NCERT Exemplar Class 12 Maths Solutions Chapter 5 Continuity and Differentiability 2021

Q: What is the difference between continuity and differentiability?

Continuity and differentiability are closely related but distinct concepts in calculus. A function is continuous at a point if there is no break, jump, or hole, meaning the graph flows smoothly through that point. On the other hand, a function is differentiable at a point if it has a defined derivative there, implying the graph has a well-defined tangent and no sharp corners or cusps. All differentiable functions are continuous, but not all continuous functions are differentiable. For example, the absolute value function is continuous everywhere but not differentiable at x=0 due to a sharp corner.

Q: Can a function be differentiable but not continuous?

No, a function cannot be differentiable without being continuous. Differentiability implies continuity. If a function is differentiable at a point, it must also be continuous at that point. This is because finding a derivative involves calculating a limit, and for that limit to exist, the function must not break or jump at that point. However, the reverse is not always true, a function can be continuous but not differentiable, such as the absolute value function at x=0, which is continuous but has a sharp corner, making it non-differentiable at that poin

Q: What is the significance of Rolle’s theorem in differentiability?

Rolle's Theorem is significant in understanding the behavior of differentiable functions. It states that if a function is continuous on [a, b], differentiable on (a, b), and f(a)=f(b), then there exists at least one point c in (a, b) where f'(c)=0. This means the function has a horizontal tangent (slope zero) at some point between a and b. Rolle's Theorem helps in proving other important results like the Mean Value Theorem and is useful in analyzing turning points, ensuring the existence of critical points in various real-life applications involving motion or optimization.

Q: What is the concept of logarithmic differentiation?

Logarithmic differentiation is a technique used to differentiate complex functions, especially those involving products, quotients, or variables raised to variable powers. The process involves taking the natural logarithm (ln) of both sides of the function, simplifying using logarithmic identities, and then differentiating implicitly. This method is particularly useful when dealing with functions like y=x x or y=(x 2 +1) x , where standard rules are difficult to apply directly. Logarithmic differentiation simplifies calculations, making it easier to handle functions with multiple terms or exponents, and is a powerful tool in both calculus and applied mathematics.

Q: Why is every differentiable function always continuous, but not vice versa?

Every differentiable function is always continuous because differentiability requires the function's limit to exist and match the function's value at a point. In other words, for a function to be differentiable, it must first be smooth and unbroken-i.e., continuous. However, not all continuous functions are differentiable. A function can be continuous without having a defined derivative at some points, especially where there are sharp turns. For example, the absolute value function |x| is continuous everywhere but not differentiable at x=0, since it has a sharp corner at that point.

NCERT Exemplar Class 12 Maths Solutions Chapter 5 Continuity and Differentiability

Updated on Apr 02, 2025 11:47 PM IST | #CBSE Class 12th

Limits and differentiability are fundamental concepts in calculus that help us understand change and continuity in mathematical functions. You can have a question about what the limit tells us. A limit describes the value that a function approaches as the input (variable) approaches a certain point, while differentiability indicates whether a function is differentiable at that point. These concepts are essential for analyzing motion, growth, and change in various fields. For example, in real life, the speed of a moving car at a specific instant is found using differentiation by calculating the limit of the average speed as the time interval approaches zero. The NCERT Class 12 Mathematics book covers everything that will come in the board exams and helps in creating a base of the topics.

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Class 12 Maths Chapter 5 exemplar solutions Exercise: 5.3
Page number: 107-116
Total questions: 106

Question:1

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

Answer:

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

For continuity at $x = 1$

$\begin{aligned} ∴ R.H.L. = lim_{x \to 1^{+}} f (x) \\ = lim_{h \to 0} f (1 + h) \\ = lim_{h \to 0} [(1 + h)^{3} + 2 (1 + h)^{2} - 1] = 2 \end{aligned}$

And L.H.L. $= lim_{x \to 1^{-}} f (x)$

$\begin{aligned} = lim_{h \to 0} f (1 - h) \\ = lim_{h \to 0} [(1 - h)^{3} + 2 (1 - h)^{2} - 1] = 2 \end{aligned}$

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

Thus $lim_{x \to 1^{+}} f (x) = lim_{x \to 1^{-}} f (x) = f (1)$

Thus $f (x)$ is continuous at $x = 1$

Question:2

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

Answer:

Given,
$f (x) = {\begin{array}{cl} 3 x + 5, & if x \geq 2 \\ x^{2}, & if x < 2 \end{array}$
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
$lim_{h \to 0} f (c - h) = lim_{h \to 0} f (c + h) = f (c)$
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 -
$\begin{aligned} lim_{h \to 0} f (2 - h) = lim_{h \to 0} f (2 + h) = f (2) \\ Now we can see that, \\ L H L = lim_{h \to 0} f (2 - h) = lim_{h \to 0} (2 - h)^{2} Using equation \\ ∴ L H L = (2 - 0)^{2} = 4 \dots (2) \end{aligned}$
Similarly, we proceed with RHL-
$\begin{aligned} lim_{RHL} = \underset{h \to 0}{f (2 + h)} = lim_{h \to 0} {3 (2 + h) + 5} \\ ∴ RHL = 3 (2 + 0) + 5 = 11 \dots (3) \\ then, \\ f (2) = 3 (2) + 5 = 11 \dots (4) \end{aligned}$
Now, from equations 2, 3, and 4, we can conclude that
$\begin{aligned} lim_{h \to 0} f (2 - h) \neq lim_{h \to 0} f (2 + h) \\ ∴ f (x) is discontinuous at x = 2 \end{aligned}$

Question:3

Find which of the functions is continuous or discontinuous at the indicated points:
$f (x) = {\begin{array}{cl} \frac{1 - \cos 2 x}{x^{2}}, & if x \neq 0 \\ 5, & if x = 0 \end{array} at x = 0$

Answer:

Given,
$f (x) = {\begin{array}{cl} \frac{1 - \cos 2 x}{x^{2}}, & if x \neq 0 \\ 5, & if x = 0 \end{array}$
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
$lim_{h \to 0} f (c - h) = lim_{h \to 0} f (c + h) = f (c)$

Where $h$ is a very small number very close to $0 (h \to 0)$ Now according to above theory- $f (x)$ is continuous at $x = 0$ if

$lim_{h \to 0} f (0 - h) = lim_{h \to 0} f (0 + h) = f (0)$
$\begin{aligned} then, \\ lim_{h HL = h \to 0} f (- h) = lim_{h \to 0} \frac{1 - \cos 2 (- h)}{(- h)^{2}} \underset{⏟}{{using equation 1}} \\ As we know \cos (- θ) = \cos θ \\ \Rightarrow LHL = lim_{h \to 0} \frac{1 - \cos 2 h}{h^{2}} \\ ∵ 1 - \cos 2 x = 2 \sin^{2} x \\ ∴ \lim_{h \to 0} \frac{2 \sin^{2} h}{h^{2}} = 2 lim_{h \to 0} {(\frac{\sin h}{h})}^{2} \end{aligned}$
This limit can be evaluated directly by putting the value of h because it is taking the indeterminate form (0/0)
As we know,
$\begin{aligned} lim_{x \to 0} \frac{\sin x}{x} = 1 \\ ∴ LHL = 2 \times 1^{2} = 2 \dots (2) \\ Similarly, we proceed for RHL- \\ lim_{RHL} h \to 0^{+} f (h) = lim_{h \to 0} \frac{1 - \cos 2 (h)}{(h)^{2}} \\ \Rightarrow RHL = lim_{h \to 0} \frac{1 - \cos 2 h}{h^{2}} \\ \Rightarrow RHL = lim_{h \to 0} \frac{2 \sin^{2} h}{h^{2}} = 2 lim_{h \to 0} {(\frac{\sin h}{h})}^{2} \end{aligned}$
Again, using the sandwich theorem, we get -
$R H L = 2 \times 1^{2} = 2. . . (3)$
And,
f (0) = 5 …(4)
From equations 2, 3, and 4 we can conclude that
$lim_{h \to 0} f (0 - h) = lim_{h \to 0} f (0 + h) \neq f (0)$
∴ f(x) is discontinuous at x = 0

Question:4

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

Answer:

Given,
$f (x) = {\begin{aligned} \frac{2 x^{2} - 3 x - 2}{x - 2}, if & x \neq 2 \\ 5, & if x = 2 \end{aligned}$
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
$lim_{h \to 0} f (c - h) = lim_{h \to 0} f (c + h) = f (c)$
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 -
$lim_{h \to 0} f (2 - h) = lim_{h \to 0} f (2 + h) = f (2)$
Then,
$\begin{aligned} lim_{h \to 0} f (2 - h) = lim_{h \to 0} \frac{2 (2 - h)^{2} - 3 (2 - h) - 2}{(2 - h) - 2} {using equation 1} \\ \Rightarrow LHL = lim_{h \to 0} \frac{2 (4 + h^{2} - 4 h) - 6 + 3 h - 2}{- h} \\ \Rightarrow \lim \frac{8 + 2 h^{2} - 8 h - 6 + 3 h - 2}{- h} = lim_{h \to 0} \frac{2 h^{2} - 5 h}{- h} \\ \Rightarrow LHL = lim_{h \to 0} \frac{- h (5 - 2 h)}{- h} = lim_{h \to 0} (5 - 2 h) \\ ∴ LHL = 5 - 2 (0) = 5 \dots (2) \end{aligned}$
Similarly, we proceed with RHL-

$lim_{RHL = h \to 0} f (2 + h) = lim_{h \to 0} \frac{2 (2 + h)^{2} - 3 (2 + h) - 2}{(2 + h) - 2} [using equation 1}$
$\Rightarrow RHL = lim_{h \to 0} \frac{2 (4 + h^{2} + 4 h) - 6 - 3 h - 2}{h}$ $\Rightarrow RHL = lim_{h \to 0} \frac{8 + 2 + 8 h - 6 - 3 h - 2}{- h} = lim_{h \to 0} \frac{2 h^{2} + 5 h}{h}$

$\Rightarrow lim_{h \to 0} \frac{h (5 + 2 h)}{h} = lim_{h \to 0} (5 + 2 h)$

$∴ RHL = 5 + 2 (0) = 5 \dots (3)$

And, $f (2) = 5 {$ using equation 1 $} \dots (4)$
From the above equations 2, 3, and 4, we can say that

$lim_{h \to 0} f (2 - h) = lim_{h \to 0} f (2 + h) = f (2) = 5$
∴ f(x) is continuous at x = 2

Question:5

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

Answer:

Given,
$f (x) = {\begin{array}{cl} \frac{| x - 4 |}{2 (x - 4)}, if & x \neq 4 \\ 0, & if x = 4 \end{array}$ ...(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
$lim_{h \to 0} f (c - h) = lim_{h \to 0} f (c + h) = f (c)$
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 -
$lim_{h \to 0} f (4 - h) = lim_{h \to 0} f (4 + h) = f (4)$
Clearly,

