NCERT Exemplar Class 12 Maths Solutions Chapter 5 Continuity and Differentiability

NCERT Exemplar Class 12 Maths Solutions Chapter 5 Continuity and Differentiability

Question 1

Examine the continuity of the function
$f(x) = x^3 + 2x - 1 $at x = 1$

Answer:

We have, $f(x)=x^3+2 x^2-1$

For continuity at $\mathrm{x}=1$

$\begin{aligned} & \therefore \text { R.H.L. }=\lim _{x \rightarrow 1^{+}} f(x) \\ & =\lim _{h \rightarrow 0} f(1+\mathrm{h}) \\ & =\lim _{\mathrm{h} \rightarrow 0}\left[(1+\mathrm{h})^3+2(1+\mathrm{h})^2-1\right]=2\end{aligned}$

And L.H.L. $=\lim _{x \rightarrow 1^{-}} \mathrm{f}(x)$

$\begin{aligned} & =\lim _{h \rightarrow 0} f(1-h) \\ & =\lim _{h \rightarrow 0}\left[(1-h)^3+2(1-h)^2-1\right]=2\end{aligned}$

Also $f(1)=1+2-1=2$

Thus $\lim _{x \rightarrow 1^{+}} \mathrm{f}(x)=\lim _{x \rightarrow 1^{-}} \mathrm{f}(x)=\mathrm{f}(1)$

Thus $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=1$

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 the 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 with 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 equations 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) $
$\begin{aligned} &\text { then, }\\ &\lim _{\mathrm{h} \mathrm{HL}=\mathrm{h} \rightarrow 0} \mathrm{f}(-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \frac{1-\cos 2(-\mathrm{h})}{(-\mathrm{h})^{2}} \underbrace{\{\text { using equation } 1\}}\\ &\text { As we know } \cos (-\theta)=\cos \theta\\ &\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0} \frac{1-\cos 2 \mathrm{~h}}{\mathrm{~h}^{2}}\\ &\because 1-\cos 2 x=2 \sin ^{2} x\\ &\therefore \mathrm{lim}_{\mathrm{h} \rightarrow 0} \frac{2 \sin ^{2} \mathrm{~h}}{\mathrm{~h}^{2}}=2 \lim _{\mathrm{h} \rightarrow 0}\left(\frac{\sin \mathrm{h}}{\mathrm{h}}\right)^{2} \end{aligned}$
This limit can be evaluated directly by putting the value of h because it is taking the 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 the sandwich theorem, we get -
$RHL = 2 \times 1^2 = 2...(3)$
And,
f (0) = 5 …(4)
From equations 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 the 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 with 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 equation 1$\} \ldots(4)$
From the above equations 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 the 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 with 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 equations 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:

We have, $\begin{cases}|x| \cos \frac{1}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0\end{cases}$

At $\mathrm{x}=0$

$\begin{aligned} & \text { L.H.L. }=\lim _{x \rightarrow 0^{-}}|x| \cos \frac{1}{x} \\ & =\lim _{\mathrm{h} \rightarrow 0}|0-\mathrm{h}| \cos \frac{1}{0-\mathrm{h}} \\ & =\lim _{\mathrm{h} \rightarrow 0} \mathrm{~h} \cos \frac{1}{\mathrm{~h}} \\ & =0 \times[\text { an oscillating number between }-1 \text { and } 1]=0\end{aligned}$
$\begin{aligned} & \text { R.H.L. }=\lim _{x \rightarrow 0^{+}}|x| \cos \frac{1}{x} \\ & =\lim _{\mathrm{h} \rightarrow 0}|0+\mathrm{h}| \cos \frac{1}{0+\mathrm{h}} \\ & =\lim _{\mathrm{h} \rightarrow 0} \mathrm{~h} \cos \frac{1}{\mathrm{~h}} \\ & =0 \times[\text { an oscillating number between }-1 \text { and } 1]=0\end{aligned}$
Also $\mathrm{f}(0)=0 \quad \ldots$. (Given)

Thus, L.H.L. $=$ R.H.L. $=f(0)$

So, $f(x)$ 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:

We have, $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}|x-\mathrm{a}| \sin \frac{1}{x-\mathrm{a}}, & \text { if } x \neq 0 \\ 0, & \text { if } x=\mathrm{a}\end{array}\right.$ at $\mathrm{x}=\mathrm{a}$ At $\mathrm{x}=\mathrm{a}$

$\begin{aligned} & \text { L.H.L. }=\lim _{x \rightarrow \mathrm{a}^{-}}|x-\mathrm{a}| \sin \frac{1}{x-\mathrm{a}} \\ & =\lim _{\mathrm{h} \rightarrow 0}|\mathrm{a}-\mathrm{h}-\mathrm{a}| \sin \left(\frac{1}{\mathrm{a}-\mathrm{h}-\mathrm{a}}\right) \\ & =\lim _{\mathrm{h} \rightarrow 0}-\mathrm{h} \sin \frac{1}{\mathrm{~h}} \\ & =0 \times[\text { an oscillating number between }-1 \text { and } 1]=0\end{aligned}$

$\begin{aligned} & \text { R.H.L. }=\lim _{x \rightarrow a^{+}}|x-a| \sin \left(\frac{1}{x-a}\right) \\ & =\lim _{\mathrm{h} \rightarrow 0}|\mathrm{a}+\mathrm{h}-\mathrm{a}| \sin \left(\frac{1}{\mathrm{a}+\mathrm{h}-\mathrm{a}}\right) \\ & =\lim _{\mathrm{h} \rightarrow 0} \mathrm{~h} \sin \frac{1}{\mathrm{~h}} \\ & =0 \times[\text { an oscillating number between }-1 \text { and } \mathrm{l}]=0\end{aligned}$

Also $f(a)=0 \quad \ldots($ Given $)$

Thus L.H.L. $=$ R.H.L. $=\mathrm{f}(\mathrm{a})$

So, $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=\mathrm{a}$.

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 the 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 equations 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:

We have, $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}\frac{x^2}{2}, & \text { if } 0 \leq x \leq 1 \\ 2 x^2-3 x+\frac{3}{2}, & \text { if } 1<x \leq 2\end{array}\right.$ at $\mathrm{x}=1$

At $\mathrm{x}=1$

L.H.L. $=\lim _{x \rightarrow 1^{-}} \frac{x^2}{2}$

$\begin{aligned} & =\lim _{\mathrm{h} \rightarrow 0} \frac{(1-\mathrm{h})^2}{2} \\ & =\lim _{\mathrm{h} \rightarrow 0} \frac{1+\mathrm{h}^2-2 \mathrm{~h}}{2} \\ & =\frac{1}{2} \\ & \text { R.H.L. }=\lim _{x \rightarrow 1^{+}}\left(2 x^2-3 x+\frac{3}{2}\right) \\ & =\lim _{\mathrm{h} \rightarrow 0}\left[2(1+\mathrm{h})^2-3(1+\mathrm{h})+\frac{3}{2}\right] \\ & =2-3+\frac{3}{2} \\ & =\frac{1}{2}\end{aligned}$

Also $f(1)=\frac{1^2}{2}=\frac{1}{2}$ Thus L.H.L. $=$ R.H.L. $=f(1)$

Hence, $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=1$.

Question 10

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

Answer:

We have, $\mathrm{f}(\mathrm{x})=|\mathrm{x}|+|\mathrm{x}-1|$ at $\mathrm{x}=1$ At $x=1$

$\begin{aligned} & \text { L.H.L. }=\lim _{x \rightarrow 1^{-}}[|x|+|x-1|] \\ & =\lim _{\mathrm{h}-? 0^{-}}[|1-\mathrm{h}|+|1-\mathrm{h}-1|] \\ & =1+0 \\ & =1\end{aligned}$

$\begin{aligned} & \text { And R.H.L. }=\lim _{x \rightarrow+}[|x|+x-1 \mid] \\ & =\lim _{\mathrm{h} \rightarrow 0}[|1+\mathrm{h}|+|1+\mathrm{h}-1|] \\ & =1+0 \\ & =1\end{aligned}$

Also $f(1)=|1|+|0|=1$

Thus, L.H.L. $=$ R.H.L $=f(1)$

Hence, $f(x)$ 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=\frac 72$

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:

We have, $\mathrm{f}(\mathrm{x})= \begin{cases}\frac{2^{x+2}-16}{4^x-16}, & \text { if } x \neq 2 \\ \mathrm{k}, & \text { if } x=2\end{cases}$ Since, $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=2$

$\begin{aligned} & \therefore \mathrm{f}(2)=\lim _{x \rightarrow 2} \mathrm{f}(x) \\ & \therefore \mathrm{k}=\lim _{x \rightarrow 2} \frac{2^{x+2}-16}{4^x-16} \\ & =\lim _{x \rightarrow 2} \frac{4\left(2^x-4\right)}{\left(2^x-4\right)\left(2^x+4\right)} \\ & =\lim _{x \rightarrow 2} \frac{4}{2^x+4} \\ & =\frac{4}{4+4} \\ & =\frac{1}{2}\end{aligned}$

Question 13

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

Answer:

Given,

$\mathrm{f}(\mathrm{x})= \begin{cases}\frac{\sqrt{1+\mathrm{k} x}-\sqrt{1-\mathrm{k} x}}{x}, & \text { if }-1 \leq x<0 \\ \frac{2 x+1}{x-1}, & \text { if } 0 \leq x \leq 1\end{cases}$

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 the 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)
$\begin{array}{l} \lim _{h \rightarrow 0} f(-h)=f(0) \\ \quad \lim _{h \rightarrow 0}\left\{\frac{1-\cos k(-h)}{(-h) \sin (-h)}\right\}=\frac{1}{2}\{u \text { ing equation } 1\} \end{array}$
$\begin{array}{l} \because \cos (-x)=\cos x \text { and } \sin (-x)=-\sin x \\ \quad \lim _{h \rightarrow 0}\left\{\frac{1-\cos k(h)}{(h) \sin (h)}\right\}=\frac{1}{2} \\ \text { Also, } 1-\cos x=2 \sin ^{2}(x / 2) \\ \quad \lim _{h \rightarrow 0}\left\{\frac{2 \sin ^{2}\left(\frac{k h}{2}\right)}{(h) \sin (h)}\right\}=\frac{1}{2} \end{array}$
This limit can be evaluated directly by putting the value of h because it is taking an indeterminate form (0/0)
Thus, we use the 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 the 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 of the choice of k.

