}-4|}{2(4-\mathrm{h}-4)}\{\text { using equation } 1\}\\ &\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{|-\mathrm{h}|}{-2 \mathrm{~h}}\\ &\because \mathrm{h}>0 \text { as defined above. }\\ &\therefore|-h|=h\\ &\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{h}}{-2 \mathrm{~h}}=-\frac{1}{2} \lim _{\mathrm{h} \rightarrow 0} 1\\ &\therefore \mathrm{LHL}=-1 / 2 \ldots(2) \end{aligned}$
Similarly, we proceed for RHL-
$\begin{aligned} &{\mathrm{RHL}}=\lim _{\mathrm{h}\rightarrow 0}{\mathrm{f}}(4+\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \frac{|4+\mathrm{h}-4|}{2(4+\mathrm{h}-4)}\{\text { using equation } 1\}\\ &\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{|\mathrm{h}|}{2(\mathrm{~h})}\\ &\because \mathrm{h}>0 \text { as defined above. }\\ &\therefore|h|=h \end{aligned}$
$\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{h}}{2 \mathrm{~h}}=\frac{1}{2} \lim _{\mathrm{h} \rightarrow 0} 1$
$\therefore \mathrm{RHL}=1 / 2 \ldots(3)$ \\And, $f(4)=0\{$ using eqn 1$\} \ldots(4)$ \\From equation 2,3 and 4 we can conclude that \\$$ \lim _{h \rightarrow 0} f(4-h) \neq \lim _{h \rightarrow 0} f(4+h) \neq f(4) $$$
∴ f(x) is discontinuous at x = 4

Question:6

$f(x)=\left\{\begin{array}{cl} |x| \cos \frac{1}{x}, \text { if } & x \neq 0 \\ 0, & \text { if } x=0 \end{array}\right.$
at x = 0

Answer:

Given,

$f(x)=\left\{\begin{array}{cl} |x| \cos \frac{1}{x}, \text { if } & x \neq 0 \\ 0, & \text { if } x=0 \end{array}\right.$
We need to check its continuity at x = 0
A function f(x) is said to be continuous at x = c if,
Left hand limit (LHL at x = c) = Right hand limit (RHL at x = c) = f(c).
Mathematically we can represent it as-

$\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)$
Where h is a very small number very close to 0 (h→0)
Now according to above theory-
f(x) is continuous at x = 4 if -
$\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(0+h)=f(0)$
$then, \\\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}\left\{|-\mathrm{h}| \cos \frac{1}{(-\mathrm{h})}\right\}_{\{\mathrm{using} \text { equation } 1\}}$ \\$\because \mathrm{h}>0$ as defined above.\\ \\$\therefore|-h|=h$ \\$\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0}\left\{\mathrm{~h} \cos \left(\frac{1}{\mathrm{~h}}\right)\right\}$$
As cos (1/h) is going to be some finite value from -1 to 1 as h→0
∴ LHL = 0 × (finite value) = 0 …(2)
Similarly, we proceed for RHL-
$\lim _ {\mathrm{h} \rightarrow 0}{\mathrm{f}}(\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}\left\{|\mathrm{~h}| \cos \frac{1}{(\mathrm{~h})}\right\}_{\{\text {using equation } 1\}}$
$\because \mathrm{h}>0$ as defined above. \\$\therefore|h|=h$ \\$\Rightarrow \lim _{\mathrm{RHL}}=\lim _{h \rightarrow 0}\left\{\mathrm{~h} \cos \left(\frac{1}{\mathrm{~h}}\right)\right\}$ \\As $\cos (1 / \mathrm{h})$ is going to be some finite value from -1 to 1 as $\mathrm{h} \rightarrow 0$ \\$\therefore \mathrm{RHL}=0 \times($ finite value $)=0 \ldots(3)$$
And,
f(0) = 0 {using eqn 1} …(4)
From the above equation 2 , 3 and 4 we can conclude that
$\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(0+h)=f(4)=0$
∴ f(x) is continuous at x = 0

Question:7

Find which of the functions is continuous or discontinuous at the indicated points:
Check continuity at x =a $f(x)=\left\{\begin{array}{cl} |x-a| \sin \frac{1}{x-a}, \text { if } & x \neq a \\ 0, & \text { if } x=a \end{array}\right.$

Answer:

Given,
$f(x)=\left\{\begin{array}{cl} |x-a| \sin \frac{1}{x-a}, \text { if } & x \neq a \\ 0, & \text { if } x=a \end{array}\right.$
We need to check its continuity at x = a
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
$\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)$
Where h is a very small number very close to 0 (h→0)
Now according to above theory-
f(x) is continuous at x = a if -
$\lim _{h \rightarrow 0} f(a-h)=\lim _{h \rightarrow 0} f(a+h)=f(a)$
Clearly,
$\begin{aligned} &\left.\lim _{\mathrm{h} \mathrm{HL}} \mathrm{f}(\mathrm{a}-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}\left\{|\mathrm{a}-\mathrm{h}-\mathrm{a}| \sin \frac{1}{(\mathrm{a}-\mathrm{h}-\mathrm{a})}\right\}_{\{\text {using }} \operatorname{egn} 1\right\}\\ &\Rightarrow \mathrm{LHL}=\lim _{h \rightarrow 0}\left\{|-\mathrm{h}| \sin \frac{1}{(-\mathrm{h})}\right\}\\ &\because \mathrm{h}>0 \text { as defined above. }\\ &\therefore|-h|=h\\ &\Rightarrow \mathrm{LHL}=\lim _{h \rightarrow 0}\left\{\mathrm{~h} \sin \left(-\frac{1}{\mathrm{~h}}\right)\right\} \end{aligned}$
As sin (-1/h) is going to be some finite value from -1 to 1 as h→0
∴ LHL = 0 × (finite value) = 0 …(2)
Similarly we proceed for RHL-
$\lim _{\mathrm{RHL}}=\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(\mathrm{a}+\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}\left\{|\mathrm{a}+\mathrm{h}-\mathrm{a}| \sin \frac{1}{(\mathrm{a}+\mathrm{h}-\mathrm{a})}\right\}_{\{\text {using eqn } 1\}}$ .
h > 0 as defined above.
∴ |h| = h
$\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0}\left\{\mathrm{~h} \sin \left(\frac{1}{\mathrm{~h}}\right)\right\}$
As sin (1/h) is going to be some finite value from -1 to 1 as h→0
∴ RHL = 0 × (finite value) = 0 …(3)
And,
f(a) = 0 {using eqn 1} …(4)
From equation 2, 3 and 4 we can conclude that
$\begin{aligned} &\lim _{h \rightarrow 0} f(a-h)=\lim _{h \rightarrow 0} f(a+h)=f(a)=0\\ &\therefore f(x) \text { is continuous at } x=a \end{aligned}$

Question:8

Find which of the functions is continuous or discontinuous at the indicated points:
$f(x)=\left\{\begin{array}{cl} \frac{e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}, \text { if } & x \neq 0 \\ 0, & \text { if } x=0 \end{array}\right.$
at x = 0
Answer:

Given,
$f(x)=\left\{\begin{array}{cl} \frac{e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}, \text { if } & x \neq 0 \\ 0, & \text { if } x=0 \end{array}\right.$
We need to check its continuity at x = 0
A function f(x) is said to be continuous at x = c if,
Left hand limit (LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
$\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)$
Where h is a very small number very close to 0 (h→0)
Now according to above theory-
f(x) is continuous at x = 4 if -

$\begin{aligned} &\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(0+h)=f(0)\\ &\text { Clearly, }\\ &\lim _{h \rightarrow 0} f(-h)=\lim _{h \rightarrow 0}\left\{\frac{e^{\frac{1}{-h}}}{1+e^{-\frac{1}{-h}}}\right\}_{\{u \operatorname{sing} \text { equation } 1\}}\\ &\Rightarrow \mathrm{LHL}=\frac{\mathrm{e}^{\frac{1}{-0}}}{1+\mathrm{e}^{-\frac{1}{-0}}}=\frac{\mathrm{e}^{-\infty}}{1+\mathrm{e}^{-\infty}}=\frac{0}{1+0}=0\\ &\therefore \mathrm{LHL}=0 . . .(2) \end{aligned}$
Similarly, we proceed for RHL-
$\begin{array}{l} \lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}\left\{\frac{\mathrm{e}^{\frac{1}{\mathrm{~h}}}}{1+\mathrm{e}^{\frac{1}{\mathrm{~h}}}}\right\}_{\{\text {using equation } 1\}} \\ \quad \quad \lim _{\mathrm{h} \rightarrow 0}\left\{\frac{\mathrm{e}^{\frac{1}{\mathrm{~h}}}}{\mathrm{e}^{\frac{1}{\mathrm{~h}}\left(1+\mathrm{e}^{-\frac{1}{\mathrm{~h}}}\right)}}\right\}=\lim _{\mathrm{h} \rightarrow 0}\left\{\frac{1}{\left(1+\mathrm{e}^{-\frac{1}{\mathrm{~h}}}\right)}\right\} \end{array}$
$\begin{array}{l} \Rightarrow \mathrm{RHL}=\frac{1}{1+\mathrm{e}-0}=\frac{1}{1+\mathrm{e}^{-\infty}}=\frac{1}{1+0}=1 \\ \therefore \mathrm{RHL}=1 \ldots(3) \end{array}$
And,
f(0) = 0 {using eqn 1} …(4)
from equation 2 , 3 and 4 we can conclude that
$\lim _{h \rightarrow 0} \mathrm{f}(0-\mathrm{h}) \neq \lim _{h \rightarrow 0} \mathrm{f}(0+\mathrm{h})$
∴ f (x) is discontinuous at x = 0

Question:9

Find which of the functions is continuous or discontinuous at the indicated points:
$f(x)=\left\{\begin{array}{cc} \frac{x^{2}}{2}, \text { if } & 0 \leq x \leq 1 \\ 2 x^{2}-3 x+\frac{3}{2}, & \text { if } 1<x \leq 2 \end{array}\right.$

Answer:

Given,
$f(x)=\left\{\begin{array}{cc} \frac{x^{2}}{2}, \text { if } & 0 \leq x \leq 1 \\ 2 x^{2}-3 x+\frac{3}{2}, & \text { if } 1<x \leq 2 \end{array}\right.$
We need to check its continuity at x = 1
A function f(x) is said to be continuous at x = c if,
Left hand limit (LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
$\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)$
Where h is a very small number very close to 0 (h→0)
Now according to above theory-
f(x) is continuous at x = 1 if -
$\begin{aligned} &\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0} f(1+h)=f(1)\\ &\text { then, }\\ &\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(1-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \frac{(1-\mathrm{h})^{2}}{2} \quad\{\text { using equation } 1\}\\ &\therefore \mathrm{LHL}=(1-0)^{2} / 2=1 / 2 \ldots(2) \end{aligned}$
Similarly, we proceed for RHL-
$\begin{array}{l} \lim _{\mathrm{RHL}=} \lim _{h \rightarrow 0} \mathrm{f}(1+\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}\left(2(1+\mathrm{h})^{2}-3(1+\mathrm{h})+\frac{3}{2}\right)_{\{\text {using eqn } 1\}} \\ \Rightarrow \mathrm{RHL}=\lim _{h \rightarrow 0}\left(2\left(\mathrm{~h}^{2}+2 \mathrm{~h}+1\right)-3-3 \mathrm{~h}+\frac{3}{2}\right) \\ \Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0}\left(2 \mathrm{~h}^{2}+4 \mathrm{~h}+2-3-3 \mathrm{~h}+\frac{3}{2}\right) \\ \Rightarrow \mathrm{RHL}=\lim _{h \rightarrow 0}\left(2 \mathrm{~h}^{2}+\mathrm{h}+\frac{1}{2}\right) \\ \therefore \mathrm{RHL}=2(0)^{2}+0+1 / 2=1 / 2 \ldots(3) \end{array}$
And,
$\\f(1)=1^{2} / 2=1 / 2\{$ using egn 1$\} \ldots(4)\\$ From the equation 2,3 and 4 \\we can conclude that \\$\lim _{h \rightarrow 0} \mathrm{f}(1-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(1+\mathrm{h})=\mathrm{f}(1)=\frac{1}{2}$ \\$\therefore f(x)$ is continuous at $x=1$$

Question:10

Find which of the functions is continuous or discontinuous at the indicated points:
$f(x) = |x| + |x - 1|$ at x = 1

Answer:

Given,
$f(x) = |x| + |x - 1|$
We need to check its continuity at x = 1
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
$\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)$
Where h is a very small number very close to 0 (h→0)
Now according to above theory-
f(x) is continuous at x = 1 if -
$\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0} f(1+h)=f(1)$
Then,
$\begin{aligned} &\left.\lim _{\mathrm{h} \mathrm{HL}} \mathrm{f}(1-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}\{|1-\mathrm{h}|+|1-\mathrm{h}-1|\}_{\{\text {using }} \operatorname{egn} 1\right\}\\ &\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0}\{|1-\mathrm{h}|+|-\mathrm{h}|\}\\ &\because \mathrm{h}>0 \text { as defined above and } \mathrm{h} \rightarrow 0\\ &\therefore|-h|=h\\ &\text { And }(1-h)>0 \end{aligned}$
$\begin{array}{l} \therefore|1-h|=1-h \\ \Rightarrow \quad L H L=h \rightarrow 0\{(1-h)+h\}=\lim _{h \rightarrow 0} 1 \\ \therefore L H L=1 \ldots(2) \end{array}$
Similarly, we proceed for RHL-
$\begin{aligned} &\lim _{\mathrm{RHL}} \mathrm{h} \rightarrow 0^{\mathrm{f}}(1+\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}\{|1+\mathrm{h}|+|1+\mathrm{h}-1|\}_{\{\text {using eqn } 1\}}\\ &\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0}\{|1+\mathrm{h}|+|\mathrm{h}|\}\\ &\because \mathrm{h}>0 \text { as defined above and } \mathrm{h} \rightarrow 0\\ &\therefore|h|=h \end{aligned}$
$\begin{aligned} &\text { And }(1+h)>0\\ &\therefore|1+h|=1+h\\ &\Rightarrow \lim _{h \rightarrow 0}\{(1+h)+h\}=\lim _{h \rightarrow 0}(1+2 h)\\ &\therefore \mathrm{RHL}=1+2(0)=1 \ldots(3)\\ &\text { And, }\\ &f(1)=|1|+|1-1|=1\{\text { using egn } 1\} \ldots(4) \end{aligned}$
From equation 2,3 and 4 we can conclude that
$\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0} f(1+h)=f(1)=1$$
$\therefore$ F(x) is continuous at x=1

Question:11

Find the value of k so that the function f is continuous at the indicated point:
$f(x)=\begin{array}{cl} 3 x-8, & \text { if } x \leq 5 \\ 2 k, & \text { if } x>5 \end{array} \text { at } x=5$

Answer:

Given,
$f(x)=\begin{array}{cl} 3 x-8, & \text { if } x \leq 5 \\ 2 k, & \text { if } x>5 \end{array} \text { at } x=5$
We need to find the value of k such that f(x) is continuous at x = 5.
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
$\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)$
Where h is a very small number very close to 0 (h→0)
Now, let’s assume that f(x) is continuous at x = 5.
$\lim _{h \rightarrow 0} f(5-h)=\lim _{h \rightarrow 0} f(5+h)=f(5)$
As we have to find k so pick out a combination so that we get k in our equation.
In this question we take LHL = f(5)
$\\ \therefore \lim _{h \rightarrow 0} f(5-h)=f(5) \\ \Rightarrow \lim _{h \rightarrow 0}\{3(5-h)-8\}=2 k \\ \Rightarrow 3(5-0)-8=2 k \\ \Rightarrow 15-8=2 k \\ \Rightarrow 2 k=7 \\ \therefore k=7 / 2$

Question:12

Find the value of k so that the function f is continuous at the indicated point:
$f(x)=\frac{2^{x+2}-16}{4^{x}-16}, \quad$ if $x \neq 2$ if $x=2$ . at $x=2$$

Answer:

Given,
$f(x)=\frac{2^{x+2}-16}{4^{x}-16}, \quad$ if $x \neq 2$ if $x=2$ . at $x=2$$
We need to find the value of k such that f(x) is continuous at x = 2.
A function f(x) is said to be continuous at x = c if,
Left hand limit (LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
$\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)$
Where h is a very small number very close to 0 (h→0)
Now, let’s assume that f(x) is continuous at x = 2.
$\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} f(2+h)=f(2)$
Now to find k we have to pick out a combination so that we get k in our equation.
In this question we take LHL = f(5)
$\\\therefore \lim _{h \rightarrow 0} f(2-h)=f(2) \\ \Rightarrow \lim _{h \rightarrow 0}\left\{\frac{2^{(2-h)+2}-16}{4^{2-h}-16}\right\}=\mathrm{k}_{\{\text {using equation } 1\}} \\ \Rightarrow \lim _{h \rightarrow 0}\left\{\frac{2^{4-h}-16}{4^{2-h}-16}\right\}=\mathrm{k} \\ \Rightarrow \lim _{h \rightarrow 0}\left\{\frac{2^{4}\left(2^{-h}-1\right)}{4^{2}\left(4^{-h}-1\right)}\right\}=\mathrm{k}$
$\Rightarrow \lim _{h \rightarrow 0}\left\{\frac{\left(2^{-h}-1\right)}{\left(4^{-h}-1\right)}\right\}=k$
As the limit can't be evaluated directly as it is taking 0 / 0 form.
So, use the formula:
$\lim _{x \rightarrow 0} \frac{a^{x}-1}{x}=\log a$$
Divide the numerator and denominator by -h to match with the form in formula-
$\lim _{h \rightarrow 0}\left\{\frac{\frac{\left(2^{-h}-1\right)}{-h}}{\frac{\left(4^{-h}-1\right)}{-h}}\right\}=k$$
$\begin{aligned} &\text { Using algebra of limits, we get, }\\ &\therefore \mathrm{k}=\frac{\log 2}{\log 4}=\frac{\log 2}{2 \log 2}=\frac{1}{2} \end{aligned}$

Question:13

Find the value of k so that the function f is continuous at the indicated point:
$\begin{array}{c} f(x )= \frac{\sqrt{1+\mathrm{kx}}-\sqrt{1-\mathrm{kx}}}{\mathrm{x}}, \quad \text { if }-1 \leq \mathrm{x}<0 \\\\ \frac{2 \mathrm{x}+1}{\mathrm{x}-1}, \quad \text { if } 0 \leq \mathrm{x} \leq 1 \end{array}$ at x= 0

Answer:

Given,

$\begin{array}{c} f(x )= \frac{\sqrt{1+\mathrm{kx}}-\sqrt{1-\mathrm{kx}}}{\mathrm{x}}, \quad \text { if }-1 \leq \mathrm{x}<0 \\\\ \frac{2 \mathrm{x}+1}{\mathrm{x}-1}, \quad \text { if } 0 \leq \mathrm{x} \leq 1 \end{array}$
We need to find the value of k such that f(x) is continuous at x = 0.
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
$\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)$
Where h is a very small number very close to 0 (h→0)
Now, let’s assume that f(x) is continuous at x = 0.
$\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(0+h)=f(0)$
Now to find k pick out a combination using which we get k in our equation.
In this question we take LHL = f(0)
$\begin{array}{l} \quad \lim _{h \rightarrow 0} f(-h)=f(0) \\ \Rightarrow \lim _{h \rightarrow 0}\left\{\frac{\sqrt{1+k(-h)}-\sqrt{1-k(-h)})}{-h}\right\}=\frac{2(0)+1}{(0)-1}\{\text { using eqn } 1\} \\ \Rightarrow \lim _{h \rightarrow 0}\left\{\frac{\sqrt{1+k h}-\sqrt{1-k h}}{h}\right\}=-1 \end{array}$
We can’t find the limit directly, because it is taking 0/0 form.
thus, we will rationalize it.
$\begin{aligned} &\Rightarrow \lim _{h \rightarrow 0}\left\{\frac{\sqrt{1+k h}-\sqrt{1-k h}}{h} \times \frac{\sqrt{1+k h}+\sqrt{1-k h}}{\sqrt{1+k h}+\sqrt{1-k h}}\right\}=-1\\ &\text { Using }(a+b)(a-b)=a^{2}-b^{2}, \text { we have }-\\ &\lim _{h \rightarrow 0}\left\{\frac{(\sqrt{1+k h})^{2}-(\sqrt{1-k h)})^{2}}{h(\sqrt{1+k h}+\sqrt{1-k h})}\right\}=-1\\ &\Rightarrow \lim _{h \rightarrow 0}\left\{\frac{2 k h}{h(\sqrt{1+k h}+\sqrt{1-k h})}\right\}=-1\\ &\lim _{h \rightarrow 0}\left\{\frac{2 k}{(\sqrt{1+k h}+\sqrt{1-k h})}\right\}=\\ &\Rightarrow \frac{2 k}{\sqrt{1+k(0)}+\sqrt{1-k(0)}}=-1\\ &\therefore 2 k / 2=-1 \\&\therefore k =-1 \end{aligned}$

Question:14

Find the value of k so that the function f is continuous at the indicated point:
$\begin{array}{c} f(x)=\frac{1-\cos \mathrm{kx}}{\mathrm{x} \sin \mathrm{x}}, \quad \text { if } \mathrm{x} \neq 0 \\ \frac{1}{2}, \quad \text { if } \mathrm{x}=0 \end{array}$ at x= 0

Answer:

Given,
$\begin{array}{c} f(x)=\frac{1-\cos \mathrm{kx}}{\mathrm{x} \sin \mathrm{x}}, \quad \text { if } \mathrm{x} \neq 0 \\ \frac{1}{2}, \quad \text { if } \mathrm{x}=0 \end{array}$
We need to find the value of k such that f(x) is continuous at x = 0.
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically, we can represent it as-
$\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)$
Where h is a very small number very close to 0 (h→0)
Now, let’s assume that f(x) is continuous at x = 0.
$\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(0+h)=f(0)$
to find k we have to pick out a combination so that we get k in our equation.
In this question we take LHL = f(0)
$\begin{array}{l} \lim _{h \rightarrow 0} f(-h)=f(0) \\ \quad \lim _{h \rightarrow 0}\left\{\frac{1-\cos k(-h)}{(-h) \sin (-h)}\right\}=\frac{1}{2}\{u \text { ing equation } 1\} \end{array}$
$\begin{array}{l} \because \cos (-x)=\cos x \text { and } \sin (-x)=-\sin x \\ \quad \lim _{h \rightarrow 0}\left\{\frac{1-\cos k(h)}{(h) \sin (h)}\right\}=\frac{1}{2} \\ \text { Also, } 1-\cos x=2 \sin ^{2}(x / 2) \\ \quad \lim _{h \rightarrow 0}\left\{\frac{2 \sin ^{2}\left(\frac{k h}{2}\right)}{(h) \sin (h)}\right\}=\frac{1}{2} \end{array}$
As this limit can be evaluated directly by putting value of h because it is taking indeterminate form (0/0)
Thus, we use sandwich or squeeze theorem according to which -
$\lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \\ \Rightarrow {\lim_{h\rightarrow 0} }\left\{\frac{\sin ^{2}\left(\frac{k h}{2}\right)}{(h) \sin (h)}\right\}=\frac{1}{4}$
Dividing and multiplying by $(kh/2)^2$ to match the form in formula we have-
$\begin{aligned} &\lim _{h \rightarrow 0}\left\{\frac{\sin ^{2}\left(\frac{\mathrm{kh}}{2}\right)}{(\mathrm{h}) \sin (\mathrm{h}) \times\left(\frac{\mathrm{kh}}{2}\right)^{2}} \times\left(\frac{\mathrm{kh}}{2}\right)^{2}\right\}=\frac{1}{4}\\ &\text { Using algebra of limits we get - }\\ &\lim _{h \rightarrow 0}\left(\frac{\sin \frac{k h}{2}}{\frac{k h}{2}}\right)^{2} \times \lim _{h \rightarrow 0} \frac{k^{2}}{4}\left(\frac{h}{\sin h}\right)=\frac{1}{4} \end{aligned}$
$\begin{aligned} &\text { Applying the formula- }\\ &\Rightarrow 1 \times\left(\mathrm{k}^{2} / 4\right)=(1 / 4)\\ &\Rightarrow \mathrm{k}^{2}=1\\ &\Rightarrow(k+1)(k-1)=0\\ &\therefore \mathrm{k}=1 \text { or } \mathrm{k}=-1 \end{aligned}$

Question:15

Prove that the function f defined by

$f(x)=\left\{\begin{array}{cl} \frac{x}{|x|+2 x^{2}}, & x \neq 0 \\ k, & x=0 \end{array}\right.$
remains discontinuous at x=0, regardless the choice of k.