$\begin{aligned} lim_{h HL} f (4 - h) = lim_{h \to 0} \frac{| 4 - h - 4 |}{2 (4 - h - 4)} {using equation 1} \\ \Rightarrow LHL = lim_{h \to 0} \frac{| - h |}{- 2 h} \\ ∵ h > 0 as defined above. \\ ∴ | - h | = h \\ \Rightarrow LHL = lim_{h \to 0} \frac{h}{- 2 h} = - \frac{1}{2} lim_{h \to 0} 1 \\ ∴ LHL = - 1 / 2 \dots (2) \end{aligned}$
Similarly, we proceed with RHL-
$\begin{aligned} RHL = lim_{h \to 0} f (4 + h) = lim_{h \to 0} \frac{| 4 + h - 4 |}{2 (4 + h - 4)} {using equation 1} \\ \Rightarrow RHL = lim_{h \to 0} \frac{| h |}{2 (h)} \\ ∵ h > 0 as defined above. \\ ∴ | h | = h \end{aligned}$
$\Rightarrow RHL = lim_{h \to 0} \frac{h}{2 h} = \frac{1}{2} lim_{h \to 0} 1$
$∴ RHL = 1 / 2 \dots (3)$ And, $f (4) = 0 {$ using eqn 1 $} \dots (4)$

From equations 2, 3, and 4, we can conclude that

$lim_{h \to 0} f (4 - h) \neq lim_{h \to 0} f (4 + h) \neq f (4)$

∴ f(x) is discontinuous at x = 4

Question: 6

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

Answer:

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

At $x = 0$

$\begin{aligned} L.H.L. = lim_{x \to 0^{-}} | x | \cos \frac{1}{x} \\ = lim_{h \to 0} | 0 - h | \cos \frac{1}{0 - h} \\ = lim_{h \to 0} h \cos \frac{1}{h} \\ = 0 \times [an oscillating number between - 1 and 1] = 0 \end{aligned}$
$\begin{aligned} R.H.L. = lim_{x \to 0^{+}} | x | \cos \frac{1}{x} \\ = lim_{h \to 0} | 0 + h | \cos \frac{1}{0 + h} \\ = lim_{h \to 0} h \cos \frac{1}{h} \\ = 0 \times [an oscillating number between - 1 and 1] = 0 \end{aligned}$
Also $f (0) = 0 \dots$ . (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 $f (x) = {\begin{array}{cl} | x - a | \sin \frac{1}{x - a}, if & x \neq a \\ 0, & if x = a \end{array}$

Answer:

We have, $f (x) = {\begin{cases} | x - a | \sin \frac{1}{x - a}, & if x \neq 0 \\ 0, & if x = a \end{cases}$ at $x = a$ At $x = a$

$\begin{aligned} L.H.L. = lim_{x \to a^{-}} | x - a | \sin \frac{1}{x - a} \\ = lim_{h \to 0} | a - h - a | \sin (\frac{1}{a - h - a}) \\ = lim_{h \to 0} - h \sin \frac{1}{h} \\ = 0 \times [an oscillating number between - 1 and 1] = 0 \end{aligned}$

$\begin{aligned} R.H.L. = lim_{x \to a^{+}} | x - a | \sin (\frac{1}{x - a}) \\ = lim_{h \to 0} | a + h - a | \sin (\frac{1}{a + h - a}) \\ = lim_{h \to 0} h \sin \frac{1}{h} \\ = 0 \times [an oscillating number between - 1 and l] = 0 \end{aligned}$

Also $f (a) = 0 \dots ($ Given $)$

Thus L.H.L. $=$ R.H.L. $= f (a)$

So, $f (x)$ is continuous at $x = a$ .

Question:8 Find which of the functions is continuous or discontinuous at the indicated points:
$f (x) = {\begin{array}{cl} \frac{e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}}, if & x \neq 0 \\ 0, & if x = 0 \end{array}$
at x = 0
Answer:

Given,
$f (x) = {\begin{array}{cl} \frac{e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}}, if & x \neq 0 \\ 0, & if x = 0 \end{array}$
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
$lim_{h \to 0} f (c - h) = lim_{h \to 0} f (c + h) = f (c)$
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 -

$\begin{aligned} lim_{h \to 0} f (0 - h) = lim_{h \to 0} f (0 + h) = f (0) \\ Clearly, \\ lim_{h \to 0} f (- h) = lim_{h \to 0} {\frac{e^{\frac{1}{- h}}}{1 + e^{- \frac{1}{- h}}}}_{{u sing equation 1}} \\ \Rightarrow LHL = \frac{e^{\frac{1}{- 0}}}{1 + e^{- \frac{1}{- 0}}} = \frac{e^{- \infty}}{1 + e^{- \infty}} = \frac{0}{1 + 0} = 0 \\ ∴ LHL = 0 . . . (2) \end{aligned}$
Similarly, we proceed for RHL-
$\begin{array}{l} lim_{h \to 0} f (h) = lim_{h \to 0} {\frac{e^{\frac{1}{h}}}{1 + e^{\frac{1}{h}}}} {using equation 1} \\ lim_{h \to 0} {\frac{e^{\frac{1}{h}}}{e^{\frac{1}{h} (1 + e^{- \frac{1}{h}})}}} = lim_{h \to 0} {\frac{1}{(1 + e^{- \frac{1}{h}})}} \end{array}$
$\begin{array}{l} \Rightarrow RHL = \frac{1}{1 + e - 0} = \frac{1}{1 + e^{- \infty}} = \frac{1}{1 + 0} = 1 \\ ∴ RHL = 1 \dots (3) \end{array}$
And,
f(0) = 0 {using eqn 1} …(4)
From equations 2, 3, and 4, we can conclude that
$lim_{h \to 0} f (0 - h) \neq lim_{h \to 0} f (0 + h)$
∴ f (x) is discontinuous at x = 0

Question:9

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

Answer:

We have, $f (x) = {\begin{cases} \frac{x^{2}}{2}, & if 0 \leq x \leq 1 \\ 2 x^{2} - 3 x + \frac{3}{2}, & if 1 < x \leq 2 \end{cases}$ at $x = 1$

At $x = 1$

L.H.L. $= lim_{x \to 1^{-}} \frac{x^{2}}{2}$

$\begin{aligned} = lim_{h \to 0} \frac{(1 - h)^{2}}{2} \\ = lim_{h \to 0} \frac{1 + h^{2} - 2 h}{2} \\ = \frac{1}{2} \\ R.H.L. = lim_{x \to 1^{+}} (2 x^{2} - 3 x + \frac{3}{2}) \\ = lim_{h \to 0} [2 (1 + h)^{2} - 3 (1 + h) + \frac{3}{2}] \\ = 2 - 3 + \frac{3}{2} \\ = \frac{1}{2} \end{aligned}$

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

Hence, $f (x)$ is continuous at $x = 1$ .

Question:10

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

Answer:

We have, $f (x) = | x | + | x - 1 |$ at $x = 1$ At $x = 1$

$\begin{aligned} L.H.L. = lim_{x \to 1^{-}} [| x | + | x - 1 |] \\ = lim_{h - ? 0^{-}} [| 1 - h | + | 1 - h - 1 |] \\ = 1 + 0 \\ = 1 \end{aligned}$

$\begin{aligned} And R.H.L. = lim_{x \to +} [| x | + x - 1 ∣] \\ = lim_{h \to 0} [| 1 + h | + | 1 + h - 1 |] \\ = 1 + 0 \\ = 1 \end{aligned}$

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:
$f (x) = \begin{array}{cl} 3 x - 8, & if x \leq 5 \\ 2 k, & if x > 5 \end{array} at x = 5$

Answer:

Given,
$f (x) = \begin{array}{cl} 3 x - 8, & if x \leq 5 \\ 2 k, & if x > 5 \end{array} at x = 5$
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
$lim_{h \to 0} f (c - h) = lim_{h \to 0} f (c + h) = f (c)$
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.
$lim_{h \to 0} f (5 - h) = lim_{h \to 0} f (5 + h) = f (5)$
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)
$∴ lim_{h \to 0} f (5 - h) = f (5) \Rightarrow lim_{h \to 0} {3 (5 - h) - 8} = 2 k \Rightarrow 3 (5 - 0) - 8 = 2 k \Rightarrow 15 - 8 = 2 k \Rightarrow 2 k = 7 ∴ k = 7 / 2$

Question:12

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

Answer:

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

$\begin{aligned} ∴ f (2) = lim_{x \to 2} f (x) \\ ∴ k = lim_{x \to 2} \frac{2^{x + 2} - 16}{4^{x} - 16} \\ = lim_{x \to 2} \frac{4 (2^{x} - 4)}{(2^{x} - 4) (2^{x} + 4)} \\ = lim_{x \to 2} \frac{4}{2^{x} + 4} \\ = \frac{4}{4 + 4} \\ = \frac{1}{2} \end{aligned}$

Question:13

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

Answer:

Given,

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

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
$lim_{h \to 0} f (c - h) = lim_{h \to 0} f (c + h) = f (c)$
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.
$lim_{h \to 0} f (0 - h) = lim_{h \to 0} f (0 + h) = f (0)$
Now, to find k, pick out a combination using which we get k in our equation.
In this question we take LHL = f(0)
$\begin{array}{l} lim_{h \to 0} f (- h) = f (0) \\ \Rightarrow lim_{h \to 0} {\frac{\sqrt{1 + k (- h)} - \sqrt{1 - k (- h)})}{- h}} = \frac{2 (0) + 1}{(0) - 1} {using eqn 1} \\ \Rightarrow lim_{h \to 0} {\frac{\sqrt{1 + k h} - \sqrt{1 - k h}}{h}} = - 1 \end{array}$
We can’t find the limit directly, because it is taking the 0/0 form.
Thus, we will rationalize it.
$\begin{aligned} \Rightarrow lim_{h \to 0} {\frac{\sqrt{1 + k h} - \sqrt{1 - k h}}{h} \times \frac{\sqrt{1 + k h} + \sqrt{1 - k h}}{\sqrt{1 + k h} + \sqrt{1 - k h}}} = - 1 \\ Using (a + b) (a - b) = a^{2} - b^{2}, we have - \\ lim_{h \to 0} {\frac{(\sqrt{1 + k h})^{2} - (\sqrt{1 - k h)})^{2}}{h (\sqrt{1 + k h} + \sqrt{1 - k h})}} = - 1 \\ \Rightarrow lim_{h \to 0} {\frac{2 k h}{h (\sqrt{1 + k h} + \sqrt{1 - k h})}} = - 1 \\ lim_{h \to 0} {\frac{2 k}{(\sqrt{1 + k h} + \sqrt{1 - k h})}} = \\ \Rightarrow \frac{2 k}{\sqrt{1 + k (0)} + \sqrt{1 - k (0)}} = - 1 \\ ∴ 2 k / 2 = - 1 \\ ∴ k = - 1 \end{aligned}$

Question:14

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

Answer:

Given,
$\begin{matrix} f (x) = \frac{1 - \cos kx}{x \sin x}, if x \neq 0 \\ \frac{1}{2}, if x = 0 \end{matrix}$
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
$lim_{h \to 0} f (c - h) = lim_{h \to 0} f (c + h) = f (c)$
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.
$lim_{h \to 0} f (0 - h) = lim_{h \to 0} f (0 + h) = f (0)$
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)
$\begin{array}{l} lim_{h \to 0} f (- h) = f (0) \\ lim_{h \to 0} {\frac{1 - \cos k (- h)}{(- h) \sin (- h)}} = \frac{1}{2} {u ing equation 1} \end{array}$
$\begin{array}{l} ∵ \cos (- x) = \cos x and \sin (- x) = - \sin x \\ lim_{h \to 0} {\frac{1 - \cos k (h)}{(h) \sin (h)}} = \frac{1}{2} \\ Also, 1 - \cos x = 2 \sin^{2} (x / 2) \\ lim_{h \to 0} {\frac{2 \sin^{2} (\frac{k h}{2})}{(h) \sin (h)}} = \frac{1}{2} \end{array}$
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 -
$lim_{x \to 0} \frac{\sin x}{x} = 1 \Rightarrow lim_{h \to 0} {\frac{\sin^{2} (\frac{k h}{2})}{(h) \sin (h)}} = \frac{1}{4}$
Dividing and multiplying by $(k h / 2)^{2}$ to match the form in the formula we have-
$\begin{aligned} lim_{h \to 0} {\frac{\sin^{2} (\frac{kh}{2})}{(h) \sin (h) \times {(\frac{kh}{2})}^{2}} \times {(\frac{kh}{2})}^{2}} = \frac{1}{4} \\ Using algebra of limits we get - \\ lim_{h \to 0} {(\frac{\sin \frac{k h}{2}}{\frac{k h}{2}})}^{2} \times lim_{h \to 0} \frac{k^{2}}{4} (\frac{h}{\sin h}) = \frac{1}{4} \end{aligned}$
$\begin{aligned} Applying the formula- \\ \Rightarrow 1 \times (k^{2} / 4) = (1 / 4) \\ \Rightarrow k^{2} = 1 \\ \Rightarrow (k + 1) (k - 1) = 0 \\ ∴ k = 1 or k = - 1 \end{aligned}$

Question:15

Prove that the function f defined by

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

Answer:

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

At $x = 0$

$\begin{aligned} L.H.L. = lim_{x \to 0^{+}} \frac{(0 - h)}{| 0 - h | + 2 (0 - h)^{2}} \\ = lim_{h \to 0} \frac{- h}{h + 2 h^{2}} \\ = lim_{h \to 0} \frac{- 1}{1 + 2 h} \\ = - 1 \\ R.H.L. = lim_{x \to 0^{+}} \frac{x}{| x | + 2 x^{2}} \\ = lim_{h \to 0} \frac{0 + h}{| 0 + h | + 2 (0 + h)^{2}} \\ = lim_{h \to 0} \frac{h}{h + 2 h^{2}} \\ = lim_{h \to 0} \frac{1}{1 + 2 h} \\ = 1 \end{aligned}$

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
$f (x) = {\begin{cases} \frac{x - 4}{| x - 4 |} + a, & if x < 4 \\ a + b & , if x = 4 \\ \frac{x - 4}{| x - 4 |} + b, & if x > 4 \end{cases}$
is a continuous function at x = 4.

Answer:

Given,
$f (x) = {\begin{cases} \frac{x - 4}{| x - 4 |} + a, & if x < 4 \\ a + b & , if x = 4 \\ \frac{x - 4}{| x - 4 |} + b, & if x > 4 \end{cases}$ …(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
$lim_{h \to 0} f (c - h) = lim_{h \to 0} f (c + h) = f (c)$
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.
$∴ lim_{h \to 0} f (4 - h) = lim_{h \to 0} f (4 + h) = f (4)$
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)
$∴ lim_{h \to 0} f (4 - h) = f (4)$
$\Rightarrow lim_{h \to 0} {\frac{4 - h - 4}{| 4 - h - 4 |} + a} = a + b$ {using equation 1}
$\Rightarrow lim_{h \to 0} {\frac{- h}{| - h |} + a} = a + b$
$∵$ h > 0 as defined in the theory above.
$∴ | - h | = h$
$\Rightarrow lim_{h \to 0} {\frac{- h}{h} + a} = a + b$
$\Rightarrow lim_{h \to 0} {a - 1} = a + b$
$\Rightarrow$ a - 1 = a + b
$∴$ b = -1
Now, taking another combination,
RHL = f(4)
$lim_{h \to 0} f (4 + h) = f (4)$
$\Rightarrow lim_{h \to 0} {\frac{4 + h - 4}{| 4 + h - 4 |} + b} = a + b$ {using equation 1}
$lim_{h \to 0} {\frac{h}{| h |} + b} = a + b$
$∵$ h > 0 as defined in the theory above.
$∴$ |h| = h
$\Rightarrow lim_{h \to 0} {\frac{h}{h} + b} = a + b$
$\Rightarrow lim_{h \to 0} {b + 1} = a + b$
⇒ 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 discontinuity of composite function f(f(x))

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

$∴$ x = -2 is a point of discontinuity
$∵ f (f (x)) = f (\frac{1}{x + 2}) = \frac{1}{\frac{1}{x + 2} + 2} = \frac{x + 2}{2 x + 5}$
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 $f (t) = \frac{1}{t^{2} + t - 2}, t = \frac{1}{x - 1}$ .

Answer:

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

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

$\begin{aligned} ∴ f (t) = \frac{1}{{(\frac{1}{x - 1})}^{2} + \frac{1}{x - 1} - 2} \\ = \frac{(x - 1)^{2}}{1 + (x - 1) - 2 (x - 1)^{2}} \\ = \frac{(x - 1)^{2}}{- (2 x^{2} - 5 x + 2)} \\ = \frac{(x - 1)^{2}}{(2 x - 1) (2 - x)} \end{aligned}$

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

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

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

Question:19

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

Answer:

Given,

$f (x) = | \sin x + \cos x | \underset{―}{\dots} (1)$

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

$lim_{h \to 0} f (c - h) = lim_{h \to 0} f (c + h) = f (c)$

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 -

$lim_{h \to 0} f (π - h) = lim_{h \to 0} f (π + h) = f (π)$

Now,

LHL = $lim_{h \to 0} f (π - h)$

⇒ LHL = $lim_{h \to 0} {| \sin (π - h) + \cos (π - h) |}$ {using eqn 1}

$∵ \sin (π - x) = \sin x & \cos (π - x) = - \cos x$

$\Rightarrow LHL = lim_{h \to 0} | \sinh - \cosh |$

$\Rightarrow LHL = | \sin 0 - \cos 0 | = | 0 - 1 |$

$∴ LHL = 1 \underset{―}{\dots (2)}$

Similarly, we proceed for RHL-

$RHL = lim_{h \to 0} f (π + h)$

$\Rightarrow RHL = lim_{h \to 0} {| \sin (π + h) + \cos (π + h) |}$ {using eqn 1}

$∵ \sin (π + x) = - \sin x & \cos (π + x) = - \cos x$

$\Rightarrow RHL = lim_{h \to 0} | - \sin h - \cosh |$

$\Rightarrow RHL = | - \sin 0 - \cos 0 | = | 0 - 1 |$

$∴ RHL = 1 \dots (3)$

$Also, f (π) = | \sin π + \cos π | = | 0 - 1 | = 1 \underset{―}{\dots (4)}$

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

$lim_{h \to 0} f (π - h) = lim_{h \to 0} f (π + h) = f (π) = 1$

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

Question:20

Examine the differentiability of f, where f is defined by

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

Answer:

Given,
$f (x) = {\begin{array}{cc} x [x], & if 0 \leq x < 2 \\ (x - 1) x, & if 2 \leq x < 3 \end{array} at x = 2$ …(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
$lim_{h \to 0} f (c - h) = lim_{h \to 0} f (c + h) = f (c)$
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
$lim_{x \to c^{-}} \frac{f (x) - f (c)}{x - c} = lim_{x \to c^{+}} \frac{f (x) - f (c)}{x - c}$
$lim_{h \to 0} \frac{f (c - h) - f (c)}{c - h - c} = lim_{h \to 0} \frac{f (c + h) - f (c)}{c + h - c}$
Finally, we can state that for a function to be differentiable at x = c

$lim_{h \to 0} \frac{f (c - h) - f (c)}{- h} = lim_{h \to 0} \frac{f (c + h) - f (c)}{h}$
Checking for the continuity:
Now according to the above theory-
f(x) is continuous at x = 2 if -
$lim_{h \to 0} f (2 - h) = lim_{h \to 0} f (2 + h) = f (2)$
$∴ LHL = lim_{h \to 0} f (2 - h)$
$\Rightarrow LHL = lim_{h \to 0} (2 - h) [2 - h]_{{using equation 1}}$
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 = $lim_{h \to 0} f (2 + h)$
⇒ RHL = $lim_{h \to 0} (2 + h - 1) (2 + h)_{{using equation 1}}$
$∴ RHL = (1 + 0) (2 + 0) = 2 \dots (3)$
And, f(2) = (2-1)(2) = 2 …(4) {using equation 1}
From equation 2,3 and 4, we observe that:
$lim_{h \to 0} f (2 - h) = lim_{h \to 0} f (2 + h) = f (2) = 2$
∴ 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 -
$lim_{h \to 0} \frac{f (2 - h) - f (2)}{- h} = lim_{h \to 0} \frac{f (2 + h) - f (2)}{h}$
$∴$ LHD = $lim_{h \to 0} \frac{f (2 - h) - f (2)}{- h}$
⇒ LHD = $lim_{h \to 0} \frac{(2 - h) [2 - h] - (2 - 1) (2)}{- h} {using equation 1}$
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 = $lim_{h \to 0} \frac{(2 - h) \times 1 - 2}{- h}$
⇒ LHD = $lim_{h \to 0} \frac{- h}{- h} = lim_{h \to 0} 1$
$∴ LHD = 1 \underset{―}{\dots} (5)$
Now,
RHD = $lim_{h \to 0} \frac{f (2 + h) - f (2)}{h}$
⇒ RHD = $lim_{h \to 0} \frac{(2 + h - 1) (2 + h) - (2 - 1) (2)}{h} {using equation 1}$
⇒ RHD = $lim_{h \to 0} \frac{(1 + h) (2 + h) - 2}{h} = lim_{h \to 0} \frac{2 + h^{2} + 3 h - 2}{h}$
∴ RHD = $lim_{h \to 0} \frac{h (h + 3)}{h} = lim_{h \to 0} (h + 3)$
$\Rightarrow RHD = 0 + 3 = 3 \underset{―}{\dots} (6)$
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
$f (x) = {\begin{array}{cl} x^{2} \sin \frac{1}{x}, & if x \neq 0 \\ 0, & if x = 0 \end{array} at x = 0$

Answer:

Given,
$f (x) = {\begin{array}{cl} x^{2} \sin \frac{1}{x}, & if x \neq 0 \\ 0, & if x = 0 \end{array} at x = 0$
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
$lim_{h \to 0} f (c - h) = lim_{h \to 0} f (c + h) = f (c)$
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
$lim_{x \to c^{-}} \frac{f (x) - f (c)}{x - c} = lim_{x \to c^{+}} \frac{f (x) - f (c)}{x - c}$
$lim_{h \to 0} \frac{f (c - h) - f (c)}{c - h - c} = lim_{h \to 0} \frac{f (c + h) - f (c)}{c + h - c}$
Finally, we can state that for a function to be differentiable at x = c
$lim_{h \to 0} \frac{f (c - h) - f (c)}{- h} = lim_{h \to 0} \frac{f (c + h) - f (c)}{h}$
Checking for the continuity:
Now according to the above theory-
f(x) is continuous at x = 0 if -
$lim_{h \to 0} f (0 - h) = lim_{h \to 0} f (0 + h) = f (0) ∴ L H L = h \to 0 \Rightarrow L H L = lim_{h \to 0} {(- h)^{2} \sin (\frac{1}{- h})}_{{u sing} equation 1}$
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,
$lim_{RHL} = lim_{h \to 0} f (h) \Rightarrow RHL = lim_{h \to 0} {(h)^{2} \sin (\frac{1}{h})}_{{using equation 1}}$
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:
$lim_{h \to 0} f (0 - h) = lim_{h \to 0} f (0 + h) = f (0)$
∴ 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 -
$lim_{h \to 0} \frac{f (0 - h) - f (0)}{- h} = lim_{h \to 0} \frac{f (0 + h) - f (0)}{h} ∴ L H D = lim_{h \to 0} \frac{f (- h) - f (0)}{- h} \Rightarrow L H D = lim_{h \to 0} \frac{(- h)^{2} \sin (\frac{1}{- h}) - 0}{- h} {using equation 1} \Rightarrow L H D = lim_{h \to 0} \frac{h^{2} \sin (\frac{1}{- h})}{- h} = lim_{h \to 0} {h \sin (\frac{1}{h})}$
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,
$RHD = lim_{h \to 0} \frac{f (h) - f (0)}{h} \Rightarrow R H D = h \to 0 \frac{(h)^{2} \sin (\frac{1}{h}) - 0}{h} {using equation 1}$
$\Rightarrow RHD = lim_{h \to 0} {h \sin (\frac{1}{h})}$
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
$f (x) = {\begin{cases} 1 + x, & if x \leq 2 \\ 5 - x, & if x > 2 \end{cases} at x = 2$

Answer:

Given,
$f (x) = {\begin{cases} 1 + x, & if x \leq 2 \\ 5 - x, & if x > 2 \end{cases} at x = 2$
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
$lim_{h \to 0} f (c - h) = lim_{h \to 0} f (c + h) = f (c)$ .
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
$lim_{x \to c^{-}} \frac{f (x) - f (c)}{x - c} = lim_{x \to c^{+}} \frac{f (x) - f (c)}{x - c} lim_{h \to 0} \frac{f (c - h) - f (c)}{c - h - c} = lim_{h \to 0} \frac{f (c + h) - f (c)}{c + h - c}$
Finally, we can state that for a function to be differentiable at x = c
$lim_{h \to 0} \frac{f (c - h) - f (c)}{- h} = lim_{h \to 0} \frac{f (c + h) - f (c)}{h}$
Checking for the continuity:
Now according to the above theory-
f(x) is continuous at x = 2 if -
$lim_{h \to 0} f (2 - h) = lim_{h \to 0} f (2 + h) = f (2) ∴ L H L = lim_{h \to 0} f (2 - h)$
$\begin{aligned} \Rightarrow LHL = lim_{h \to 0} {1 + (2 - h)}_{{using equation 1}} \\ \Rightarrow LHL = lim_{h \to 0} {3 - h} \\ ∴ L H L = (3 - h) = 3 \\ ∴ LHL = 3 \dots (2) \\ Similarly, \\ lim_{RHL} = h_{h \to 0} f (2 + h) \\ \Rightarrow RHL = lim_{h \to 0} {5 - (2 + h)}_{{using equation 1}} \\ \Rightarrow RHL = lim_{h \to 0} {3 + h} \\ ∴ RHL = 3 + 0 = 3 . .0 (3) \end{aligned}$
And, f(2) = 1 + 2 = 3 {using equation 1} …(4)
From equation 2,3 and 4, we observe that:
$lim_{h \to 0} f (2 - h) = lim_{h \to 0} f (2 + h) = f (2) = 3$
∴ 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 -
$\begin{array}{l} lim_{h \to 0} \frac{f (2 - h) - f (2)}{- h} = lim_{h \to 0} \frac{f (2 + h) - f (2)}{h} \\ ∴ L H D = lim_{h \to 0} \frac{f (2 - h) - f (2)}{- h} \end{array}$
$\begin{aligned} \Rightarrow LHD = lim_{h \to 0} \frac{1 + (2 - h) - (1 + 2)}{- h} {using equation 1} \\ \Rightarrow LHD = lim_{h \to 0} \frac{3 - h - 3}{- h} = lim_{h \to 0} 1 \\ ∴ LHD = 1 . .0 (5) \\ Now, \\ lim_{RHD} = lim_{h \to 0} \frac{f (2 + h) - f (2)}{h} \\ \Rightarrow RHD = lim_{h \to 0} \frac{5 - (2 + h) - 3}{h} {using equation 1} \\ \Rightarrow lim_{RHD} = \underset{h \to 0}{- h} = lim_{h \to 0} - 1 \\ ∴ RHD = - 1 \dots (6) \end{aligned}$
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 $f (x) = | x - 5 |$

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

$\begin{aligned} L.H.L. lim_{h \to 5^{-}} f (x) = - (x - 5) \\ = lim_{h \to 0} - (5 - h - 5) \\ = lim_{h \to 0} h = 0 \\ R.H.L. lim_{x \to 5^{+}} f (x) = x - 5 \\ = lim_{h \to 0} (5 + h - 5) \\ = lim_{h \to 0} h = 0 \end{aligned}$

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

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

Now, for differentiability

$\begin{aligned} {Lf}^{'} (5) = lim_{h \to 0} \frac{f (5 - h) - f (5)}{- h} \\ = lim_{h \to 0} \frac{- (5 - h - 5) - (5 - 5)}{- h} \\ = lim_{h \to 0} \frac{h}{- h} \\ = - 1 \\ {Rf}^{'} (5) = lim_{h \to 0} \frac{f (5 + h) - f (5)}{h} \\ = lim_{h \to 0} \frac{(5 + h - 5) - (5 - 5)}{h} \\ = lim_{h \to 0} \frac{h - 0}{h} \\ = 1 \\ ∵ {Lf}^{'} (5) \neq Rd (5) \end{aligned}$

Hence, $f (x)$ is not differentiable at $x = 5$ .

Question:24 A function $f : R \to R$ 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, $f : R \to R$ 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 $x = 0$ at which the function $f (x)$ is differentiable.

$\begin{aligned} ∴ f^{'} (0) = lim_{h \to 0} \frac{f (0 + h) - f (0)}{h} \\ 2 = lim_{h \to 0} \frac{f (0) \cdot f (h) - f (0)}{h} . \\ \Rightarrow 2 = lim_{h \to 0} \frac{f (0) [f (h) - 1]}{h} \end{aligned}$

Now $f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

$\begin{aligned} = lim_{h \to 0} \frac{f (x) \cdot f (h) - f (x)}{h} \\ = lim_{h \to 0} \frac{f (x) [f (h) - 1]}{h} \\ = 2 f (x) \end{aligned}$

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

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,
$\begin{aligned} \log y = \log 2^{\cos^{2} x} \\ \log y = \cos^{2} x \cdot \log 2 \\ Now, Differentiate w.r.t x \\ \frac{1}{y} \frac{d y}{d x} = \frac{d}{d x} [\cos^{2} x \cdot \log 2] \\ \frac{1}{y} \frac{d y}{d x} = [2 \cos x \cdot (- \sin x) \cdot \log 2] \\ \frac{d y}{d x} = y [2 \cos x \cdot (- \sin x) \cdot \log 2] \end{aligned}$
Now, substitute the value of y
$\frac{dy}{dx} = - 2^{\cos^{2} x} [2 \cos x \cdot (\sin x) \cdot \log 2]$

$Hence, \frac{dy}{dx} = - 2^{\cos^{2} x} [2 \cos x \cdot (\sin x) \cdot \log 2]$

Question:26 Differentiate each of the following w.r.t. x

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

Answer:

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

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

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

Differentiating both sides w.r.t. X

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

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

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

Answer:

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

Differentiating both sides w.r.t. x

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

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 $y = \log [\log (\log x^{5})]$

Differentiating both sides w.r.t. x

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

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

Question:29

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

Answer:

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

Differentiating both sides w.r.t. x

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

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

Question:30

Differentiate each of the following w.r.t. x
$s i n^{n} (a x^{2} + b x + c)$

Answer:

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

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 $\tan \sqrt{x + 1} = v$
$So, y = \cos v$
Now, differentiate w.r.t v
$\frac{dy}{d v} = (- \sin v)$
And, $v =$ tan $w$
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,
$\frac{dy}{dx} = \frac{dy}{dv} \times \frac{dv}{dw} \times \frac{d w}{dx}$ \ $\frac{d y}{d x} = (- \sin v) \times \sec^{2} w \times \frac{1}{2 \sqrt{x + 1}}$
Substitute the value of v and w
$\frac{dy}{dx} = (- \sin (\tan \sqrt{x + 1}) \times \sec^{2} \sqrt{x + 1} \times \frac{1}{2 \sqrt{x + 1}}) .$
Question:32 Differentiate each of the following w.r.t. x $s i n x^{2} + s i n^{2} x + s i n^{2} (x^{2})$

Answer:

Let us Assume $y = s i n x^{2} + s i n^{2} x + s i n^{2} (x^{2})$
$\begin{aligned} ∴ \frac{dy}{dx} = \frac{d}{dx} \sin (x^{2}) + \frac{d}{dx} (\sin x)^{2} + \frac{d}{dx} {(\sin x^{2})}^{2} \\ = \cos (x^{2}) \frac{d}{dx} (x^{2}) + 2 \sin x \cdot \frac{d}{dx} (\sin x) + 2 \sin x^{2} \frac{d}{dx} (\sin x^{2}) \\ = 2 x \cos x^{2} + 2 \sin x \cos x + 2 \sin x^{2} \cos x^{2} \frac{d}{dx} x^{2} \\ = 2 x \cos x^{2} + 2 \sin x \cos x + 2 \sin x^{2} \cos x^{2} \times 2 x \\ = 2 x \cos (x^{2}) + \sin 2 x + 2 x \sin 2 (x^{2}) \end{aligned}$

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

Answer:

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

$\begin{aligned} ∴ \frac{dy}{dx} = \frac{d}{dx} (\sin^{- 1} \frac{1}{\sqrt{x + 1}}) \\ = \frac{1}{\sqrt{1 - {(\frac{1}{\sqrt{x + 1}})}^{2}}} \cdot \frac{d}{dx} \frac{1}{(x + 1)^{2}} \\ = \frac{1}{\sqrt{\frac{x + 1 - 1}{x + 1}}} \cdot \frac{d}{dx} (x + 1)^{2} \\ = \sqrt{\frac{x + 1}{x}} \cdot \frac{- 1}{2} (x + 1)^{\frac{- 3}{2}} \\ = \frac{- 1}{2 \sqrt{x}} \cdot (\frac{1}{x + 1}) \end{aligned}$

Question:34 Differentiate each of the following w.r.t. x
$(s i n x)^{c o s x}$

Answer:

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

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

Question:35

Differentiate each of the following w.r.t. x
$s i n^{m} x . c o s^{n} x$

Answer:

It is given $s i n^{m} x . c o s^{n} x$
$y = s i n^{m} x . c o s^{n} x$
$\begin{aligned} ∴ \frac{dy}{dx} = \frac{d}{dx} [(\sin x)^{m} \cdot (\cos x)^{n}] \\ = (\sin x)^{m} \frac{d}{dx} (\cos x)^{n} + (\cos x)^{n} \frac{d}{dx} (\sin x)^{m} \\ = (\sin x)^{m} n (\cos x)^{n - 1} \frac{d}{dx} (\cos x) + (\cos x)^{n} m (\sin x)^{m - 1} \frac{d}{dx} (\sin x) \\ = (\sin x)^{m} n (\cos x)^{n - 1} (- \sin x) + (\cos x)^{n} m (\sin x)^{m - 1} \cos x \\ = \sin^{m} x \cos^{n} x [- n \tan x + m \cot x] \end{aligned}$

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

Answer:

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

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

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

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

Question:37

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

Answer:

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

Question:38

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

Answer:

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

$\frac{d y}{d x} = {\begin{matrix} - \frac{1}{2}, - \frac{π}{4} < x < 0 \\ \frac{1}{2}, 0 \leq x < \frac{π}{4} \end{matrix}$

Question:39

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

Answer:

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

Differentiating both sides w.r.t. x

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

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

Question:40

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

Answer:

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

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

Differentiating both sides concerning x

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

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

Question:41

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

Answer:

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

Put $x = \cos θ$

$\begin{aligned} ∴ θ = \cos^{- 1} x \\ y = \sec^{- 1} (\frac{1}{4 \cos^{3} θ - 3 \cos θ}) \\ \Rightarrow y = \sec^{- 1} (\frac{1}{\cos 3 θ}) \dots \dots [∵ \cos 3 θ = 4 \cos^{3} θ - 3 \cos θ] \\ \Rightarrow y = \sec^{- 1} (\sec 3 θ) \\ \Rightarrow y = 3 θ \\ y = 3 \cos^{- 1} x \end{aligned}$

Differentiating both sides w.r.t. X

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

Hence, $\frac{d y}{d x} = \frac{- 3}{\sqrt{1 - x^{2}}}$ .

Question:42 Differentiate each of the following w.r.t. x
$\tan^{- 1} \frac{3 a^{2} x - x^{3}}{a^{3} - 3 {ax}^{2}}, \frac{- 1}{\sqrt{3}} < \frac{x}{\sqrt{3}} < \frac{x}{a} < \frac{1}{\sqrt{3}}$

Answer:

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

Question:43 Differentiate each of the following w.r.t. X $\tan^{- 1} (\frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}), 1 < x < 1, x \neq 0$

Answer:

$\begin{aligned} We have given \\ \tan^{- 1} (\frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}) \\ y = \tan^{- 1} (\frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}) \\ Put x^{2} = \cos 2 θ \\ So \end{aligned}$
$y = \tan^{- 1} (\frac{\sqrt{1 + \cos 2 θ} + \sqrt{1 - \cos 2 θ}}{\sqrt{1 + \cos 2 θ - \sqrt{1 - \cos 2 θ}}}) y = \tan^{- 1} (\frac{\sqrt{2 \cos^{2} θ} + \sqrt{2 \sin^{2} θ}}{\sqrt{2 \cos^{2} θ} - \sqrt{2 \sin^{2} θ}}) y = \tan^{- 1} (\frac{\sqrt{2} \cos θ + \sqrt{2} \sin θ}{\sqrt{2} \cos θ - \sqrt{2} \sin θ}) y = \tan^{- 1} (\frac{\cos θ + \sin θ}{\cos θ - \sin θ})$
$y = \tan^{- 1} (\frac{\frac{\cos θ}{\cos θ} + \frac{\sin θ}{\cos θ}}{\frac{\cos θ}{\cos θ} - \frac{\sin θ}{\cos θ}}) y = \tan^{- 1} (\frac{1 + \tan θ}{1 - \tan θ}) y = \tan^{- 1} (\frac{\tan \frac{π}{4} + \tan θ}{1 - \tan \frac{π}{4} \tan θ}) y = \tan^{- 1} \tan (\frac{π}{4} + θ)$
$\begin{aligned} y = \frac{π}{4} + θ \\ Differentiate, y w.r.t x \\ \frac{d y}{d x} = \frac{π}{4} + \frac{1}{2} \frac{d}{d x} \cos^{- 1} x^{2} \\ \frac{d y}{d x} = 0 + \frac{1}{2} \cdot \frac{- 1}{\sqrt{1 - x^{4}}} \frac{d}{d x} (x^{2}) \\ \frac{d y}{d x} = \frac{1}{2} \cdot \frac{- 2 x}{\sqrt{1 - x^{4}}} \\ Hence, \frac{dy}{dx} = \frac{- x}{\sqrt{1 - x^{4}}} \end{aligned}$

Question:44 Find $\frac{d y}{d x}$ 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, $\frac{d}{d x} (\frac{1}{x}) = - \frac{1}{x^{2}}$ 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,
$\begin{aligned} \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} \\ \frac{d y}{d x} = \frac{1 + \frac{1}{t^{2}}}{1 - \frac{1}{t^{2}}} \\ Hence, \\ \frac{dy}{dx} = \frac{t^{2} + 1}{t^{2} - 1} \end{aligned}$

Question:45

Find $\frac{d y}{d x}$ of each of the functions expressed in parametric form
$x = e^{θ} (θ + \frac{1}{θ}), y = e^{- θ} (θ - \frac{1}{θ})$

Answer:

Given that, $x = e^{θ} (θ + \frac{1}{θ}), y = e^{- θ} (θ - \frac{1}{θ})$

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

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

$\begin{aligned} \Rightarrow e^{- θ} (\frac{θ^{2} + 1 - θ^{3} + θ}{θ^{2}}) \\ = e^{- θ} \frac{(- θ^{3} + θ^{2} + θ + 1)}{θ^{2}} \\ ∴ \frac{dy}{dx} = \frac{\frac{dy}{d θ}}{\frac{d x}{d θ}} \\ = \frac{e^{- θ} (\frac{- θ^{3} + θ^{2} + θ + 1}{θ^{2}})}{e^{θ} (\frac{θ^{3} + θ^{2} + θ + 1}{θ^{2}})} \\ = e^{- 2 θ} (\frac{- θ^{3} + θ^{2} + θ + 1}{θ^{3} + θ^{2} + θ - 1}) \end{aligned}$

Hence, $\frac{dy}{dx} = e^{- 2 θ} (\frac{- θ^{3} + θ^{2} + θ + 1}{θ^{3} + θ^{2} + θ - 1})$

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

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

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

$\begin{aligned} \frac{dx}{d θ} = - 3 \sin θ - 6 \cos^{2} θ \cdot \frac{d}{d θ} (\cos θ) \\ = - 3 \sin θ - 6 \cos^{2} θ \cdot (- \sin θ) \\ = - 3 \sin θ + 6 \cos^{2} θ \cdot \sin θ \\ \frac{dy}{d θ} = 3 o s θ - 6 \sin^{2} θ \cdot \frac{d}{d θ} (\sin θ) \\ == 3 \cos θ - 6 \sin^{2} θ \cdot \cos θ \\ ∴ \frac{dy}{dx} = \frac{\frac{d y}{d θ}}{\frac{dx}{d θ}} \end{aligned}$

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

$∴ \frac{dy}{dx} = \cot θ$

Question:47

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

Question:48 Find dy/dx of each of the functions expressed in parametric form
$x = \frac{1 + \log t}{t^{2}}, y = \frac{3 + 2 \log t}{t}$

Answer:

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

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

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

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

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

Question:49

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

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

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

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

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

Question:50

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

Answer:

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

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

Question:51

If , find $\frac{dy}{dx} at t = \frac{π}{3}$

Answer:

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

$\begin{aligned} \frac{d x}{d t} = \frac{d}{d t} (3 \sin t - \sin 3 t) \\ \frac{d x}{d t} = \frac{d}{d t} (3 \sin t) - \frac{d}{d t} (\sin 3 t) \\ \frac{d x}{d t} = 3 \cos t - 3 \cos 3 t \\ Now, for y \\ y = 3 cost - \cos 3 t \end{aligned}$
$\frac{dy}{dt} = \frac{d}{dt} (3 \cos t - \cos 3 t) \frac{dy}{dt} = 3 (- \sin t) - \frac{d}{dt} (\cos 3 t) \frac{dy}{dt} = 3 (- \sin t) - 3 (- \sin 3 t) \frac{dy}{dt} = - 3 \sin t + 3 \sin 3 t \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{d x}{dt}}$
$\frac{d y}{d x} = \frac{- 3 \sin t + 3 \sin 3 t}{3 \cos t - 3 \cos 3 t} At t = π / 3 \frac{d y}{d x} = \frac{- 3 \sin \frac{π}{3} + 3 \sin \frac{3 π}{3}}{3 \cos \frac{π}{3} - 3 \cos \frac{3 π}{3}} \frac{d y}{d x} = \frac{- \sin \frac{π}{3} + \sin π}{\cos \frac{π}{3} - \cos π} \frac{d y}{d x} = \frac{- \frac{\sqrt{3}}{2} + 0}{\frac{1}{2} - (- 1)}$
$\frac{dy}{dx} = \frac{- \frac{\sqrt{3}}{2}}{\frac{1}{2} + 1} \frac{dy}{dx} = \frac{- \frac{\sqrt{3}}{2}}{\frac{3}{2}} \frac{dy}{dx} = - \frac{\sqrt{3}}{2} \times \frac{2}{3} \frac{dy}{dx} = \frac{- 1}{\sqrt{3}}$

Question:52

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

Answer:

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

Question:53

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

Answer:

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

Question:54

Find $\frac{d y}{d x}$ 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
$\frac{d}{d x} [\frac{f (x)}{g (x)}] = \frac{g (x) f^{'} (x) - f (x) g^{'} (x)}{[g (x)]^{2}}$
On differentiating both the sides concerning x, we get