Answer:

we have $\mathrm{f}(\mathrm{x})= \begin{cases}\frac{x}{|x|+2 x^2}, & x \neq 0 \\ \mathrm{k} & x=0\end{cases}$

At $x=0$

$\begin{aligned} & \text { L.H.L. }=\lim _{x \rightarrow 0^{+}} \frac{(0-\mathrm{h})}{|0-\mathrm{h}|+2(0-\mathrm{h})^2} \\ & =\lim _{\mathrm{h} \rightarrow 0} \frac{-\mathrm{h}}{\mathrm{h}+2 \mathrm{~h}^2} \\ & =\lim _{\mathrm{h} \rightarrow 0} \frac{-1}{1+2 \mathrm{~h}} \\ & =-1 \\ & \text { R.H.L. }=\lim _{x \rightarrow 0^{+}} \frac{x}{|x|+2 x^2} \\ & =\lim _{\mathrm{h} \rightarrow 0} \frac{0+\mathrm{h}}{|0+\mathrm{h}|+2(0+\mathrm{h})^2} \\ & =\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{~h}}{\mathrm{~h}+2 \mathrm{~h}^2} \\ & =\lim _{\mathrm{h} \rightarrow 0} \frac{1}{1+2 \mathrm{~h}} \\ & =1\end{aligned}$

Since, L.H.L. $\neq$ R.H.L. for any value of $k$.

Hence, $f(x)$ is discontinuous at $x=0$ regardless of the choice 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 the 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 another 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 the 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 the points of discontinuity of the 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 the 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:

We have, $\mathrm{f}(\mathrm{t})=\frac{1}{\mathrm{t}^2+\mathrm{t}-2}$

Where $\mathrm{t}=\frac{1}{x-1}$

$\begin{aligned} & \therefore \mathrm{f}(\mathrm{t})=\frac{1}{\left(\frac{1}{x-1}\right)^2+\frac{1}{x-1}-2} \\ & =\frac{(x-1)^2}{1+(x-1)-2(x-1)^2} \\ & =\frac{(x-1)^2}{-\left(2 x^2-5 x+2\right)} \\ & =\frac{(x-1)^2}{(2 x-1)(2-x)}\end{aligned}$

So, $\mathrm{f}(\mathrm{t})$ is discontinuous at $2 \mathrm{x}-1=0$

$\begin{aligned} & \Rightarrow \mathrm{x}=\frac{1}{2} \text { and } 2-\mathrm{x}=0 \\ & \Rightarrow \mathrm{x}=2\end{aligned}$

Also $\mathrm{f}(\mathrm{t})$ is discontinuous at $\mathrm{x}=1$, where $\mathrm{t}=\frac{1}{x-1}$ is discontinuous.

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 the 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 equations 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 continuity:
Now, according to the 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 the greatest integer function which gives the 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 the 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 the greatest integer function which gives the 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 equations 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 the 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 the 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 equations 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
LLeft-handderivative(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 the 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 the 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 equations 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:

We have $f(x)=|x-5|$

$\Rightarrow \mathrm{f}(\mathrm{x})= \begin{cases}-(x-5), & \text { if } x-5<0 \text { or } x<5 \\ x-5, & \text { if } x-5>0 \text { or } x>5\end{cases}$

$\begin{aligned} & \text { L.H.L. } \lim _{h \rightarrow 5^{-}} f(x)=-(x-5) \\ & =\lim _{h \rightarrow 0}-(5-h-5) \\ & =\lim _{h \rightarrow 0} h=0 \\ & \text { R.H.L. } \lim _{x \rightarrow 5^{+}} f(x)=x-5 \\ & =\lim _{h \rightarrow 0}(5+h-5) \\ & =\lim _{h \rightarrow 0} h=0\end{aligned}$

L.H.L. = R.H.L.

So, $f(x)$ is continuous at $x=5$

Now, for differentiability

$\begin{aligned} & \operatorname{Lf}^{\prime}(5)=\lim _{h \rightarrow 0} \frac{f(5-h)-f(5)}{-h} \\ & =\lim _{h \rightarrow 0} \frac{-(5-h-5)-(5-5)}{-h} \\ & =\lim _{h \rightarrow 0} \frac{h}{-h} \\ & =-1 \\ & \operatorname{Rf}^{\prime}(5)=\lim _{h \rightarrow 0} \frac{f(5+h)-f(5)}{h} \\ & =\lim _{h \rightarrow 0} \frac{(5+h-5)-(5-5)}{h} \\ & =\lim _{h \rightarrow 0} \frac{h-0}{h} \\ & =1 \\ & \because \operatorname{Lf}^{\prime}(5) \neq \operatorname{Rd}(5)\end{aligned}$

Hence, $\mathrm{f}(\mathrm{x})$ is not differentiable at $\mathrm{x}=5$.

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 that, $f: R \rightarrow R$ satisfies the equation $f(x+y)=f(x) f(y)$ for all $x, y \in R, f(x) \neq 0$.

Let us take any point $\mathrm{x}=0$ at which the function $\mathrm{f}(\mathrm{x})$ is differentiable.

$\begin{aligned} & \therefore \mathrm{f}^{\prime}(0)=\lim _{h \rightarrow 0} \frac{\mathrm{f}(0+\mathrm{h})-\mathrm{f}(0)}{\mathrm{h}} \\ & 2=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0) \cdot \mathrm{f}(\mathrm{h})-\mathrm{f}(0)}{\mathrm{h}} . \\ & \Rightarrow 2=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0)[\mathrm{f}(\mathrm{h})-1]}{\mathrm{h}}\end{aligned}$

Now $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$\begin{aligned} & =\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(x) \cdot \mathrm{f}(\mathrm{h})-\mathrm{f}(x)}{\mathrm{h}} \\ & =\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(x)[\mathrm{f}(\mathrm{h})-1]}{\mathrm{h}} \\ & =2 \mathrm{f}(\mathrm{x})\end{aligned}$

Hence, $f^{\prime}(x)=2 f(x)$.

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:

Let $\mathrm{y}=\frac{8^x}{x^8}$

Taking $\log$ on both sides, we get,

$\begin{aligned} & \log \mathrm{y}=\log \frac{8^x}{x^8} \\ & \Rightarrow \log \mathrm{y}=\log 8^x-\log x^8 \\ & \Rightarrow \log \mathrm{y}=\mathrm{x} \log 8-8 \log \mathrm{x}\end{aligned}$

Differentiating both sides w.r.t. X

$\begin{aligned} & \Rightarrow \frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}=\log 8.1-\frac{8}{x} \\ & \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=y\left[\log 8-\frac{8}{x}\right]\end{aligned}$

Hence, $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{8^x}{x^8}\left[\log 8-\frac{8}{x}\right]$

Question 27 Differentiate each of the following w.r.t. x
$\log\left ( x+\sqrt{x^2 +a}\right )$

Answer:

Let $\mathrm{y}=\log \left(x+\sqrt{x^2+\mathrm{a}}\right)$

Differentiating both sides w.r.t. x

$\begin{aligned} & \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}} \log \left(x+\sqrt{x^2+\mathrm{a}}\right) \\ & =\frac{1}{x+\sqrt{x^2+\mathrm{a}}} \cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(x+\sqrt{x^2+\mathrm{a}}\right) \\ & =\frac{1}{x+\sqrt{x^2+\mathrm{a}}} \cdot\left[1+\frac{1}{2 \sqrt{x^2+\mathrm{a}}} \times \frac{\mathrm{d}}{\mathrm{dx}}\left(x^2+\mathrm{a}\right)\right] \\ & =\frac{1}{x+\sqrt{x^2+\mathrm{a}}} \cdot\left[1+\frac{1}{2 \sqrt{x^2+\mathrm{a}}} \cdot 2 x\right] \\ & =\frac{1}{x+\sqrt{x^2+\mathrm{a}}} \cdot\left[1+\frac{x}{\sqrt{x^2+\mathrm{a}}}\right] \\ & =\frac{1}{x+\sqrt{x^2+\mathrm{a}}} \cdot\left(\frac{\sqrt{x^2+\mathrm{a}+x}}{\sqrt{x^2+\mathrm{a}}}\right) \\ & =\frac{1}{\sqrt{x^2+\mathrm{a}}}\end{aligned}$

Hence. $\frac{d y}{d x}=\frac{1}{\sqrt{x^2+a}}$

Question 28

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

Answer:

Let $\mathrm{y}=\log \left[\log \left(\log x^5\right)\right]$

Differentiating both sides w.r.t. x

$\begin{aligned} & \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}} \log \left[\log \left(\log x^5\right)\right] \\ & =\frac{1}{\log \left(\log x^5\right)} \times \frac{\mathrm{d}}{\mathrm{dx}} \log \left(\log x^5\right) \\ & =\frac{1}{\log \left(\log x^5\right)} \times \frac{1}{\log \left(x^5\right)} \times \frac{\mathrm{d}}{\mathrm{dx}} \log x^5 \\ & =\frac{1}{\log \left(\log x^5\right)} \cdot \frac{1}{\log \left(x^5\right)} \cdot \frac{1}{x^5} \cdot \frac{\mathrm{~d}}{\mathrm{dx}} x^5 \\ & =\frac{1}{\log \left(\log x^5\right)} \cdot \frac{1}{\log \left(x^5\right)} \cdot \frac{1}{x^5} \cdot 5 x^4 \\ & =\frac{5}{x \log \left(x^5\right) \cdot \log \left(\log x^5\right)}\end{aligned}$

Hence, $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{5}{x \log \left(x^5\right) \cdot \log \left(\log x^5\right)}$

Question 29

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

Answer:

Let $\mathrm{y}=\sin \sqrt{x}+\cos ^2 \sqrt{x}$

Differentiating both sides w.r.t. x

$\begin{aligned} & \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}(\sin \sqrt{x})+\frac{\mathrm{d}}{\mathrm{dx}}\left(\cos ^2 \sqrt{x}\right) \\ & =\cos \sqrt{x} \cdot \frac{\mathrm{~d}}{\mathrm{dx}}(\sqrt{x})+2 \cos \sqrt{x} \cdot \frac{\mathrm{~d}}{\mathrm{dx}}(\cos \sqrt{x}) \\ & =\cos \sqrt{x} \cdot \frac{1}{2 \sqrt{x}}+2 \cos \sqrt{x}(-\sin \sqrt{x}) \cdot \frac{\mathrm{d}}{\mathrm{dx}} \sqrt{x} \\ & =\frac{1}{2 \sqrt{x}} \cdot \cos \sqrt{x}-2 \cos \sqrt{x} \cdot \sin \sqrt{x} \cdot \frac{1}{2 \sqrt{x}} \\ & =\frac{\cos \sqrt{x}}{2 \sqrt{x}}-\frac{\sin 2 \sqrt{x}}{2 \sqrt{x}}\end{aligned}$

Hence, $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\cos \sqrt{x}}{2 \sqrt{x}}-\frac{\sin 2 \sqrt{x}}{2 \sqrt{x}}$.