Answer:

Given,

$f(x)=\left\{\begin{array}{cl} \frac{x}{|x|+2 x^{2}}, & x \neq 0 \\ k, & x=0 \end{array}\right.$
We need to prove that f(x) is discontinuous at x = 0 irrespective of the value of k.
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically, we can represent it as-
$\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)....(1)$
Where h is a very small number very close to 0 (h→0)
Now, we need to prove that f(x) is discontinuous at x = 0 irrespective of the value of k
If we show that,
$\lim _{h \rightarrow 0} f(0-h) \neq \lim _{h \rightarrow 0} f(0+h)$
Then there will not be involvement of k in the equation & we can easily prove it.
So let’s take LHL first -
$\begin{aligned} &{L H L}=\lim _{h \rightarrow 0} f(0-h)\\ &\Rightarrow \mathrm{LHL}=\lim _{h \rightarrow 0} \frac{(0-\mathrm{h})}{|0-\mathrm{h}|+2(0-\mathrm{h})^{2}}\\ &\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{-\mathrm{h}}{|-\mathrm{h}|+2 \mathrm{~h}^{2}}\\ &\because \mathrm{h}>0 \text { as defined in theory above. }\\ &\therefore|-h|=h \end{aligned}$
$\\\therefore\lim _{ \mathrm{h} \rightarrow 0} \frac{-\mathrm{h}}{\mathrm{h}+2 \mathrm{~h}^{2}}=\lim _{\mathrm{h} \rightarrow 0} \frac{-\mathrm{h}}{\mathrm{h}(1+2 \mathrm{~h})}$ \\$\Rightarrow \mathrm{LHL}=\lim _{h \rightarrow 0} \frac{-1}{(1+2 \mathrm{~h})}$ \\$\therefore \mathrm{LHL}=\frac{-1}{1+2(0)}=-1.....(2)$$
Now Let's find RHL,
$\lim _{\mathrm{RHL}}=\operatorname{h}_{\rightarrow 0} \mathrm{f}(0+\mathrm{h})$
$\begin{aligned} &\Rightarrow \mathrm{RHL}=\lim _{h \rightarrow 0} \frac{(0+\mathrm{h})}{|0+\mathrm{h}|+2(0+\mathrm{h})^{2}}\\ &\Rightarrow \mathrm{RHL}=\lim _{h \rightarrow 0} \frac{\mathrm{h}}{|\mathrm{h}|+2 \mathrm{~h}^{2}}\\ &\because \mathrm{h}>0 \text { as defined in theory above. }\\ &\therefore|h|=h\\ &\therefore\mathrm{RHL}=\lim _{ h\rightarrow 0}=\frac{\mathrm{h}}{\mathrm{h}+2 \mathrm{~h}^{2}}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{h}}{\mathrm{h}(1+2 \mathrm{~h})} \end{aligned}$
$\begin{aligned} &\Rightarrow \mathrm{RHL}=\lim _{h \rightarrow 0} \frac{1}{(1+2 \mathrm{~h})}\\ &\therefore \mathrm{RHL}=\frac{1}{1+2(0)}=1 ....(3)\end{aligned}$
From the equation 2 and 3, conclude that
LHL ≠ RHL
Hence,
f(x) is discontinuous at x = 0 irrespective of the value of k.

Question:16

Find the values of a and b such that the function f defined by
$f(x)=\left\{\begin{array}{ll} \frac{x-4}{|x-4|}+a, & \text { if } x<4 \\ a+b & , \text { if } x=4 \\ \frac{x-4}{|x-4|}+b, & \text { if } x>4 \end{array}\right.$
is a continuous function at x = 4.

Answer:

Given,
$f(x)=\left\{\begin{array}{ll} \frac{x-4}{|x-4|}+a, & \text { if } x<4 \\ a+b & , \text { if } x=4 \\ \frac{x-4}{|x-4|}+b, & \text { if } x>4 \end{array}\right.$ …(1)
We need to find the value of a & b such that f(x) is continuous at x = 4.
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
$\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)$
Where h is a very small number very close to 0 (h→0)
Now, let’s assume that f(x) is continuous at x = 4.
$\therefore \lim _{h \rightarrow 0} \mathrm{f}(4-\mathrm{h})=\lim _{h \rightarrow 0} \mathrm{f}(4+\mathrm{h})=\mathrm{f}(4)$
to find a & b, we have to pick out a combination so that we get a or b in our equation.
In this question first we take LHL = f(4)
$\therefore \lim _{h \rightarrow 0} f(4-h)=f(4)$
$\Rightarrow \lim _{h \rightarrow 0}\left\{\frac{4-h-4}{|4-h-4|}+a\right\}=a+b$ {using equation 1}
$\Rightarrow \lim _{h \rightarrow 0}\left\{\frac{-h}{|-h|}+a\right\}=a+b$
$\because$ h > 0 as defined in theory above.
$\therefore|-h|=h$
$\Rightarrow \lim _{h \rightarrow 0}\left\{\frac{-h}{h}+a\right\}=a+b$
$\Rightarrow \lim _{h \rightarrow 0}\{a-1\}=a+b$
$\Rightarrow$ a - 1 = a + b
$\therefore$ b = -1
Now, taking other combination,
RHL = f(4)
$\lim _{h \rightarrow 0} \mathrm{f}(4+\mathrm{h})=\mathrm{f}(4)$
$\Rightarrow \lim _{h \rightarrow 0}\left\{\frac{4+h-4}{|4+h-4|}+b\right\}=a+b$ {using equation 1}
$\underset{h \rightarrow 0}{\lim }\left\{\frac{h}{|h|}+b\right\}=a+b$
$\because$ h > 0 as defined in theory above.
$\therefore$ |h| = h
$\Rightarrow \lim _{h \rightarrow 0}\left\{\frac{h}{h}+b\right\}=a+b$
$\Rightarrow \lim _{h \rightarrow 0}\{b+1\}=a+b$
⇒ b + 1 = a + b
∴ a = 1
Hence,
a = 1 and b = -1

Question:17

Given the function $f(x)=\frac{1}{x+2}$ . Find the point of discontinuity of the composite function y = f(f(x)).

Answer:

Given,
$F(x)=\frac{1}{x+2}$
we have to find: Points discontinuity of composite function f(f(x))

As f(x) is not defined at x = -2 as denominator becomes 0, at x = -2.

$\therefore$ x = -2 is a point of discontinuity
$\because f(f(x))=f\left(\frac{1}{x+2}\right)=\frac{1}{\frac{1}{x+2}+2}=\frac{x+2}{2 x+5}$
And f(f(x)) is not defined at x = -5/2 as denominator becomes 0, at x = -5/2.
∴ x = -5/2 is another point of discontinuity
Thus f (f(x)) has 2 points of discontinuity at x = -2 and x = -5/2

Question:18

Find all points of discontinuity of the function $f(t)=\frac{1}{t^{2}+t-2}, \quad t=\frac{1}{x-1}$ .

Answer:

$f(t)=\frac{1}{t^{2}+t-2}$
We have to find: Points discontinuity of function f(t) where $\mathrm{t}=\frac{1}{\mathrm{x}-1}$
As t is not defined at x = 1 as denominator becomes 0, at x = 1.
$\therefore$ x = 1 is a point of discontinuity
$\because$ f(t) = $f\left(\frac{1}{x-1}\right)=\frac{1}{\left(\frac{1}{x-1}\right)^{2}+\frac{1}{x-1}-2}=\frac{(x-1)^{2}}{1+x-1-2(x-1)^{2}}$
$\Rightarrow f(t)=\frac{(x-1)^{2}}{1+x-1-2\left(x^{2}-2 x+1\right)}=\frac{(x-1)^{2}}{-2 x^{2}+5 x-2}$
The f(t) is not going to be defined whenever denominator is 0 and thus will give a point of discontinuity.

∴ Solution of the following equation gives other points of discontinuities.
$\\ -2x\textsuperscript{2} + 5x - 2 = 0 \\ $ \Rightarrow $ 2x\textsuperscript{2} - 5x + 2 = 0 \\ $ \Rightarrow $ 2x\textsuperscript{2} - 4x - x + 2 = 0 \\ $ \Rightarrow $ 2x(x - 2) - (x - 2) = 0 \\ $ \Rightarrow $ (2x - 1)(x - 2) = 0 \\ $ \therefore $ x = 2 or x = 1/2 \\$
Hence,
f(t) is discontinuous at x = 1, x = 2 and x = 1/2

Question:19

Show that the function f(x) = |sin x + cos x| is continuous at x = $\pi$ .

Answer:

Given,

$f(x)=|\sin x+\cos x| \underline{\ldots}(1)$

We need to prove that f(x) is continuous at x = π

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

$\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)$

Where h is a very small number very close to 0 (h→0)

Now according to above theory-

f(x) is continuous at x = π if -

$\lim _{h \rightarrow 0} f(\pi-h)=\lim _{h \rightarrow 0} f(\pi+h)=f(\pi)$

Now,

LHL = $\lim _{h \rightarrow 0} f(\pi-h)$

⇒ LHL = $\lim _{h \rightarrow 0}\{|\sin (\pi-h)+\cos (\pi-h)|\}$ {using eqn 1}

$\because \sin (\pi-x)=\sin x \& \cos (\pi-x)=-\cos x$

$\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0}|\sinh -\cosh |$

$\Rightarrow \mathrm{LHL}=|\sin 0-\cos 0|=|0-1|$

$\therefore \mathrm{LHL}=1 \underline{\ldots(2)}$

Similarly, we proceed for RHL-

$\operatorname{RHL}=\lim _{h \rightarrow 0} f(\pi+h)$

$\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0}\{|\sin (\pi+\mathrm{h})+\cos (\pi+\mathrm{h})|\}$ {using eqn 1}

$\because \sin (\pi+x)=-\sin x \& \cos (\pi+x)=-\cos x$

$\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0}|-\sin \mathrm{h}-\cosh |$

$\Rightarrow \mathrm{RHL}=|-\sin 0-\cos 0|=|0-1|$

$\therefore \mathrm{RHL}=1 \ldots(3)$

$\text { Also, } f(\pi)=|\sin \pi+\cos \pi|=|0-1|=1 \underline{\ldots(4)}$

Now from equation 2, 3 and 4 we can conclude that

$\lim _{h \rightarrow 0} \mathrm{f}(\pi-\mathrm{h})=\lim _{h \rightarrow 0} \mathrm{f}(\pi+\mathrm{h})=\mathrm{f}(\pi)=1$

∴ f(x) is continuous at x = π is proved

Question:20

Examine the differentiability of f, where f is defined by

$f(x)=\left\{\begin{array}{cc} x[x], & \text { if } 0 \leq x<2 \\ (x-1) x, & \text { if } 2 \leq x<3 \end{array}\right. \text { at } x=2$ .

Answer:

Given,
$f(x)=\left\{\begin{array}{cc} x[x], & \text { if } 0 \leq x<2 \\ (x-1) x, & \text { if } 2 \leq x<3 \end{array}\right. \text { at } x=2$ …(1)
We need to check whether f(x) is continuous and differentiable at x = 2
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
$\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)$
Where h is a very small number very close to 0 (h→0)
And a function is said to be differentiable at x = c if it is continuous there and
Left hand derivative (LHD at x = c) = Right hand derivative (RHD at x = c) = f(c).
Mathematically we can represent it as-
$\lim _{x \rightarrow c^{-}} \frac{f(x)-f(c)}{x-c}=\lim _{x \rightarrow c^{+}} \frac{f(x)-f(c)}{x-c}$
$\lim _{h \rightarrow 0} \frac{f(c-h)-f(c)}{c-h-c}=\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{c+h-c}$
Finally, we can state that for a function to be differentiable at x = c

$\lim _{h \rightarrow 0} \frac{f(c-h)-f(c)}{-h}=\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}$
Checking for the continuity:
Now according to above theory-
f(x) is continuous at x = 2 if -
$\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} f(2+h)=f(2)$
$\therefore \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(2-\mathrm{h})$
$\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0}(2-\mathrm{h})[2-\mathrm{h}]_{\{\text {using equation } 1\}}$
Note: As [.] represents greatest integer function which gives greatest integer less than the number inside [.].
E.g. [1.29] = 1; [-4.65] = -5 ; [9] = 9
[2-h] is just less than 2 say 1.9999 so [1.999] = 1
⇒ LHL = (2-0) ×1
∴ LHL = 2 …(2)
Similarly,
RHL = $\lim _{h \rightarrow 0} f(2+h)$
⇒ RHL = $\lim _{h \rightarrow 0}(2+h-1)(2+h)_{\{\text {using equation } 1\}}$
$\therefore \mathrm{RHL}=(1+0)(2+0)=2 \ldots(3)$
And, f(2) = (2-1)(2) = 2 …(4) {using equation 1}
From equation 2,3 and 4 we observe that:
$\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} f(2+h)=f(2)=2$
∴ f(x) is continuous at x = 2. So we will proceed now to check the differentiability.
Checking for the differentiability:
Now according to above theory-
f(x) is differentiable at x = 2 if -
$\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h}=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}$
$\therefore$ LHD = $\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h}$
⇒ LHD = $\lim _{h \rightarrow 0} \frac{(2-h)[2-h]-(2-1)(2)}{-h} \quad\{\text { using equation } 1\}$
Note: As [.] represents greatest integer function which gives greatest integer less than the number inside [.].
E.g. [1.29] = 1 ; [-4.65] = -5 ; [9] = 9
[2-h] is just less than 2 say 1.9999 so [1.999] = 1

⇒ LHD = $\lim _{h \rightarrow 0} \frac{(2-h) \times 1-2}{-h}$
⇒ LHD = $\lim _{h \rightarrow 0} \frac{-h}{-h}=\lim _{h \rightarrow 0} 1$
$\therefore \mathrm{LHD}=1 \underline{\ldots}(5)$
Now,
RHD = $\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}$
⇒ RHD = $\lim _{h \rightarrow 0} \frac{(2+h-1)(2+h)-(2-1)(2)}{h} \quad\{\text { using equation } 1\}$
⇒ RHD = $\lim _{h \rightarrow 0} \frac{(1+h)(2+h)-2}{h}=\lim _{h \rightarrow 0} \frac{2+h^{2}+3 h-2}{h}$
∴ RHD = $\lim _{h \rightarrow 0} \frac{h(h+3)}{h}=\lim _{h \rightarrow 0}(h+3)$
$\Rightarrow \mathrm{RHD}=0+3=3 \underline{\ldots}(6)$
Clearly from equation 5 and 6, we can conclude that-
(LHD at x=2) ≠ (RHD at x = 2)
∴ f(x) is not differentiable at x = 2

Question:21

Examine the differentiability of f, where f is defined by
$f(x)=\left\{\begin{array}{cl} x^{2} \sin \frac{1}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0 \end{array}\right. \text { at } x=0$

Answer:

Given,
$f(x)=\left\{\begin{array}{cl} x^{2} \sin \frac{1}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0 \end{array}\right. \text { at } x=0$
We need to check whether f(x) is continuous and differentiable at x = 0
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
$\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)$
Where h is a very small number very close to 0 (h→0)
And a function is said to be differentiable at x = c if it is continuous there and
Left hand derivative(LHD at x = c) = Right hand derivative(RHD at x = c) = f(c).
Mathematically we can represent it as-
$\lim _{x \rightarrow c^{-}} \frac{f(x)-f(c)}{x-c}=\lim _{x \rightarrow c^{+}} \frac{f(x)-f(c)}{x-c}$
$\lim _{h \rightarrow 0} \frac{f(c-h)-f(c)}{c-h-c}=\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{c+h-c}$
Finally, we can state that for a function to be differentiable at x = c
$\lim _{h \rightarrow 0} \frac{f(c-h)-f(c)}{-h}=\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}$
Checking for the continuity:
Now according to above theory-
f(x) is continuous at x = 0 if -
$\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(0+h)=f(0) \\ \therefore L H L=h \rightarrow 0 \\ \Rightarrow L H L=\lim _{h \rightarrow 0}\left\{(-h)^{2} \sin \left(\frac{1}{-h}\right)\right\}_{\{u \operatorname{sing}} \text { equation } \left.1\right\}$
As sin (-1/h) is going to be some finite value from -1 to 1 as h→0
∴ LHL = $0^2$ × (finite value) = 0
∴ LHL = 0 …(2)
Similarly,
$\lim _{\mathrm{RHL}}=\lim _{h \rightarrow 0} \mathrm{f}(\mathrm{h}) \\ \Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0}\left\{(\mathrm{~h})^{2} \sin \left(\frac{1}{\mathrm{~h}}\right)\right\}_{\{\text {using equation } 1\}}$
As sin (1/h) is going to be some finite value from -1 to 1 as h→0
∴ RHL = $0^2$ (finite value) = 0 …(3)
And, f(0) = 0 {using equation 1} …(4)
From equation 2,3 and 4 we observe that:
$\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(0+h)=f(0)$
∴ f(x) is continuous at x = 0. So we will proceed now to check the differentiability.
Checking for the differentiability:
Now according to above theory-
f(x) is differentiable at x = 0 if -
$\\ \lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} \\ \therefore L H D=\lim_{h \rightarrow 0 }\frac{f(-h)-f(0)}{-h} \\ \Rightarrow L H D=\lim_{h \rightarrow 0 }\frac{(-h)^{2} \sin \left(\frac{1}{-h}\right)-0}{-h} \quad\{\text { using equation } 1\} \\ \Rightarrow L H D=\lim _{h \rightarrow 0} \frac{h^{2} \sin \left(\frac{1}{-h}\right)}{-h}=\lim _{h \rightarrow 0}\left\{h \sin \left(\frac{1}{h}\right)\right\}$
As sin (1/h) is going to be some finite value from -1 to 1 as h→0
∴ LHD = 0×(some finite value) = 0
∴ LHD = 0 …(5)
Now,
$\operatorname{RHD}=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h} \\ \Rightarrow R H D=h \rightarrow 0 \frac{(h)^{2} \sin \left(\frac{1}{h}\right)-0}{h} \quad\{\text { using equation } 1\} \\$
$\Rightarrow \mathrm{RHD}=\lim _{\mathrm{h} \rightarrow 0}\left\{\mathrm{~h} \sin \left(\frac{1}{\mathrm{~h}}\right)\right\}$
As sin (1/h) is going to be some finite value from -1 to 1 as h→0
∴ RHD = 0×(some finite value) = 0
∴ RHD = 0 …(6)
Clearly from equation 5 and 6, we can conclude that-
(LHD at x=0) = (RHD at x = 0)
∴ f(x) is differentiable at x = 0

Question:22

Examine the differentiability of f, where f is defined by
$f(x)=\left\{\begin{array}{ll} 1+x, & \text { if } x \leq 2 \\ 5-x, & \text { if } x>2 \end{array}\right. \text { at } x=2$

Answer:

Given,
$f(x)=\left\{\begin{array}{ll} 1+x, & \text { if } x \leq 2 \\ 5-x, & \text { if } x>2 \end{array}\right. \text { at } x=2$
We need to check whether f(x) is continuous and differentiable at x = 2
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
$\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)$ .
Where h is a very small number very close to 0 (h→0)
And a function is said to be differentiable at x = c if it is continuous there and
Left hand derivative(LHD at x = c) = Right hand derivative(RHD at x = c) = f(c).
Mathematically we can represent it as-
$\\ \lim _{x \rightarrow c^{-}} \frac{f(x)-f(c)}{x-c}=\lim _{x \rightarrow c^{+}} \frac{f(x)-f(c)}{x-c} \\ \lim _{h \rightarrow 0} \frac{f(c-h)-f(c)}{c-h-c}=\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{c+h-c}$
Finally, we can state that for a function to be differentiable at x = c
$\lim _{h \rightarrow 0} \frac{f(c-h)-f(c)}{-h}=\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}$
Checking for the continuity:
Now according to above theory-
f(x) is continuous at x = 2 if -
$\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} f(2+h)=f(2) \\ \therefore L H L=\lim _{h \rightarrow 0} f(2-h)$
$\begin{aligned} &\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0}\{1+(2-\mathrm{h})\}_{\{\text {using equation } 1\}}\\ &\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0}\{3-\mathrm{h}\}\\ &\therefore L H L=(3-h)=3\\ &\therefore \mathrm{LHL}=3 \ldots(2)\\ &\text { Similarly, }\\ &\lim _{\mathrm{RHL}}=\operatorname{h}_{h \rightarrow 0} \mathrm{f}(2+\mathrm{h})\\ &\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0}\{5-(2+\mathrm{h})\}_{\{\text {using equation } 1\}}\\ &\Rightarrow \mathrm{RHL}=\lim _{h \rightarrow 0}\{3+\mathrm{h}\}\\ &\therefore \mathrm{RHL}=3+0=3 . .0(3) \end{aligned}$
And, f(2) = 1 + 2 = 3 {using equation 1} …(4)
From equation 2,3 and 4 we observe that:
$\lim _{h \rightarrow 0} \mathrm{f}(2-\mathrm{h})=\lim _{h \rightarrow 0} \mathrm{f}(2+\mathrm{h})=\mathrm{f}(2)=3$
∴ f(x) is continuous at x = 2. So we will proceed now to check the differentiability.
Checking for the differentiability:
Now according to above theory-
f(x) is differentiable at x = 2 if -
$\begin{array}{l} \lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h}=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h} \\ \therefore L H D=\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h} \end{array}$
$\begin{aligned} &\Rightarrow \mathrm{LHD}=\lim _{h \rightarrow 0} \frac{1+(2-\mathrm{h})-(1+2)}{-\mathrm{h}} \quad\{\text { using equation } 1\}\\ &\Rightarrow \mathrm{LHD}=\lim _{\mathrm{h} \rightarrow 0} \frac{3-\mathrm{h}-3}{-\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} 1\\ &\therefore \mathrm{LHD}=1 . .0(5)\\ &\text { Now, }\\ &\lim _{\mathrm{RHD}}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(2+\mathrm{h})-\mathrm{f}(2)}{\mathrm{h}}\\ &\Rightarrow \mathrm{RHD}=\lim _{\mathrm{h} \rightarrow 0} \frac{5-(2+\mathrm{h})-3}{\mathrm{~h}} \quad\{\text { using equation } 1\}\\ &\Rightarrow \lim _{\mathrm{RHD}}=\underset{\mathrm{h} \rightarrow 0}{-\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0}-1\\ &\therefore \mathrm{RHD}=-1 \ldots(6) \end{aligned}$
Clearly from equation 5 and 6,we can conclude that-
(LHD at x=2) ≠ (RHD at x = 2)
∴ f(x) is not differentiable at x = 2

Question:23

Show that f(x) = |x - 5| is continuous but not differentiable at x = 5.