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

Question:55

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

$s e c (x + y) = x y$

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:
$\begin{aligned} \frac{d}{d x} [f (x) g (x)] = f (x) g^{'} (x) + g (x) f^{'} (x) \\ On differentiating both sides with respect to x, we get \\ \sec (x + y) \tan (x + y) \frac{d}{d x} (x + y) = y + x \frac{d}{d x} y \\ [∵ \frac{d}{d x} \sec (x) = \sec x \tan x] \\ \Rightarrow \sec (x + y) \tan (x + y) [1 + \frac{d y}{d x}] = y + x \frac{dy}{dx} \end{aligned}$
$\Rightarrow \sec (x + y) \tan (x + y) + \sec (x + y) \tan (x + y) \frac{d y}{d x} = y + x \frac{d y}{d x} \Rightarrow \sec (x + y) \tan (x + y) \frac{d y}{d x} - x \frac{d y}{d x} = y - \sec (x + y) \tan (x + y) \Rightarrow \frac{d y}{d x} [\sec (x + y) \tan (x + y) - x] = y - \sec (x + y) \tan (x + y) ∴ \frac{d y}{d x} = \frac{y - \sec (x + y) \tan (x + y)}{\sec (x + y) \tan (x + y) - x}$

Question:58

If $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$ , then show that $\frac{d y}{d x} \cdot \frac{d x}{d y} = 1$

Answer:

Given: $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$
Differentiating the above concerning x, we get
$2 ax + 2 h [x \frac{dy}{dx} + y] + b \frac{d}{dx} (y^{2}) + 2 g + 2 f \frac{d}{dx} y = 0 \Rightarrow 2 ax + 2 hx \frac{dy}{dx} + 2 hy + 2 by \frac{dy}{dx} + 2 g + 2 f \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} [2 hx + 2 by + 2 f] = - 2 ax - 2 hy - 2 g \Rightarrow \frac{dy}{dx} = \frac{- [2 ax + 2 hy + 2 g]}{2 hx + 2 by + 2 f} \dots (ii)$
Now, we again differentiate eq (i) concerning y, we get,

$a \frac{d}{d y} (x^{2}) + 2 h [x + y \frac{d}{d y} x] + 2 y b + 2 g \frac{d x}{d y} + 2 f = 0 \Rightarrow 2 a x \frac{d x}{d y} + 2 h x + 2 h y \frac{d x}{d y} + 2 b y + 2 g \frac{d x}{d y} + 2 f = 0 \Rightarrow \frac{d x}{d y} [2 a x + 2 h y + 2 g] = - 2 h x - 2 b y - 2 f \Rightarrow \frac{d x}{d y} = \frac{- [2 h x + 2 b y + 2 f]}{2 a x + 2 h y + 2 g} \dots (i i i)$
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 $\frac{d y}{d x} = \frac{x - y}{x \log x}$

Answer:

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

Taking $\log$ on both the sides,

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

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

Differentiating both sides w.r.t. X

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

Hence, $^{'}$ "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: $y^{x} = e^{y - x}$

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

$\begin{aligned} \Rightarrow x \log y = (y - x) \log e \\ \Rightarrow x \log y = y - x \dots . . [∵ \log e = 1] \\ \Rightarrow x \log y + x = y \\ \Rightarrow x (\log y + 1) = y \\ \Rightarrow x = \frac{y}{\log y + 1} \end{aligned}$

Differentiating both sides w.r.t. Y

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

We know that

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

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

Question:61

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

Answer:

Given that $y = (\cos x)^{(\cos x)^{(\cos x) \dots \infty}}$ ,

$\Rightarrow y = (\cos x)^{y} \dots . . [y = (\cos x)^{(\cos x)^{(\cos x) \dots \infty}]}]$

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

Differentiating both sides w.r.t. x

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

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

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

$\begin{aligned} \Rightarrow \frac{d x}{d y} = - \sin a \cdot \frac{d}{d y} \cot (a + y) \\ \Rightarrow \frac{d x}{d y} = - \sin a [- {cosec}^{2} (a + y) \\ \Rightarrow \frac{d x}{d y} = \frac{\sin a}{\sin^{2} (a + y)} \\ ∴ \frac{d y}{d x} = \frac{1}{\frac{d x}{d y}} \\ = \frac{1}{\frac{\sin a}{\sin^{2} (a + y)}} \end{aligned}$

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

Hence proved.

Question:63

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

Answer:

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

Put $x = \sin θ$ and $y = \sin Φ$ .

$\begin{aligned} ∴ θ = \sin^{- 1} x and Φ = \sin^{- 1} y \\ \sqrt{1 - \sin^{2} θ} + \sqrt{1 - \sin^{2} ϕ} = a (\sin θ - \sin Φ) \\ \Rightarrow \sqrt{\cos^{2} θ} + \sqrt{\cos^{2} ϕ} = a (\sin θ - \sin Φ) \\ \Rightarrow \cos θ + \cos Φ = a (\sin θ - \sin Φ) \\ \Rightarrow \frac{\cos θ + \cos ϕ}{\sin θ - \sin ϕ} = a \end{aligned}$

$\begin{aligned} \Rightarrow \frac{2 \cos \frac{θ + ϕ}{2} \cdot \cos \frac{θ - ϕ}{2}}{2 \cos \frac{θ + ϕ}{2} \cdot \sin \frac{θ - ϕ}{2}} = a \\ \Rightarrow \frac{\cos (\frac{θ - ϕ}{2})}{\sin (\frac{θ - ϕ}{2})} = a \\ \Rightarrow \cot (\frac{θ - ϕ}{2}) = a \\ \Rightarrow \frac{θ - ϕ}{2} = \cot^{- 1} a \\ \Rightarrow θ - Φ = 2 \cot^{- 1} a \\ \Rightarrow \sin^{- 1} x - \sin^{- 1} y = 2 \cot^{- 1} a \end{aligned}$

Differentiating both sides w.r.t. X

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

Question:64 If $y = t a n^{- 1} x$ , find $\frac{d^{2} y}{d x^{2}}$ in terms of y alone.

Answer:

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

$\Rightarrow x = \tan y$

Differentiating both sides w.r.t. Y

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

$\frac{dy}{d x} = \frac{1}{\sec^{2} y} = \cos^{2} y$

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

$\begin{aligned} \Rightarrow \frac{d}{dx} (\frac{dy}{dx}) = \frac{d}{dx} (\cos^{2} y) \\ \Rightarrow \frac{d^{2} y}{{dx}^{2}} = 2 \cos y \cdot \frac{d}{dx} (\cos y) \\ \Rightarrow \frac{d^{2} y}{{dx}^{2}} = 2 \cos y (- \sin y) \cdot \frac{dy}{dx} \\ \Rightarrow \frac{d^{2} y}{{dx}^{2}} = - 2 \sin y \cos y \cdot \cos^{2} y \\ ∴ \frac{d^{2} y}{{dx}^{2}} = - 2 \sin y \cos^{3} y \end{aligned}$

Question:65

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

Answer:

We have, $f (x) = x (x - 1)^{2}$ 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 $f (1)$

$\Rightarrow f (0) = f (1)$

f satisfies the conditions of Rolle's theorem.

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

$\begin{aligned} \Rightarrow 3 c^{2} - 4 c + 1 = 0 \\ \Rightarrow (3 c - 1) (c - 1) = 0 \\ \Rightarrow c = \frac{1}{3} \in (0, 1) \end{aligned}$
Thus, Rolle’s theorem is verified.

Question:66

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

Answer:

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

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

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

Now $f (0) = 0 + 1 = 1$ and $f (\frac{π}{2}) = 1 + 0 = 1$

$\Rightarrow f (0) = f (\frac{π}{2})$

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

Hence, there exists at least one $c \in (0, \frac{π}{2})$ such that $f^{'} (c) = 0$

$\begin{aligned} ∴ 4 \sin^{3} c \cos c - 4 \cos^{3} c \sin c = 0 \\ \Rightarrow 4 \sin c \cos c (\sin^{2} c - \cos^{2} c) = 0 \\ \Rightarrow 4 \sin c \cos c (- \cos 2 c) = 0 \\ \Rightarrow - 2 \sin 2 c \cdot \cos 2 c = 0 \\ \Rightarrow \sin 4 c = 0 \\ \Rightarrow 4 c = π \\ \Rightarrow c = \frac{π}{4} \in (0, \frac{π}{2}) \end{aligned}$

Thus, Rolle’s theorem is verified.

Question:67

Verify Rolle’s theorem for each of the functions
$f (x) = l o g (x^{2} + 2) - l o g 3$ in [- 1, 1].

Answer:

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

We know that $x^{2} + 2$ and the logarithmic function are continuous and differentiable $∴ f (x) = \log (x^{2} + 2) - \log 3$ 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 $c \in (- 1, 1)$ such that $f^{'} (c) = 0$

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

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: $f (x) = x (x + 3) e^{- x / 2}$
$\Rightarrow f (x) = (x^{2} + 3 x) e^{\frac{- x}{2}}$
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:
$\Rightarrow f (x) = (x^{2} + 3 x) e^{\frac{- x}{2}}$
Since f(x) is the multiplication of algebra and exponential function,n and is defined everywhere in its domain.
$\Rightarrow f (x) = (x^{2} + 3 x) e^{\frac{- x}{2}}$ is continuous at x ∈ [-3,0]
Hence, condition 1 is satisfied.
Condition 2: $\Rightarrow f (x) = (x^{2} + 3 x) e^{\frac{- x}{2}}$
On differentiating f(x) concerning x, we get,

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

Question:69 Verify Rolle’s theorem for each of the functions
$f (x) = \sqrt{4 - x^{2}} in [- 2, 2]$

Answer:

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

Since $(4 - x^{2})$ 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 $c \in (- 2, 2)$ such that $f^{'} (c) = 0$ .

Now $f^{'} (x) = \frac{1}{2} {(4 - x^{2})}^{\frac{- 1}{2}} (- 2 x)$

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

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

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

Hence, Rolle's theorem has been verified.

Question:70

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

Answer:

Given: $f (x) = \begin{array}{ll} x^{2} + 1, & if 0 \leq x \leq 1 \\ 3 - x, & if 1 \leq x \leq 2 \end{array}$
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
$lim_{LHL} (x^{2} + 1) = 1 + 1 = 2$ \ $lim_{RHL} = lim_{x \to 1^{+}} (3 - x) = 3 - 1 = 2$ \ $∵ LHL = RHL = 2$
and $f (1) = 3 - x = 3 - 1 = 2$
$∴ f (x)$ is continuous at $x = 1$
Hence, condition 1 is satisfied.
Condition 2:
Now, we have to check f(x) is differentiable
$f (x) = {\begin{cases} x^{2} + 1, if 0 \leq x \leq 1 \\ 3 - x, if 1 \leq x \leq 2 \end{cases}$
On differentiating concerning x, we get
$\Rightarrow f^{'} (x) = {\begin{matrix} 2 x + 0, if 0 < x < 1 \\ 0 - 1, if 1 < x < 2 \end{matrix}$ or
$f^{'} (x) = {\begin{cases} 2 x, if 0 < x < 1 \\ - 1, if 1 < x < 2 \end{cases}$
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, $2 π$ ], where the tangent is parallel to the x-axis.