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 the 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
$\frac{\mathrm{dy}}{\mathrm{dx}}=\left(-\sin (\tan \sqrt{\mathrm{x}+1}) \times \sec ^{2} \sqrt{\mathrm{x}+1} \times \frac{1}{2 \sqrt{\mathrm{x}+1}}\right).$
Question 32 Differentiate each of the following w.r.t. x $sinx^2 + sin^2x + sin^2 (x^2)$

Answer:

Let us Assume $y=sinx^2 + sin^2x + sin^2 (x^2)$
$\begin{aligned} & \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}} \sin \left(x^2\right)+\frac{\mathrm{d}}{\mathrm{dx}}(\sin x)^2+\frac{\mathrm{d}}{\mathrm{dx}}\left(\sin x^2\right)^2 \\ & =\cos \left(x^2\right) \frac{\mathrm{d}}{\mathrm{dx}}\left(x^2\right)+2 \sin x \cdot \frac{\mathrm{~d}}{\mathrm{dx}}(\sin x)+2 \sin x^2 \frac{\mathrm{~d}}{\mathrm{dx}}\left(\sin x^2\right) \\ & =2 x \cos x^2+2 \sin x \cos x+2 \sin x^2 \cos x^2 \frac{\mathrm{~d}}{\mathrm{dx}} x^2 \\ & =2 x \cos x^2+2 \sin x \cos x+2 \sin x^2 \cos x^2 \times 2 x \\ & =2 \mathrm{x} \cos \left(\mathrm{x}^2\right)+\sin 2 \mathrm{x}+2 \mathrm{x} \sin 2\left(\mathrm{x}^2\right)\end{aligned}$

Question 33 Differentiate each of the following w.r.t. x
$\sin ^{-1}\left(\frac{1}{\sqrt{x+1}}\right)$

Answer:

Let $\mathrm{y}=\sin ^{-1} \frac{1}{\sqrt{x+1}}$

$\begin{aligned} & \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\sin ^{-1} \frac{1}{\sqrt{x+1}}\right) \\ & =\frac{1}{\sqrt{1-\left(\frac{1}{\sqrt{x+1}}\right)^2}} \cdot \frac{\mathrm{~d}}{\mathrm{dx}} \frac{1}{(x+1)^2} \\ & =\frac{1}{\sqrt{\frac{x+1-1}{x+1}}} \cdot \frac{\mathrm{~d}}{\mathrm{dx}}(x+1)^2 \\ & =\sqrt{\frac{x+1}{x}} \cdot \frac{-1}{2}(x+1)^{\frac{-3}{2}} \\ & =\frac{-1}{2 \sqrt{x}} \cdot\left(\frac{1}{x+1}\right)\end{aligned}$

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 sides 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 $sin^mx. cos^nx$
$y=sin^mx . cos^nx$
$\begin{aligned} & \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[(\sin x)^{\mathrm{m}} \cdot(\cos x)^{\mathrm{n}}\right] \\ & =(\sin x)^{\mathrm{m}} \frac{\mathrm{d}}{\mathrm{dx}}(\cos x)^{\mathrm{n}}+(\cos x)^{\mathrm{n}} \frac{\mathrm{d}}{\mathrm{dx}}(\sin x)^{\mathrm{m}} \\ & =(\sin x)^{\mathrm{m}} \mathrm{n}(\cos x)^{\mathrm{n}-1} \frac{\mathrm{~d}}{\mathrm{dx}}(\cos x)+(\cos x)^{\mathrm{n}} \mathrm{m}(\sin x)^{\mathrm{m}-1} \frac{\mathrm{~d}}{\mathrm{dx}}(\sin x) \\ & =(\sin x)^{\mathrm{m}} \mathrm{n}(\cos x)^{\mathrm{n}-1}(-\sin x)+(\cos x)^{\mathrm{n}} \mathrm{m}(\sin x)^{\mathrm{m}-1} \cos x \\ & =\sin ^{\mathrm{m}} \mathrm{x} \cos ^{\mathrm{n}} \mathrm{x}[-\mathrm{n} \tan \mathrm{x}+\mathrm{m} \cot \mathrm{x}]\end{aligned}$

Question 36 Differentiate each of the following w.r.t. x $(x + 1)^2 (x + 2)^3 (x + 3)^4$

Answer:

Let $y=(x+1)^2(x+2)^3(x+3)^4$

$\begin{aligned} & \therefore \log y=\log \left[(x+1)^2 \cdot(x+2)^3(x+3)^4\right] \\ & =2 \log (x+1)+3 \log (x+2)+4 \log (x+3)\end{aligned}$

Differentiating w.r.t. x both sides, we get

$\begin{aligned} & \frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2}{x+1}+\frac{3}{x+2}+\frac{4}{x+3} \\ & \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=y\left[\frac{2}{x+1}+\frac{3}{x+2}+\frac{4}{x+3}\right] \\ & =(x+1)^2 \cdot(x+2)^3 \cdot(x+3)^4\left[\frac{2}{(x+1)}+\frac{3}{(x+2)}+\frac{4}{(x+3)}\right] \\ & =(x+1)^2 \cdot(x+2)^3 \cdot(x+3)^4 \times\left[\frac{2(x+3)(x+3)+3(x+1)(x+3)+4(x+1)(x+2)}{(x+1)(x+2)(x+3)}\right] \\ & =(\mathrm{x}+1)(\mathrm{x}+2)^2(\mathrm{x}+3)^3\left[9 \mathrm{x}^2+34 \mathrm{x}+29\right]\end{aligned}$

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:

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

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:

Let $\mathrm{y}=\tan ^{-1}(\sec \mathrm{x}+\tan \mathrm{x})$

Differentiating both sides w.r.t. x

$\begin{aligned} & \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[\tan ^{-1}(\sec x+\tan x)\right] \\ & =\frac{1}{1+(\sec x+\tan x)^2} \cdot \frac{\mathrm{~d}}{\mathrm{dx}}(\sec x+\tan x) \\ & =\frac{1}{1+\sec ^2+\tan ^2 x+2 \sec x \tan x} \cdot\left(\sec x \tan x+\sec ^2 x\right) \\ & =\frac{1}{\left(1+\tan ^2 x\right)+\sec ^2 x+2 \sec x \tan x} \cdot \sec x(\tan x+\sec x) \\ & =\frac{1}{\sec ^2 x+\sec ^2 x+2 \sec x \tan x} \cdot \sec x(\tan x+\sec x) \\ & =\frac{1}{2 \sec ^2 x+2 \sec x \tan x} \cdot \sec x(\tan x+\sec x) \\ & =\frac{1}{2 \sec x(\sec x+\tan x)} \cdot \sec x(\tan x+\sec x) \\ & =\frac{1}{2}\end{aligned}$

Hence, $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}$

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:

Let $\mathrm{y}=\tan ^{-1}\left(\frac{\mathrm{a} \cos x-\mathrm{b} \sin x}{\mathrm{~b} \cos x-\mathrm{a} \sin x}\right)$

$\begin{aligned} & \Rightarrow \mathrm{y}=\tan ^{-1}\left[\frac{\frac{\mathrm{a} \cos x}{\mathrm{~b} \cos x}-\frac{\mathrm{b} \sin x}{\mathrm{~b} \cos x}}{\frac{\mathrm{~b} \cos x}{\mathrm{~b} \cos x}+\frac{\mathrm{a} \sin x}{\mathrm{~b} \cos x}}\right] \\ & \Rightarrow \mathrm{y}=\tan ^{-1}\left[\frac{\frac{\mathrm{a}}{\mathrm{b}}-\tan x}{1+\frac{\mathrm{a}}{\mathrm{b}} \tan x}\right] \\ & \Rightarrow \mathrm{y}=\tan ^{-1} \frac{\mathrm{a}}{\mathrm{b}}-\tan ^{-1}(\tan x) \\ & \Rightarrow \mathrm{y}=\tan ^{-1} \frac{\mathrm{a}}{\mathrm{b}}-x\end{aligned}$

Differentiating both sides concerning x

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\tan ^{-1} \frac{\mathrm{a}}{\mathrm{b}}\right)-\frac{\mathrm{d}}{\mathrm{dx}}(x)=0-1=-1$

Hence, $\frac{\mathrm{dy}}{\mathrm{dx}}=-1$.