Answer:

Given,
f(x) = |x - 5| …(1)
We need to prove that f(x) is continuous but not differentiable at x = 5
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
$\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)$
Where h is a very small number very close to 0 (h→0)
And a function is said to be differentiable at x = c if it is continuous there and
Left hand derivative(LHD at x = c) = Right hand derivative(RHD at x = c) = f(c).
Mathematically we can represent it as-
$\\\lim _{x \rightarrow c^{-}} \frac{f(x)-f(c)}{x-c}=\lim _{x \rightarrow c^{+}} \frac{f(x)-f(c)}{x-c} \\\lim _{h \rightarrow 0} \frac{f(c-h)-f(c)}{c-h-c}=\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{c+h-c}$
Finally, we can state that for a function to be differentiable at x = c
$\lim _{h \rightarrow 0} \frac{f(c-h)-f(c)}{-h}=\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}$
Checking for the continuity:
Now according to above theory-
f(x) is continuous at x = 5 if -
$\lim _{h \rightarrow 0} f(5-h)=\lim _{h \rightarrow 0} f(5+h)=f(5) \\ \therefore L H L=h \rightarrow 0 \\ \Rightarrow L H L=\lim _{h \rightarrow 0}|5-h-5|$
$\begin{aligned} &\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0}|-\mathrm{h}|\\ &\therefore \mathrm{LHL}=|-0|=0\\ &\therefore \mathrm{LHL}=0 \ldots(2)\\ &\text { Similarly, }\\ &\lim _{\mathrm{RHL}}=\lim _{h \rightarrow 0} \mathrm{f}(5+\mathrm{h})\\ &\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0}|5+\mathrm{h}-5|_{\{\text {using equation } 1\}}\\ &\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0}|\mathrm{~h}|\\ &\therefore \mathrm{RHL}=|0|=0 \ldots(3) \end{aligned}$
And, f(5) = |5-5| = 0 {using equation 1} …(4)
From equation 2,3 and 4 we observe that:
$\lim _{h \rightarrow 0} f(5-h)=\lim _{h \rightarrow 0} f(5+h)=f(5) =0$
∴ f(x) is continuous at x = 5. So we will proceed now to check the differentiability.
Checking for the differentiability:
Now according to above theory
f(x) is differentiable at x = 2 if -
$\begin{array}{l} \lim _{h \rightarrow 0} \frac{f(5-h)-f(5)}{-h}=\lim _{h \rightarrow 0} \frac{f(5+h)-f(5)}{h} \\ \therefore L H D=h \rightarrow 0 \quad \frac{|5-h-5|-0}{-h} \\ \Rightarrow L H D=h \rightarrow 0 \frac{|-h|}{-h}\{\text { using equation } 1\} \end{array}$
As h > 0 as defined in theory above.
∴ |-h| = h
$\Rightarrow \mathrm{LHD}=\lim _{h \rightarrow 0} \frac{\mathrm{h}}{-\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0}-1 \\\therefore \mathrm{LHD}=-1 . .(5)$
Now,
$\\\operatorname{RHD}=\lim _{h \rightarrow 0} \frac{\mathrm{f}(5+h)-\mathrm{f}(5)}{\mathrm{h}}$ \\$\Rightarrow \lim _{h \rightarrow 0} \frac{|5+h-5|-0}{h}=\lim _{h \rightarrow 0} \frac{|h|}{h}\{u sin g$ equation 1$\}$$
As h>0 as defined in theory above.
$\\\therefore|h|=h$ \\$\Rightarrow \lim _{\mathrm{RHD}}=\operatorname{h} \rightarrow 0^{\frac{\mathrm{h}}{\mathrm{h}}}=\lim _{\mathrm{h} \rightarrow 0} 1$ \\$\therefore \mathrm{RHD}=1 \ldots(6)$$
Clearly from equation 5 and 6,we can conclude that-
(LHD at x=5) ≠ (RHD at x = 5)
∴ f(x) is not differentiable at x = 5 but continuous at x = 5.
Hence proved.

Question:24

A function $f:R\rightarrow R$ satisfies the equation f(x + y) = f(x) f(y) for all x, y $\in R,f(x)\neq 0$ . Suppose that the function is differentiable at x = 0 and f’(0) = 2. Prove that f’(x) = 2f(x).

Answer:

Given f(x) is differentiable at x = 0 and f(x) ≠ 0
And f(x + y) = f(x)f(y) also f’(0) = 2
To prove: f’(x) = 2f(x)
As we know that,
$\\ f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ \text { as } f(x+h)=f(x) f(h) \\ \therefore f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x) f(h)-f(x)}{h} \\ \Rightarrow f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x)(f(h)-1\}}{h}=f(x) \lim _{h \rightarrow 0} \frac{f(h)-1}{h}$
As
f(x + y) = f(x)f(y)
put x = y = 0
$\\ \therefore f(0+0) = f(0)f(0) \\ \Rightarrow $ f(0) = $ \{ $ f(0)$ \} ^2$
$\\$ \therefore $ f(0) = 1 $ \{ $ $\because$ f(x) $ \neq $ 0 $ \ldots $ .given$ \\$ \therefore $ equation 1 is deduced as$
$\begin{aligned} &f^{\prime}(x)=f(x) \lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}\\ &\Rightarrow f^{\prime}(x)=f(x) \lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}\\ &\Rightarrow f^{\prime}(x)=f(x) f^{\prime}(0)\{\text { using formula of derivative }\}\\ &\therefore f^{\prime}(x)=2 f(x) \ldots \text { proved }\left\{\because \text { it is given that } f^{\prime}(0)=2\right\} \end{aligned}$

Question:25

Differentiate each of the following w.r.t. x
$2 ^{\cos ^2 x}$

Answer:

Given: $2 ^{\cos ^2 x}$
Let Assume $y=2 ^{\cos ^2 x}$
Now, Taking Log on both sides we get,
$\begin{aligned} &\log y=\log 2^{\cos ^{2} x}\\ &\log \mathrm{y}=\cos ^{2} \mathrm{x} \cdot \log 2\\ &\text { Now, Differentiate w.r.t } x\\ &\frac{1}{y} \frac{d y}{d x}=\frac{d}{d x}\left[\cos ^{2} x \cdot \log 2\right]\\ &\frac{1}{y} \frac{d y}{d x}=[2 \cos x \cdot(-\sin x) \cdot \log 2]\\ &\frac{d y}{d x}=y[2 \cos x \cdot(-\sin x) \cdot \log 2] \end{aligned}$
Now, substitute the value of y
$\\ \frac{\mathrm{dy}}{\mathrm{dx}}=-2^{\cos ^{2} x}[2 \cos \mathrm{x} \cdot(\sin \mathrm{x}) \cdot \log 2] \\ \text { Hence, } \frac{\mathrm{dy}}{\mathrm{dx}}=-2^{\cos ^{2} x}[2 \cos \mathrm{x} \cdot(\sin \mathrm{x}) \cdot \log 2]$

Question:26

Differentiate each of the following w.r.t. x

$\frac{8^x}{x^8}$

Answer:

We have been given $\frac{8^x}{x^8}$
Let us Assume $y=\frac{8^x}{x^8}$
Now, taking log on both sides, we get
$\log y=\log \frac{8^{x}}{x^{8}}$$
since we know, $\log \frac{\mathrm{m}}{\mathrm{n}}=\log \mathrm{m}-\log \mathrm{n}$$
$\log y=\log 8^{x}-\log x^{8}$$
since we know, $\log \mathrm{a}^{n}=n \log a$$
$\log y=x \log 8-8 \log {x}$$
Now, Differentiate w.r.t x
$\\\frac{1}{y} \frac{d y}{d x}=\frac{d}{d x}(x \log 8-8 \log x)$ \\$\frac{d y}{d x}=y\left(1 \times \log 8-8 \times \frac{1}{x}\right)$$
$\frac{d y}{d x}=\frac{8^{x}}{x^{8}}\left(\log 8-\frac{8}{x}\right)$$
Hence, $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{8^{\mathrm{x}}}{\mathrm{x}^{8}}\left(\log 8-\frac{8}{\mathrm{x}}\right)$$

Question:27

Differentiate each of the following w.r.t. x
$\log\left ( x+\sqrt{x^2 +a}\right )$

Answer:

Given:
$\log \left(x+\sqrt{x^{2}+a}\right)$$
$\mathrm{y}=\log \mathrm{t}$$
Now, Differentiate w.r.t t
$\\\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\log \mathrm{t})$ \\$\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{1}{\mathrm{t}}$$
And, $t=x+\sqrt{x^{2}+a}$$
Now, differentiate w.r.t x
$\\\frac{d t}{d x}=\frac{d}{d x}\left(x+\sqrt{x^{2}+a}\right)$ \\$\frac{d t}{d x}=\left[1+\frac{d}{d x}\left(\left(x^{2}+a\right)^{\frac{1}{2}}\right)\right.$$
$\\\frac{d t}{d x}=\left[1+\frac{1}{2}\left(x^{2}+a\right)^{-\frac{1}{2}} \cdot \frac{d}{d x}\left(x^{2}\right)\right]$ \\\\$\frac{d t}{d x}=\left[1+\frac{1}{2}\left(x^{2}+a\right)^{-\frac{1}{2}} \cdot(2 x)\right]$$
Now, using chain rule, we get
$\\\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}} \times \frac{\mathrm{dt}}{\mathrm{dx}}$ \\\\$\frac{d y}{d x}=\frac{1}{t} \times\left[1+\frac{1}{2}\left(x^{2}+a\right)^{-\frac{1}{2}} \cdot(2 x)\right]$$
Substitute the value of t
$\frac{d y}{d x}=\frac{1}{x+\sqrt{x^{2}+a}} \times\left[1+\frac{1}{2}\left(x^{2}+a\right)^{-\frac{1}{2}} \cdot(2 x)\right]$ \\$=\frac{1}{x+\sqrt{x^{2}+a}} \times\left[1+\frac{x}{ \sqrt{x^{2}+a}}\right]$ \\$=\frac{\sqrt{x^{2}+a}+x}{\left(x+\sqrt{x^{2}+a}\right) \sqrt{x^{2}+a}}$$
$=\frac{1}{\sqrt{x^{2}+a}} \\$

Question:28

Differentiate each of the following w.r.t. x
$\log [\log (\log x^5)]$

Answer:

Differentiate the given function w.r.t x
Let Assume $y=\log [\log (\log x^5)]$
$y= log [log (log x\textsuperscript{5})]$
Let $log(log x\textsuperscript{5}) = u$
Let Assume $log x\textsuperscript{5}=v$
Let Assume $x\textsuperscript{5}=w$
Differentiate both side w.r.t x
$\\ \frac{\mathrm{dy}}{\mathrm{du}}=\frac{\mathrm{d}}{\mathrm{du}}[\log \mathrm{u}] \\ \frac{\mathrm{dy}}{\mathrm{du}}=\frac{1}{\mathrm{u}} \\ \frac{\mathrm{d} \mathrm{w}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}\left(\mathrm{x}^{5}\right) \\ \frac{\mathrm{d} \mathrm{w}}{\mathrm{dx}}=5 \mathrm{x}^ 4 \\ \frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{w}}=\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}(\log \mathrm{w}) \\ \frac{\mathrm{d} \mathrm{v}}{\mathrm{dw}}=\frac{1}{\mathrm{w}} \frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}(\mathrm{w})$
Now, Bu using chain rule we get, Differentiation of $\log [\log (\log x^5)]$
$\begin{aligned} &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dv}} \times \frac{\mathrm{d} \mathrm{v}}{\mathrm{dw}} \times \frac{\mathrm{d} \mathrm{w}}{\mathrm{dx}}\\ &\frac{d y}{d x}=\frac{1}{u} \times \frac{1}{v} \times \frac{1}{w} \times 5 x^{4}\\ &\text { Now, Substitute the value of } u_{\ell} v \text { and } w \text { then, we get }\\ &\frac{d y}{d x}=\frac{1}{\log \left(\log x^{5}\right)} \times \frac{1}{\log x^{5}} \times \frac{1}{x^{5}} \times 5 x^{4}\\ &=\frac{5}{x \cdot \log \left(\log x^{5}\right) \log x^{5}} \end{aligned}$
Hence, This the differentiation of given function.

Question:29

Differentiate each of the following w.r.t. x
$\sin \sqrt{x}+\cos ^{2} \sqrt{x}$

Answer:

Given:
$\sin \sqrt{x}+\cos ^{2} \sqrt{x}$
We have $\sin \sqrt{x}+\cos ^{2} \sqrt{x}$
$y=\sin \sqrt{x}+\cos ^{2} \sqrt{x}$
Differentiate w.r.t x
$\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}\left(\sin (\mathrm{x})^{\frac{1}{2}}+\left(\cos \mathrm{x} ^\frac{1}{2}\right)^{2}\right) \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\left(\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}\left(\sin (\mathrm{x})^{\frac{1}{2}}\right)+\frac{\mathrm{d}}{\mathrm{dx}}\left(\left(\cos \mathrm{x}^{\frac{1}{2}}\right)^{2}\right)\right) \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\left(\frac{1}{2 \sqrt{\mathrm{x}}} \cdot \cos \sqrt{\mathrm{x}}+2 \cos \sqrt{\mathrm{x}}\left[-\sin \sqrt{\mathrm{x}} \cdot \frac{1}{2 \sqrt{\mathrm{x}}}\right]\right) \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2 \sqrt{\mathrm{x}}}[\cos (\sqrt{\mathrm{x}})-\sin (2 \sqrt{\mathrm{x}})]$

Question:30

Differentiate each of the following w.r.t. x
$sin^n (ax^2 + bx + c)$

Answer:

We have $sin^n (ax^2 + bx + c)$
$\begin{aligned} &y=\sin^n \left(a x^{2}+b x+c\right)\\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\sin^n\left(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right)\right)\\ &\text { since, we know, } x^{n}=n x^{n-1}\\ &\frac{d y}{d x}=n \cdot \sin^{n-1} \left(a x^{2}+b x+c\right) \cdot \frac{d}{d x} \sin \left(a x^{2}+b x+c\right)\\ &\frac{d y}{d x}=n \cdot \sin^{n-1} \left(a x^{2}+b x+c\right) \cdot \cos \left(a x^{2}+b x+c\right) \cdot \frac{d}{d x}\left(a x^{2}+b x+c\right)\\ &\frac{d y}{d x}=n \cdot \sin^{n-1} \left(a x^{2}+b x+c\right) \cdot \cos \left(a x^{2}+b x+c\right) \cdot(2 a x \cdot+b)\\ &\mathrm{y}^{\prime}=\mathrm{n}(2 \mathrm{ax} \cdot+\mathrm{b}) \cdot \sin ^{\mathrm{n}-1}\left(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right) \cdot \cos \left(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right) \end{aligned}$

Question:31

Differentiate each of the following w.r.t. x
$\cos (\tan \sqrt{x+1})$

Answer:

We have given $\cos (\tan \sqrt{x+1})$
Let us Assume $\sqrt{x+1}=w$
And $\tan \sqrt{x+1}=v$
$\mathrm{So}, \mathrm{y}=\cos \mathrm{v}$
Now, differentiate w.r.t v
$\frac{\mathrm{dy}}{\mathrm{d} \mathrm{v}}=(-\sin \mathrm{v})$$
And, $\mathrm{v}=$ tan $\mathrm{w}$
Now, again differentiate w.r.t. w
$\frac{d v}{d w}=\sec ^{2} w$$
And, we know, $\sqrt{x+1}=w$
So, differentiate w w.r.t. x we get
$\frac{d w}{d x}=\frac{1}{2 \sqrt{x+1}}$$
Now, using chain rule we get,
$\\\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dv}} \times \frac{\mathrm{dv}}{\mathrm{dw}} \times \frac{\mathrm{d} \mathrm{w}}{\mathrm{dx}}$ \\$\frac{d y}{d x}=(-\sin v) \times \sec ^{2} w \times \frac{1}{2 \sqrt{x+1}}$$
Substitute the value of v and w
$\frac{\mathrm{dy}}{\mathrm{dx}}=\left(-\sin (\tan \sqrt{\mathrm{x}+1}) \times \sec ^{2} \sqrt{\mathrm{x}+1} \times \frac{1}{2 \sqrt{\mathrm{x}+1}}\right).$$
Hence, dy/dy is the differentiation of function.

Question:32

Differentiate each of the following w.r.t. x $sinx^2 + sin^2x + sin^2 (x^2)$

Answer:

Let us Assume $y=sinx^2 + sin^2x + sin^2 (x^2)$
Now, differentiate y w.r.t x
$\\\frac{d y}{d x}=\frac{d}{d x}\left(\sin x^{2}+\sin ^{2} x+\sin ^{2}(x)^{2}\right)$ \\We know, $\frac{d}{d x}(\sin x)=\cos x$ and $x^{n}=n x^{n-1}$ \\$\frac{d y}{d x}=\frac{d}{d x}\left(\sin x^{2}\right)+\frac{d}{d x}\left(\sin ^{2} x\right)+\frac{d}{d x}\left(\sin ^{2}(x)^{2}\right)$ \\$\frac{d y}{d x}=\cos x^{2} \cdot \frac{d}{d x}\left(x^{2}\right)+2 \sin x \cdot \frac{d}{d x}(\sin x)+2 \cdot \sin x^{2} \frac{d}{d x}\left(\sin x^{2}\right)$ \\$\frac{d y}{d x}=\cos x^{2} \cdot 2 x+2 \sin x \cdot \cos x+2 \cdot \sin x^{2} \cdot \cos x^{2} \frac{d}{d x}\left(x^{2}\right)$ \\$\frac{\mathrm{dy}}{\mathrm{dx}}=\cos \mathrm{x}^{2} \cdot 2 \mathrm{x}+2 \sin \mathrm{x} \cdot \cos \mathrm{x}+4 \mathrm{x} \cdot \sin \mathrm{x}^{2} \cdot \cos \mathrm{x}^{2}$ \\Hence, $\frac{d y}{d x}=2 x \cdot \cos x^{2}+\sin 2 x+4 x \cdot \sin ^x{2} \cos x^{2}$$

Question:33

Differentiate each of the following w.r.t. x
$\sin ^{-1}\left(\frac{1}{\sqrt{x+1}}\right)$

Answer:

Given $\sin ^{-1}\left(\frac{1}{\sqrt{x+1}}\right)$
Let us assume $t=\left(\frac{1}{\sqrt{x+1}}\right)$
$\\So, y=\sin ^{-1} t$ \\Now, differentiate y w.r.t t \\$\frac{d y}{d t}=\frac{1}{\sqrt{1-t^{2}}}$ \\And, differentiate t w.r.t x \\$\frac{d t}{d x}=\frac{-1}{2 ({x+1})^\frac{3}{2}}$ \\Now, using chain we get $\mathrm{dy} / \mathrm{dx}$ \\$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}} \times \frac{\mathrm{dt}}{\mathrm{dx}}$ \\$\frac{d y}{d x}=\frac{1}{\sqrt{1-t^{2}}} \times-\frac{1}{2 (x+1)^\frac{3}{2}}$$
Substitute the value of t
$\\ \frac{d y}{d x}=\frac{1}{\sqrt{1-\left(\frac{1}{\sqrt{x+1}}\right)^{2}}} \times-\frac{1}{2 \sqrt{x+1}} \times \frac{1}{x+1} \\ \frac{d y}{d x}=\frac{1}{\sqrt{1-\frac{1}{x+1}}} \times-\frac{1}{2 \sqrt{x+1}} \times \frac{1}{x+1}$
$\\ \frac{d y}{d x}=\frac{1}{\sqrt{\frac{x+1-1}{x+1}}} \times-\frac{1}{2 \sqrt{x+1}} \times \frac{1}{x+1} \\ \frac{d y}{d x}=\frac{\sqrt{x+1}}{\sqrt{x}} \times-\frac{1}{2 \sqrt{x+1}} \times \frac{1}{x+1} \\ \frac{d y}{d x}=\frac{-1}{2 \sqrt{x}}\left(\frac{1}{x+1}\right)$
Hence, this is the differentiation of $\sin ^{-1}\left(\frac{1}{\sqrt{x+1}}\right)$ .

Question:34

Differentiate each of the following w.r.t. x
$(sin x)^{cos x}$

Answer:

Given: $(sin x)^{cos x}$
To Find: Differentiate w.r.t x
We have $(sin x)^{cos x}$
Let $y=(sin x)^{cos x}$
Now, Taking Log on both sides, we get
Log y = cos x.log(sin x)
Now, Differentiate both side w.r.t. x

$\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=\cos x \cdot \frac{d}{d x}(\log (\sin x))+\log (\sin x) \cdot \frac{d}{d x}(\cos x)\\ &\text { By using product rule of differentiation }\\ &\frac{1}{y} \frac{d y}{d x}=\cos x \cdot \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x)+\log (\sin x) \cdot(-\sin x)\\ &\frac{1}{y} \frac{d y}{d x}=\cos x \cdot \frac{1}{\sin x}(\cos x)+\log (\sin x) \cdot(-\sin x)\\ &\frac{1}{y} \frac{d y}{d x}=\cos x \cdot \cot x-\log (\sin x) \cdot(\sin x)\\ &\frac{d y}{d x}=y[\cos x \cdot \cot x-\log (\sin x) \cdot(\sin x)] \end{aligned}$
Substitute the value of y, we get
$\begin{array}{l} y^{\prime}=(\sin x)^{\cos x}[\cos x \cdot \cot x-\sin x(\log \sin x)] \\ \text { Hence, } y^{\prime}=(\sin x)^{\cos x}[\cos x \cdot \cot x-\sin x(\log \sin x)] \end{array}$

Question:35

Differentiate each of the following w.r.t. x
$sin^mx . cos^nx$

Answer:

It is given $sin^mx . cos^nx$
$y=sin^mx . cos^nx$
Taking log both side, we get
$\\$\log \mathrm{y}=\log \sin ^{\mathrm{m}} \mathrm{x} \cdot \cos ^{\mathrm{n}} \mathrm{x}$ \\$\log \mathrm{y}=\log \sin ^{\mathrm{m}} \mathrm{x}+\log \cos ^{\mathrm{n}} \mathrm{x}$ \\$\log y=m \log \sin x+n \log \cos x$$
Differentiate w.r.t x
$\frac{1}{y} \frac{d y}{d x}=m \cdot \frac{1}{\sin x} \frac{d}{d x}(\sin x)+n \frac{1}{\cos x} \frac{d}{d x}(\cos x)$$
$\begin{aligned} &\frac{1}{y} \frac{d y}{d x}=m \cdot \frac{\cos x}{\sin x}+n \cdot \frac{-\sin x}{\cos x}\\ &\frac{1}{y} \frac{d y}{d x}=m \cdot \cot x-n \cdot \tan x\\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}[\mathrm{m} \cdot \cot \mathrm{x}-\mathrm{n} \cdot \tan \mathrm{x}]\\ &\text { Substitute the value of } y \text { we get, }\\ &\frac{d y}{d x}=\sin ^{m} x \cdot \cos ^{n} x[m \cdot \cot x-n \cdot \tan x] \end{aligned}$

Question:36

Differentiate each of the following w.r.t. x $(x + 1)^2 (x + 2)^3 (x + 3)^4$

Answer:

We have given, $(x + 1)^2 (x + 2)^3 (x + 3)^4$
Let us Assume, $y=(x + 1)^2 (x + 2)^3 (x + 3)^4$
Taking log both side
$\\Log y=log [(x + 1)\textsuperscript{2} (x + 2)\textsuperscript{3} (x + 3)\textsuperscript{4}] \\Log y=log(x + 1)\textsuperscript{2} +log(x + 2)\textsuperscript{3} +log(x + 3)\textsuperscript{4} \\Log y=2log(x + 1) +3log(x + 2) +4log(x + 3)\\$
Differentiate w.r.t x
$\\ \frac{1}{y} \cdot \frac{d y}{d x}=\frac{2}{x+1}+\frac{3}{x+2}+\frac{4}{(x+3)} \\ \frac{d y}{d x}=y\left[\frac{2}{x+1}+\frac{3}{x+2}+\frac{4}{(x+3)}\right] \\ \frac{d y}{d x}=(x+1)^{2} \cdot(x+2)^{3} \cdot(x+3)^{4}\left[\frac{2}{x+1}+\frac{3}{x+2}+\frac{4}{x+3}\right]$
$\\ =(x+1)^{2} \cdot(x+2)^{3} \cdot(x +3)^{4}\left[\frac{2(x+2)(x+3)+3(x+1)(x+3)+4(x+1)(x+2)}{(x+1)(x+2)(x+3)}\right] \\ =(x+1)(x+2)^{2}(x+3)^{3}\left[9 x^{2}+34 x+29\right]$

Question:37

Differentiate each of the following w.r.t. x
$\cos ^{-1}\left(\frac{\sin \mathrm{x}+\cos \mathrm{x}}{\sqrt{2}}\right), \frac{-\pi}{4}<\mathrm{x}<\frac{\pi}{4}$

Answer:

$\begin{aligned} &\cos ^{-1}\left(\frac{\sin x+\cos x}{\sqrt{2}}\right)\\ &=\cos ^{-1}\left(\sin \mathrm{x} \frac{1}{\sqrt{2}}+\cos \mathrm{x} \frac{1}{\sqrt{2}}\right)\\ &\text { As we know } \sin \pi / 4=\cos \pi / 4=1 / \mathrm{v} 2\\ &=\cos ^{-1}\left(\sin x \sin \frac{\pi}{4}+\cos x \cos \frac{\pi}{4}\right)\\ &\text { We know } \cos (a-b)=\sin a \sin b+\cos a \cos b\\ &=\cos ^{-1}\left(\cos \left(\frac{\pi}{4}-\mathrm{x}\right)\right)\\ &=\left(\frac{\pi}{4}-\mathrm{x}\right) \end{aligned}$
$\begin{aligned} &\text { Now, }\\ &\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\pi}{4}-x\right)\\ &=-1 \end{aligned}$

Question:38

Differentiate each of the following w.r.t. x
$\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right),-\frac{\pi}{4}<x<\frac{\pi}{4}$

Answer:

We have $\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right),-\frac{\pi}{4}<x<\frac{\pi}{4}$
$\begin{aligned} &\mathrm{y}=\tan ^{-1} \sqrt{\left(\frac{1-\cos \mathrm{x}}{1+\cos \mathrm{x}}\right)}\\ &\text { We know, }\\ &\cos x=\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}\\ &\mathrm{y}=\tan ^{-1} \sqrt{\left(\frac{\sin ^{2} \frac{\mathrm{x}}{2}+\cos ^{2} \frac{\mathrm{x}}{2}-\cos ^{2} \frac{\mathrm{x}}{2}+\sin ^{2} \frac{\mathrm{x}}{2}}{\sin ^{2} \frac{\mathrm{x}}{2}+\cos ^{2} \frac{\mathrm{x}}{2}+\cos ^{2} \frac{\mathrm{x}}{2}-\sin ^{2} \frac{\mathrm{x}}{2}}\right)}\\ &\mathrm{y}=\tan ^{-1} \sqrt{\left(\frac{2 \sin ^{2} \frac{\mathrm{x}}{2}}{2 \cos ^{2} \frac{\mathrm{x}}{2}}\right)}\\ &\mathrm{y}=\tan ^{-1} \sqrt{\left(\tan ^{2} \frac{\mathrm{x}}{2}\right)}\\ &y=\tan ^{-1}\left(\tan \frac{x}{2}\right) \end{aligned}$
$\begin{aligned} &\text { As the interval is }\\ &-\frac{\pi}{4}<x<\frac{\pi}{4}\\ &=\left\{\begin{array}{r} \tan ^{-1}\left(\tan \frac{x}{2}\right),-\frac{\pi}{4}<x<0 \\ \tan ^{-1}\left(\tan \frac{x}{2}\right), 0 \leq x<\frac{\pi}{4} \end{array}\right.\\ &=\left\{\begin{array}{c} -\frac{\mathrm{x}}{2},-\frac{\mathrm{\pi}}{4}<\mathrm{x}<0 \\ \frac{\mathrm{x}}{2}, 0 \leq \mathrm{x}<\frac{\pi}{4} \end{array}\right. \end{aligned}$
Differentiate w.r.t. x

$\frac{d y}{d x}=\left\{\begin{array}{c} -\frac{1}{2},-\frac{\pi}{4}<x<0 \\ \frac{1}{2}, 0 \leq x<\frac{\pi}{4} \end{array}\right.$

Question:39

Differentiate each of the following w.r.t. x
$\tan ^{-1}(\sec x+\tan x),-\frac{\pi}{4}<x<\frac{\pi}{2}$

Answer:

We have given $\tan ^{-1}(\sec x+\tan x),-\frac{\pi}{4}<x<\frac{\pi}{2}$
Let us Assume (sec x +tan x) =t
So, $y = tan^{-1} t$
Now, differentiate w.r.t t
$\\\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{1}{1+\mathrm{t}^{2}}$ \\since, $\frac{d}{d x} \tan ^{-1} x=\frac{1}{1+x^{2}}$ \\And, $t=(\sec x+\tan x)$ \\Differentiate t w.r.t $x$ \\$\frac{d t}{d x}=\sec x \cdot \tan x+\sec ^{2} x$ \\Therefore, $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}} \times \frac{\mathrm{dt}}{\mathrm{dx}}$$
$\begin{aligned} &\frac{d y}{d x}=\frac{1}{1+t^{2}} \times\left(\sec x \tan x+\sec ^{2} x\right)\\ &\text { When, substitute the value of t we get }\\ &\frac{d y}{d x}=\frac{1}{1+(\sec x+\tan x)^{2}} \times\left(\sec x \tan x+\sec ^{2} x\right)\\ &\frac{d y}{d x}=\frac{1}{1+\sec ^{2} x+\tan ^{2} x+2 \sec x \tan x} \times \sec x \cdot(\sec x+\tan x)\\ &\frac{d y}{d x}=\frac{1}{2 \sec x(\tan x+\sec x)} \times \sec x \cdot(\sec x+\tan x)\\ &\frac{d y}{d x}=\frac{1}{2}\\ &\text { Hence, } \mathrm{dy} / \mathrm{dx}=1 / 2 \end{aligned}$

Question:40

Differentiate each of the following w.r.t. x
$\tan ^{-1}\left(\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right),-\frac{\pi}{2}<x<\frac{\pi}{2} \text { and } \frac{a}{b} \tan x>-1$

Answer:

We have given $\tan ^{-1}\left(\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right)$
$y=\tan ^{-1}\left(\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right)$
Now, divide by cos x in both numerator and denominator
$\\ y=\tan ^{-1}\left(\frac{\frac{\operatorname{acos} x}{\cos x}-\frac{b \sin x}{\cos x}}{\frac{b \cos x}{\cos x}+\frac{\operatorname{asin} x}{\cos x}}\right) \\ y=\tan ^{-1}\left(\frac{a-b \tan x}{b+a \tan x}\right) \\ y=\tan ^{-1} \frac{\left(\frac{a}{b}-\tan x\right)}{\left(1+\frac{a}{b} \tan x\right)}$
$\\$since, we know,$ tan ^{1} x-\tan ^{-1} y={ }^{\tan ^{-1} \frac{x-y}{1+x y}}\\$ So, $\mathrm{y}=\tan ^{-1} \frac{\mathrm{a}}{\mathrm{b}}-\tan ^{-1}(\tan \mathrm{x})$ \\$\mathrm{y}=\tan ^{-1} \frac{\mathrm{a}}{\mathrm{b}}-\mathrm{x}$ $\\\frac{d y}{d x}=0-1=-1$

Question:41

Differentiate each of the following w.r.t. x
$\sec ^{-1}\left(\frac{1}{4 x^{3}-3 x}\right), 0<x<\frac{1}{\sqrt{2}}$

Answer:

$\\We have given, \sec ^{-1}\left(\frac{1}{4 x^{3}-3 x}\right)$ \\Let us assume, $\mathrm{y}=\sec ^{-1}\left(\frac{1}{4 \mathrm{x}^{3}-3 \mathrm{x}}\right)$ \\Put $x=\cos \theta$ then $\theta=\cos ^{1} x$ \\So, $y=\sec ^{-1}\left(\frac{1}{4 \cos ^{3} \theta-3 \cos \theta}\right)$ \\And, $4 \cos ^{3} \theta-3 \cos \theta=\cos 3 \theta$ \\Therefore, \\$\mathrm{y}=\sec ^{-1} \frac{1}{\cos 3 \theta}$ \\$\mathrm{y}=\sec ^{-1}(\sec 3 \theta)$$
$y=3 \theta$
$\\$Substitute the value of $ \theta$ \\$\mathrm{y}=3 \cos ^{-1} \mathrm{x}$ \\Now, Differentiate w.r.t $x$ \\$\frac{d y}{d x}=\frac{-3}{\sqrt{1-x^{2}}}$ \\since, $\frac{d}{d x}\left(\cos ^{-1} x\right)=\frac{-1}{\sqrt{1-x^{2}}}$ \\Hence, $\frac{d y}{d x}=\frac{-3}{\sqrt{1-x^{2}}}$$

Question:42

Differentiate each of the following w.r.t. x
$\tan ^{-1} \frac{3 \mathrm{a}^{2} \mathrm{x}-\mathrm{x}^{3}}{\mathrm{a}^{3}-3 \mathrm{ax}^{2}}, \frac{-1}{\sqrt{3}}<\frac{\mathrm{x}}{\sqrt{3}}<\frac{\mathrm{x}}{\mathrm{a}}<\frac{1}{\sqrt{3}}$

Answer:

$\\y=\tan ^{-1}\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right)\\=\tan ^{-1}\left(\frac{3 \frac{x}{a}-\left(\frac{x}{a}\right)^{3}}{1-3\left(\frac{x}{a}\right)^{2}}\right)$
Let $x=a \tan \theta \Rightarrow \theta=\tan ^{-1} \frac{x}{a}$
$\\y=\tan ^{-1}\left[\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\right]\\=\tan ^{-1}(\tan 3 \theta)\\ =3 \theta\\=3 \tan ^{-1} \frac{x}{a} \\ \frac{d y}{d x}=3 \frac{d}{d x} \tan ^{-1} \frac{x}{a} \\ \quad=3\left[\frac{1}{1+\frac{x^{2}}{a^{2}}}\right] \cdot \frac{d}{d x}\left(\frac{x}{a}\right)\\=\frac{3 a^{2}}{a^{2}+x^{2}} \cdot \frac{1}{a}\\=\frac{3 a}{a^{2}+x^{2}}$

Question:43

Differentiate each of the following w.r.t. x
$\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right), 1<x<1, x \neq 0$

Answer:

$\begin{aligned} &\text { We have given }\\ &\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right)\\ &y=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right)\\ &\text { Put } x^{2}=\cos 2 \theta\\ &\mathrm{SO} \end{aligned}$
$\\ \mathrm{y}=\tan ^{-1}\left(\frac{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta-\sqrt{1-\cos 2 \theta}}}\right) \\ \mathrm{y}=\tan ^{-1}\left(\frac{\sqrt{2 \cos ^{2} \theta}+\sqrt{2 \sin ^{2} \theta}}{\sqrt{2 \cos ^{2} \theta}-\sqrt{2 \sin ^{2} \theta}}\right) \\ \mathrm{y}=\tan ^{-1}\left(\frac{\sqrt{2} \cos \theta+\sqrt{2} \sin \theta}{\sqrt{2} \cos \theta-\sqrt{2} \sin \theta}\right) \\ \mathrm{y}=\tan ^{-1}\left(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\right)$
$\\ \mathrm{y}=\tan ^{-1}\left(\frac{\frac{\cos \theta}{\cos \theta}+\frac{\sin \theta}{\cos \theta}}{\frac{\cos \theta}{\cos \theta}-\frac{\sin \theta}{\cos \theta}}\right) \\ \mathrm{y}=\tan ^{-1}\left(\frac{1+\tan \theta}{1-\tan \theta}\right) \\ \mathrm{y}=\tan ^{-1}\left(\frac{\tan \frac{\pi}{4}+\tan \theta}{1-\tan \frac{\pi}{4} \tan \theta}\right) \\ \mathrm{y}=\tan ^{-1} \tan \left(\frac{\pi}{4}+\theta\right)$
$\begin{aligned} &\mathrm{y}=\frac{\pi}{4}+\theta\\ &\text { Differentiate, y w.r.t x }\\ &\frac{d y}{d x}=\frac{\pi}{4}+\frac{1}{2} \frac{d}{d x} \cos ^{-1} x^{2}\\ &\frac{d y}{d x}=0+\frac{1}{2} \cdot \frac{-1}{\sqrt{1-x^{4}}} \frac{d}{d x}\left(x^{2}\right)\\ &\frac{d y}{d x}=\frac{1}{2} \cdot \frac{-2 x}{\sqrt{1-x^{4}}}\\ &\text { Hence, } \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{x}}{\sqrt{1-\mathrm{x}^{4}}} \end{aligned}$

Question:44

Find $\frac{dy}{dx}$ of each of the functions expressed in parametric form in
$x=t+\frac{1}{t}\: \: ,\: \: y=t-\frac{1}{t}$

Answer:

We have given, two parametric equation,
$x=t+\frac{1}{t}\: \: ,\: \: y=t-\frac{1}{t}$
Now, differentiate both equation w.r.t x
We know, $\frac{d}{d x}\left(\frac{1}{x}\right)=-\frac{1}{x^{2}}$ and $\frac{d}{d x}(x)=1$$
So,
$\frac{d x}{d t}=1-\frac{1}{t^{2}}$$
and,
$\frac{d y}{d x}=1+\frac{1}{t^{2}}$$
Now,
$\begin{aligned} &\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\\ &\frac{d y}{d x}=\frac{1+\frac{1}{t^{2}}}{1-\frac{1}{t^{2}}}\\ &\text { Hence, }\\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{t}^{2}+1}{\mathrm{t}^{2}-1} \end{aligned}$

Question:45

Find $\frac{dy}{dx}$ of each of the functions expressed in parametric form
$\mathrm{x}=\mathrm{e}^{\theta}\left(\theta+\frac{1}{\theta}\right), \mathrm{y}=\mathrm{e}^{-\theta}\left(\theta-\frac{1}{\theta}\right)$

Answer:

We have two equations $\mathrm{x}=\mathrm{e}^{\theta}\left(\theta+\frac{1}{\theta}\right), \mathrm{y}=\mathrm{e}^{-\theta}\left(\theta-\frac{1}{\theta}\right)$
Now, differentiate w.r.t θ
So,
$\\ \mathrm{x}=\mathrm{e}^{\theta}\left(\theta+\frac{1}{\theta}\right) \\ \frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \theta}=\frac{\mathrm{d}}{\mathrm{d} \theta}\left[\mathrm{e}^{\theta}\left(\theta+\frac{1}{\theta}\right)\right] \\ \frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \theta}=\mathrm{e}^{\theta} \frac{\mathrm{d}}{\mathrm{d} \theta}\left(\theta+\frac{1}{\theta}\right)+\left(\theta+\frac{1}{\theta}\right) \frac{\mathrm{d}}{\mathrm{d} \theta}\left(\mathrm{e}^{\theta}\right) \\ \frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \theta}=\mathrm{e}^{\theta}\left(1-\frac{1}{\theta^{2}}\right)+\left(\theta+\frac{1}{\theta}\right)\left(\mathrm{e}^{\theta}\right) \\ \frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \theta}=\mathrm{e}^{\theta}\left(1-\frac{1}{\theta^{2}}+\theta+\frac{1}{\theta}\right) \\ \frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \theta}=\mathrm{e}^{\theta}\left(\frac{\theta^{2}-1+\theta^{3}+\theta}{\theta^{2}}\right)...(i)$
Also,
$\\ \mathrm{y}=\mathrm{e}^{-\theta}\left(\theta-\frac{1}{\theta}\right) \\ \frac{\mathrm{dy}}{\mathrm{d} \theta}=\frac{\mathrm{d}}{\mathrm{d} \theta}\left[\mathrm{e}^{-\theta}\left(\theta-\frac{1}{\theta}\right)\right] \\ \frac{\mathrm{d} y}{\mathrm{~d} \theta}=\mathrm{e}^{-\theta} \frac{\mathrm{d}}{\mathrm{d} \theta}\left(\theta-\frac{1}{\theta}\right)+\left(\theta-\frac{1}{\theta}\right) \frac{\mathrm{d}}{\mathrm{d} \theta}\left(\mathrm{e}^{-\theta}\right) \\ \frac{\mathrm{d} y}{\mathrm{~d} \theta}=\mathrm{e}^{-\theta}\left(1+\frac{1}{\theta^{2}}\right)-\left(\theta-\frac{1}{\theta}\right)\left(\mathrm{e}^{-\theta}\right) \\ \frac{\mathrm{d} y}{\mathrm{~d} \theta}=\mathrm{e}^{-\theta}\left(\frac{\theta^{2}+1-\theta^{3}+\theta}{\theta^{2}}\right) \\ \frac{\mathrm{d} y}{\mathrm{~d} \theta}=\mathrm{e}^{-\theta}\left(\frac{\theta^{2}+1-\theta^{3}+\theta}{\theta^{2}}\right)_{-\mathrm{ }(\mathrm{ii})}$
$\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}} \times \frac{\mathrm{dt}}{\mathrm{dx}} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{e}^{-\theta}\left(\frac{\theta^{2}+1-\theta^{3}+\theta}{\theta^{2}}\right)}{\mathrm{e}^{\theta}\left(\frac{\theta^{2}-1+\theta^{3}+\theta}{\theta^{2}}\right)} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{-2 \theta}\left(\frac{-\theta^{3}+\theta^{2}+\theta+1}{\theta^{3}+\theta^{2}+\theta-1}\right)$

Question:46

Find dy/dx of each of the functions expressed in parametric form in
$x=3 \cos \theta-2 \cos ^{3} \theta, y=3 \sin \theta-2 \sin ^{3} \theta$
Answer:

We have given, $x=3 \cos \theta-2 \cos ^{3} \theta, y=3 \sin \theta-2 \sin ^{3} \theta$
Now, differentiate both the equation w.r.t. x then we get
$x=3 \cos \theta-2 \cos ^{3} \theta$$
$\frac{d x}{d \theta}=\left(-3 \sin \theta-2 \times 3 \cos ^{2} \theta \cdot \frac{d}{d \theta}(\cos \theta)\right)$
$\\\frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \theta}=\left(-3 \sin \theta+6 \cos ^{2} \theta \cdot(\sin \theta)\right)$ \\And, for $y=3 \sin \theta-2 \sin ^{3} \theta$$
$\begin{aligned} &\frac{\mathrm{dy}}{\mathrm{d} \theta}=\left(3 \cos \theta-2 \times 3 \sin ^{2} \theta \cdot \frac{\mathrm{d}}{\mathrm{d} \theta}(\sin \theta)\right)\\ &\frac{\mathrm{dy}}{\mathrm{d} \theta}=\left(3 \cos \theta-6 \sin ^{2} \theta \cdot \cos \theta\right)\\ &\text { Therefore, }\\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{d} \theta} \times \frac{\mathrm{d} \theta}{\mathrm{d} \mathrm{x}}\\ &\frac{\mathrm{dy}}{\mathrm{dx}}=\left(3 \cos \theta-6 \sin ^{2} \theta \cdot \cos \theta\right) \times \frac{1}{\left(-3 \sin \theta+6 \cos ^{2} \theta \cdot(\sin \theta)\right.} \end{aligned}$
$\\ \frac{d y}{d x}=\frac{3 \cos \theta\left(1-2 \sin ^{2} \theta\right)}{3 \sin \theta\left(-1+2 \cos ^{2} \theta\right)} \\ \frac{d y}{d x}=\frac{\cos 2 \theta}{\cos 2 \theta} \times \cot \theta \\ \frac{d y}{d x}=\cot \theta \\ \text { Hence, } d y / d x=\cot \theta$

Question:47

Find dy/dx of each of the functions expressed in parametric form in
$\sin \mathrm{x}=\frac{2 \mathrm{t}}{1+\mathrm{t}^{2}}, \tan \mathrm{y}=\frac{2 \mathrm{t}}{1-\mathrm{t}^{2}}$

Answer:

We have given two parametric equation: $\sin \mathrm{x}=\frac{2 \mathrm{t}}{1+\mathrm{t}^{2}}, \tan \mathrm{y}=\frac{2 \mathrm{t}}{1-\mathrm{t}^{2}}$
Let us Assume t= tan θ
$So, \frac{2 t}{1+t^{2}} \Rightarrow \frac{2 \tan \theta}{1+\tan ^{2} \theta} \\\Rightarrow \sin 2 \theta$ \\Therefore, \\$\sin x=\sin 2 \theta$ \\$x=2 \theta-(i)$ \\Also, \\tan $y=\tan 2 \theta$ \\$\mathrm{y}=2 \theta-\mathrm{c}(\mathrm{ii})$ \\From equation (i) and (ii) \\$y=x$ \\now, differentiate w.r.t x \\$\frac{\mathrm{dy}}{\mathrm{dx}}=1$ \\Hence, $\mathrm{dy} / \mathrm{dx}=1$$

Question:48

Find dy/dx of each of the functions expressed in parametric form
$\mathrm{x}=\frac{1+\log \mathrm{t}}{\mathrm{t}^{2}}, \mathrm{y}=\frac{3+2 \log \mathrm{t}}{\mathrm{t}}$

Answer:

We have given, $\mathrm{x}=\frac{1+\log \mathrm{t}}{\mathrm{t}^{2}}, \mathrm{y}=\frac{3+2 \log \mathrm{t}}{\mathrm{t}}$
Now, differentiate w.r.t t
$\\ \frac{d x}{d t}=\frac{t^{2} \cdot \frac{d}{d t}(1+\log t)-(1+\log t) \cdot \frac{d}{d t}\left(t^{2}\right)}{\left(t^{2}\right)^{2}} \\ \frac{d x}{d t}=\frac{t^{2} \cdot \frac{1}{t}-(1+\log t) \cdot 2 t}{t^{4}} \\ \frac{d x}{d t}=\frac{t-(1+\log t) \cdot 2 t}{t^{4}} \\ \frac{d x}{d t}=\frac{-1-2 \log t}{t^{3}}$
Also,
$\\ \frac{d y}{d t}=\frac{t \cdot \frac{d}{d t}(3+2 \log t)-(3+2 \log t) \cdot \frac{d}{d t}(t)}{t^{2}} \\ \frac{d y}{d t}=\frac{t \cdot \frac{2}{t}-(3+2 \log t)}{t^{2}} \\ \frac{d y}{d t}=\frac{-1-2 \log t}{t^{2}} \\ \frac{d y}{d t}=\frac{-1-2 \log t}{t^{2}} \quad \ldots(i)$
$\\$Now$, \\\frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}$ \\$\frac{d y}{d x}=\frac{-1-2 \log t}{t^{2}} \times \frac{t^{3}}{-1-2 \log t}$ \\Hence, $\frac{\mathrm{dy}}{\mathrm{dx}}=t$$

Question:49

If $\mathrm{x}=e^{\cos 2 t}$ and $\mathrm{y}=\mathrm{e}^{ \sin 2 t}$ , prove that $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{y} \log \mathrm{x}}{\mathrm{x} \log \mathrm{y}} \text { . }$

Answer:

$\\x = e\textsuperscript{cos2t} $and $y = e\textsuperscript{sin2t} \\$Now x = e\textsuperscript{cos2t},$
Taking log on both sides to get,
$\\\log x = cos 2t \\$For y$ = e\textsuperscript{sin2t}$
Taking log on both sides we get,
log y = sin 2t
$\\\therefore $ cos\textsuperscript{2}2t + sin\textsuperscript{2}2t = (log x)\textsuperscript{2} + (log y)\textsuperscript{2} \\ 1 = (log x)\textsuperscript{2} + (log y)\textsuperscript{2}$
Differentiating w.r.t x,
$\\\Rightarrow 0=2 \log{x} \frac{1}{x}+2 \log y \frac{1}{y} \\ \\\frac{d y}{d x}=-\frac{y \log x}{x \log y}$

Question:50

If $x = a\sin 2t (1 + \cos2t) $and y$ = b\cos2t(1 - \cos2t)$ , show that $\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{at} \mathrm{t}=\frac{\pi}{4}}=\frac{\mathrm{b}}{\mathrm{a}}$

Answer:

$x = a\sin 2t (1 + \cos2t) $and y$ = b\cos2t(1 - \cos2t)$
Differentiate w.r.t t
$\\ x=a \sin 2 t(1+\cos 2 t) \\ \frac{d x}{d t}=\frac{d}{d t}[\operatorname{asin} 2 t(1+\cos 2 t)] \\ \frac{d x}{d t}=a \frac{d}{d t}[\sin 2 t(1+\cos 2 t)] \\ \frac{d x}{d t}=a\left[\sin 2 t \frac{d}{d t}(1+\cos 2 t)+(1+\cos 2 t) \frac{d}{d t}(\sin 2 t)\right] \\ \frac{d x}{d t}=a[\sin 2 t \cdot(-2 \sin 2 t)+(1+\cos 2 t) \cdot 2 \cos 2 t] \\ \frac{d x}{d t}=-2 a\left[\sin ^{2} 2 t-\cos 2 t(1+\cos 2 t)\right]$
Also,
y = bcos2t
$\\ \frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{b} \cos 2 \mathrm{t}) \\ \frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{b} \frac{\mathrm{d}}{\mathrm{dt}}(\cos 2 \mathrm{t}) \\ \frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{b}\left[-\sin 2 \mathrm{t} \cdot \frac{\mathrm{d}}{\mathrm{dt}}(2 \mathrm{t})\right] \\ \frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{b}[-2 \sin 2 \mathrm{t}]$
Now, for dy/dx

$\\ \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \\ \frac{d y}{d x}=\frac{b[-2 \sin 2 t]}{-2 a\left[\sin ^{2} 2 t-\cos 2 t(1+\cos 2 t)\right]} \\ \frac{d y}{d x}=\frac{b[\sin 2 t]}{a\left[\sin ^{2} 2 t-\cos 2 t(1+\cos 2 t)\right]} \\ \frac{d y}{d x} \text { at } t=\frac{\pi}{4}$
$\\ \frac{d y}{d x}=\frac{b\left[\sin \frac{2 \pi}{4}\right]}{a\left[\sin ^{2} \frac{2 \pi}{4}-\cos \frac{2 \pi}{4}\left(1+\cos \frac{2 \pi}{4}\right)\right]} \\ \frac{d y}{d x}=\frac{b\left[\sin \frac{\pi}{2}\right]}{a\left[\sin ^{2} \frac{\pi}{2}-\cos \frac{\pi}{2}\left(1+\cos \frac{\pi}{2}\right)\right]} \\ \frac{d y}{d x}=\frac{b[1]}{a[1-0(1+0)]} \\ \frac{d y}{d x}=\frac{b}{a}$
Hence Proved.