Answer:

We have $y = \cos x - 1$

$∴ \frac{dy}{dx} = - \sin x$

For the tangent to be parallel to the x-axis

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

$\begin{aligned} ∴ - \sin x = 0 \\ ∴ x = π \in [0, 2 π] \end{aligned}$

$y (π) = \cos π - 1 = - 2$

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

Question:72

Using Rolle’s theorem, find the point on the curve y = x(x - 4), $x \in [0, 4]$ where the tangent is parallel to x-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 $c \in (0, 4)$ such that $f^{'} (c) = 0$

$\Rightarrow 2 c - 4 = 0$

$\Rightarrow c = 2$

$\Rightarrow x = 2$ 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
$f (x) = \frac{1}{4 x - 1} in [1, 4]$

Answer:

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

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

Also, $f^{'} (x) = - \frac{4}{(4 x - 1)^{2}}$ , 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 $c \in (1, 4)$ such that

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

Hence mean value theorem has been verified.

Question:74

Verify the mean value theorem for each of the functions given
$f (x) = x^{3} - 2 x^{2} - x + 3$ 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 $c \in (0, 1)$ such that

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

Hence, the mean value theorem has been verified.

Question:75

Verify the mean value theorem for each of the functions given
$f (x) = s i n x - s i n 2 x$ in $[0, π]$

Answer:

We have, $f (x) = \sin x - \sin 2 x$ in $[0, π]$

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 $c \in (0, π)$ such that,

$\begin{aligned} f^{'} (c) = \frac{f (π) - f (0)}{π - 0} \\ \Rightarrow \cos c - 2 \cos 2 c = \frac{\sin π - \sin 2 π - \sin 0 + \sin 0}{π - 0} \\ \Rightarrow 2 \cos 2 c - \cos c = 0 \\ \Rightarrow 2 (2 \cos^{2} c - 1) - \cos c = 0 \\ \Rightarrow 4 \cos^{2} c - \cos c - 2 = 0 \\ \Rightarrow \cos c = \frac{1 \pm \sqrt{1 + 32}}{8} \\ = \frac{1 \pm \sqrt{33}}{8} \\ \Rightarrow c = \cos^{- 1} (\frac{1 \pm \sqrt{33}}{8}) \in (0, π) \end{aligned}$

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, $f (x) = \sqrt{25 - x^{2}}$ 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 $c \in (1, 5)$ such that,

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

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, $y = (x - 3)^{2}$ , 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 $c \in (3, 4)$ such that,

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

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

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

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

So, $(\frac{7}{2}, \frac{1}{4})$ 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 $y = 2 x^{2} - 5 x + 3$ 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, $y = 2 x^{2} - 5 x + 3$ , 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 $c \in (1, 2)$ such that,

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

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

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

Question:79

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

Answer:

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

$L f^{'} (1) = lim_{x \to 1^{-}} \frac{f (x) - f (1)}{x - 1} = lim_{h \to 0} \frac{f (1 - h) - f (1)}{(1 - h) - 1} = lim_{h \to 0} \frac{[{(1 - h)^{2} + 3 (1 - h) + p] - (1 + 3 + p)]}{(1 - h) - 1} (∵ f (x) = x^{2} + 3 x + p, if x \leq 1) = lim_{h \to 0} \frac{[(1 + h^{2} - 2 h + 3 - 3 h + p) - (4 + p)]}{- h} = h m \frac{[h^{2} - 5 h + p + 4 - 4 - p]}{- h} = lim_{h \to 0} \frac{[h^{2} - 5 h]}{- h} = lim_{h \to 0} \frac{h (h - 5)}{- h} = lim_{h \to 0} h (5 - h)$
$= 5 lim_{{Rf}^{'} (1)} = lim_{x \to 1} \frac{f (x) - f (1)}{x - 1} = lim_{h \to 0} \frac{f (1 + h) - f (1)}{(1 + h) - 1} = lim_{h \to 0} \frac{[(q (1 + h) + 2) - (q + 2)]}{(1 + h) - 1} (v f (x) = q x + 2, if x > 1) = lim_{h \to 0} \frac{[(q + q h + 2) - (q + 2)]}{h} = lim_{h \to 0} \frac{[q + q h + 2 - q - 2]}{h} = q$
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 $\frac{d y}{d x} = \frac{y}{x}$
B. If $x^{m} . y^{n} = (x + y)^{m + n}$ prove that $\frac{d^{2} y}{d x^{2}} = 0$

Answer:

A. We have,
$x^{m} . y^{n} = (x + y)^{m + n}$
Taking logs on both sides, we get

$\log (x^{m} y^{n}) = \log (x + y)^{m + n} \Rightarrow m \log x + n \log y = (m + n) \log (x + y)$

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

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

Hence proved.

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

Question:82

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

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

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

Now, $v = \sqrt{\frac{x^{2} + 1}{2}}$

$\Rightarrow v = {(\frac{x^{2} + 1}{2})}^{1 / 2}$

Differentiating w.r.t x, we get

$\Rightarrow \frac{d v}{d x} = \frac{1}{2} {(\frac{x^{2} + 1}{2})}^{- 1 / 2} \cdot \frac{2 x}{2} \Rightarrow \frac{d v}{d x} = \frac{x}{2} {(\frac{2}{x^{2} + 1})}^{1 / 2} = \frac{x}{2} \sqrt{\frac{2}{x^{2} + 1}}$

Now, $y = u + v$

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

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 ${(\frac{f (x)}{g (x)}), when g (x) \neq 0}$ 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 $f (x) = | s i n x |$ . Then
A. f is everywhere differentiable
B. f is everywhere continuous but not differentiable $x = n π, n \in Z$
C. f is everywhere continuous but not differentiable at $x = (2 n + 1) \frac{π}{2}, n \in Z$
D. None of these

Answer:
Given that, $f (x) = | s i n x |$
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.
$\begin{aligned} L H D = lim_{0^{-}} \frac{v (x) - v (0)}{x - 0} \\ lim_{h \to 0} \frac{v (0 - h) - v (0)}{(0 - h) - 0} \\ lim_{= h \to 0} \frac{| 0 - h | - | 0 |}{- h} (∵ v (x) = | x |) \\ lim_{h \to 0} \frac{| - h |}{- h} \\ = lim_{h \to 0} \frac{h}{- h} \\ = lim_{h \to 0} - 1 = - 1 \end{aligned}$
$R H D = lim_{h \to 0} \frac{v (0 + h) - v (0)}{(0 + h) - 0}$

$= lim_{h \to 0} \frac{| 0 + h | - | 0 |}{h} (∵ v (x) = | x |) lim_{h \to 0} \frac{| h |}{h} = lim_{h \to 0} \frac{h}{h} = lim_{h \to 0} 1 = 1 \Rightarrow {LY}^{'} (0) \neq {RC}^{'} (0) \Rightarrow | x |$ is not differentiable at $x = 0 . \Rightarrow h (x)$ is not differentiable at $x = 0$ .

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

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

Hence, $f (x)$ is everywhere continuous but not differentiable $x = n π, n \in Z$ .

Hence, the correct answer is option (B).

Question:91

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

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

Answer:

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

Hence, the correct answer is option (B).

Question:92

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

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

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

Squaring both sides, we get

$y^{2} = \sin x + y$

Differentiating w.r.t y, we get

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

Hence, the correct answer is option (A).

Question:93

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

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

Answer:

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

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

$\begin{aligned} u = \cos^{- 1} (2 \cos^{2} v - 1) [\cdot v = \cos^{- 1} x \to \cos v = x] \\ u = \cos^{- 1} (\cos 2 v) [∵ 2 \cos^{2} x - 1 = \cos 2 x] \\ \Rightarrow u = 2 v \\ \frac{d u}{d v} = 2 \end{aligned}$

Hence, the correct answer is option (A).

Question:94

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

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

Hence, the correct answer is option (B).

Question:95

The value of c in Rolle’s theorem for the function $f (x) = x^{3} - 3 x$ 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

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

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

B. $\sqrt{3}$
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
$f^{'} (c) = \frac{f (b) - f (a)}{b - a} We have, f (x) = x + \frac{1}{x}$
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
$f^{'} (c) = \frac{f (b) - f (a)}{b - a} \Rightarrow 1 - \frac{1}{c^{2}} = \frac{(3 + \frac{1}{3}) - (1 + 1)}{3 - 1} [∵ f^{'} (x) = 1 + \frac{1}{x^{2}}] \Rightarrow \frac{c^{2} - 1}{c^{2}} = \frac{(\frac{10}{3}) - (2)}{2} \Rightarrow \frac{c^{2} - 1}{c^{2}} = \frac{(\frac{10}{3}) - (2)}{2} = \frac{\frac{10 - 6}{3}}{2} = \frac{2}{3} \Rightarrow 3 (c^{2} - 1) = 2 c^{2} \Rightarrow 3 c^{2} - 2 c^{2} = 3 \Rightarrow c^{2} = 3 \Rightarrow c = \pm \sqrt{3} \Rightarrow c = \sqrt{3} \in (1, 3)$

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|
$\begin{array}{l} f (x) = {\begin{matrix} - (x - 1) - (x - 2), if x < 1 \\ (x - 1) - (x - 2), if 1 \leq x < 2 \\ (x - 1) + (x - 2), if x \geq 2 \end{matrix} \\ f (x) = {\begin{matrix} - 2 x + 3, if x < 1 \\ 1, if 1 \leq x < 2 \\ 2 x - 3, if x \geq 2 \end{matrix} \end{array}$
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
$\begin{array}{l} f (x) = {\begin{matrix} - (x - 1) - (x - 2), if x < 1 \\ (x - 1) - (x - 2), if 1 \leq x < 2 \\ (x - 1) + (x - 2), if x \geq 2 \end{matrix} \\ f (x) = {\begin{matrix} - 2 x + 3, if x < 1 \\ 1, if 1 \leq x < 2 \\ 2 x - 3, if x \geq 2 \end{matrix} \end{array}$
$L H D = lim_{x \to 1^{-}} \frac{f (x) - f (1)}{x - 1} = lim_{h \to 0} \frac{f (1 - h) - f (1)}{(1 - h) - 1} = lim_{h \to 0} \frac{[(- 2 (1 - h) + 3) - (- 2 + 3)]}{(1 - h) - 1} (∵ f (x) = - 2 x + 3, if x < 1)$
$\begin{aligned} = lim_{h \to 0} \frac{[- 2 + 2 h + 3 - 1]}{- h} \\ lim_{= h \to 0} \frac{[2 h]}{- h} = lim_{h \to 0} 2 = 2 \\ R L D = lim_{x \to 1^{+}} \frac{f (x) - f (1)}{x - 1} \\ = lim_{x \to 1^{+}} \frac{1 - 1}{1 - 1} (∵ f (x) = 1, if 1 \leq x < 2) \\ = 0 \\ \Rightarrow ⌊ f^{'} (1) \neq {Rf}^{'} (1) \\ \Rightarrow f (x) is not differentiable at x = 1 . \\ L f^{'} (2) = lim_{x \to 2^{-}} \frac{f (x) - f (2)}{x - 2} \end{aligned}$
$\begin{aligned} = lim_{x \to 2} \frac{1 - 1}{2 - 2} (: f (x) = 1, if 1 \leq x < 2 and f (2) = 2 \times 2 - 3 = 1) \\ = 0 \\ R f^{'} (2) = lim_{x \to 2^{+}} \frac{f (x) - f (2)}{x - 2} \\ = lim_{h \to 0} \frac{f (1 + h) - f (2)}{(1 + h) - 2} \\ = lim_{h \to 0} \frac{[(2 (1 + h) - 3) - (2 \times 2 - 3)]}{(1 + h) - 2} (∵ f (x) = 2 x - 3, if x \geq 2) \\ = lim_{h \to 0} \frac{[2 + 2 h - 3 - 1]}{h - 1} \\ = lim_{h \to 0} \frac{[2 h - 2]}{h - 1} = lim_{h \to 0} \frac{2 (h - 1)}{h - 1} = 2 \\ \Rightarrow {Lf}^{'} (2) \neq {Rf}^{'} (2) \end{aligned}$
⇒ 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:

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

Question:99 Fill in the blanks in each of the
If $f (x) = | c o s x |$ , then ’ $f^{'} (\frac{π}{4})$ =.