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:

Let $\mathrm{y}=\sec ^{-1}\left(\frac{1}{4 x^3-3 x}\right)$

Put $x=\cos \theta$

$\begin{aligned} & \therefore \theta=\cos ^{-1} x \\ & y=\sec ^{-1}\left(\frac{1}{4 \cos ^3 \theta-3 \cos \theta}\right) \\ & \Rightarrow y=\sec ^{-1}\left(\frac{1}{\cos 3 \theta}\right) \ldots \ldots\left[\because \cos 3 \theta=4 \cos ^3 \theta-3 \cos \theta\right] \\ & \Rightarrow y=\sec ^{-1}(\sec 3 \theta) \\ & \Rightarrow y=3 \theta \\ & y=3 \cos ^{-1} x\end{aligned}$

Differentiating both sides w.r.t. X

$\begin{aligned} & \frac{d y}{d}=3 \cdot \frac{d}{d x} \cos ^{-1} x \\ & =3\left(\frac{-1}{\sqrt{1-x^2}}\right) \\ & =\frac{-3}{\sqrt{1-x^2}}\end{aligned}$

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} $
$=3\left[\frac{1}{1+\frac{x^{2}}{a^{2}}}\right] \cdot \frac{d}{d x}\left(\frac{x}{a}\right)$
$=\frac{3 a^{2}}{a^{2}+x^{2}} \cdot \frac{1}{a}\\=\frac{3 a}{a^{2}+x^{2}}$

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 equations,
$x=t+\frac{1}{t}\: \: ,\: \: y=t-\frac{1}{t}$
Now, differentiate both equations 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:

Given that, $\mathrm{x}=\mathrm{e}^\theta\left(\theta+\frac{1}{\theta}\right), \mathrm{y}=\mathrm{e}^{-\theta}\left(\theta-\frac{1}{\theta}\right)$

Differentiating both the parametric functions w.r.t. $\theta$

$\begin{aligned} & \frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{e}^\theta\left(1-\frac{1}{\theta^2}\right)+\left(\theta+\frac{1}{\theta}\right) \cdot \mathrm{e}^\theta \\ & \frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{e}^\theta\left(1-\frac{1}{\theta^2}+\theta+\frac{1}{\theta}\right) \\ & \Rightarrow \mathrm{e}^\theta\left(\frac{\theta^2-1+\theta^3+\theta}{\theta^2}\right) \\ & =\frac{\mathrm{e}^\theta\left(\theta^3+\theta^2+\theta-1\right)}{\theta^2} \\ & \mathrm{y}=\mathrm{e}^{-\theta}\left(\theta-\frac{1}{\theta}\right) \\ & \frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{e}^{-\theta}\left(1+\frac{1}{\theta^2}\right)+\left(\theta-\frac{1}{\theta}\right) \cdot\left(-\mathrm{e}^{-\theta}\right) \\ & \frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{e}^{-\theta}\left(1+\frac{1}{\theta^2}-\theta+\frac{1}{\theta}\right)\end{aligned}$

$\begin{aligned} & \Rightarrow \mathrm{e}^{-\theta}\left(\frac{\theta^2+1-\theta^3+\theta}{\theta^2}\right) \\ & =\mathrm{e}^{-\theta} \frac{\left(-\theta^3+\theta^2+\theta+1\right)}{\theta^2} \\ & \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{d} \theta}}{\frac{\mathrm{d} x}{\mathrm{~d} \theta}} \\ & =\frac{\mathrm{e}^{-\theta}\left(\frac{-\theta^3+\theta^2+\theta+1}{\theta^2}\right)}{\mathrm{e}^\theta\left(\frac{\theta^3+\theta^2+\theta+1}{\theta^2}\right)} \\ & =\mathrm{e}^{-2 \theta}\left(\frac{-\theta^3+\theta^2+\theta+1}{\theta^3+\theta^2+\theta-1}\right)\end{aligned}$

Hence, $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{-2 \theta}\left(\frac{-\theta^3+\theta^2+\theta+1}{\theta^3+\theta^2+\theta-1}\right)$

Question 46 Find dy/dx of each of the functions expressed in the parametric form in
$x=3 \cos \theta-2 \cos ^{3} \theta, y=3 \sin \theta-2 \sin ^{3} \theta$
Answer:

Given that: $\mathrm{x}=3 \cos \theta-2 \cos ^3 \theta$ and $\mathrm{y}=3 \sin \theta-2 \sin ^3 \theta$.

Differentiating both the parametric functions w.r.t. $\theta$

$\begin{aligned} & \frac{\mathrm{dx}}{\mathrm{d} \theta}=-3 \sin \theta-6 \cos ^2 \theta \cdot \frac{\mathrm{~d}}{\mathrm{~d} \theta}(\cos \theta) \\ & =-3 \sin \theta-6 \cos ^2 \theta \cdot(-\sin \theta) \\ & =-3 \sin \theta+6 \cos ^2 \theta \cdot \sin \theta \\ & \frac{\mathrm{dy}}{\mathrm{d} \theta}=3 o s \theta-6 \sin ^2 \theta \cdot \frac{\mathrm{~d}}{\mathrm{~d} \theta}(\sin \theta) \\ & ==3 \cos \theta-6 \sin ^2 \theta \cdot \cos \theta \\ & \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{d y}{\mathrm{~d} \theta}}{\frac{\mathrm{dx}}{\mathrm{d} \theta}}\end{aligned}$

$\begin{aligned} & =\frac{3 \cos \theta-6 \sin ^2 \theta \cos \theta}{-3 \sin \theta+6 \cos ^2 \theta \cdot \sin \theta} \\ & \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\cos \theta\left(3-6 \sin ^2 \theta\right)}{\sin \theta\left(-3+6 \cos ^2 \theta\right)} \\ & =\frac{\cos \theta\left[3-6\left(1-\cos ^2 \theta\right)\right]}{\sin \theta\left[-3+6 \cos ^2 \theta\right]} \\ & =\cot \theta\left(\frac{3-6+6 \cos ^2 \theta}{-3+6 \cos ^2 \theta}\right) \\ & =\cot \theta\left(\frac{-3+6 \cos ^2 \theta}{-3+6 \cos ^2 \theta}\right) \\ & =\cot \theta\end{aligned}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\cot \theta$

Question 47

Find the dy/dx of each of the functions expressed in the 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:

Given that $\sin \mathrm{x}=\frac{2 \mathrm{t}}{1+\mathrm{t}^2}$ and $\tan \mathrm{y}=\frac{2 \mathrm{t}}{1-\mathrm{t}^2}$

$\therefore$ Taking $\sin \mathrm{x}=\frac{2 \mathrm{t}}{1+\mathrm{t}^2}$

Differentiating both sides w.r.t t, we get

$\begin{aligned} & \cos x \cdot \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\left(1+\mathrm{t}^2\right) \cdot \frac{\mathrm{d}}{\mathrm{dt}}(2 \mathrm{t})-2 \mathrm{t} \cdot \frac{\mathrm{d}}{\mathrm{dt}}\left(1+\mathrm{t}^2\right)}{\left(1+\mathrm{t}^2\right)^2} \\ & \Rightarrow \cos x \cdot \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{2\left(1+\mathrm{t}^2\right)-2 \mathrm{t} \cdot 2 \mathrm{t}}{\left(1+\mathrm{t}^2\right)^2} \\ & \Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{2+2 \mathrm{t}^2-4 \mathrm{t}^2}{\left(1-\mathrm{t}^2\right)^2} \times \frac{1}{\cos x} \\ & \Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{2-2 \mathrm{t}^2}{\left(1+\mathrm{t}^2\right)^2} \times \frac{1}{\sqrt{1-\sin ^2 x}} \\ & \Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{2\left(1-\mathrm{t}^2\right)}{\left(1+\mathrm{t}^2\right)^2} \times \frac{1}{\sqrt{1-\left(\frac{2 \mathrm{t}}{1+\mathrm{t}^2}\right)^2}}\end{aligned}$

$\begin{aligned} & \Rightarrow \frac{d x}{d t}=\frac{2\left(1-t^2\right)}{\left(1+t^2\right)^2} \times \frac{1}{\frac{\sqrt{\left(1+t^2\right)^2-4 t^2}}{\left(1+t^2\right)^2}} \\ & \Rightarrow \frac{d x}{d t}=\frac{2\left(1-t^2\right)}{\left(1+t^2\right)^2} \times \frac{1+t^2}{\sqrt{1+t^4+2 t^2-4 t^2}} \\ & \Rightarrow \frac{d x}{d t}=\frac{2\left(1-t^2\right)}{\left(1+t^2\right)^2} \times \frac{1}{\sqrt{1+t^4-2 t^2}} \\ & \Rightarrow \frac{d x}{d t}=\frac{2\left(1-t^2\right)}{\left(1+t^2\right)^2} \times \frac{1}{\sqrt{\left(1-t^2\right)^2}}\end{aligned}$

$\begin{aligned} & \Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{2\left(1-\mathrm{t}^2\right)}{\left(1+\mathrm{t}^2\right)^2} \times \frac{1}{\left(1-\mathrm{t}^2\right)} \\ & \Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{2}{1+\mathrm{t}^2}\end{aligned}$

Now taking, $\tan \mathrm{y}=\frac{2}{1-\mathrm{t}^2}$

Differentiating both sides w.r.t, t , we get

$\begin{aligned} & \frac{\mathrm{d}}{\mathrm{dt}}(\tan y)=\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{2 \mathrm{t}}{1-\mathrm{t}^2}\right) \\ & \Rightarrow \sec ^2 y \frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\left(1-\mathrm{t}^2\right) \cdot \frac{\mathrm{d}}{\mathrm{dt}}(2 \mathrm{t})-2 \mathrm{t} \cdot \frac{\mathrm{d}}{\mathrm{dt}}\left(1-\mathrm{t}^2\right)}{\left(1-\mathrm{t}^2\right)^2} \\ & \Rightarrow \sec ^2 y \frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\left(1-\mathrm{t}^2\right) \cdot 2-2 \mathrm{t} \cdot(-2 \mathrm{t})}{\left(1-\mathrm{t}^2\right)^2} \\ & \Rightarrow \sec ^2 y \frac{\mathrm{dy}}{\mathrm{dt}}=\frac{2-2 \mathrm{t}^2+4 \mathrm{t}^2}{\left(1-\mathrm{t}^2\right)^2} \\ & \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=\frac{2+2 \mathrm{t}^2}{\left(1-\mathrm{t}^2\right)^2} \times \frac{1}{\sec ^2 y}\end{aligned}$