Question:51

If , find $\frac{\mathrm{dy}}{\mathrm{dx}} \text { at } \mathrm{t}=\frac{\pi}{3}$

Answer:

$x = 3\sin t - \sin 3t, y = 3\cos t - \cos 3t$
Differentiate w.r.t t in both equation
x = 3sint - sin3t

$\begin{aligned} &\frac{d x}{d t}=\frac{d}{d t}(3 \sin t-\sin 3 t)\\ &\frac{d x}{d t}=\frac{d}{d t}(3 \sin t)-\frac{d}{d t}(\sin 3 t)\\ &\left.\frac{d x}{d t}=3 \cos t-3 \cos 3 t\right.\\ &\text { Now, for y }\\ &y=3 \operatorname{cost}-\cos 3 t \end{aligned}$
$\\ \frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(3 \cos \mathrm{t}- \cos 3 \mathrm{t}) \\ \frac{\mathrm{dy}}{\mathrm{dt}}=3(-\sin \mathrm{t})- \frac{\mathrm{d}}{\mathrm{dt}}(\cos 3 \mathrm{t}) \\ \frac{\mathrm{dy}}{\mathrm{dt}}=3(-\sin \mathrm{t})-3(-\sin 3 \mathrm{t}) \\ \frac{\mathrm{dy}}{\mathrm{dt}}=-3 \sin \mathrm{t}+3 \sin 3 \mathrm{t} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}}$
$\\ \frac{d y}{d x}=\frac{-3 \sin t+3 \sin 3 t}{3 \cos t-3 \cos 3 t} \\ \text { At } t=\pi / 3 \\ \frac{d y}{d x}=\frac{-3 \sin \frac{\pi}{3}+3 \sin \frac{3 \pi}{3}}{3 \cos \frac{\pi}{3}-3 \cos \frac{3 \pi}{3}} \\ \frac{d y}{d x}=\frac{-\sin \frac{\pi}{3}+\sin \pi}{\cos \frac{\pi}{3}-\cos \pi} \\ \frac{d y}{d x}=\frac{-\frac{\sqrt{3}}{2}+0}{\frac{1}{2}-(-1)}$
$\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}+1} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\frac{\sqrt{3}}{2}}{\frac{3}{2}} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\sqrt{3}}{2} \times \frac{2}{3} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-1}{\sqrt{3}}$

Question:52

Differentiate $\frac{x}{\sin x}$ w.r.t. sinx.

Answer:

Let us Assume,
$\begin{aligned} &\mathrm{u}=\frac{\mathrm{x}}{\sin \mathrm{x}}, \mathrm{v}=\sin \mathrm{x}\\ &\text { Now, differentiate w.r.t } x\\ &\frac{d u}{d x}=\frac{\sin x \cdot \frac{d}{d x}(x)-x \cdot \frac{d}{d x}(\sin x)}{(\sin x)^{2}}\\ &\frac{d u}{d x}=\frac{\sin x-x \cdot \cos x}{(\sin x)^{2}}\\ &\text { And, } v=\sin x\\ &\frac{\mathrm{d} \mathrm{v}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{x}) \end{aligned}$
$\begin{aligned} &\frac{d v}{d x}=\cos x\\ &\text { Now, }\\ &\frac{\mathrm{du}}{\mathrm{dv}}=\frac{\frac{\mathrm{du}}{\mathrm{dx}}}{\frac{\mathrm{dv}}{\mathrm{dx}}}\\ &\frac{\mathrm{du}}{\mathrm{dv}}=\frac{\sin \mathrm{x} \cdot-\mathrm{x} \cdot \cos \mathrm{x}}{(\sin \mathrm{x})^{2}} \times \frac{1}{\cos \mathrm{x}}\\ &\frac{d u}{d v}=\frac{\tan x-x}{\sin ^{2} x} \end{aligned}$

Question:53

Differentiate $\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$ w.r.t. $tan^{-1}x$ when $x\neq 0$

Answer:

We have $\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$
Let us Assume,
$\begin{aligned} &\mathrm{p}=\tan ^{-1}\left(\frac{\sqrt{1+\mathrm{x}^{2}}-1}{\mathrm{x}}\right) \text { and } \theta=\tan ^{-1}{ \mathrm{x}}\\ &\text { And, put } x=\tan \theta\\ &p=\tan ^{-1} \frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}\\ &p=\tan ^{-1} \frac{\sqrt{\sec ^{2} \theta}-1}{\tan \theta}\\ &p=\tan ^{-1} \frac{\sec \theta-1}{\tan \theta} \end{aligned}$
$\\ \mathrm{p}=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right) \\ \mathrm{p}=\tan ^{-1}\left(\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}}\right) \\ \mathrm{p}=\tan ^{-1}\left(\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\right) \\ \mathrm{p}=\tan ^{-1}\left(\tan \frac{\theta}{2}\right) \\ \mathrm{p}=\frac{\theta}{2}$
$\frac{dp}{d\theta} = \frac{1}{2}$
Hence Differentiation of $\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)$ w.r.t. $\tan^{-1}x$ is $\frac{1}{2}$ .

Question:54

Find $\frac{dy}{dx}$ when x and y are connected by the relation given

$\sin (x y)+\frac{x}{y}=x^{2}-y$

Answer:

We have,
$\sin (x y)+\frac{x}{y}=x^{2}-y$
Use chain rule and quotient rule to get:
Chain Rule
f(x) = g(h(x))
f’(x) = g’(h(x))h’(x)
By Quotient Rule
$\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}\left[\frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}\right]=\frac{\mathrm{g}(\mathrm{x}) \mathrm{f}^{\prime}(\mathrm{x})-\mathrm{f}(\mathrm{x}) \mathrm{g}^{\prime}(\mathrm{x})}{[\mathrm{g}(\mathrm{x})]^{2}}$
On differentiating both the sides with respect to x, we get

$\begin{aligned} &\cos x y \times \frac{d}{d x}(x y)+\frac{y \frac{d}{d x}(x)-x \frac{d}{d x}(y)}{y^{2}}=2 x-\frac{d y}{d x}\\ &\text { By product rule: }\\ &\frac{\mathrm{d}}{\mathrm{dx}}[\mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x})]=\mathrm{f}(\mathrm{x}) \mathrm{g}^{\prime}(\mathrm{x})+\mathrm{g}(\mathrm{x}) \mathrm{f}^{\prime}(\mathrm{x})\\ &\left[\because \frac{d}{d x} \sin x=\cos x\right] \end{aligned}$
$\begin{aligned} &\Rightarrow \cos (x y)\left[x \frac{d y}{d x}+y \times(1)\right]+\frac{y \times 1-x \frac{d y}{d x}}{y^{2}}=2 x-\frac{d y}{d x}\\ &\text { Multiplying by } y^{2} \text { to both the sides, we get }\\ &\Rightarrow \cos (x y)\left[x y^{2} \frac{d y}{d x}+y^{3}\right]+y-x \frac{d y}{d x}=2 x y^{2}-y^{2} \frac{d y}{d x}\\ &\Rightarrow x y^{2} \cos (x y) \frac{d y}{d x}+y^{3} \cos (x y)+y-x \frac{d y}{d x}=2 x y^{2}-y^{2} \frac{d y}{d x}\\ &\Rightarrow \mathrm{xy}^{2} \cos (\mathrm{xy}) \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{xy}^{2}-\mathrm{y}^{3} \cos (\mathrm{xy})-\mathrm{y}\\ &\Rightarrow \frac{d y}{d x}\left[x y^{2} \cos (x y)-x+y^{2}\right]=2 x y^{2}-y^{3} \cos (x y)-y\\ &\Rightarrow \frac{d y}{d x}=\frac{2 x y^{2}-y^{3} \cos (x y)-y}{x y^{2} \cos (x y)-x+y^{2}} \end{aligned}$

Question:55

Find $\frac{dy}{dx}$ when x and y are connected by the relation given

$sec (x + y) = xy$

Answer:

We have,
sec(x + y) = xy
By the rules given below:
Chain Rule
f(x) = g(h(x))
f’(x) = g’(h(x))h’(x)
Product rule:
$\begin{aligned} &\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}[\mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x})]=\mathrm{f}(\mathrm{x}) \mathrm{g}^{\prime}(\mathrm{x})+\mathrm{g}(\mathrm{x}) \mathrm{f}^{\prime}(\mathrm{x})\\ &\text { On differentiating both sides with respect to } x, \text { we get }\\ &\sec (x+y) \tan (x+y) \frac{d}{d x}(x+y)=y+x \frac{d}{d x} y\\ &\left[\because \frac{d}{d x} \sec (x)=\sec x \tan x\right]\\ &\Rightarrow \sec (\mathrm{x}+\mathrm{y}) \tan (\mathrm{x}+\mathrm{y})\left[1+\frac{\mathrm{d} y}{\mathrm{~d} \mathrm{x}}\right]=\mathrm{y}+\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}} \end{aligned}$
$\\ \Rightarrow \sec (x+y) \tan (x+y)+\sec (x+y) \tan (x+y) \frac{d y}{d x}=y+x \frac{d y}{d x} \\ \Rightarrow \sec (x+y) \tan (x+y) \frac{d y}{d x}-x \frac{d y}{d x}=y-\sec (x+y) \tan (x+y) \\ \Rightarrow \frac{d y}{d x}[\sec (x+y) \tan (x+y)-x]=y-\sec (x+y) \tan (x+y) \\ \therefore \frac{d y}{d x}=\frac{y-\sec (x+y) \tan (x+y)}{\sec (x+y) \tan (x+y)-x}$

Question:58

If $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ , then show that $\frac{d y}{d x} \cdot \frac{d x}{d y}=1$

Answer:

Given: $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$
Differentiating the above with respect to x, we get
$\\ 2 \mathrm{ax}+2 \mathrm{~h}\left[\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}\right]+\mathrm{b} \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{y}^{2}\right)+2 \mathrm{~g}+2 \mathrm{f} \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{y}=0 \\ \Rightarrow 2 \mathrm{ax}+2 \mathrm{hx} \frac{\mathrm{dy}}{\mathrm{dx}}+2 \mathrm{hy}+2 \mathrm{by} \frac{\mathrm{dy}}{\mathrm{dx}}+2 \mathrm{~g}+2 \mathrm{f} \frac{\mathrm{dy}}{\mathrm{dx}}=0 \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}[2 \mathrm{hx}+2 \mathrm{by}+2 \mathrm{f}]=-2 \mathrm{ax}-2 \mathrm{hy}-2 \mathrm{~g} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-[2 \mathrm{ax}+2 \mathrm{hy}+2 \mathrm{~g}]}{2 \mathrm{hx}+2 \mathrm{by}+2 \mathrm{f}} \ldots(\mathrm{ii})$
Now, we again differentiate eq (i) with respect to y, we get,

$\\ a \frac{d}{d y}\left(x^{2}\right)+2 h\left[x+y \frac{d}{d y} x\right]+2 y b+2 g \frac{d x}{d y}+2 f=0 \\ \Rightarrow 2 a x \frac{d x}{d y}+2 h x+2 h y \frac{d x}{d y}+2 b y+2 g \frac{d x}{d y}+2 f=0 \\ \Rightarrow \frac{d x}{d y}[2 a x+2 h y+2 g]=-2 h x-2 b y-2 f \\ \Rightarrow \frac{d x}{d y}=\frac{-[2 h x+2 b y+2 f]}{2 a x+2 h y+2 g} \ldots(i i i)$
Now, multiplying Eq. (ii) and (iii), we get

$\\ \Rightarrow \frac{d y}{d x} \times \frac{d x}{d y}=\frac{-[2 a x+2 h y+2 g]}{2 h x+2 b y+2 f} \times \frac{-[2 h x+2 b y+2 f]}{2 a x+2 h y+2 g} =1$
Hence Proved

Question:59

If $x=e^{x/y}$ prove that $\frac{dy}{dx}= \frac{x-y}{x\log x}$

Answer:

Given: $x=e^{x/y}$
Taking log on both the sides, we get
$\begin{aligned} &\log \mathrm{x}=\log \mathrm{e}^{\frac{x}{y}}\\ &\Rightarrow \log \mathrm{x}=\frac{\mathrm{x}}{\mathrm{y}}\\ &\text { or } y \log x=x\\ &\text { On differentiating above with respect to } x, \text { we get }\\ &\mathrm{y} \times \frac{1}{\mathrm{x}}+\log \mathrm{x} \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{y}=1 \end{aligned}$
$\\ \Rightarrow \frac{y}{x}+\log x \frac{d y}{d x}=1\left[\because \frac{d}{d x}(\log x)=\frac{1}{x}\right] \\ \Rightarrow \log x \frac{d y}{d x}=1-\frac{y}{x} \\ \Rightarrow \frac{d y}{d x}=\frac{1}{\log x}\left[\frac{x-y}{x}\right] \\ \Rightarrow \frac{d y}{d x}=\frac{x-y}{x \log x}$
Hence Proved.

Question:60

If $y^x = e^{y-x}$ , prove that $\frac{d y}{d x}=\frac{(1+\log y)^{2}}{\log y}$

Answer:

Given:.
$y^x = e^{y-x}$
Taking log on both the sides, we get

$\begin{aligned} &\log y^x=\log e^{y-x} \\ &\Rightarrow x \log y=y-x\\ \end{aligned}$
$\log y=\frac{y}{x}-1$
$\\ \Rightarrow \frac{x d y}{y d x}-\frac{d y}{d x}=-1-\log y \\ \Rightarrow \frac{d y}{d x}\left[\frac{x}{y}-1\right]=-[1+\log y] \\ \Rightarrow \frac{d y}{d x}=\frac{-[1+\log y]}{\frac{x}{y}-1} \\ \Rightarrow \frac{d y}{d x}=\frac{\log y+1}{1-\frac{x}{y}}$
$\\ \Rightarrow \frac{d y}{d x}=\frac{\log y+1}{\frac{y-x}{y}} \\ \Rightarrow \frac{d y}{d x}=\frac{y[\log y+1]}{y-x} \\ \Rightarrow \frac{d y}{d x}=\frac{y[\log y+1]}{x \log y} \\ \Rightarrow \frac{d y}{d x}=\frac{[\log y+1] (\log y+1)}{\log y} \quad[u \sin g(i)] \\ \Rightarrow \frac{d y}{d x}=\frac{(\log y+1)^{2}}{\log y}$
Hence Proved

Question:61

If $\mathrm{y}=(\cos \mathrm{x})^{(\cos \mathrm{x})^{(\cos \mathrm{x}) \ldots . \infty}}$ Show that $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}^{2} \tan \mathrm{x}}{\mathrm{y} \log \cos \mathrm{x}-1}$

Answer:

Given: $\mathrm{y}=(\cos \mathrm{x})^{(\cos \mathrm{x})^{(\cos \mathrm{x}) \ldots . \infty}}$
⇒ y = (cos x)^y
Taking log both the sides, we get
log y = y log(cos x)
On differentiating both the sides, we get
$\\ \frac{1}{y} \frac{d y}{d x}=y \frac{d}{d x} \log (\cos x)+\log \cos x \frac{d}{d x} y \\ \Rightarrow \frac{1}{y} \frac{d y}{d x}=y \times \frac{1}{\cos x} \times(-\sin x)+\log \cos x \frac{d y}{d x} \\ \Rightarrow \frac{d y}{d x}\left[\frac{1}{y}-\log \cos x\right]=-y \tan x \because\left[\frac{\sin x}{\cos x}=\tan x\right] \\ \Rightarrow \frac{d y}{d x}=\frac{-y \tan x}{\frac{1}{y}-\log \cos x} \\ \Rightarrow \frac{d y}{d x}=\frac{y \tan x}{\log \cos x-\frac{1}{y}} \\ \Rightarrow \frac{d y}{d x}=\frac{y^{2} \tan x}{y \log \cos x-1}$
Hence Proved

Question:62

If $x \sin (a + y) + \sin a \cos (a + y) = 0$ , prove that $\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}$ .

Answer:

Given: $x \sin (a + y) + \sin a \cos (a + y) = 0$
$\begin{aligned} &\Rightarrow x=\frac{-\sin a \cos (a+y)}{\sin (a+y)}\\ &\Rightarrow x=-\sin a \cot (a+y)\left[\because \frac{\cos x}{\sin x}=\cot x\right]\\ &\text { Differentiating with respect to } y \text { , we get }\\ &\frac{d x}{d y}=\frac{d}{d y}[-\sin a \cot (a+y)]\\ &\Rightarrow \frac{d x}{d y}=-\sin a \frac{d}{d y} \cot (a+y) \end{aligned}$
$\\ \Rightarrow \frac{d x}{d y}=-\sin a\left[-\operatorname{cosec}^{2}(a+y)\right] \because\left[\frac{d}{d y} \cot (x)=\operatorname{cosec}^{2} x\right] \\ \Rightarrow \frac{d x}{d y}=\sin a\left[\frac{1}{\sin ^{2}(a+y)}\right] \\ \Rightarrow \frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}$
Hence Proved.

Question:63

If $\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)$ prove that $\frac{d y}{d x}=\sqrt{\frac{1-y^{2}}{1-x^{2}}}$

Answer:

Given: $\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)$
$\begin{array}{l} \text { Put } x=\sin \alpha \text { and } y=\sin \beta \ldots(i) \\ \sqrt{1-(\sin \alpha)^{2}}+\sqrt{1-(\sin \beta)^{2}}=a(\sin \alpha-\sin \beta) \end{array}$
$\Rightarrow \sqrt{1-\sin ^{2} \alpha}+\sqrt{1-\sin ^{2} \beta}=a(\sin \alpha-\sin \beta)$$
Now, we know that $\sin ^{2} \theta+\cos ^{2} \theta=1$$
$\\\Rightarrow \sqrt{\cos ^{2} \alpha}+\sqrt{\cos ^{2} \beta}=a(\sin \alpha-\sin \beta)$ \\$\Rightarrow \cos \alpha+\cos \beta=a(\sin \alpha-\sin \beta) \ldots(i i)$$
Now, we use some trigonometry formulas,
$\\ \cos \alpha+\cos \beta=2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)$
$\\ \sin \alpha-\sin \beta=2 \cos \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)$$
$\begin{aligned} &\text { So, eq.(ii) become }\\ &2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)=2 \mathrm{a} \cos \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)\\ &\Rightarrow \cos \left(\frac{\alpha-\beta}{2}\right)=\mathrm{a} \sin \left(\frac{\alpha-\beta}{2}\right)\\ &\Rightarrow \mathrm{a}=\frac{\cos \left(\frac{\alpha-\beta}{2}\right)}{\sin \left(\frac{\alpha-\beta}{2}\right)}\\ &\Rightarrow \mathrm{a}=\cot \left(\frac{\alpha-\beta}{2}\right) \end{aligned}$
$\\\Rightarrow \cot ^{-1} \mathrm{a}=\frac{\alpha-\beta}{2}$ \\$\Rightarrow 2 \cot ^{-1} a=\alpha-\beta....(i i)$ \\Now, from eq.(i), we have \\$\alpha=\sin ^{-1} x$ and $\beta=\sin ^{-1} y$ \\Now, put value of $\alpha$ and $\beta$ in eq. (iii), we get \\$2 \cot ^{-1} a=\sin ^{-1} x-\sin ^{-1} y$ \\or $\sin ^{-1} x-\sin ^{-1} y=2 \cot ^{-1} a$ \\On differentiating above with respect to $x,$ we get \\$\frac{d}{d x}\left(\sin ^{-1} x-\sin ^{-1} y\right)=\frac{d}{d x}\left(2 \cot ^{-1} a\right)$ \\$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}} \sin ^{-1} \mathrm{x}-\frac{\mathrm{d}}{\mathrm{dx}} \sin ^{-1} \mathrm{y}=2 \frac{\mathrm{d}}{\mathrm{dx}} \cot ^{-1} \mathrm{a}$$
$\\ \Rightarrow \frac{1}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-y^{2}}} \frac{d y}{d x}=0\left[\because \frac{d}{d x} \sin ^{-1} x=\frac{1}{\sqrt{1-x^{2}}}\right] \\ \Rightarrow \frac{d y}{d x}=\sqrt{\frac{1-y^{2}}{1-x^{2}}}$
Hence Proved

Question:64

If $y = tan^{-1}x$ , find $\frac{d^2y}{dx^2}$ in terms of y alone.

Answer:

Given: $y = tan^{-1}x$
$\\\Rightarrow \tan y=x$ \\On differentiating Eq. (i) with respect to $x,$ we get \\$\frac{d y}{d x}=\frac{1}{1+x^{2}}$ \\Now, again differentiating the above with respect to $x,$ we get \\$\frac{d^{2} y}{d x^{2}}=\frac{\left(1+x^{2}\right) \times \frac{d}{d x}(1)-1 \times \frac{d}{d x}\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}}$ \\$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{0-2 \mathrm{x}}{\left(1+\mathrm{x}^{2}\right)^{2}}$ \\$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{-2 \tan \mathrm{y}}{\left(1+\tan ^{2} \mathrm{y}\right)^{2}}$ \\$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{-2 \tan\mathrm{y}}{\left(\mathrm{sec}^{2} \mathrm{y}\right)^{2}}$$
$\\ \Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{-2 \sin y}{\cos y} \times \cos ^{4} y \\ \Rightarrow \frac{d^{2} y}{d x^{2}}=-2 \sin y \cos ^{3} y$

Question:65

Verify the Rolle’s theorem for each of the functions
$f(x) = x (x - 1)^2$ in [0, 1].

Answer:

Given: $f(x) = x (x - 1)^2$
$\\ \Rightarrow f(x)=x\left(x^{2}+1-2 x\right) \\ -f(x)=x^{3}+x-2 x^{2} \text { in }[0,1]$
Now, we have to show that f(x) verify the Rolle’s Theorem
First of all, Conditions of Rolle’s theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
c) f(a) = f(b)
If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0
Condition 1:

On expanding $f(x) = x(x - 1)\textsuperscript{2}, we get f(x) = x\textsuperscript{3} + x - 2x\textsuperscript{2\\ }$
Since, f(x) is a polynomial and we know that, every polynomial function is continuous for all $x $ \in $ R$
$$ \Rightarrow $ f(x) = x\textsuperscript{3} + x - 2x\textsuperscript{2}$ is continuous at x $ \in $ [0,1]$
Hence, condition 1 is satisfied.
Condition 2:
$f(x) = x\textsuperscript{3} + x - 2x\textsuperscript{2\\ }$
Since, f(x) is a polynomial and every polynomial function is differentiable for all $x $ \in $ R$
$$ \Rightarrow $ f(x) $is differentiable at [0,1]$
Hence, condition 2 is satisfied.
Condition 3:
$\\f(x) = x\textsuperscript{3} + x - 2x\textsuperscript{2} \\f(0) = 0 \\f(1) = (1)\textsuperscript{3} + (1) - 2(1)\textsuperscript{2} = 1 + 1 - 2 = 0$
Hence, f(0) = f(1)
Hence, condition 3 is also satisfied.
Now, let us show that $c $ \in $ (0,1)$ such that f’(c) = 0
$f(x) = x\textsuperscript{3} + x - 2x\textsuperscript{2}$
On differentiating above with respect to x, we get
$f'(x) = 3x\textsuperscript{2} + 1 - 4x$
Put x = c in above equation, we get
$f'(c) = 3c\textsuperscript{2} + 1 - 4c$
$\because$ , all the three conditions of Rolle’s theorem are satisfied
f’(c) = 0
$3c\textsuperscript{2} + 1 - 4c = 0$
On factorising, we get
$\\$ \Rightarrow $ 3c\textsuperscript{2} - 3c - c + 1 = 0 \\$ \Rightarrow $ 3c(c - 1) - 1(c - 1) = 0 \\$ \Rightarrow $ (3c - 1) (c - 1) = 0 \\$ \Rightarrow $ (3c - 1) = 0 or (c - 1) = 0$
$\begin{aligned} &\Rightarrow c=\frac{1}{3} \text { or } c=1\\ &\text { So, value of }\\ &c=\frac{1}{3} \in(0,1) \end{aligned}$
Thus, Rolle’s theorem is verified.

