If $f\left( x \right)$ has a derivative at $x = a,$ then $\mathop {\lim }\limits_{x \to a} \frac{{xf\left( a \right) - af\left( x \right)}}{{x - a}}$ is equal to

$af\left( a \right) - f'\left( a \right)$

$f\left( a \right) + f'\left( a \right)$

$af\left( a \right) + f'\left( a \right)$

Let $f:R \to R$ be differentiable at $c \in R$ and $f(c) = 0.$ If $g(x) = \left| {f(x)} \right|,$ then at $x = c,$ $g$ is

Differentiable if $f'(c) \ne 0$

Not differentiable if $f'(c) = 0$

Let $f(x)=\left\{\begin{array}{cc}a x, & x<2 \\ a x^2+b x+3, & x \geq 2\end{array}\right.$ If $f$ is differentiable for all $x$, then the value of ( $a$, b) is equal to

$\left(\frac{3}{2}, \frac{9}{2}\right)$

$\left(\frac{3}{4},-\frac{9}{2}\right)$

If $\tan ^{-1} y-y+x=0$ then $\frac{d^2 y}{d x^2}$ is equal to

$\frac{2\left(1+y^2\right)}{y^4}$

The function $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {1 - 2x + 3{x^2} - 4{x^3} + \ldots {\rm{to}}\,\,\,\,\infty ,x \ne - 1}\\ {1,x = - 1} \end{array}} \right.$ is

Continuous and derivable at $x = - 1$

Continuous but not derivable at $x = - 1$

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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 601-630 of 932 questions

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601 Easy

601 of 932

Application +4 | -1

f(x)=\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)

, then

\frac{d}{d x}(f(x))

is equal to

View answer and explanation

Correct option: C

\frac{3}{1+x^2}

for all

x \in R-\left\{-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\}

Option C is correct because it correctly simplifies the inverse tangent function and finds its derivative, considering the domain restrictions.

602 Medium

602 of 932

Analyzing +4 | -1

f\left( x \right)

has a derivative at

x = a,

then

\lim\limits_{x \to a} \frac{{xf\left( a \right) - af\left( x \right)}}{{x - a}}

is equal to

View answer and explanation

Correct option: A

f\left( a \right) - af'\left( a \right)

Option A (

f\left( a \right) - af'\left( a \right)

) is correct because by adding and subtracting

af(a)

in the numerator and applying L'Hopital's rule, the limit simplifies to this expression.

603 Easy

603 of 932

Understanding +4 | -1

At how many points between the interval

(-\infty, \infty)

is the function

f(x)=\sin x

is not differentiable.

View answer and explanation

Correct option: A

Option A (0) is correct because the sine function, sin(x), is differentiable for all real numbers.

604 Easy

604 of 932

Application +4 | -1

Let

f: R \rightarrow R

be a function given by

f(x+y)=f(x) f(y)

for all

x, y \in R

f(x)=1+x g(x) \cdot \log _e 2

, where

\lim\limits _{x \rightarrow 0} g(x)=1

. Then,

f^{\prime}(x)=

View answer and explanation

Correct option: A

\log _e 2^{f(x)}

Option A is correct because using the given condition

f(x)=1+x g(x) \cdot \log _e 2

, where

\lim\limits _{x \rightarrow 0} g(x)=1

, and the functional equation, we find

f^{\prime}(x)=\log _e 2^{f(x)}

605 Medium

605 of 932

Analyzing +4 | -1

Let

f: R \rightarrow R

be a twice differentiable function such that

\lim \limits_{{x \rightarrow \infty}} f(x)=a(a \neq 0)

, then

\lim \limits_{{x \rightarrow \infty}} x^2 f^{\prime \prime}(x)

is equal to

View answer and explanation

Correct option: D

None of these

Option B (

\frac{{1}}{{16}}

) is correct because it represents the maximum value obtained by completing the square and integrating.

606 Medium

606 of 932

Application +4 | -1

y=a \cos (\log x)-b \sin (\log x)

, then the value of

x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}+y

View answer and explanation

Correct option: A

Option A (0) is correct because substituting the given function and its derivatives into the expression results in zero.

607 Easy

607 of 932

Application +4 | -1

If the function

f(x)=x^3+e^{x / 2}

and

g(x)=f^{-1}(x)

, then the value of

g^{\prime}(1)

View answer and explanation

Correct option: B

Option B (2) is correct because it uses the derivative of an inverse function and the given values.