$\begin{aligned} & \Rightarrow \frac{d y}{d t}=\frac{2\left(1+t^2\right)}{\left(1-t^2\right)^2} \times \frac{1}{1+\tan ^2 y} \\ & \Rightarrow \frac{d y}{d t}=\frac{2\left(1+\mathrm{t}^2\right)}{\left(1-\mathrm{t}^2\right)^2} \times \frac{1}{1+\left(\frac{2 \mathrm{t}}{1-\mathrm{t}^2}\right)^2} \\ & \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=\frac{2\left(1+\mathrm{t}^2\right)}{\left(1-\mathrm{t}^2\right)^2} \times \frac{1}{\frac{\left(1-\mathrm{t}^2\right)^2+4 \mathrm{t}^2}{\left(1-\mathrm{t}^2\right)^2}} \\ & \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=\frac{2\left(1+\mathrm{t}^2\right)}{\left(1-\mathrm{t}^2\right)^2} \times \frac{\left(1-\mathrm{t}^2\right)^2}{1+\mathrm{t}^2+2 \mathrm{t}^2+4 \mathrm{t}^2} \\ & \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=\frac{2\left(1+\mathrm{t}^2\right)}{\left(1-\mathrm{t}^2\right)^2} \times \frac{\left(1-\mathrm{t}^2\right)^2}{1+\mathrm{t}^4+2 \mathrm{t}^2}\end{aligned}$

$\begin{aligned} & \Rightarrow \frac{d y}{d t}=\frac{2\left(1+t^2\right)}{\left(1-t^2\right)^2} \times \frac{\left(1-t^2\right)^2}{\left(1+t^2\right)^2} \\ & \Rightarrow \frac{d y}{d t}=\frac{2}{1+t^2} \\ & \therefore \frac{d y}{d t}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \\ & =\frac{\frac{2}{1+t^2}}{\frac{2}{1+t^2}} \\ & =1\end{aligned}$

Hence $\frac{\mathrm{dy}}{\mathrm{dt}}=1$

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:

Given that: $\mathrm{x}=\frac{1+\log \mathrm{t}}{\mathrm{t}^2}, \mathrm{y}=\frac{3+2 \log \mathrm{t}}{\mathrm{t}}$

Differentiating both the parametric functions w.r.t. t

$\begin{aligned} & \frac{d x}{d t}=\frac{t^2 \cdot \frac{d}{d t}(1+\log t)-(1+\log t) \cdot \frac{d}{d t}\left(t^2\right)}{t^4} \\ & =\frac{t^2 \cdot\left(\frac{1}{t}\right)-(1+\log t) \cdot 2 t}{t^4} \\ & =\frac{t-(1+\log t) \cdot 2 t}{t^4} \\ & =\frac{t[1-2-2 \log t]}{t^4} \\ & =\frac{-(1+2 \log t)}{t^3} \\ & y=\frac{3+2 \log t}{t}\end{aligned}$

$\begin{aligned} & \frac{d y}{d t}=\frac{t \cdot \frac{d}{d t}(3+2 \log t)-(3+2 \log t) \cdot \frac{d}{d t}(t)}{t^2} \\ & =\frac{t\left(\frac{2}{t}\right)-(3+2 \log t) \cdot 1}{t^2} \\ & =\frac{2-3-2 \log t}{t^2} \\ & =\frac{-(1+2 \log t)}{t^2} \\ & \therefore \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \\ & =\frac{-(1+2 \log t)}{t^2} \\ & \frac{-(1+2 \log t)}{t^3} \\ & =\frac{t^3}{t^2} \\ & =t\end{aligned}$

Hence, $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{t}$.

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:

Given that: $e^{\cos 2 t}$ and $y=e^{\sin 2 t}$

$\Rightarrow \cos 2 t=\log x$ and $\sin 2 t=\log y$

Differentiating both the parametric functions w.r.t. T

$\begin{aligned} & \frac{d x}{d t}=e^{\cos 2 t} \cdot \frac{d}{d t}(\cos 2 t) \\ & =e^{\cos 2 t}(-\sin 2 t) \cdot \frac{d}{d t}(2 t) \\ & =-e^{\cos 2 t} \cdot \sin 2 t \cdot 2 \\ & =2 e^{\cos 2 t} \cdot \sin 2 t\end{aligned}$

Now $y=e^{\sin 2 t}$

$\begin{aligned} & \frac{d y}{d t}=e^{\sin 2 t} \cdot \frac{d}{d t}(\sin 2 t) \\ & =e^{\sin 2 t} \cdot \cos 2 t \cdot \frac{d}{d t}(2 t) \\ & =e^{\sin 2 t} \cdot \cos 2 t \cdot 2 \\ & =2 e^{\sin 2 t} \cdot \cos 2 t \\ & \therefore \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \\ & =\frac{2 e^{\sin 2 t} \cdot \cos 2 t}{-2 e^{\cos 2 t} \cdot \sin 2 t} \\ & =\frac{e^{\sin 2 t} \cdot \cos 2 t}{-e^{\cos 2 t} \cdot \sin 2 t} \\ & =\frac{y \cos 2 t}{-x \sin 2 t}\end{aligned}$

$=\frac{y \log x}{-x \log y}$

Hence, $\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{y \log x}{x \log y}$.

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 the 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 concerning 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 concerning 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) concerning 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, by 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 that: $\mathrm{x}=\mathrm{e}^{\frac{x}{y}}$

Taking $\log$ on both the sides,

$\begin{aligned} & \log \mathrm{x}=\log \mathrm{e}^{\frac{x}{y}} \\ & \Rightarrow \log \mathrm{x}=\frac{x}{y} \log \mathrm{e}\end{aligned}$

$\Rightarrow \log \mathrm{x}=\frac{x}{y}$

Differentiating both sides w.r.t. X

$\begin{aligned} & \frac{\mathrm{d}}{\mathrm{dx}} \log x=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{x}{y}\right) \\ & \Rightarrow \frac{1}{x}=\frac{y \cdot 1-x \cdot \frac{\mathrm{dy}}{\mathrm{dx}}}{y^2} \\ & \Rightarrow \mathrm{y}^2=x y-x^2 \cdot \frac{\mathrm{dy}}{\mathrm{dx}} \\ & \Rightarrow x^2 \cdot \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{xy}-\mathrm{y}^2 \\ & \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{y(x-y)}{x^2} \\ & \Rightarrow \frac{\mathrm{dy}}{\mathrm{d}}=\frac{y}{x} \cdot\left(\frac{x-y}{x}\right) \\ & \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\log x} \cdot\left(\frac{x-y}{x}\right)\end{aligned}$

Hence, ${ }^{\prime}$ "dy"/"dx" = (x - y)/(xlogx).

Question 60

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

Answer:

Given that: $\mathrm{y}^{\mathrm{x}}=\mathrm{e}^{\mathrm{y}-\mathrm{x}}$

Taking $\log$ on both sides $\log \mathrm{y}^{\mathrm{x}}=\log \mathrm{e}^{\mathrm{y}-\mathrm{x}}$

$\begin{aligned} & \Rightarrow \mathrm{x} \log \mathrm{y}=(\mathrm{y}-\mathrm{x}) \log \mathrm{e} \\ & \Rightarrow \mathrm{x} \log \mathrm{y}=\mathrm{y}-\mathrm{x} \quad \ldots . .[\because \log \mathrm{e}=1] \\ & \Rightarrow \mathrm{x} \log \mathrm{y}+\mathrm{x}=\mathrm{y} \\ & \Rightarrow \mathrm{x}(\log \mathrm{y}+1)=\mathrm{y} \\ & \Rightarrow \mathrm{x}=\frac{y}{\log y+1}\end{aligned}$

Differentiating both sides w.r.t. Y

$\begin{aligned} & \frac{\mathrm{dx}}{\mathrm{dy}}=\frac{\mathrm{d}}{\mathrm{dy}}\left(\frac{y}{\log y+1}\right) \\ & =\frac{(\log y+1) \cdot 1-y \cdot \frac{\mathrm{~d}}{\mathrm{dy}}(\log y+1)}{(\log y+1)^2} \\ & =\frac{\log y+1-y \cdot \frac{1}{2}}{(\log y+1)^2} \\ & =\frac{\log y}{(\log y+1)^2}\end{aligned}$

We know that

$\begin{aligned} & \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\frac{\mathrm{dx}}{\mathrm{dy}}} \\ & =\frac{1}{\frac{\log y}{(\log y+1)^2}} \\ & =\frac{(\log y+1)^2}{\log y}\end{aligned}$

Hence, $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{(\log y+1)^2}{\log y}$.

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 that $\mathrm{y}=(\cos x)^{(\cos x)^{(\cos x) \ldots \infty}}$,

$\Rightarrow \mathrm{y}=(\cos \mathrm{x})^{\mathrm{y}} \ldots . .\left[y=(\cos x)^{\left.(\cos x)^{(\cos x) \ldots \infty}\right]}\right]$

Taking $\log$ on both sides $\log y=y \cdot \log (\cos x)$

Differentiating both sides w.r.t. x

$\begin{aligned} & \frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}=y \cdot \frac{\mathrm{~d}}{\mathrm{dx}} \log (\cos x)+\log (\cos x) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} \\ & \Rightarrow \frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}=y \cdot \frac{1}{\cos x} \cdot \frac{\mathrm{~d}}{\mathrm{dx}}(\cos x)+\log (\cos x) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} \\ & \Rightarrow \frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}=y \cdot \frac{1}{\cos x} \cdot(-\sin x)+\log (\cos x) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} \\ & \Rightarrow \frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}-\log (\cos x) \frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{y} \tan \mathrm{x} \\ & \Rightarrow\left[\frac{1}{y}-\log (\cos x)\right] \frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{y} \tan \mathrm{x} \\ & \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-y \tan x}{\frac{1}{y}-\log (\cos x)} \\ & =\frac{y^2 \tan x}{y \log \cos x-1}\end{aligned}$

Hence, $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{y^2 \tan x}{y \log \cos x-1}$.