Answer:

$\begin{aligned} We have, f (x) = | \cos x | \\ Foc. 0 < x < \frac{π}{2}, \cos x > 0 \\ \begin{matrix} ∴ f (x) = \cos x \\ \Rightarrow f^{'} (x) = - \sin x \\ \Rightarrow f^{'} (\frac{π}{4}) = - \sin \frac{π}{4} = - \frac{1}{\sqrt{2}} \end{matrix} \\ Hence, \\ f^{'} (\frac{π}{4}) = - \frac{1}{\sqrt{2}} \end{aligned}$

Question:100 Fill in the blanks in each of the
If $f (x) = | c o s x - s i n x |$ , then $f^{'} \frac{π}{3} =$ .

Answer:

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

$\begin{aligned} ∴ f (x) = \sin x - \cos x \\ \Rightarrow f^{'} (x) = \cos x - (- \sin x) = \cos x + \sin x \\ \Rightarrow f^{'} (\frac{π}{3}) = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2} \end{aligned}$

Hence, $f^{'} (\frac{π}{3}) = \frac{1 + \sqrt{3}}{2}$

Question:101 Fill in the blanks in each of the
For the curve $\sqrt{x} + \sqrt{y} = 1, \frac{d y}{d x} a t (\frac{1}{4}, \frac{1}{4})$ 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, ${[\frac{d y}{d x}]}_{(\frac{1}{4}, \frac{1}{4})} = - \frac{\sqrt{\frac{1}{4}}}{\sqrt{\frac{1}{4}}} = - 1$

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,
$\begin{aligned} f (x) = {\begin{matrix} - (x - 1), if x < 1 \\ (x - 1), if x > 1 \end{matrix} \\ lim_{x \to 1^{-}} \frac{f (x) - f (1)}{x - 1} \\ lim_{h \to 0} \frac{f (1 - h) - f (1)}{(1 - h) - 1} \\ lim_{h \to 0} \frac{[1 - (1 - h) - 0]}{(1 - h) - 1} (∵ f (x) = 1 - x, if x < 1) \\ lim_{n \to 0} \frac{[h]}{- h} \\ lim_{= h \to 0} \frac{[h]}{- h} = lim_{h \to 0} - 1 = - 1 \\ lim_{{Rf}^{'} (1)} = lim_{x \to 1^{+}} \frac{f (x) - f (1)}{x - 1} \\ lim_{h \to 0} \frac{f (1 + h) - f (1)}{(1 + h) - 1} \\ lim_{h \to 0} \frac{[(1 + h) - 1 - 0]}{(1 + h) - 1} (∵ f (x) = x - 1, if 1 < x) \\ lim_{= h \to 0} \frac{[h]}{h} = lim_{h \to 0} 1 = 1 \end{aligned}$
⇒ 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.
$\begin{aligned} lim_{x \to a} f (x) = f (a) \\ Now, \\ lim_{x \to a} | f | (x) = lim_{x \to a} | f (x) | [∵: | f | (x) =∣ f (x) L] \\ lim_{x \to a} | f | (x) = | lim_{x \to a} f (x) | lim_{x \to a} | f | (x) = | f (a) | = | f | (a) \end{aligned}$

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 $x = 0$ .

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.

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NCERT Exemplar Class 12 Maths Solutions Chapter

NCERT Exemplar for Class 12 Mathematics Chapter 1: Relations and Functions

NCERT Exemplar Class 12 Mathematics Chapter 2: Inverse Trigonometric Functions

NCERT Exemplar Class 12 Mathematics Chapter 3: Matrices

NCERT Exemplar Class 12 Mathematics Chapter 4: Determinants

NCERT Exemplar Class 12 Mathematics Chapter 5: Continuity and Differentiability

NCERT Exemplar Class 12 Mathematics Chapter 6: Applications of derivatives

NCERT Exemplar Class 12 Mathematics Chapter 7: Integrals

NCERT Exemplar Class 12 Mathematics Chapter 8: Applications of Integrals

NCERT Exemplar Class 12 Mathematics Chapter 9: Differential equations

NCERT Exemplar Class 12 Mathematics Chapter 10: Vector Algebra

NCERT Exemplar Class 12 Mathematics Chapter 11: Three-DimensionalGeometry

NCERT Exemplar Class 12 Mathematics Chapter 12: Linear Programming

NCERT Exemplar Class 12 Mathematics Chapter 13: Probability

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NCERT Exemplar Class 12 Solutions

NCERT Exemplar Class 12 Chemistry Solutions

NCERT Exemplar Class 12 Physics Solutions

NCERT Exemplar Class 12 Biology Solutions

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

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

Must Read NCERT Solution subject-wise

Here are the subject-wise links for the NCERT solutions of class 12:

Also, check NCERT Books and NCERT Syllabus here

Here are some useful links for NCERT books and the NCERT syllabus for class 12:

Frequently Asked Questions (FAQs)

1. What is the difference between continuity and differentiability?

Continuity and differentiability are closely related but distinct concepts in calculus. A function is continuous at a point if there is no break, jump, or hole, meaning the graph flows smoothly through that point. On the other hand, a function is differentiable at a point if it has a defined derivative there, implying the graph has a well-defined tangent and no sharp corners or cusps. All differentiable functions are continuous, but not all continuous functions are differentiable. For example, the absolute value function is continuous everywhere but not differentiable at x=0 due to a sharp corner.

2. Can a function be differentiable but not continuous?

No, a function cannot be differentiable without being continuous. Differentiability implies continuity. If a function is differentiable at a point, it must also be continuous at that point. This is because finding a derivative involves calculating a limit, and for that limit to exist, the function must not break or jump at that point. However, the reverse is not always true, a function can be continuous but not differentiable, such as the absolute value function at x=0, which is continuous but has a sharp corner, making it non-differentiable at that poin

3. What is the significance of Rolle’s theorem in differentiability?

Rolle's Theorem is significant in understanding the behavior of differentiable functions. It states that if a function is continuous on [a, b], differentiable on (a, b), and f(a)=f(b), then there exists at least one point c in (a, b) where f'(c)=0. This means the function has a horizontal tangent (slope zero) at some point between a and b. Rolle's Theorem helps in proving other important results like the Mean Value Theorem and is useful in analyzing turning points, ensuring the existence of critical points in various real-life applications involving motion or optimization.

4. What is the concept of logarithmic differentiation?

Logarithmic differentiation is a technique used to differentiate complex functions, especially those involving products, quotients, or variables raised to variable powers. The process involves taking the natural logarithm (ln) of both sides of the function, simplifying using logarithmic identities, and then differentiating implicitly. This method is particularly useful when dealing with functions like y=x^x or y=(x²+1)^x, where standard rules are difficult to apply directly. Logarithmic differentiation simplifies calculations, making it easier to handle functions with multiple terms or exponents, and is a powerful tool in both calculus and applied mathematics.

5. Why is every differentiable function always continuous, but not vice versa?

Every differentiable function is always continuous because differentiability requires the function's limit to exist and match the function's value at a point. In other words, for a function to be differentiable, it must first be smooth and unbroken-i.e., continuous. However, not all continuous functions are differentiable. A function can be continuous without having a defined derivative at some points, especially where there are sharp turns. For example, the absolute value function |x| is continuous everywhere but not differentiable at x=0, since it has a sharp corner at that point.

Also Read

NCERT Solutions for Class 12 Physics Chapter 7 - Alternating Current

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

NCERT Solutions for Miscellaneous Exercise Chapter 7 Class 12 - Integrals

NCERT Solutions for Exercise 4.1 Class 11 Maths Chapter 4 - Principle of Mathematical Induction

Quadratic Equations Class 10th Notes - Free NCERT Class 10 Maths Chapter 4 Notes - Download PDF

NCERT Class 12th Maths Chapter 5 Continuity and Differentiability Notes

NCERT Class 12 Physics Chapter 4 Notes, Moving Charges and Magnetism Class 12 Chapter 4 Notes

NCERT Class 12 Physics Chapter 3 Notes, Current Electricity Class 12 Chapter 3 Notes

Atoms Class 12th Notes - Free NCERT Class 12 Physics Chapter 12 Notes - Download PDF

NCERT Class 12 Physics Chapter 6 Notes, Electromagnetic Induction Class 12 Chapter 6 Notes

NCERT Exemplar Class 12 Physics Solutions Chapter 8 Electromagnetic Waves

NCERT Syllabus 2025-26 for Class 6-12 - Download All Subjects PDF Free

NCERT Book Class 12 Physics 2025-26 FREE PDF Download

NCERT Book Class 12 English 2025-26 Free PDF Download

NCERT Books for Class 12 Hindi 2025-26; Download Chapter Wise PDF here

NCERT Book Class 12 Maths 2025-26 PDF Free Download

NCERT Books for Class 12 Biology 2025-26: Download Free PDF

NCERT Books for Class 12 - Download All Subjects PDFs Here

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Can I change my board from CBSE to CHSE Odisha in Class 12th?

Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.

Read Complete Answer

Manasvini Gupta 30 January,2025

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer

Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

Read Complete Answer

Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer

Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer

Ashvadha 12 July,2024

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Central Board of Secondary Education Class 12th Examination

JEE Test Series

Applications Open

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

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

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

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

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

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

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

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

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

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NCERT Exemplar Class 12 Maths Solutions Chapter 5 Continuity and Differentiability

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JEE Main Important Mathematics Formulas

NCERT Exemplar Class 12 Maths Solutions Chapter

NCERT Exemplar Class 12 Solutions

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

Must Read NCERT Solution subject-wise

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Also, check NCERT Books and NCERT Syllabus here

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