Question:66

Verify the Rolle’s theorem for each of the functions
$f(x)=\sin ^{4} x+\cos ^{4} x \text { in }\left[0, \frac{\pi}{2}\right]$

Answer:

Given: $f(x)=\sin ^{4} x+\cos ^{4} x \text { in }\left[0, \frac{\pi}{2}\right]$
Now, we have to show that f(x) verify the Rolle’s Theorem
First of all, Conditions of Rolle’s theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
c) f(a) = f(b)
If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0
Condition 1:
$f(x)=\sin ^{4} x+\cos ^{4} x$
Since, f(x) is a trigonometric function and trigonometric function is continuous everywhere
$\\$ \Rightarrow $ f(x) = sin\textsuperscript{4}x + cos\textsuperscript{4}x $ is continuous at x \in \left [ 0,\pi/2 \right ]$
Hence, condition 1 is satisfied.
Condition 2:
$f(x)=\sin ^{4} x+\cos ^{4} x$
On differentiating above with respect to x, we get
$f'(x) = 4 $ \times $ sin\textsuperscript{3} (x) $ \times $ cos x + 4 $ \times $ cos\textsuperscript{3} x $ \times $ (- sin x)$
$\because\left[\frac{\mathrm{d}}{\mathrm{dx}} \sin \mathrm{x}=\cos \mathrm{x} \& \frac{\mathrm{d}}{\mathrm{dx}} \cos \mathrm{x}=-\sin \mathrm{x}\right]$
$\Rightarrow f'(x) = 4sin\textsuperscript{3} x cos x - 4 cos\textsuperscript{3} x sinx \\ \Rightarrow f'(x) = 4sin x cos x [sin\textsuperscript{2}x - cos\textsuperscript{2} x] \\ \Rightarrow f'(x) = 2 sin2x [sin\textsuperscript{2}x - cos\textsuperscript{2} x] [\because 2 sin x cos x = sin 2x] \\ \Rightarrow f'(x) = 2 sin 2x [- cos 2x] [\because cos\textsuperscript{2} x - sin\textsuperscript{2} x = cos 2x] \\ \Rightarrow f'(x) = - 2 sin 2x cos 2x$
$\Rightarrow$ f(x) is differentiable at $\left [ 0, \pi/2 \right ]$
Hence, condition 2 is satisfied.
Condition 3:
$\\f(x) = sin\textsuperscript{4}x + cos\textsuperscript{4}x \\\\f(0) = sin\textsuperscript{4}(0) + cos\textsuperscript{4}(0) = 1$
$\begin{aligned} \mathrm{f}\left(\frac{\pi}{2}\right)=& \sin ^{4}\left(\frac{\pi}{2}\right)+\cos ^{4}\left(\frac{\pi}{2}\right)=1 \\ & \therefore \mathrm{f}(0)=\mathrm{f}\left(\frac{\pi}{2}\right) \end{aligned}$
Hence, condition 3 is also satisfied.
Now, let us show that c ∈ $\left [ 0, \pi/2 \right ]$ such that f’(c) = 0
$\\f(x) = sin\textsuperscript{4}x + cos\textsuperscript{4}x \\ $ \Rightarrow $ f'(x) = - 2 sin 2x cos 2x$
Put x = c in above equation, we get
$\Rightarrow $ f'(c) = - 2 sin 2c cos 2c$
$\because$ all the three conditions of Rolle’s theorem are satisfied
f’(c) = 0
$\\ \Rightarrow $ - 2 sin 2c cos 2c = 0 \\$ \Rightarrow $ sin 2c cos 2c = 0 \\$ \Rightarrow $ sin 2c = 0 \\$ \Rightarrow $ 2c = 0\\ \\$ \Rightarrow $ c = 0$
or Now, cos 2c = 0
$\\ \Rightarrow \cos 2 c=\cos \frac{\pi}{2} \\ \Rightarrow 2 c=\frac{\pi}{2} \\ \Rightarrow c=\frac{\pi}{4} \\ \Rightarrow c=\left\{0, \frac{\pi}{4}\right\} \in\left[0, \frac{\pi}{2}\right]$
Thus, Rolle’s theorem is verified.

Question:67

Verify the Rolle’s theorem for each of the functions
$f(x) = log (x^2 + 2) - log3$ in [- 1, 1].

Answer:

Given: $f(x) = log (x^2 + 2) - log3$
Now, we have to show that f(x) verify the Rolle’s Theorem
First of all, Conditions of Rolle’s theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
c) f(a) = f(b)
If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0
Condition 1:
$f(x) = log (x^2 + 2) - log3$
Since, f(x) is a logarithmic function and logarithmic function is continuous for all values of x.
$$ \Rightarrow $ f(x) = log (x\textsuperscript{2} + 2) - log3 $ is continuous at x $ \in $ [-1,1]$
Hence, condition 1 is satisfied.
Condition 2:
$f(x) = log (x\textsuperscript{2} + 2) - log3$
$\begin{aligned} &\text { Chain Rule }\\ &f(x)=g(h(x))\\ &f^{\prime}(x)=g^{\prime}(h(x)) h^{\prime}(x) \end{aligned}$
$\begin{aligned} &\Rightarrow f(x)=\log \left(\frac{x^{2}+2}{3}\right)\left[\because \log m-\log n=\log \frac{m}{n}\right]\\ &\text { On differentiating above with respect to } x \text { , we get }\\ &f^{\prime}(x)=\frac{1}{\frac{x^{2}+2}{3}} \times \frac{d}{d x}\left(\frac{x^{2}+2}{3}\right)\left[\because \frac{d}{d x} \log x=\frac{1}{x}\right]\\ &\Rightarrow f^{\prime}(x)=\frac{3}{x^{2}+2} \times \frac{d}{d x}\left(\frac{x^{2}}{3}+\frac{2}{3}\right)\\ &\Rightarrow f^{\prime}(x)=\frac{3}{x^{2}+2} \times\left(\frac{2 x}{3}+0\right)\\ &\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{3}{\mathrm{x}^{2}+2} \times \frac{2 \mathrm{x}}{3} \end{aligned}$
$\\\Rightarrow f^{\prime}(x)=\frac{2 x}{x^{2}+2}$ \\$\Rightarrow f(x)$ is differentiable at [-1,1]$
Hence, condition 2 is satisfied.
Condition 3:
$\\f(x)=\log \left(\frac{x^{2}+2}{3}\right) \\f(-1)=\log \left(\frac{(-1)^{2}+2}{3}\right)=\log \frac{1+2}{3}=\log 1=0 \\f(1)=\log \left(\frac{(1)^{2}+2}{3}\right)=\log \frac{1+2}{3}=\log 1=0 \\\therefore f(-1)=f(1)$
Hence, condition 3 is also satisfied.
Now, let us show that $c \in(-1,1)$ such that f(c)=0
$\\ f(x)=\log \left(\frac{x^{2}+2}{3}\right)$ \\$\Rightarrow f^{\prime}(x)=\frac{2 x}{x^{2}+2}$$
Put x=c in above equation, we get
$\\\Rightarrow f^{\prime}(c)=\frac{2 c}{c^{2}+2}=0\Rightarrow c=0\epsilon (-1,1)$ \\$\because,$ all the three conditions of Rolle's theorem are satisfied $f(c)=0$$

Question:68

Verify the Rolle’s theorem for each of the functions
$f(x) = x(x + 3)e^{-x/2}$ in [-3, 0].

Answer:

Given: $f(x) = x(x + 3)e^{-x/2}$
$\Rightarrow f(x)=\left(x^{2}+3 x\right) e^{\frac{-x}{2}}$
Now, we have to show that f(x) verify the Rolle’s Theorem
First of all, Conditions of Rolle’s theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
c) f(a) = f(b)
If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0
Condition 1:
$\Rightarrow f(x)=\left(x^{2}+3 x\right) e^{\frac{-x}{2}}$
Since, f(x) is multiplication of algebra and exponential function and is defined everywhere in its domain.
$\Rightarrow f(x)=\left(x^{2}+3 x\right) e^{\frac{-x}{2}}$ is continuous at x ∈ [-3,0]
Hence, condition 1 is satisfied.
Condition 2: $\Rightarrow f(x)=\left(x^{2}+3 x\right) e^{\frac{-x}{2}}$
On differentiating f(x) with respect to x, we get,

$\\ f^{\prime}(x)=e^{\frac{-x}{2}} \frac{d}{d x}\left(x^{2}+3 x\right)+\left(x^{2}+3 x\right) \frac{d}{d x} e^{-\frac{x}{2}} \text { [by product rule }] \\ \Rightarrow f^{\prime}(x)=e^{-\frac{x}{2}}[2 x+3]+\left(x^{2}+3 x\right) \times\left(-\frac{1}{2} e^{-\frac{x}{2}}\right) \\ \Rightarrow f^{\prime}(x)=2 x e^{-\frac{x}{2}}+3 e^{-\frac{x}{2}}-\frac{x^{2}}{2} e^{-\frac{x}{2}}-\frac{3 x}{2} \mathrm{e}^{-\frac{x}{2}} \\ \Rightarrow f^{\prime}(x)=e^{-\frac{x}{2}}\left[2 x+3-\frac{x^{2}}{2}-\frac{3 x}{2}\right] \\ \Rightarrow f^{\prime}(x)=\frac{e^{-\frac{x}{2}}}{2}\left[x+6-x^{2}\right] \\ \Rightarrow f^{\prime}(x)=-\frac{e^{-\frac{x}{2}}}{2}\left[x^{2}-x-6\right] \\ \Rightarrow \quad(x)=-\frac{e^{-\frac{x}{2}}}{2}\left[x^{2}-3 x+2 x-6\right]$
$\\ \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=-\frac{\mathrm{e}^{-\frac{\mathrm{x}}{2}}}{2}[\mathrm{x}(\mathrm{x}-3)+2(\mathrm{x}-3)] \\ \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=-\frac{\mathrm{e}^{-\frac{\mathrm{x}}{2}}}{2}[(\mathrm{x}-3)(\mathrm{x}+2)]$
⇒ f(x) is differentiable at [-3,0]
Hence, condition 2 is satisfied.
Condition 3:
$\\f(x)=\left(x^{2}+3 x\right) e^{\frac{-x}{2}}$ \\$f(-3)=\left[(-3)^{2}+3(-3)\right] e^{\frac{-(-3)}{2}}$ \\$=[9-9] \mathrm{e}^{2 / 2}$ \\$=0$ \\$f(0)=\left[(0)^{2}+3(0)\right] e^{\frac{-0}{2}}$ \\$=0$ \\Hence, $f(-3)=f(0)$ \\Hence, condition 3 is also satisfied. \\Now, let us show that $c \in(0,1)$ such that $f^{\prime}(c)=0$ \\$f(x)=\left(x^{2}+3 x\right) e^{\frac{-x}{2}}$$
$$On differentiating above with respect to x , we get $\\ f^{\prime}(x)=-\frac{e^{-\frac{x}{2}}}{2}[(x-3)(x+2)]$
$\\ Put$ x=c in above equation, we get$\\ \mathrm{f}^{\prime}(\mathrm{c})=-\frac{\mathrm{e}^{-\frac{\mathrm{c}}{2}}}{2}[(\mathrm{c}-3)(\mathrm{c}+2)]$
all the three conditions of Rolle’s theorem are satisfied
f’(c) = 0
$\begin{aligned} &-\frac{e^{-\frac{c}{2}}}{2}[(c-3)(c+2)]=0\\ &\because-\frac{e^{-\frac{c}{2}}}{2} c a n^{\prime} t \text { be zero }\\ &\Rightarrow(c-3)(c+2)=0\\ &\Rightarrow c-3=0 \text { or } c+2=0\\ &\Rightarrow c=3 \text { or } c=-2\\ &\text { So, value of } c=-2,3\\ &c=-2 \in(-3,0) \text { but } c=3 \in(-3,0)\\ &\therefore c=-2 \end{aligned}$
Thus, Rolle’s theorem is verified.

Question:69

Verify the Rolle’s theorem for each of the functions
$f(x)=\sqrt{4-x^{2}} \text { in }[-2,2]$

Answer:

Given: $f(x)=\sqrt{4-x^{2}}$
Now, we have to show that f(x) verify the Rolle’s Theorem
First of all, Conditions of Rolle’s theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
c) f(a) = f(b)
If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0
Condition 1:
Firstly, we have to show that f(x) is continuous.
Here, f(x) is continuous because f(x) has a unique value for each x ∈ [-2,2]
Condition 2:
Now, we have to show that f(x) is differentiable
$\\ f(x)=\sqrt{4-x^{2}}$ \\$\Rightarrow f(x)=\left(4-x^{2}\right)^{\frac{1}{2}} \\ \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2}\left(4-\mathrm{x}^{2}\right)^{-\frac{1}{2}} \times(-2 \mathrm{x})$ [using chain rule $]$ \\$\Rightarrow f^{\prime}(x)=\frac{-x}{\sqrt{4-x^{2}}}$$
$\\\therefore \mathrm{f}^{\prime}(\mathrm{x})$ exists for all $\mathrm{x} \in(-2,2)$
So, f(x) is differentiable on (-2,2)
Hence, Condition 2 is satisfied.
Condition 3:
$f(x)=\sqrt{4-x^{2}}$$
Now, we have to show that f(a) = f(b)
so, f(a) = f(-2)
$f(-2)=\sqrt{4-(-2)^{2}}=\sqrt{4-4}=0$$
and $f(b)=f(2)$$
$\\ f(2)=\sqrt{4-(2)^{2}}=\sqrt{4-4}=0$ \\$\therefore f(-2)=f(2)=0$$
Hence, condition 3 is satisfied
Now, let us show that
$\\ c \in(0,1)$ such that $f^{\prime}(c)=0$ \\f(x)=\sqrt{4-x^{2}}$$
On differentiating above with respect to x, we get
$f^{\prime}(x)=\frac{-x}{\sqrt{4-x^{2}}}$$
Put x=c in above equation, we get,
$f^{\prime}(c)=\frac{-c}{\sqrt{4-c^{2}}}$$
Thus, all the three conditions of Rolle's theorem is satisfied. Now we have to see that there exist $c \in(-2,2)$ such that
$\\f^{\prime}(c)=0$ \\$\Rightarrow \frac{-c}{\sqrt{4-c^{2}}}=0$ \\$\Rightarrow c=0$ \\$\because c=0 \in(-2,2)$$
Hence, Rolle’s theorem is verified.

Question:70

Discuss the applicability of Rolle’s theorem on the function given by
$f(x)=\begin{array}{ll} x^{2}+1, & \text { if } 0 \leq x \leq 1 \\ 3-x, & \text { if } 1 \leq x \leq 2 \end{array}$

Answer:

Given: $f(x)=\begin{array}{ll} x^{2}+1, & \text { if } 0 \leq x \leq 1 \\ 3-x, & \text { if } 1 \leq x \leq 2 \end{array}$
First of all, Conditions of Rolle’s theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
c) f(a) = f(b)
If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0
Condition 1:
At x = 1
$\\\lim _{\mathrm{LHL}}\left(\mathrm{x}^{2}+1\right)=1+1=2$ \\$\lim _{\mathrm{RHL}}=\lim _{x \rightarrow 1^{+}}(3-\mathrm{x})=3-1=2$ \\$\because \mathrm{LHL}=\mathrm{RHL}=2$$
and $f(1)=3-x=3-1=2$$
$\therefore f(x)$ is continuous at $x=1$$
Hence, condition 1 is satisfied.
Condition 2:
Now, we have to check f(x) is differentiable
$f(x)=\left\{\begin{array}{l}x^{2}+1, \text { if } 0 \leq x \leq 1 \\ 3-x, \text { if } 1 \leq x \leq 2\end{array}\right.$$
On differentiating with respect to x, we get
$\Rightarrow f^{\prime}(x)=\left\{\begin{array}{c}2 x+0, \text { if } 0<x<1 \\ 0-1, \text { if } 1<x<2\end{array}\right.$$ or
$\mathrm{f}^{\prime}(\mathrm{x})=\left\{\begin{array}{l} 2 \mathrm{x}, \text { if } 0<\mathrm{x}<1 \\ -1, \text { if } 1<\mathrm{x}<2 \end{array}\right.$
Now, let us consider the differentiability of f(x) at x = 1
LHD ⇒ f(x) = 2x = 2(1) = 2
RHD ⇒ f(x) = -1 = -1
LHD ≠ RHD
∴ f(x) is not differentiable at x = 1
Thus, Rolle’s theorem is not applicable to the given function.

Question:71

Find the points on the curve y = (cosx - 1) in [0, $2\pi$ ], where the tangent is parallel to x-axis.

Answer:

Given: Equation of curve, y = cos x - 1
Firstly, we differentiate the above equation with respect to x, we get
$\\ \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{y}=\frac{\mathrm{d}}{\mathrm{dx}} \cos \mathrm{x}-\frac{\mathrm{d}}{\mathrm{dx}}(1) \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\sin \mathrm{x}-0\left[\because \frac{\mathrm{d}}{\mathrm{dx}} \cos \mathrm{x}=-\sin \mathrm{x}\right] \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\sin \mathrm{x}$
Given tangent to the curve is parallel to the x - axis
This means, Slope of tangent = Slope of x - axis
$\begin{aligned} &\frac{d y}{d x}=0\\ &\Rightarrow-\sin x=0\\ &-\sin x=0\\ &\Rightarrow \mathrm{x}=\sin ^{-1}(0)\\ &-x=\pi\lfloor(0,2 \pi)\\ &\text { Put } x=\pi \text { in } y=\cos x-1, \text { we have }\\ &\mathrm{y}=\cos \pi-1=-1-1=-2[\because \cos \pi=-1] \end{aligned}$
Hence, the tangent to the curve is parallel to the x -axis at
(π, -2)

Question:72

Using Rolle’s theorem, find the point on the curve y = x(x - 4), $x\in [0,4]$ where the tangent is parallel to x-axis.

Answer:

Given: y = x(x - 4)
$y = (x^2 - 4x)$
Now, we have to show that f(x) verify the Rolle’s Theorem
First of all, Conditions of Rolle’s theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
c) f(a) = f(b)
If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0
Condition 1:
On expanding $y=x(x-4),$$ we get $y=\left(x^{2}-4 x\right)$$
since, $x^{2}-4 x$$ is a polynomial and we know that, every polynomial function is continuous for all $X\in R$
$-y=\left(x^{2}-4 x\right)$ is continuous at $x \mathbb{E}[0,4]$$
Hence, condition 1 is satisfied.
Condition 2
$\\ y=\left(x^{2}-4 x\right)$ \\y^{\prime}=2 x-4$$
$-x^{2}-4 x$$ is differentiable at [0,4]
Hence, condition 2 is satisfied.
Condition 3:
$y=x^{2}-4 x$ \\ $x=0$ $\Rightarrow y=0$ \\ $x=4$ $\Rightarrow y=(4)^{2}-4(4)=16-16=0$
Hence, condition 3 is also satisfied.
Now, there is atleast one value of c ∈ (0,4)
Given tangent to the curve is parallel to the x - axis
This means, Slope of tangent = Slope of x - axis
$\begin{aligned} &\frac{\mathrm{dy}}{\mathrm{dx}}=0\\ &2 x-4=0\\ &2 x=4\\ &x=2 \in(0,4)\\ &\text { Put } x=2 \text { in } y=x^{2}-4 x, \text { we have }\\ &y=(2)^{2}-4(2)=4-8=-4 \end{aligned}$
Hence, the tangent to the curve is parallel to the x -axis at (2, -4).

Question:73

Verify mean value theorem for each of the functions given
$f(x)=\frac{1}{4 x-1} \text { in }[1,4]$

Answer:

Given: $f(x)=\frac{1}{4 x-1} \text { in }[1,4]$
Now, we have to show that f(x) verify the Mean Value Theorem
First of all, Conditions of Mean Value theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
If both the conditions are satisfied, then there exist some ‘c’ in (a,b) such that
$f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$
Here,
$f(x)=\frac{1}{4 x-1}=(4 x-1)^{-1}$
On differentiating above with respect to x, we get
$\\f^{\prime}(x)=-1 \times(4 x-1)^{-1-1} \times 4$ \\$\Rightarrow f^{\prime}(x)=-4 \times(4 x-1)^{-2}$ \\$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{-4}{(4 \mathrm{x}-1)^{2}}$ \\$\Rightarrow f^{\prime}(x)$ exist$
Hence, f(x) is differentiable in (1,4)
We know that,
Differentiability $\Rightarrow$ Continuity
Hence, f(x) is continuous in (1,4)
Thus, Mean Value Theorem is applicable to the given function
Now,
$\\ f(x)=\frac{1}{4 x-1} x \in[1,4] \\ f(a)=f(1)=\frac{1}{4(1)-1}=\frac{1}{4-1}=\frac{1}{3} \\ f(b)=f(4)=\frac{1}{4(4)-1}=\frac{1}{16-1}=\frac{1}{15}$
Now, let us show that there exist c ∈ (0,1) such that
$\\f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$ \\$f(x)=\frac{1}{4 x-1}$ \\On differentiating above with respect to $x,$ we get \\$f^{\prime}(x)=\frac{-4}{(4 x-1)^{2}}$ \\Put $x=c$ in above equation, we get \\$f^{\prime}(c)=\frac{-4}{(4 c-1)^{2}}$ \\By Mean Value Theorem, \\$f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$ \\$f^{\prime}(c)=\frac{f(4)-f(1)}{4-1}$$
$\\ \Rightarrow \frac{-4}{(4 c-1)^{2}}=\frac{\frac{1}{15}-\frac{1}{3}}{3} \\ \Rightarrow \frac{-4}{(4 c-1)^{2}}=\frac{\frac{1-5}{15}}{3} \\ \Rightarrow \frac{-4}{(4 c-1)^{2}}=\frac{-4}{15 \times 3} \\ \Rightarrow(4 c-1)^{2}=45 \\ \Rightarrow 4 c-1=\sqrt{ 45} \\ \Rightarrow 4 c-1=\pm 3 \sqrt{5}$
$\\\Rightarrow 4 c=1 \pm 3 \sqrt{5} \\\Rightarrow c=\frac{1 \pm 3 \sqrt{5}}{4}$
but $c=\frac{1-3 \sqrt{5}}{4} \notin(1,4)$$
So, value of $\mathrm{c}=\frac{1+3 \sqrt{5}}{4} \in(1,4)$$
Thus, Mean Value Theorem is verified.

Question:74

Verify mean value theorem for each of the functions given
$f(x) = x^3 - 2x^2 - x + 3 $ in [0, 1]$

Answer:

Given: $f(x) = x^3 - 2x^2 - x + 3 $ in [0, 1]$
Now, we have to show that f(x) verify the Mean Value Theorem
First of all, Conditions of Mean Value theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
If both the conditions are satisfied, then there exist some ‘c’ in (a,b) such that
$f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$
Condition 1:
$f(x) = x^3 - 2x^2 - x + 3$
Since, f(x) is a polynomial and we know that, every polynomial function is continuous for all x ∈ R
$\Rightarrow f(x) = x^3 - 2x^2 - x + 3$ is continuous at x ∈ [0,1]
Hence, condition 1 is satisfied.
Condition 2:
$\Rightarrow f(x) = x^3 - 2x^2 - x + 3$
Since, f(x) is a polynomial and every polynomial function is differentiable for all x ∈ R
$f^{\prime}(x)=3 x^{2}-4 x-1$
⇒ f(x) is differentiable at [0,1]
Hence, condition 2 is satisfied.
Thus, Mean Value Theorem is applicable to the given function
Now,
$\begin{aligned} &f(x)=x^{3}-2 x^{3}-x+3 x \in[0,1]\\ &f(a)=f(0)=3\\ &f(b)=f(1)=(1)^{3}-2(1)^{3}-1+3\\ &=1-2-1+3\\ &=4-3\\ &=1\\ &\text { Now, let us show that there exist } c \in(0,1) \text { such that }\\ &f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \end{aligned}$
$f(x)=x^{3}-2 x^{2}-x+3$$
On differentiating above with respect to x, we get
$f^{\prime}(x)=3 x^{2}-4 x-1$$
Put x=c in above equation, we get
$f^{\prime}(c)=3 c^{2}-4 c-1 \ldots(i)$$
By Mean Value Theorem,
$\\f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$ \\$f^{\prime}(c)=\frac{f(1)-f(0)}{1-0}$ \\$\Rightarrow f^{\prime}(c)=\frac{1-3}{1-0}$ \\$\Rightarrow f^{\prime}(c)=-\frac{2}{1}$$
$\begin{aligned} &\Rightarrow f^{\prime}(c)=-2\\ &\Rightarrow 3 c^{2}-4 c-1=-2[\text { from }(j)]\\ &\begin{array}{l} \Rightarrow 3 c^{2}-4 c-1+2=0 \\ \Rightarrow 3 c^{2}-4 c+1=0 \end{array}\\ &\text { On factorising, we get }\\ &\Rightarrow 3 c^{2}-3 c-c+1=0\\ &\begin{array}{l} \Rightarrow 3 c(c-1)-1(c-1)=0 \\ \Rightarrow(3 c-1)(c-1)=0 \\ \Rightarrow(3 c-1)=0 \text { or }(c-1)=0 \end{array} \end{aligned}$
$\begin{aligned} &\Rightarrow c=\frac{1}{3} \text { or } c=1\\ &\text { So, value of }\\ &c=\frac{1}{3} \in(0,1) \end{aligned}$
Thus, Mean Value Theorem is verified.