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 that: $x \sin (a+y)+\sin a \cos (a+y)=0$

$\begin{aligned} & \Rightarrow x \sin (a+y)=-\sin a \cos (a+y) \\ & \Rightarrow x=\frac{-\sin a \cdot \cos (a+y)}{\sin (a+y)} \\ & \Rightarrow x=-\sin a \cdot \cot (a+y)\end{aligned}$

Differentiating both sides w.r.t. Y

$\begin{aligned} & \Rightarrow \frac{d x}{d y}=-\sin a \cdot \frac{d}{d y} \cot (a+y) \\ & \Rightarrow \frac{d x}{d y}=-\sin a\left[-\operatorname{cosec}^2(a+y)\right. \\ & \Rightarrow \frac{d x}{d y}=\frac{\sin a}{\sin ^2(a+y)} \\ & \therefore \frac{d y}{d x}=\frac{1}{\frac{d x}{d y}} \\ & =\frac{1}{\frac{\sin a}{\sin ^2(a+y)}}\end{aligned}$

Hence, $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sin ^2(\mathrm{a}+y)}{\sin \mathrm{a}}$.

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 that: $\sqrt{1-x^2}+\sqrt{1-y^2}=\mathrm{a}(x-y)$

Put $\mathrm{x}=\sin \theta$ and $\mathrm{y}=\sin \Phi$.

$\begin{aligned} & \therefore \theta=\sin ^{-1} \mathrm{x} \text { and } \Phi=\sin ^{-1} \mathrm{y} \\ & \sqrt{1-\sin ^2 \theta}+\sqrt{1-\sin ^2 \phi}=\mathrm{a}(\sin \theta-\sin \Phi) \\ & \Rightarrow \sqrt{\cos ^2 \theta}+\sqrt{\cos ^2 \phi}=\mathrm{a}(\sin \theta-\sin \Phi) \\ & \Rightarrow \cos \theta+\cos \Phi=\mathrm{a}(\sin \theta-\sin \Phi) \\ & \Rightarrow \frac{\cos \theta+\cos \phi}{\sin \theta-\sin \phi}=\mathrm{a}\end{aligned}$

$\begin{aligned} & \Rightarrow \frac{2 \cos \frac{\theta+\phi}{2} \cdot \cos \frac{\theta-\phi}{2}}{2 \cos \frac{\theta+\phi}{2} \cdot \sin \frac{\theta-\phi}{2}}=\mathrm{a} \\ & \Rightarrow \frac{\cos \left(\frac{\theta-\phi}{2}\right)}{\sin \left(\frac{\theta-\phi}{2}\right)}=\mathrm{a} \\ & \Rightarrow \cot \left(\frac{\theta-\phi}{2}\right)=\mathrm{a} \\ & \Rightarrow \frac{\theta-\phi}{2}=\cot ^{-1} \mathrm{a} \\ & \Rightarrow \theta-\Phi=2 \cot ^{-1} \mathrm{a} \\ & \Rightarrow \sin ^{-1} \mathrm{x}-\sin ^{-1} \mathrm{y}=2 \cot ^{-1} \mathrm{a}\end{aligned}$

Differentiating both sides w.r.t. X

$\begin{aligned} & \frac{\mathrm{d}}{\mathrm{dx}}\left(\sin ^{-1} x\right)-\frac{\mathrm{d}}{\mathrm{dx}}\left(\sin ^{-1} x\right)=2 \cdot \frac{\mathrm{~d}}{\mathrm{dx}} \cot ^{-1} \mathrm{a} \\ & \Rightarrow \frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}=0 \\ & \Rightarrow \frac{1}{\sqrt{1-y^2}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\sqrt{1-x^2}} \\ & \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}} .\end{aligned}$

Question 64 If $y = tan^{-1}x$ , find $\frac{d^2y}{dx^2}$ in terms of y alone.

Answer:

Given that: $\mathrm{y}=\tan ^{-1} \mathrm{x}$

$\Rightarrow x=\tan y$

Differentiating both sides w.r.t. Y

$\frac{d x}{d y}=\sec ^2 y$

$\frac{\mathrm{dy}}{dx}=\frac{1}{\sec ^2 y}=\cos ^2 y$

Again, differentiating both sides w.r.t. x

$\begin{aligned} & \Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\cos ^2 y\right) \\ & \Rightarrow \frac{\mathrm{d}^2 y}{\mathrm{dx}^2}=2 \cos y \cdot \frac{\mathrm{~d}}{\mathrm{dx}}(\cos y) \\ & \Rightarrow \frac{\mathrm{d}^2 y}{\mathrm{dx}^2}=2 \cos y(-\sin y) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} \\ & \Rightarrow \frac{\mathrm{d}^2 y}{\mathrm{dx}^2}=-2 \sin y \cos y \cdot \cos ^2 y \\ & \therefore \frac{\mathrm{~d}^2 y}{\mathrm{dx}^2}=-2 \sin \mathrm{y} \cos ^3 \mathrm{y}\end{aligned}$

Question 65

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

Answer:

We have, $\mathrm{f}(\mathrm{x})=\mathrm{x}(\mathrm{x}-1)^2$ in $[0,1]$

Since, $f(x)=x(x-1)^2$ is a polynomial function it is continuous in $[0,1]$ and differentiable in $(0,1)$ Now, $f(0)=0$ and $f(1)$

$\Rightarrow \mathrm{f}(0)=\mathrm{f}(1)$

f satisfies the conditions of Rolle's theorem.

Hence, by Rolle's theorem there exists at least one $\mathrm{c} \in(0,1)$ such that $\mathrm{f}^{\prime}(\mathrm{c})=0$

$\begin{aligned} & \Rightarrow 3 \mathrm{c}^2-4 \mathrm{c}+1=0 \\ & \Rightarrow(3 \mathrm{c}-1)(\mathrm{c}-1)=0 \\ & \Rightarrow \mathrm{c}=\frac{1}{3} \in(0,1)\end{aligned}$
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:

We have, $\mathrm{f}(\mathrm{x})=\sin ^4 x+\cos ^4 x$ in $\left[0, \frac{\pi}{2}\right]$

We know that $\sin \mathrm{x}$ and $\cos \mathrm{x}$ are conditions and differentiable

$\therefore \sin 4 x$ and $\cos 4 x$ and hence $\sin 4 x+\cos 4 x$ is continuous and differentiable

Now $\mathrm{f}(0)=0+1=1$ and $\mathrm{f}\left(\frac{\pi}{2}\right)=1+0=1$

$\Rightarrow \mathrm{f}(0)=\mathrm{f}\left(\frac{\pi}{2}\right)$

So, the conditions of Rolle's theorem are satisfied.

Hence, there exists at least one $\mathrm{c} \in\left(0, \frac{\pi}{2}\right)$ such that $\mathrm{f}^{\prime}(\mathrm{c})=0$

$\begin{aligned} & \therefore 4 \sin ^3 \mathrm{c} \cos \mathrm{c}-4 \cos ^3 \mathrm{c} \sin \mathrm{c}=0 \\ & \Rightarrow 4 \sin \mathrm{c} \cos \mathrm{c}\left(\sin ^2 \mathrm{c}-\cos ^2 \mathrm{c}\right)=0 \\ & \Rightarrow 4 \sin \mathrm{c} \cos \mathrm{c}(-\cos 2 \mathrm{c})=0 \\ & \Rightarrow-2 \sin 2 \mathrm{c} \cdot \cos 2 \mathrm{c}=0 \\ & \Rightarrow \sin 4 \mathrm{c}=0 \\ & \Rightarrow 4 \mathrm{c}=\pi \\ & \Rightarrow \mathrm{c}=\frac{\pi}{4} \in\left(0, \frac{\pi}{2}\right)\end{aligned}$

Thus, Rolle’s theorem is verified.

Question 67

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

Answer:

We have, $\mathrm{f}(\mathrm{x})=\log \left(\mathrm{x}^2+2\right)-\log 3$

We know that $\mathrm{x}^2+2$ and the logarithmic function are continuous and differentiable $\therefore \mathrm{f}(\mathrm{x})=\log \left(\mathrm{x}^2+2\right)-\log 3$ is also continuous and differentiable.

Now $f(-1)=f(1)=\log 3-\log 3=0$

So, the conditions of Rolle's theorem are satisfied.

Hence, there exists at least one $\mathrm{c} \in(-1,1)$ such that $\mathrm{f}^{\prime}(\mathrm{c})=0$

$\begin{aligned} & f(x)=\frac{2 c}{c^2+2}-0=0 \\ & \Rightarrow c=0 \in(-1,1)\end{aligned}$

Hence, Rolle's theorem has been verified.

Question 68

Verify 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) verifies Rolle’s Theorem
First of all, the 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 exists 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 the multiplication of algebra and exponential function,n 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) concerning 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 the above equation, we get
$ \mathrm{f}^{\prime}(\mathrm{c})=-\frac{\mathrm{e}^{-\frac{\mathrm{c}}{2}}}{2}[(\mathrm{c}-3)(\mathrm{c}+2)]$
All 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 Rolle’s theorem for each of the functions
$f(x)=\sqrt{4-x^{2}} \text { in }[-2,2]$

Answer:

We have, $\sqrt{4-x^2}=\left(4-x^2\right)^{\frac{1}{2}}$

Since $\left(4-x^2\right)$ and the square root function are continuous and differentiable in their domain, given function $f(x)$ is also continuous and differentiable in $[-2,2]$

Also $f(-2)=f(2)=0$

So, the conditions of Rolle's theorem are satisfied.

Hence, there exists a real number $\mathrm{c} \in(-2,2)$ such that $\mathrm{f}^{\prime}(\mathrm{c})=0$.

Now $f^{\prime}(\mathrm{x})=\frac{1}{2}\left(4-x^2\right)^{\frac{-1}{2}}(-2 x)$

$=-\frac{x}{\sqrt{4-x^2}}$

So, $f^{\prime}(c)=0$

$\begin{aligned} & \Rightarrow \frac{c}{\sqrt{4-c^2}}=0 \\ & \Rightarrow c=0 \in(-2,2)\end{aligned}$

Hence, Rolle's theorem has been 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, the 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 exists 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 concerning 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 does not apply 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 the x-axis.

Answer:

We have $y=\cos x-1$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=-\sin \mathrm{x}$

For the tangent to be parallel to the x-axis

We must have $\frac{d y}{d x}=0$

$\begin{aligned} & \therefore-\sin \mathrm{x}=0 \\ & \therefore \mathrm{x}=\pi \in[0,2 \pi]\end{aligned}$

$y(\pi)=\cos \pi-1=-2$

Hence, the required point on the curve, where the tangent drawn is parallel to the x-axis $(\pi,-2)$

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-the axis.