Question:75

Verify mean value theorem for each of the functions given
$f(x) = sinx - sin2x$ in $[0,\pi]$

Answer:

Given:
f(x) = sinx - sin2x in [0,π]
Now, we have to show that f(x) verify the Mean Value Theorem
First of all, Conditions of Mean Value theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
If both the conditions are satisfied, then there exist some ‘c’ in (a,b) such that
$f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$
Condition 1:
f(x) = sinx - sin 2x
Since, f(x) is a trigonometric function and we know that, sine function are defined for all real values and are continuous for all x ∈ R
⇒ f(x) = sinx - sin 2x is continuous at x ∈ [0,π]
Hence, condition 1 is satisfied.
Condition 2:
f(x) = sinx - sin 2x
f’(x) = cosx - 2 cos2x
$\left[\because \frac{d}{d x} \sin x=\cos x\right]$
⇒ f(x) is differentiable at [0,π]
Hence, condition 2 is satisfied.
Thus, Mean Value Theorem is applicable to the given function
Now,
$\\ f(x)=\sin x-\sin 2 x x \in[0, \pi] \\ f(a)=f(0)=\sin (0)-\sin 2(0)=0\left[\because \sin \left(0^{\circ}\right)=0\right] \\ f(b)=f(\pi)=\sin (\pi)-\sin 2(\pi)=0-0=0 \\ {[\because \sin \pi=0 \& \sin 2 \pi=0]}$
Now, let us show that there exist c ∈ (0,1) such that
$\\f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \\f(x)=\sin x-\sin 2 x$
On differentiating above with respect to x, we get
$f^{\prime}(x)=\cos x-2 \cos 2 x$$
Put x=c in above equation, we get
f’(c) = cos(c) - 2cos2c …(i)
By Mean Value Theorem,

$\begin{aligned} &f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\\ &\mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(\mathrm{\pi})-\mathrm{f}(0)}{\pi-0}\\ &\Rightarrow \cos c-2 \cos 2 c=\frac{0-0}{\pi-0}\\ &\Rightarrow \cos c-2 \cos 2 c=0\\ &\Rightarrow \cos c-2\left(2 \cos ^{2} c-1\right)=0\left[\because \cos 2 x=2 \cos ^{2} x-1\right]\\ &\Rightarrow \cos c-4 \cos ^{2} c+2=0\\ &\Rightarrow 4 \cos ^{2} c-\cos c-2=0\\ &\text { Now, let } \cos c=x\\ &\Rightarrow 4 x^{2}-x-2=0 \end{aligned}$
Now, to find the factors of the above equation, we use
$\\ x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ \Rightarrow x=\frac{-(-1) \pm \sqrt{(-1)^{2}-4 \times 4 \times(-2)}}{2 \times 4} \\ \Rightarrow x=\frac{1 \pm \sqrt{1+32}}{8} \\ \Rightarrow x=\frac{1 \pm \sqrt{33}}{8}$
$\begin{aligned} &\Rightarrow \cos c=\frac{1 \pm \sqrt{33}}{8} \text { [above we let } \left.\cos c=x\right]\\ &\Rightarrow c=\cos ^{-1}\left(\frac{1 \pm \sqrt{33}}{8}\right)\\ &\text { So, value of }\\ &\mathrm{c}=\cos ^{-1}\left(\frac{1 \pm \sqrt{33}}{8}\right) \in(0, \pi) \end{aligned}$
Thus, Mean Value Theorem is verified.

Question:76

Verify mean value theorem for each of the functions given
$f(x)=\sqrt{25-x^2}$ in [1,5]

Answer:

Given: $f(x)=\sqrt{25-x^2}$
Now, we have to show that f(x) verify the Mean Value Theorem
First of all, Conditions of Mean Value theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
If both the conditions are satisfied, then there exist some ‘c’ in (a,b) such that

$f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$$
Condition 1:
Firstly, we have to show that f(x) is continuous.
Here, f(x) is continuous because f(x) has a unique value for each x ∈ [1,5]
Condition 2:
Now, we have to show that f(x) is differentiable
$\\ f(x)=\sqrt{25-x^{2}} \\ \Rightarrow f(x)=\left(25-x^{2}\right)^{\frac{1}{2}} \\ f^{\prime}(x)=\frac{1}{2}\left(25-x^{2}\right)^{-\frac{1}{2}} \times(-2 x) \\ \Rightarrow f^{\prime}(x)=\frac{-x}{\sqrt{25-x^{2}}}$
∴ f’(x) exists for all x ∈ (1,5)
So, f(x) is differentiable on (1,5)
Hence, Condition 2 is satisfied.
Thus, mean value theorem is applicable to given function.
Now,
$f(x)=\sqrt{25-x^{2}}$
Now, we will find f(a) and f(b)
$so,$ f(a)=f(1)$
$f(1)=\sqrt{25-(1)^{2}}=\sqrt{25-1}=\sqrt{24} $$$
and $f(b)=f(5)$$
$f(5)=\sqrt{25-(5)^{2}}=\sqrt{25-25}=0$$
Now, let us show that $c \in(1,5)$$ such that
$f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$$
$f(x)=\sqrt{25-x^{2}}$$
On differentiating above with respect to x, we get
$f^{\prime}(x)=\frac{-x}{\sqrt{25-x^{2}}}$$
Put x=c in above equation, we get
$f^{\prime}(c)=\frac{-c}{\sqrt{25-c^{2}}}$$
By Mean Value theorem,
$\\f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$ \\$\Rightarrow \frac{-c}{\sqrt{25-c^{2}}}=\frac{0-\sqrt{24}}{5-1}$ \\$\Rightarrow \frac{c}{\sqrt{25-c^{2}}}=\frac{\sqrt{24}}{4}$ \\$\Rightarrow 4 c=\sqrt{24} \times \sqrt{25-c^{2}}$$
$\begin{aligned} &\text { Squaring both sides, we get }\\ &\Rightarrow 16 c^{2}=24 \times\left(25-c^{2}\right)\\ &\Rightarrow 16 c^{2}=600-24 c^{2}\\ &\Rightarrow 24 \mathrm{c}^{2}+16 \mathrm{c}^{2}=600\\ &\Rightarrow 40 c^{2}=600\\ &\Rightarrow c^{-}=15\\ &\Rightarrow c=\sqrt{1} 5 \in(,5) \end{aligned}$
Hence, Mean Value Theorem is verified.

Question:77

Find a point on the curve $y = (x - 3)^2$ ,where the tangent is parallel to the chord joining the points (3, 0) and (4, 1).

Answer:

Given: Equation of curve, $y = (x - 3)^2$
Firstly, we differentiate the above equation with respect to x, we get
$\begin{aligned} &\frac{d}{d x} y=\frac{d}{d x}(x-3)^{2}\\ &\begin{array}{l} \text { Chain Rule } \\ f(x)=g(h(x)) \\ f^{\prime}(x)=g^{\prime}(h(x)) h^{\prime}(x) \end{array} \end{aligned}$
$\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=2 \times(\mathrm{x}-3) \times(1-0) \quad$ [using chain rule $]$ \\$\Rightarrow \frac{d y}{d x}=2 x-6$$
Given tangent to the curve is parallel to the chord joining the points (3, Q) and (4,1)
i.e $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}_{1}-\mathrm{y}_{2}}{\mathrm{x}_{1}-\mathrm{x}_{2}}$$
$\\\Rightarrow 2 x-6=\frac{0-1}{3-4}$ \\$\Rightarrow 2 x-6=\frac{-1}{-1}$ \\$\Rightarrow 2 x-6=1$ \\$\Rightarrow 2 x=7$$
$\Rightarrow \mathrm{x}=\frac{7}{2}=3.5 \in(3,4)$$
Put $x=\frac{7}{2}$$ in $y=(x-3)^{2}$ , we have
$y=\left(\frac{7}{2}-3\right)^{2}=\left(\frac{7-6}{2}\right)^{2}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$$
Hence, the tangent to the curve is parallel to chord joining the points (3,0) and (4,1) at $\left(\frac{7}{2}, \frac{1}{4}\right)$$

Question:78

Using mean value theorem, prove that there is point on the curve $y = 2x^2 - 5x + 3$ between the points A(1, 0) and B(2, 1), where tangent is parallel to the chord AB. Also, find that point.

Answer:

Given: $y = 2x^2 - 5x + 3$ in [1,2]
Now, we have to show that f(x) verify the Mean Value Theorem
First of all, Conditions of Mean Value theorem are:
a) f(x) is continuous at (a,b)
b) f(x) is derivable at (a,b)
If both the conditions are satisfied, then there exist some ‘c’ in (a,b) such that
$f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$
Condition 1:
$y=2 x^{2}-5 x+3$
since, f(x) is a polynomial and we know that, every polynomial function is continuous for all $x \in R$
$\Rightarrow y=2 x^{2}-5 x+3$ is continuous at $x \in[1,2]$
Hence, condition 1 is satisfied.
Condition 2
$y=2 x^{2}-5 x+3$$
since, f(x) is a polynomial and every polynomial function is differentiable for all $x \in R$
$y^{\prime}=4 x-5$$
$\Rightarrow \mathrm{y}=2 \mathrm{x}^{3}-5 \mathrm{x}+3$$ is differentiable at [1, 2]
Hence, condition 2 is satisfied.
Thus, Mean Value Theorem is applicable to the given function.
Now,
$\\f(x)=y=2 x^{2}-5 x+3 x \in[1,2]$ \\$f(a)=f(1)=2(1)^{2}-5(1)+3=2-5+3=0$$
$f(b)=f(2)=2(2)^{2}-5(2)+3=8-10+3=1$$
Then, there exist $c \in(0,1)$ such that
$\mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})}{\mathrm{b}-\mathrm{a}}$$
Put x=c in equation, we get
$y^{\prime}=4 c-5 \ldots(i)$$
By Mean Value Theorem,
$\\\mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})}{\mathrm{b}-\mathrm{a}}$ \\$4 c-5=\frac{f(2)-f(1)}{2-1}$ \\$\Rightarrow 4 c-5=\frac{1-0}{1}$ \\$\Rightarrow 4 c-5=1$ \\$\Rightarrow 4 c=6$ \\$\Rightarrow c=\frac{6}{4}=\frac{3}{2}$$
So, value of $c=\frac{3}{2} \in(1,2)$$
$\Rightarrow x=\frac{3}{2}$$
Thus, Mean Value Theorem is verified.
Put $x=\frac{3}{2}$$ in given equation $y=2 x^{2}-5 x+3,$$ we have
$\begin{aligned} &y=2\left(\frac{3}{2}\right)^{2}-5\left(\frac{3}{2}\right)+3\\ &\Rightarrow y=\frac{9}{2}-\frac{15}{2}+3\\ &\Rightarrow y=\frac{9-15+6}{2}\\ &\Rightarrow y=0\\ &\text { Hence, the tangent to the curve is parallel to the chord AB at }\\ &\left(\frac{3}{2}, 0\right) \end{aligned}$

Question:79

Find the values of p and q so that
$f(x)=\left\{\begin{array}{cl} x^{2}+3 x+p, & \text { if } x \leq 1 \\ q x+2, & \text { if } x>1 \end{array}\right.$
Is differentiable at x = 1.

Answer:

Given that,
$f(x)=\left\{\begin{array}{cl} x^{2}+3 x+p, & \text { if } x \leq 1 \\ q x+2, & \text { if } x>1 \end{array}\right.$
Is differentiable at x = 1.
We know that, f(x) is differentiable at x =1 ⇔ Lf’(1) = Rf’(1).

$\\ {L f^{\prime}(1)}=\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1} \\ =\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{(1-h)-1} \\ =\lim _{h \rightarrow 0} \frac{\left[\left\{(1-h)^{2}+3(1-h)+p\right]-(1+3+p)\right]}{(1-h)-1} \quad\left(\because f(x)=x^{2}+3 x+p, \text { if } x \leq 1\right) \\ =\lim _{h \rightarrow 0} \frac{\left[\left(1+h^{2}-2 h+3-3 h+p\right)-(4+p)\right]}{-h} \\ =h m \frac{\left[h^{2}-5 h+p+4-4-p\right]}{-h} \\ = \quad \lim _{h \rightarrow 0} \frac{\left[h^{2}-5 h\right]}{-h}=\lim _{h \rightarrow 0} \frac{h(h-5)}{-h} \\ =\lim _{h \rightarrow 0} h(5-h)$
$\\ =5 \\ \lim _{\operatorname{Rf}^{\prime}(1)}=\lim _{x \rightarrow 1} \frac{f(x)-f(1)}{x-1} \\ =\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{(1+h)-1} \\ =\lim _{h \rightarrow 0} \frac{[(q(1+h)+2)-(q+2)]}{(1+h)-1} \quad(v f(x)=q x+2, \text { if } x>1) \\ =\lim _{h \rightarrow 0} \frac{[(q+q h+2)-(q+2)]}{h} \\ =\lim _{h \rightarrow 0} \frac{[q+q h+2-q-2]}{h} \\ =q$
Since, Lf’(1) = Rf’(1)
∴ 5 = q (i)
Now, we know that if a function is differentiable at a point,it is necessarily continuous at that point.
⇒ f(x) is continuous at x = 1.
⇒ f(1-) = f(1+) = f(1)
⇒ 1+3+p = q+2 = 1+3+p
⇒ p-q = 2-4 = -2
⇒ q-p = 2
Now substituting the value of ‘q’ from (i), we get
⇒ 5-p = 2
⇒ p = 3
∴ p = 3 and q = 5

Question:80

A.If $x^m.y^n = (x+y)^{m+n}$ prove that $\frac{dy}{dx}= \frac{y}{x}$
B. If $x^m.y^n = (x+y)^{m+n}$ prove that $\frac{d^2y}{dx^2}=0$

Answer:

A.
We have,
$x^m.y^n = (x+y)^{m+n}$
Taking log on both sides, we get
Taking log on both sides, we get
$\log \left(x^{m} y^{n}\right)=\log (x+y)^{m+n}$ \\$\Rightarrow m \log x+n \log y=(m+n) \log (x+y)$$
Differentiating both sides w.r.t x, we get
$\\ \mathrm{m} \cdot \frac{1}{\mathrm{x}}+\mathrm{n} \cdot \frac{1}{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}=(\mathrm{m}+\mathrm{n}) \frac{1}{\mathrm{x}+\mathrm{y}} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+\mathrm{y})$ \\$=\frac{m}{x}+\frac{n}{y} \frac{d y}{d x}=\frac{m+n}{x+y} \cdot\left(1+\frac{d y}{d x}\right)$$
$\\ \Rightarrow\left(\frac{n}{y}-\frac{m+n}{x+y}\right) \frac{d y}{d x}=\frac{m+n}{x+y}-\frac{m}{x} \\ \Rightarrow\left(\frac{n x+n y-m y-n y}{y(x+y)}\right) \frac{d y}{d x}=\left(\frac{m x+n x-m x-m y}{x(x+y)}\right) \\ \Rightarrow \quad\left(\frac{n x-m y}{y(x+y)}\right) \frac{d y}{d x}=\left(\frac{n x-m x}{x(x+y)}\right) \\ \Rightarrow \frac{d y}{d x}=\left(\frac{n x-m x}{x(x+y)}\right)\left(\frac{y(x+y)}{n x-m y}\right) \\ \Rightarrow \frac{d y}{d x}=\frac{y}{x}$
Hence proved.
B.
We have,
$\begin{aligned} &\frac{d y}{d x}=\frac{y}{x}\\ &\text { Differentiating both sides w.r.t } x, \text { we get }\\ &\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y} \cdot 1}{\mathrm{x}^{2}}\\ &\frac{\mathrm{d}^{2} y}{\mathrm{dx}^{2}}=\frac{\mathrm{x} \frac{\mathrm{y}}{\mathrm{x}}-\mathrm{y} \cdot \mathrm{.}}{\mathrm{x}^{2}}\left(\because \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{\mathrm{x}}\right)\\ &\frac{d^{2} y}{d x^{2}}=\frac{y-y}{x^{2}}=0\\ &\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=0 \end{aligned}$
Hence Proved

Question:81

If x = sint and y = sin pt, prove that

$(1-x^2)$ $\frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}+p^{2} y=0$

Answer:

We have,
$\begin{aligned} &x=\sin t \text { and } y=\sin p t\\ &\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\cos \mathrm{t} \text { and } \frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{p} \cos \mathrm{pt}\\ &\therefore \frac{\mathrm{d} y}{\mathrm{dx}}=\frac{\mathrm{dy} / \mathrm{dt}}{\mathrm{dx} / \mathrm{dt}}=\frac{\mathrm{p} \cdot \cos \mathrm{pt}}{\cos \mathrm{t}}\\ &\Rightarrow \frac{d y}{d x}=\frac{p \cdot \cos p t}{\cos t}\\ &\text { Differentiating both sides w.r.t } x, \text { we get }\\ &\frac{\mathrm{d}^{2 y}}{\mathrm{dx}^{2}}=\frac{\operatorname{cost} \frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{p} \cdot \cos \mathrm{pt}) \frac{\mathrm{dt}}{\mathrm{dx}}-\mathrm{p} \cdot \cos \mathrm{pt} \frac{\mathrm{d}}{\mathrm{dt}} \operatorname{cost} \frac{\mathrm{dt}}{\mathrm{dx}}}{\cos ^{2} \mathrm{t}}\\ \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{d^{2 y}}{d x^{2}}=\frac{\left(-p^{2} \sin p t \cos t+p \cdot s i n t \cos p t\right) \frac{1}{\operatorname{cost}}}{\cos ^{2} t}\left(\because \frac{d x}{d t}=\cos t \Rightarrow \frac{d t}{d x}=\frac{1}{\operatorname{cost}}\right)\\ &\Rightarrow \frac{\mathrm{d}^{2 y}}{\mathrm{dx}^{2}}=\frac{-\mathrm{p}^{2} \sin \mathrm{pt} \cdot \cos t+\mathrm{p} \cdot \operatorname{sint} \cos \mathrm{pt}}{\left(1-\sin ^{2} \mathrm{t}\right) \operatorname{cost}}\\ &\Rightarrow\left(1-\sin ^{2} t\right) \frac{d^{2 y}}{d x^{2}}=\frac{-p^{2} \sin p t \cdot \cos t+p \sin t \cos p t}{\operatorname{cost}}\\ &\Rightarrow\left(1-\sin ^{2} t\right) \frac{d^{2 y}}{d x^{2}}=\frac{-p^{2} \sin p t \cdot \cos t}{\cos t}+\frac{p \cdot s i n t \cos p t}{\operatorname{cost}}\\ \end{aligned}$
$\\ \Rightarrow\left(1-x^{2}\right) \frac{d^{2 y}}{d x^{2}}=-p^{2} y+x \frac{d y}{d x}\left(\because x=\operatorname{sint}, y=\sin p t \text { and } \frac{d y}{d x}=\frac{p \cdot \cos p t}{\cos t}\right)\\ \\\Rightarrow\left(1-x^{2}\right) \frac{d^{2 y}}{d x^{2}}-x \frac{d y}{d x}+p^{2} y=0$
Hence proved.

Question:82

Find $\frac{d y}{d x}, \text { if } y=x^{\tan x}+\sqrt{\frac{x^{2}+1}{2}}$

Answer:

We have,
$y=x^{\tan x}+\sqrt{\frac{x^{2}+1}{2}}$$
Putting $x^{ \tan x}=u$ and $\sqrt{\frac{x^{2}+1}{2}}=v$$
$u=x ^{\tan x}$$
Taking log on both sides, we get
$\log u=\tan x \log x$$
Differentiating w.r.t x, we get
$\Rightarrow \frac{1}{u} \frac{d u}{d x}=\tan x \cdot \frac{1}{x}+\log x \cdot \sec ^{2} x$$
$\\\Rightarrow \frac{d u}{d x}=u\left(\tan x \cdot \frac{1}{x}+\log x \cdot \sec ^{2} x\right) \\ \Rightarrow \frac{d u}{d x}=x^{\tan x}\left(\tan x \cdot \frac{1}{x}+\log x \cdot \sec ^{2} x\right)$
Now, $\mathrm{v}=\sqrt{\frac{\mathrm{x}^{2}+1}{2}}$$
$\Rightarrow v=\left(\frac{x^{2}+1}{2}\right)^{1 / 2}$$
Differentiating w.r.t x, we get
$\Rightarrow \frac{d v}{d x}=\frac{1}{2}\left(\frac{x^{2}+1}{2}\right)^{-1 / 2} \cdot \frac{2 x}{2}$ \\$\Rightarrow \frac{d v}{d x}=\frac{x}{2}\left(\frac{2}{x^{2}+1}\right)^{1 / 2}=\frac{x}{2} \sqrt{\frac{2}{x^{2}+1}}$$
Now, y=u+v
$\\\Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$ \\$\Rightarrow \frac{d y}{d x}=x^{\tan x}\left(\tan x \cdot \frac{1}{x}+\log x \cdot \sec ^{2} x\right)+\frac{x}{2} \sqrt{\frac{2}{x^{2}+1}}$ \\$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x}^{\operatorname{tanx}}\left(\frac{\tan \mathrm{x}}{\mathrm{x}}+\log \mathrm{x} \cdot \sec ^{2} \mathrm{x}\right)+\frac{\mathrm{x}}{\sqrt{2\left(\mathrm{x}^{2}+1\right)}}$$

Question:83

If f(x) = 2x and g(x) = $\frac{x^2}{2}+1$ then which of the following can be a discontinuous function.
A. f(x) + g(x)
B. f(x) - g(x)
C. f(x) . g(x)
D. $\frac{g(x)}{f(x)}$

Answer:

We know that if two functions f(x) and g(x) are continuous then {f(x) +g(x)},{f(x)-g(x)}, {f(x).g(x)} and $\left\{\left(\frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}\right), \text { when } \mathrm{g}(\mathrm{x}) \neq 0\right\}$ are continuous.
Since, f(x) = 2x and $g(x)=$ $\frac{x^2}{2}+1$ are polynomial functions, they are continuous everywhere.
⇒ {f(x) +g(x)},{f(x)-g(x)}, {f(x).g(x)} are continuous functions.
for, $\frac{g(x)}{f(x)}=\frac{\frac{x^{2}}{2}+1}{2 x}=\frac{x^{2}+2}{4 x}$
now, f(x) = 0
⇒ 4x = 0
⇒ x = 0
∴ $\frac{g(x)}{f(x)}$ is discontinuous at x=0.

Question:90

Let $f(x) = |sinx|$ . Then
A. f is everywhere differentiable
B. f is everywhere continuous but not differentiable $x = n\pi,n\in Z$
C. f is everywhere continuous but not differentiable at $x = (2n+1)\frac{\pi}{2},n\in Z$
D. None of these

Answer:

B)
Given that, $f(x) = |sinx|$
Let g(x) = sinx and h(x) = |x|
Then, f(x) = hog(x)
We know that, modulus function and sine function are continuous everywhere.
Since, composition of two continuous functions is a continuous function.
Hence, f(x) = hog(x) is continuous everywhere.
Now, v(x)=|x| is not differentiable at x=0.
$\begin{aligned} &LHD=\lim _{0^{-}} \frac{\mathrm{v}(\mathrm{x})-\mathrm{v}(0)}{\mathrm{x}-0}\\ &\lim _{h \rightarrow 0} \frac{v(0-h)-v(0)}{(0-h)-0}\\ &\lim _{=h \rightarrow 0} \frac{|0-h|-|0|}{-h} \quad(\because v(x)=|x|)\\ &\lim _{h \rightarrow 0} \frac{|-h|}{-h}\\ &=\lim _{h \rightarrow 0} \frac{h}{-h}\\ &=\lim _{h \rightarrow 0}-1=-1\\ \end{aligned}$
$\\RHD = \lim _{h \rightarrow 0} \frac{v(0+h)-v(0)}{(0+h)-0}$ \\$=\lim _{h \rightarrow 0} \frac{|0+h|-|0|}{h} \quad(\because v(x)=|x|)$ \\$\lim _{h \rightarrow 0} \frac{|h|}{h}$ \\$=\lim _{h \rightarrow 0} \frac{h}{h}$ \\$=\lim _{h \rightarrow 0} 1=1$ \\$\Rightarrow \mathrm{LY}^{\prime}(0) \neq \mathrm{R} \mathrm{C}^{\prime}(0)$ \\$\Rightarrow|\mathrm{x}|$ is not differentiable at $\mathrm{x}=0 .$ \\$\Rightarrow \mathrm{h}(\mathrm{x})$ is not differentiable at $\mathrm{x}=0 .$$
So, f(x) is not differentiable where $\sin x=0$
We know that $\sin x=0$ at $x=\mathrm{n} \pi, n \in Z$$
Hence, f(x) is everywhere continuous but not differentiable $x=\mathrm{n} \pi, \mathrm{n} \in \mathrm{Z}$.$

Question:91

If $y=\log \left(\frac{1-x^{2}}{1+x^{2}}\right)$ then $\frac{dy}{dx}$ is equal to

$\\A.\frac{4 x^{3}}{1-x^{2}}$ \\B.$\frac{-4 x}{1-x^{4}}$ \\C. $\frac{1}{4-\mathrm{x}^{4}}$ \\D. $\frac{-4 x^{3}}{1-x^{4}}$$

Answer:

$\begin{aligned} &\text { We have, }\\ &y=\log \frac{1-x^{2}}{1+x^{2}}\\ &\Rightarrow \mathrm{y}=\log \left(1-\mathrm{x}^{2}\right)-\log \left(1+\mathrm{x}^{2}\right)\\ &\Rightarrow \frac{d y}{d x}=\frac{1}{1-x^{2}} \frac{d}{d x}\left(1-x^{2}\right)-\frac{1}{1+x^{2}} \frac{d}{d x}\left(1+x^{2}\right)\\ &\Rightarrow \frac{d y}{d x}=\frac{1}{1-x^{2}} \cdot(-2 x)-\frac{1}{1+x^{2}} \cdot(2 x)\\ &\Rightarrow \frac{d y}{d x}=\frac{-2 x}{1-x^{2}}-\frac{2 x}{1+x^{2}}\\ &\frac{d y}{d x}=\frac{-2 x\left(1+x^{2}\right)-2 x\left(1-x^{2}\right)}{\left(1-x^{2}\right)\left(1+x^{2}\right)}\\ &\frac{d y}{d x}=\frac{-2 x-2 x^{3}-2 x+2 x^{3}}{\left(1-x^{4}\right)}\\ &\frac{d y}{d x}=\frac{-4 x}{\left(1-x^{4}\right)} \end{aligned}$

Question:92

If $y=\sqrt{\sin x+y}$ then $\frac{d y}{d x}$ is equal to

$\\A. \ \frac{\cos \mathrm{x}}{2 \mathrm{y}-1}$ \\\\B. $\frac{\cos x}{1-2 y}$ \\\\C. $\frac{\sin \mathrm{x}}{1-2 \mathrm{y}}$\\\\ D. $\frac{\sin \mathrm{x}}{2 \mathrm{y}-1}$$

Answer:

A)
$\\$We have$, y=\sqrt{\sin x+y}$ \\$y^{2}=\sin x+y$ \\Differentiating both sides w.r.t $x,$ we get \\$2 y=\cos x+\frac{d y}{d x}$ \\$2 y-\frac{d y}{d x}=\cos x$ \\$(2 y-1) \frac{d y}{d x}=\cos x$ \\$\frac{d y}{d x}=\frac{\cos x}{(2 y-1)}$$

Question:93

The derivative of $\cos ^{-1}\left(2 x^{2}-1\right)$ w.r.t $\cos ^{-1} x$ is

$\\A \ 2 \\B \ \frac{-1}{2 \sqrt{1-\mathrm{x}^{2}}} \\\\C \ \frac{2}{x} \\ \\D. 1-x^{2}$

Answer:

A)
$\\ \text { Let } u=\cos ^{-1}\left(2 x^{2}-1\right) \text { and } v=\cos ^{-1} x \\ \text { Now, } u=\cos ^{-1}\left(2 x^{2}-1\right) \\ u=\cos ^{-1}\left(2 \cos ^{2} v-1\right)\left[\cdot v=\cos ^{-1} x\rightarrow \cos v=x\right] \\ u=\cos ^{-1}(\cos 2 v)\left[\because 2 \cos ^{2} x-1=\cos 2 x\right] \\ \Rightarrow u=2 v \\ \quad \frac{d u}{d v}=2$

Question:94

If $x = t^2, y = t^3$ , then $\frac{d^2y}{dx^2}$ is
A. $\frac{3}{2}$
B. $\frac{3}{4t}$
C. $\frac{3}{2t}$
D. $\frac{3}{4}$

Answer:

B)
$\begin{aligned} &\text { Given that, } x=t ^2, y=t^{3}\\ &\Rightarrow \frac{d x}{d t}=2 t \text { and } \frac{d y}{d t}=3 t^{2}\\ & \frac{d y}{d x}=\frac{d y / d t}{d x / d t}\\ &\frac{d y}{d x}=\frac{3 t^{2}}{2 t}=\frac{3 t}{2}\\ &\frac{d^{2} y}{d x^{2}}=\frac{3}{2} \frac{d t}{d x}\\ &\frac{d^{2} y}{d x^{2}}=\frac{3}{2} \cdot \frac{1}{2 t}\left(\because \frac{d x}{d t}=2 t \Rightarrow \frac{d t}{d x}=\frac{1}{2 t}\right)\\ &\frac{d^{2} y}{d x^{2}}=\frac{3}{4 t} \end{aligned}$

Question:95

The value of c in Rolle’s theorem for the function $f(x) = x^3 - 3x$ in the interval $[0,\sqrt 3]$ is
A. 1
B.-1
C. $\frac{3}{2}$
D. $\frac{1}{3}$

Answer:

A)
$\\$Rolle's Theorem states that, Let $f_{i}[a, b] \rightarrow R$ be continuous on $[a, b]$ and differentiable on (a, b), such that $f(a)=f(b),$ where a and $b$ are some real numbers. \\Then there exists some $c$ in $(a, b)$ such that $f^{\prime}(c)=0$ We have, $f(x)=x^{3}-3 x$ since, \\$f(x)$ is a polynomial function it is continuous on $[0, \sqrt{3}]_{\text {and }}$ differentiable on $(0, \sqrt{3})\\$ $f(0)=f(\sqrt{3})=0$ $\\f^{\prime}(c)=0$ $-3 c^{2}-3=0\left[\because f^{\prime}(x)=3 x^{2}-3\right]$ \\$\Rightarrow \mathrm{c}^{2}=1$ \\$\Rightarrow c=\pm 1$ $\\c=1$ \in $(0, \sqrt{3})$$

Question:96

For the function $\mathrm{f}(\mathrm{x})=\mathrm{x}+\frac{1}{\mathrm{x}}, \mathrm{x} \in[1,3]$ the value of c for mean value theorem is
A. 1

B. $\sqrt3$
C. 2
D. None of these

Answer:

B)
Mean Value Theorem states that, Let f : [a, b] → R be a continuous function on [a, b] and differentiable on (a, b). Then there exists some c in (a, b) such that
$\\ f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \\ \\\text { We have, } f(x)=x+\frac{1}{x}$
Since, f(x) is a polynomial function it is continuous on [1,3] and differentiable on (1,3).
Now, as per Mean value Theorem, there exists at least one c ∈ (1,3), such that
$\\ f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \\ \Rightarrow 1-\frac{1}{c^{2}}=\frac{\left(3+\frac{1}{3}\right)-(1+1)}{3-1}\left[\because f^{\prime}(x)=1+\frac{1}{x^{2}}\right] \\ \Rightarrow \quad \frac{c^{2}-1}{c^{2}}=\frac{\left(\frac{10}{3}\right)-(2)}{2} \\ \Rightarrow \quad \frac{c^{2}-1}{c^{2}}=\frac{\left(\frac{10}{3}\right)-(2)}{2}=\frac{\frac{10-6}{3}}{2}=\frac{2}{3} \\ \Rightarrow 3\left(c^{2}-1\right)=2 c^{2} \\ \Rightarrow 3 c^{2}-2 c^{2}=3 \\ \Rightarrow c^{2}=3 \\ \Rightarrow c=\pm \sqrt{3} \\ \Rightarrow c=\sqrt{3} \in(1,3)$

Question:97

Fill in the blanks in each of the
An example of a function which is continuous everywhere but fails to be differentiable exactly at two points is _______.

Answer:

Consider, f(x) = |x-1| + |x-2|
Let’s discuss the continuity of f(x).
We have, f(x) = |x-1| + |x-2|
$\begin{array}{l} f(x)=\left\{\begin{array}{l} -(x-1)-(x-2), \text { if } x<1 \\ (x-1)-(x-2), \text { if } 1 \leq x<2 \\ (x-1)+(x-2), \text { if } x \geq 2 \end{array}\right. \\ f(x)=\left\{\begin{array}{l} -2 x+3, \text { if } x<1 \\ 1, \text { if } 1 \leq x<2 \\ 2 x-3, \text { if } x \geq 2 \end{array}\right. \end{array}$
When x<1, we have f(x) = -2x+3, which is a polynomial function and polynomial function is continuous everywhere.
When 1≤x<2, we have f(x) = 1, which is a constant function and constant function is continuous everywhere.
When x≥2, we have f(x) = 2x-3, which is a polynomial function and polynomial function is continuous everywhere.
Hence, f(x) = |x-1| + |x-2| is continuous everywhere.
Let’s discuss the differentiability of f(x) at x=1 and x=2.
We have
$\begin{array}{l} f(x)=\left\{\begin{array}{l} -(x-1)-(x-2), \text { if } x<1 \\ (x-1)-(x-2), \text { if } 1 \leq x<2 \\ (x-1)+(x-2), \text { if } x \geq 2 \end{array}\right. \\ f(x)=\left\{\begin{array}{l} -2 x+3, \text { if } x<1 \\ 1, \text { if } 1 \leq x<2 \\ 2 x-3, \text { if } x \geq 2 \end{array}\right. \end{array}$
$\\\qquad LHD =\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1} \\ =\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{(1-h)-1} \\ =\lim _{h \rightarrow 0} \frac{[(-2(1-h)+3)-(-2+3)]}{(1-h)-1} \quad(\because f(x)=-2 x+3, \text { if } x<1)$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{[-2+2 h+3-1]}{-h}\\ &\lim _{=h \rightarrow 0} \frac{[2 h]}{-h}=\lim _{h \rightarrow 0} 2=2\\ &RLD =\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}\\ &=\lim _{x \rightarrow 1^{+}} \frac{1-1}{1-1}(\because f(x)=1, \text { if } 1 \leq x<2)\\ &=0\\ &\Rightarrow\left\lfloor f^{\prime}(1) \neq \operatorname{Rf}^{\prime}(1)\right.\\ &\Rightarrow f(x) \text { is not differentiable at } x=1 \text { . }\\ &L{ f^{\prime}(2)}=\lim _{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2} \end{aligned}$
$\begin{aligned} &=\lim _{x \rightarrow 2} \frac{1-1}{2-2}(: f(x)=1, \text { if } 1 \leq x<2 \text { and } f(2)=2 \times 2-3=1)\\ &=0\\ &{R f^{\prime}(2)}=\lim_{x \rightarrow 2^{+} }\frac{f(x)-f(2)}{x-2}\\ &=\lim _{h \rightarrow 0} \frac{f(1+h)-f(2)}{(1+h)-2}\\ &=\lim _{h \rightarrow 0} \frac{[(2(1+h)-3)-(2 \times 2-3)]}{(1+h)-2} \quad(\because f(x)=2 x-3, \text { if } x \geq 2)\\ &=\lim _{h \rightarrow 0} \frac{[2+2 h-3-1]}{h-1}\\ &=\lim _{h \rightarrow 0} \frac{[2 h-2]}{h-1}=\lim _{h \rightarrow 0} \frac{2(h-1)}{h-1}=2\\ &\Rightarrow \operatorname{Lf}^{\prime}(2) \neq \mathrm{Rf}^{\prime}(2) \end{aligned}$
⇒ f(x) is not differentiable at x=2.
Thus, f(x) = |x-1| + |X-2| is continuous everywhere but fails to be differentiable exactly at two points x=1 and x=2.

Question:98

Fill in the blanks in each of the
Derivative of $x^2$ w.r.t. $x^3$ is _______.

Answer:

$\begin{aligned} &\text { Let } u=x^{2} \text { and } v=x^{2}\\ &\Rightarrow \frac{d u}{d x}=2 x \text { and } \frac{d v}{d x}=3 x^{2}\\ &\therefore \frac{\mathrm{du}}{\mathrm{dv}}=\frac{\mathrm{du} / \mathrm{dx}}{\mathrm{dv} / \mathrm{dx}}=\frac{2 \mathrm{x}}{3 \mathrm{x}^{2}}\\ &\Rightarrow \frac{d u}{d v}=\frac{2}{3 x}\\ &\text { Hence, Derivative of }\\ &x^{2} \text { w.r.t. } x^{3} \text { is } \frac{2}{3 x} \text { . } \end{aligned}$

Question:99

Fill in the blanks in each of the
If $f(x) = |cos x|$ , then ’ $f'(\frac{\pi}{4})$ = ______.

Answer:

$\begin{aligned} &\text { We have, } f(x)=|\cos x|\\ &\text { Foc. } 0<x<\frac{\pi}{2}, \cos x>0\\ &\begin{array}{l} \therefore f(x)=\cos x \\ \Rightarrow f^{\prime}(x)=-\sin x \\ \Rightarrow f^{\prime}\left(\frac{\pi}{4}\right)=-\sin \frac{\pi}{4}=-\frac{1}{\sqrt{2}} \end{array}\\ &\text { Hence, } \\&f^{\prime}\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}\\ \end{aligned}$

Question:100

Fill in the blanks in each of the
If $f(x) = |cosx - sinx|$ , then $f'\frac{\pi}{3}=$ ____.

Answer:

$\\ \text { We have, } f(x)=|\cos x-\sin x| \\ \qquad \begin{array}{l} \text { Foc } \frac{\pi}{4}<x<\frac{\pi}{2}, \sin x>\cos x \\ \therefore f(x)=\sin x-\cos x \end{array} \\ \Rightarrow f^{\prime}(x)=\cos x-(-\sin x)=\cos x+\sin x \\ \Rightarrow f^{\prime}\left(\frac{\pi}{3}\right)=\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{1+\sqrt{3}}{2} \\ \text { Hence, } f^{\prime}\left(\frac{\pi}{3}\right)=\frac{1+\sqrt{3}}{2}$

Question:101

Fill in the blanks in each of the
For the curve $\sqrt x + \sqrt y =1 , \frac{dy}{dx}\: \: at\: \: \left ( \frac{1}{4},\frac{1}{4} \right )$ is _______.

Answer:

$\\$We have, $\sqrt{x}+\sqrt{y}=1$ \\On differentiating both sides, we get \\$\Rightarrow \frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x}=0$ \\$\Rightarrow \frac{1}{2 \sqrt{y}} \frac{d y}{d x}=-\frac{1}{2 \sqrt{x}}$ \\$\Rightarrow \frac{d y}{d x}=-\frac{2 \sqrt{y}}{2 \sqrt{x}}=-\frac{\sqrt{y}}{\sqrt{x}}$ \\$\left(\frac{d y}{d x}\right)_{\frac{1}{4}, \frac{1}{4}}=-\frac{\sqrt{\frac{1}{4}}}{\sqrt{\frac{1}{4}}}=-\frac{\frac{1}{2}}{\frac{1}{2}}=-1$ \\Hence, $\left(\frac{d y}{d x}\right)_{\frac{1}{4} \frac{1}{4}}=-1$$

Question:102

State True or False for the statements
Rolle’s theorem is applicable for the function f(x) = |x - 1| in [0, 2].

Answer:

False
As per Rolle’s Theorem, Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b), such that f(a) = f(b), where a and b are some real numbers. Then there exists some c in (a, b) such that f’(c) = 0.
We have, f(x) = |x - 1| in [0, 2].
Since, polynomial and modulus functions are continuous everywhere f(x) is continuous
Now, x-1=0
⇒ x=1
We need to check if f(x) is differentiable at x=1 or not.
We have,
$\begin{aligned} &f(x)=\left\{\begin{array}{l} -(x-1), \text { if } x<1 \\ (x-1), \text { if } x>1 \end{array}\right.\\ &\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}\\ &\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{(1-h)-1}\\ &\lim _{h \rightarrow 0} \frac{[1-(1-h)-0]}{(1-h)-1} \quad(\because f(x)=1-x, \text { if } x<1)\\ &\lim _{n \rightarrow 0} \frac{[\mathrm{h}]}{-\mathrm{h}}\\ &\lim _{=h \rightarrow 0} \frac{[\mathrm{h}]}{-\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0}-1=-1\\ &\lim _{\mathrm{Rf}^{\prime}(1)}=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}\\ &\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{(1+h)-1}\\ &\lim _{h \rightarrow 0} \frac{[(1+h)-1-0]}{(1+h)-1}(\because f(x)=x-1, \text { if } 1<x)\\ &\lim _{=h \rightarrow 0} \frac{[\mathrm{h}]}{\mathrm{h}}=\lim _{h \rightarrow 0} 1=1 \end{aligned}$
⇒ Lf’(1) ≠ Rf’(1)
⇒ f(x) is not differentiable at x=1.
Hence, Rolle’s theorem is not applicable on f(x) since it is not differentiable at x=1 ∈ (0,2).

Question:103

State True or False for the statements
If f is continuous on its domain D, then |f| is also continuous on D.

Answer:

True.
Given that, f is continuous on its domain D.
Let a be an arbitrary real number in D. Then f is continuous at a.
$\lim _{x \rightarrow a} f(x)=f(a)$
Now,
$\lim _{x \rightarrow a}|f|(x)=\lim _{x \rightarrow a}|f(x)|_{[\because|f|(x)=\mid f(x) L]}$$
$\lim _{x \rightarrow a}|f|(x)=\left|\lim _{x \rightarrow a} f(x)\right| \\ \lim _{x \rightarrow a}|f|(x)=|f(a)|=|f|(a)$
If |f| is continuous at x=a.
since a is an arbitrary point in D. Therefore |f| is continuous in D.

Question:104

State True or False for the statements
The composition of two continuous functions is a continuous function.

Answer:

True.
Let f be a function defined by f(x) = |1-x + |x||.
Let g(x) = 1-x + |x| and h(x) = |x| be two functions defined on R.
Then,
hog(x) = h(g(x))
⇒ hog(x) = h(1-x + |x|)
⇒ hog(x) = |(1-x + |x|)|
⇒ hog(x) = f(x) ∀ x ∈ R.
Since (1-x) is a polynomial function and |X| is modulus function are continuous in R.
⇒ g(x) = 1-x + |x| is everywhere continuous.
⇒ h(x) = |x| is everywhere continuous.
Hence, f = hog is everywhere continuous.

Question:105

State True or False for the statements
Trigonometric and inverse-trigonometric functions are differentiable in their respective domain.

Answer:

True

Question:106

State True or False for the statements
If f .g is continuous at x = a, then f and g are separately continuous at x = a.

Answer:

False
Let $f(x)=x$ and $g(x)=\frac{1}{x}$$
$f(x) \cdot g(x)=x \cdot \frac{1}{x}=1,$$ which is a constant function and continuous everywhere.
But, $g(x)=\frac{1}{x}$$ is discontinuous at $$x=0$.$

Students can use NCERT Exemplar Class 12 Math solutions chapter 5 PDF download, prepared by experts, for better understanding of concepts and topics of probability. The topics and subtopics are mentioned below.

Subtopics of NCERT Exemplar Class 12 Math Solutions Chapter 5 Continuity and Differentiability

The sub-topics covered in this chapter are:

Introduction
Continuity
Algebra of continuous functions
Differentiability
Derivatives of composite functions
Derivatives of implicit functions
Derivatives of inverse trigonometric functions
Logarithmic and exponential functions
Logarithmic differentiation
Derivatives of functions in parametric forms
Second-order derivative
Mean value theorem

What will you learn from NCERT Exemplar Class 12 Math Solutions Chapter 5?

In Class 12 Math NCERT exemplar solutions chapter 5, students will learn about continuity, differential, limits and their properties in detail. Other than learning about continuity and discontinuity in detail, one will also cover Heine's definition, Cauchy's theorem, and algebra of functions. In a separate subtopic of differentiability, one will learn about differentiability concept, fundamentals, derivatives and its types, relations between differentiability and continuity. NCERT exemplar solutions for Class 12 Math chapter 5 will also cover implicit functions, composite functions, inverse trigonometric functions and its derivatives.

NCERT exemplar class 12 Math solutions chapter 5 also includes logarithmic and exponential functions, their concept and their differentiability. The topics also cover logarithmic differentiation in detail and all its fundamentals. One will also get a closer look at parametric forms of functions and their derivatives and how to solve the questions. Other than all this, one will learn two of the most major advanced calculus theorems, Mean Value Theorem and Rolle's Theorem.

We have a team of highly experienced teachers, who have solved the NCERT exemplar Class 12 Math solutions chapter 5 continuity and differentiability. Our teachers have solved the questions in the simplest way; so that every student can understand the answer in a single get-go. The answers are exhaustive, meaning every step is mentioned as per the exam requirement. No shortcuts and no tricks are used in solving the questions, as per the NCERT syllabus. One will learn some of the most crucial theorems of calculus mathematics. Better understanding will help in higher education and also in getting a clear idea about advanced calculus.

NCERT Exemplar Class 12 Maths Solutions

Chapter 1 Relations and Functions

Chapter 2 Inverse Trigonometric Functions

Chapter 3 Matrices

Chapter 4 Determinants

Chapter 6 Applications of Derivatives

Chapter 7 Integrals

Chapter 8 Applications of Integrals

Chapter 9 Differential Equations

Chapter 10 Vector Algebra

Chapter 11 Three dimensional Geometry

Chapter 12 Linear Programming

Chapter 13 Probability

Important Topics To Cover for NCERT Exemplar Class 12 Math Solutions Chapter 5 Continuity and Differentiability

Continuity and differentiability is a chapter, which every student should be well prepared for. NCERT exemplar Class 12 Math chapter 5 solutions not only helps in understanding calculus better but also helps in other subjects like physics. This chapter explains the base of the continuity function and differentiability function in detail.

In NCERT exemplar Class 12 Math solutions chapter 5, students will learn that Continuity is characteristic of the function. This characteristic shows that the function can have an unbroken and continuous wave. Differentiability, on the other hand, is a function, which shows that the derivative of the domain exists at every single point. Differentiability shows that there can be a slope to the graph at each point.

NCERT Exemplar Class 12 Solutions

NCERT Exemplar Class 12 Chemistry Solutions

NCERT Exemplar Class 12 Physics Solutions

NCERT Exemplar Class 12 Biology Solutions

Also, check NCERT Solutions for questions given in the book:

Chapter 1	Relations and Functions
Chapter 2	Inverse Trigonometric Functions
Chapter 3	Matrices
Chapter 4	Determinants
Chapter 5	Continuity and Differentiability
Chapter 6	Application of Derivatives
Chapter 7	Integrals
Chapter 8	Application of Integrals
Chapter 9	Differential Equations
Chapter 10	Vector Algebra
Chapter 11	Three Dimensional Geometry
Chapter 12	Linear Programming
Chapter 13	Probability

Must Read NCERT Solution subject wise

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Question (FAQs)

1. Can I download the solutions for this chapter?

Yes, you can download the NCERT exemplar Class 12 Maths chapter 5 solutions from the link given in the page.

2. Are these solutions helpful for board examinations?

Yes, the NCERT exemplar solutions for Class 12 Maths chapter 5 are prepared to help students prepare well for board exams.

3. Do I have to make notes of this chapter?

You can prepare notes for this chapter by highlighting or underlining the important points which will make it easier for you to read for quick revision.

4. How many questions are there in this chapter?

The chapter has only 1 exercise with 107 questions in total that are all problem based with variable weightage.

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i couldn't clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

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Upasana Barman 24 August,2022

i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

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Mayankjha.student2007 23 August,2022

if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects and we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

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Nuthalapati Vyshnavi 21 August,2022

I will have my boards in Feb 2023 (Class 12) and I still haven't completed 1 chapter in any subject (Physics, chemistry and maths) , My 11th concepts are not that great but its okish, I wanted to know that If I do start studying from 15th of August Can I get above 85?

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

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khushi.thakurbbk 07 September,2022

I want to take a admission in class 12 privately.. so can I???

Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

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Parvati Solanki 02 June,2022

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Central Board of Secondary Education Class 12th Examination

JEE Test Series

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Exemplar Class 12 Maths Solutions Chapter 5 Continuity and Differentiability

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NCERT Exemplar Class 12 Maths Solutions

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