Answer:

We have, $y=x(x-4), x \in[0,4]$

Since the given function is polynomial, it is continuous and differentiable.

Also $y(0)=y(4)=0$

So, the conditions of Role's theorem are satisfied.

Hence there exists a point $\mathrm{c} \in(0,4)$ such that $\mathrm{f}^{\prime}(\mathrm{c})=0$

$\Rightarrow 2 c-4=0$

$\Rightarrow c=2$

$\Rightarrow x=2$ and $y(2)$

$\begin{aligned} & =2(2-4) \\ & =-4\end{aligned}$

Therefore, the required point on the curve, where the tangent drawn is parallel to the x-axis is $(2,-4)$.

Question 73

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

Answer:

We have, $\mathrm{f}(\mathrm{x})=\frac{1}{4 x-1}$ in $[1,4]$

Clearly $f(x)$ is continuous in $[1,4]$

Also, $\mathrm{f}^{\prime}(\mathrm{x})=-\frac{4}{(4 x-1)^2}$, which exists in $(1,4)$

So, it is differentiable in $(1,4)$

Thus conditions of the mean value theorem are satisfied.

Hence, there exists a real number $\mathrm{c} \in(1,4)$ such that

$\begin{aligned} & f(\mathrm{c})=\frac{\mathrm{f}(4)-f(1)}{4-1} \\ & \Rightarrow \frac{-4}{(4 \mathrm{c}-1)^2}=\frac{\frac{1}{16-1}-\frac{1}{4-1}}{4-1} \\ & =\frac{\frac{1}{15}-\frac{1}{3}}{3} \\ & \Rightarrow \frac{-4}{(4-1)^2}=\frac{-4}{45} \\ & \Rightarrow(4 \mathrm{c}-1)^2=45 \\ & \Rightarrow 4 \mathrm{c}-1= \pm 3 \sqrt{5} \\ & \Rightarrow \mathrm{c}=\frac{3 \sqrt{5}+1}{4} \in(1,4)\end{aligned}$

Hence mean value theorem has been verified.

Question 74

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

Answer:

We have, $f(x)=x^3-2 x^2-x+3$ in $[0,1]$

Since, $f(x)$ is a polynomial function it is continuous in $[0,1]$ and differentiable in $(0,1)$

Thus, the conditions of the mean value theorem are satisfied.

Hence, there exists a real number $\mathrm{c} \in(0,1)$ such that

$\begin{aligned} & f(\mathrm{c})=\frac{\mathrm{f}(1)-\mathrm{f}(0)}{1-0} \\ & \Rightarrow 3 \mathrm{c}^2-4 \mathrm{c}-1=\frac{[1-2-1+3]-[0+3]}{1-0} \\ & \Rightarrow 3 \mathrm{c}^2-4 \mathrm{c}-1=-2 \\ & \Rightarrow 3 \mathrm{c}^2-4 \mathrm{c}+1=0 \\ & \Rightarrow(3 \mathrm{c}-1)(\mathrm{c}-1)=0 \\ & \Rightarrow \mathrm{c}=\frac{1}{3} \in(0,1)\end{aligned}$

Hence, the mean value theorem has been verified.

Question 75

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

Answer:

We have, $\mathrm{f}(\mathrm{x})=\sin \mathrm{x}-\sin 2 \mathrm{x}$ in $[0, \pi]$

We know that all trigonometric functions are continuous and differentiable in their domain, a given function is also continuous and differentiable

So, the condition of the mean value theorem is satisfied.

Hence, there exists at least one $\mathrm{c} \in(0, \pi)$ such that,

$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(\pi)-\mathrm{f}(0)}{\pi-0} \\ & \Rightarrow \cos \mathrm{c}-2 \cos 2 \mathrm{c}=\frac{\sin \pi-\sin 2 \pi-\sin 0+\sin 0}{\pi-0} \\ & \Rightarrow 2 \cos 2 \mathrm{c}-\cos \mathrm{c}=0 \\ & \Rightarrow 2\left(2 \cos ^2 \mathrm{c}-1\right)-\cos \mathrm{c}=0 \\ & \Rightarrow 4 \cos ^2 \mathrm{c}-\cos \mathrm{c}-2=0 \\ & \Rightarrow \cos \mathrm{c}=\frac{1 \pm \sqrt{1+32}}{8} \\ & =\frac{1 \pm \sqrt{33}}{8} \\ & \Rightarrow \mathrm{c}=\cos ^{-1}\left(\frac{1 \pm \sqrt{33}}{8}\right) \in(0, \pi)\end{aligned}$

Hence, the mean value theorem has been verified.

Question 76

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

Answer:

We have, $\mathrm{f}(\mathrm{x})=\sqrt{25-x^2}$ in $[1,5]$

Since $25-x^2$ and the square root function are continuous and differentiable in their domain, given function $f(x)$ is also continuous and differentiable.

So, the condition of the mean value theorem is satisfied.

Hence, there exists at least one $\mathrm{c} \in(1,5)$ such that,

$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(5)-\mathrm{f}(1)}{5-1} \\ & \Rightarrow \frac{-\mathrm{c}}{\sqrt{25-\mathrm{c}^2}}=\frac{0-\sqrt{24}}{4} \\ & \Rightarrow 16 \mathrm{c}^2=24\left(25-\mathrm{c}^2\right) \\ & \Rightarrow 40 \mathrm{c}^2=600 \\ & \Rightarrow \mathrm{c}^2=15 \\ & \Rightarrow \mathrm{c}=\sqrt{15} \in(1,5)\end{aligned}$

Hence, the mean value theorem has been 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:

We have, $\mathrm{y}=(\mathrm{x}-3)^2$, which is polynomial function.

So it is continuous and differentiable.

Thus conditions of the mean value theorem are satisfied.

Hence, there exists at least one $\mathrm{c} \in(3,4)$ such that,

$\begin{aligned} & \mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(4)-\mathrm{f}(3)}{4-3} \\ & \Rightarrow 2(\mathrm{c}-3)=\frac{1-0}{1} \\ & \Rightarrow \mathrm{c}-3=\frac{1}{2} \\ & \Rightarrow \mathrm{c}=\frac{7}{2} \in(3,4)\end{aligned}$

$\Rightarrow \mathrm{x}=\frac{7}{2}$, where tangent is parallel to the chord joining points $(3,0)$ and $(4,1)$.

For $\mathrm{x}=\frac{7}{2}, \mathrm{y}=\left(\frac{7}{2}-3\right)^2$

$\begin{aligned} & =\left(\frac{1}{2}\right)^2 \\ & =\frac{1}{4}\end{aligned}$

So, $\left(\frac{7}{2}, \frac{1}{4}\right)$ is the point on the curve, where the tangent drawn is parallel to the chord joining the points $(3,0)$ and $(4,1)$.

Question 78

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

Answer:

We have, $\mathrm{y}=2 \mathrm{x}^2-5 \mathrm{x}+3$, which is polynomial function.

So it is continuous and differentiable.

Thus conditions of the mean value theorem are satisfied.

Hence, there exists at least one $\mathrm{c} \in(1,2)$ such that,

$\begin{aligned} & \mathrm{f}(\mathrm{c})=\frac{\mathrm{f}(2)-\mathrm{f}(1)}{2-1} \\ & \Rightarrow 4 \mathrm{c}-5=\frac{1-0}{1} \\ & \Rightarrow 4 \mathrm{c}-5=1 \\ & \therefore \mathrm{c}=\frac{3}{2} \in(1,2)\end{aligned}$

For $\mathrm{x}=\frac{3}{2}, \mathrm{y}=2\left(\frac{3}{2}\right)^2-5\left(\frac{3}{2}\right)+3=0$

Hence, $\left(\frac{3}{2}, 0\right)$ is the points on the curve $\mathrm{y}=2 \mathrm{x}^2-5 \mathrm{x}+3$ between the points $\mathrm{A}(1,0)$ and $\mathrm{B}(2,1)$, where tangent is parallel to the chord AB .

Question 79

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

Answer:

Given that,
$f(x)=\left\{\begin{array}{cl} x^{2}+3 x+p, & \text { if } x \leq 1 \\ q x+2, & \text { if } x>1 \end{array}\right.$
It 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} $
$=\lim \frac{\left[h^{2}-5 h+p+4-4-p\right]}{-h} $
$ = \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 logs 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

$\begin{aligned} & \mathrm{m} \cdot \frac{1}{\mathrm{x}}+\mathrm{n} \cdot \frac{1}{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}=(\mathrm{m}+\mathrm{n}) \frac{1}{\mathrm{x}+\mathrm{y}} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+\mathrm{y})=\frac{m}{x}+\frac{n}{y} \frac{d y}{d x}=\frac{m+n}{x+y} \cdot\left(1+\frac{d y}{d x}\right) \\ & \Rightarrow\left(\frac{n}{y}-\frac{m+n}{x+y}\right) \frac{d y}{d x}=\frac{m+n}{x+y}-\frac{m}{x} \\ & \Rightarrow\left(\frac{n x+n y-m y-n y}{y(x+y)}\right) \frac{d y}{d x}=\left(\frac{m x+n x-m x-m y}{x(x+y)}\right) \\ & \Rightarrow\left(\frac{n x-m y}{y(x+y)}\right) \frac{d y}{d x}=\left(\frac{n x-m x}{x(x+y)}\right) \\ & \Rightarrow \frac{d y}{d x}=\left(\frac{n x-m x}{x(x+y)}\right)\left(\frac{y(x+y)}{n x-m y}\right) \\ & \Rightarrow \frac{d y}{d x}=\frac{y}{x}\end{aligned}$

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 = sin t 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 logs on both sides, we get

$\log u=\tan x \log x$

Differentiating w.r.t x , we get

$\begin{aligned} & \Rightarrow \frac{1}{u} \frac{d u}{d x}=\tan x \cdot \frac{1}{x}+\log x \cdot \sec ^2 x \\ & \Rightarrow \frac{d u}{d x}=u\left(\tan x \cdot \frac{1}{x}+\log x \cdot \sec ^2 x\right) \\ & \Rightarrow \frac{d u}{d x}=x^{\tan x}\left(\tan x \cdot \frac{1}{x}+\log x \cdot \sec ^2 x\right)\end{aligned}$

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, $y=u+v$

$\begin{aligned} & \Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} \\ & \Rightarrow \frac{d y}{d x}=x^{\tan x}\left(\tan x \cdot \frac{1}{x}+\log x \cdot \sec ^2 x\right)+\frac{x}{2} \sqrt{\frac{2}{x^2+1}} \\ & \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x}^{\tan \mathrm{x}}\left(\frac{\tan \mathrm{x}}{\mathrm{x}}+\log \mathrm{x} \cdot \sec ^2 \mathrm{x}\right)+\frac{\mathrm{x}}{\sqrt{2\left(\mathrm{x}^2+1\right)}}\end{aligned}$

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.

Hence, the correct answer is option (D).

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:
Given that, $f(x) = |sinx|$
Let g(x) = sinx and h(x) = |x|
Then, f(x) = hog(x)
We know that the modulus function and sine function are continuous everywhere. Since
The imposition 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.
$\begin{aligned} &LHD=\lim _{0^{-}} \frac{\mathrm{v}(\mathrm{x})-\mathrm{v}(0)}{\mathrm{x}-0}\\ &\lim _{h \rightarrow 0} \frac{v(0-h)-v(0)}{(0-h)-0}\\ &\lim _{=h \rightarrow 0} \frac{|0-h|-|0|}{-h} \quad(\because v(x)=|x|)\\ &\lim _{h \rightarrow 0} \frac{|-h|}{-h}\\ &=\lim _{h \rightarrow 0} \frac{h}{-h}\\ &=\lim _{h \rightarrow 0}-1=-1\\ \end{aligned}$
$R H D=\lim _{h \rightarrow 0} \frac{v(0+h)-v(0)}{(0+h)-0}$

$=\lim _{h \rightarrow 0} \frac{|0+h|-|0|}{h} \quad(\because v(x)=|x|) \lim _{h \rightarrow 0} \frac{|h|}{h}=\lim _{h \rightarrow 0} \frac{h}{h}=\lim _{h \rightarrow 0} 1=1 \Rightarrow \mathrm{LY}^{\prime}(0) \neq \mathrm{RC}^{\prime}(0) \Rightarrow|\mathrm{x}|$ is not differentiable at $\mathrm{x}=0 . \Rightarrow \mathrm{h}(\mathrm{x})$ is not differentiable at $\mathrm{x}=0$.

So, $\mathrm{f}(\mathrm{x})$ is not differentiable where $\sin x=0$

We know that $\sin x=0$ at $x=n \pi, n \in Z$

Hence, $\mathrm{f}(\mathrm{x})$ is everywhere continuous but not differentiable $x=\mathrm{n} \pi, \mathrm{n} \in \mathrm{Z}$.

Hence, the correct answer is option (B).

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:

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

Hence, the correct answer is option (B).

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:
$\mathrm{y}=\sqrt{\sin \mathrm{x}+y}$

Squaring both sides, we get

$y^2=\sin x+y$

Differentiating w.r.t y, we get

$\begin{aligned} & 2 y=\cos x \frac{d x}{d y}+1 \\ & \frac{d x}{d y}=\frac{2 y-1}{\cos x} \\ & \therefore \frac{d y}{d x}=\frac{\cos x}{2 y-1}\end{aligned}$

Hence, the correct answer is option (A).

Question 93

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

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

Answer:

Let $u=\cos ^{-1}\left(2 x^2-1\right)$ and $v=\cos ^{-1} x$

Now, $u=\cos ^{-1}\left(2 x^2-1\right)$

$\begin{aligned} & u=\cos ^{-1}\left(2 \cos ^2 v-1\right)\left[\cdot v=\cos ^{-1} x \rightarrow \cos v=x\right] \\ & u=\cos ^{-1}(\cos 2 v)\left[\because 2 \cos ^2 x-1=\cos 2 x\right] \\ & \Rightarrow u=2 v \\ & \frac{d u}{d v}=2\end{aligned}$

Hence, the correct answer is option (A).

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

Hence, the correct answer is option (B).

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:

The given function is $f(x)=x^3-3 x$.

This is a polynomial function, which is continuous and derivable in $R$.

Therefore, the function is continuous on $[0, \sqrt{3}]$ and derivable on $(0, \sqrt{3})$

Differentiating the given function with respect to $x$, we get

$\begin{aligned} & f^{\prime}(x)=3 x^2-3 \\ & \Rightarrow f^{\prime}(c)=3 c^2-3 \\ & \therefore f^{\prime}(c)=0 \\ & \Rightarrow 3 c^2-3=0 \\ & \Rightarrow c^2=1 \\ & \Rightarrow c= \pm 1\end{aligned}$

Thus, $c=1 \in[0, \sqrt{3}]$ for which Rolle's theorem holds.

Hence, the correct answer is option (B).

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:
The 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 the ean 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} $
$\Rightarrow c=\sqrt{3} \in(1,3)$

Hence, the correct answer is option (B).

Question 97 Fill in the blanks in each of the
An example of a function that 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 the polynomial function is continuous everywhere.
When 1≤x<2, we have f(x) = 1, which is a constant function and the constant function is continuous everywhere.
When x≥2, we have f(x) = 2x-3, which is a polynomial function and the 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:

We have, $f(x)=|\cos x-\sin x|$ Foc $\frac{\pi}{4}<x<\frac{\pi}{2}, \sin x>\cos x$

$\begin{aligned} & \therefore f(x)=\sin x-\cos x \\ & \Rightarrow f^{\prime}(x)=\cos x-(-\sin x)=\cos x+\sin x \\ & \Rightarrow f^{\prime}\left(\frac{\pi}{3}\right)=\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{1+\sqrt{3}}{2}\end{aligned}$

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, $\sqrt{x}+\sqrt{y}=1$

Differentiating with respect to x, we get,

$\begin{aligned} & \frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x}=0 \\ & \Rightarrow \frac{1}{2 \sqrt{y}} \frac{d y}{d x}=-\frac{1}{2 \sqrt{x}} \\ & \Rightarrow \frac{d y}{d x}=-\frac{1}{2 \sqrt{x}} \times \frac{2 \sqrt{y}}{1} \\ & \Rightarrow \frac{d y}{d x}=-\frac{\sqrt{y}}{\sqrt{x}}\end{aligned}$

Now, $\left[\frac{d y}{d x}\right]_{\left(\frac{1}{4}, \frac{1}{4}\right)}=-\frac{\sqrt{\frac{1}{4}}}{\sqrt{\frac{1}{4}}}=-1$

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:
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 to f(x) since it is not differentiable at x=1 ∈ (0,2).

Hence, the statement is False.

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:
Given that, f is continuous on its domain D.
Let a be an arbitrary real number in D. Then f is continuous at a.
$\begin{aligned} & \lim _{x \rightarrow a} f(x)=f(a) \\ & \text { Now, } \\ & \lim _{x \rightarrow a}|f|(x)=\lim _{x \rightarrow a}|f(x)|[\because:|f|(x)=\mid f(x) L] \\ & \lim _{x \rightarrow a}|f|(x)=\left|\lim _{x \rightarrow a} f(x)\right| \lim _{x \rightarrow a}|f|(x)=|f(a)|=|f|(a)\end{aligned}$

If |f| is continuous at x=a.
Since a is an arbitrary point in D. Therefore |f| is continuous in D.

Hence, the statement is True.

Question 104 State True or False for the statements

The composition of two continuous functions is a continuous function.

Answer:
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 a modulus function is continuous in R.
⇒ g(x) = 1-x + |x| is everywhere continuous.
⇒ h(x) = |x| is everywhere continuous.
Hence, f = hog is everywhere continuous.

Hence, the statement is True.

Question 105 State True or False for the statements
Trigonometric and inverse-trigonometric functions are differentiable in their respective domain.

Answer:

It is an obvious statement. Trigonometric and inverse-trigonometric functions are differentiable within their respective domains. Their derivatives are well-defined and commonly used in calculus.

Hence, the statement is 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:
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 $\mathrm{x}=0$.

Hence, the statement is False.

Students can use the NCERT Exemplar Class 12 Math Solutions Chapter 5 PDF download, prepared by experts, for a 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 subtopics 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

Importance of Solving NCERT Exemplar Class 12 Maths Solutions Chapter 5

• The Exemplar problems go beyond the basics, helping students grasp more advanced concepts with greater clarity.
• Various types of questions, like MCQs, fill-in-the-blanks, true-false, short-answer type, and long-answer type, will enhance the logical and analytical skills of the students.
• These NCERT Exemplar exercises cover all the important topics and concepts so that students can be well-prepared for various exams.
• By solving these NCERT Exemplar problems, students will get to know about all the real-life applications of Continuity and Differentiability.

NCERT Exemplar Class 12 Maths Solutions Chapter-Wise

All NCERT Class 12 Maths Exemplar Solutions are gathered together on Careers360 for quick access. Click the links below to view them.

NCERT Solutions for Class 12 Maths: Chapter Wise

Given below is the chapter-wise list of the NCERT Class 12 Maths solutions with their respective links:

NCERT Notes for Class 12

Given below are the subject-wise NCERT Notes of class 12 :

• NCERT Notes for Class 12 Physics
• NCERT Notes for Class 12 Chemistry
• NCERT Notes for Class 12 Maths
• NCERT Notes for Class 12 Biology

NCERT Solutions for Class 12 Subject-wise

Here, you can find the NCERT Solutions for other subjects as well.

• NCERT solutions class 12 chemistry
• NCERT solutions for class 12 physics
• NCERT solutions for class 12 biology

Class-wise NCERT Solutions

Here, you can find the NCERT Solutions for classes 9 to 11.

• NCERT solutions for class 11
• NCERT solutions for class 10
• NCERT solutions for class 9

NCERT Books and NCERT Syllabus

Checking the updated syllabus at the start of the academic year helps students stay prepared. Below, you’ll find the syllabus links along with useful reference books.

• NCERT Books Class 12 Maths
• NCERT Syllabus Class 12 Maths
• NCERT Books Class 12
• NCERT Syllabus Class 12