608 Medium

608 of 932

Analyzing +4 | -1

Let

K

be the set of all real values of

x

, where the function

f(x) = \sin \left| x \right| - \left| x \right| + 2(x - \pi )\cos \left| x \right|

is not differentiable. Then, the set

K

is equal to

View answer and explanation

Correct option: B

\phi

(an empty set)

Option B (

\phi

) is correct because the function

f(x)

is differentiable for all real values of

x

, meaning there are no points where it's not differentiable.

609 Medium

609 of 932

Analyzing +4 | -1

f\left( x \right) = a\left| {\sin x} \right| + b{e^{\left| x \right|}} + c{\left| x \right|^3}

and if

f\left( x \right)

is differentiable at

x = 0

, then

View answer and explanation

Correct option: B

a = 0,b = 0;c \in R

Option B (

a = 0,b = 0;c \in R

) is correct because for

f(x)

to be differentiable at

x=0

, the coefficients of non-differentiable terms must be zero.

610 Easy

610 of 932

Application +4 | -1

f\left( x \right) = {\log _e}\left\{ {\frac{{u\left( x \right)}}{{\upsilon \left( x \right)}}} \right\},u\left( 1 \right) = \upsilon \left( 1 \right)

and

u'\left( 1 \right) = \upsilon '\left( 1 \right) = 2

, then

f'\left( 1 \right)

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because when

u(1) = v(1)

and

u'(1) = v'(1) = 2

, the derivative

f'(1)

becomes zero.

611 Medium

611 of 932

Application +4 | -1

y=\sec \left(\tan ^{-1} x\right)

, then

\frac{d y}{d x}

x=1

is equal to

View answer and explanation

Correct option: D

\frac{1}{\sqrt{2}}

The correct answer is D (

\frac{1}{\sqrt{2}}

) because it is the value of the derivative of y with respect to x at x=1, after simplifying the function and differentiating.

612 Medium

612 of 932

Application +4 | -1

x=\sin ^{-1}\left(t^2-1\right), y=\cos ^{-1}(2 t)

then

\frac{d y}{d x}

t=0

View answer and explanation

Correct option: A

-\sqrt{2}

Option A (

-\sqrt{2}

) is correct because it is the value of dy/dx at t=0, calculated using parametric differentiation.

613 Medium

613 of 932

Application +4 | -1

Let

f

be a differentiable function satisfying

[f(x)]^n=f(n x)

for all

x \in R

.

Then,

f^{\prime}(x) f(n x)

View answer and explanation

Correct option: C

f(x) f^{\prime}(n x)

Option C (

f(x) f^{\prime}(n x)

) is correct because it is derived by differentiating the given equation and manipulating the result.

614 Medium

614 of 932

Application +4 | -1

Let

f(x) = \sin x,\mathrm{g}\left( x \right) = {x^2}

and

\,h\left( x \right) = {\log _e}x

. If

F\left( x \right) = \left( {hog\,\,of} \right)\left( x \right)

, then

F''\left( x \right)

is equal to

View answer and explanation

Correct option: D

- 2\mathrm{cose}{\mathrm{c}^2}x

Option D (

- 2\mathrm{cose}{\mathrm{c}^2}x

) is correct because after finding F(x) and differentiating twice, the result is

- 2\mathrm{cose}{\mathrm{c}^2}x

615 Hard

615 of 932

Analyzing +4 | -1

Let

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {{x^P}\sin \frac{1}{x},x \ne 0}\\ {0,x = 0} \end{array}} \right.

then

f\left( x \right)

is continuous but not differentiable at

x = 0

, if

View answer and explanation

Correct option: A

0 < p \le 1

Option A (

0 < p \le 1

) is correct because it ensures continuity but not differentiability at

x=0

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs

616 Easy

616 of 932

Application +4 | -1

f\left( x \right) = 10\cos x + \left( {13 + 2x} \right)\sin x,

then

f''\left( x \right) + f\left( x \right)

is equal to

View answer and explanation

Correct option: B

4\cos x

Option B (

4\cos x

) is correct because after finding the second derivative and adding it to the original function, the terms simplify to

4\cos x

617 Easy

617 of 932

Analyzing +4 | -1

Let

f:R \to R

be differentiable at

c \in R

and

f(c) = 0.

g(x) = \left| {f(x)} \right|,

then at

x = c,

g

View answer and explanation

Correct option: D

Differentiable if

f'(c) = 0

Option D (Differentiable if

f'(c) = 0

) is correct because if

f'(c) = 0

, then the limit defining

g'(c)

exists and is equal to 0, making

g

differentiable.

618 Medium

618 of 932

Application +4 | -1

The derivative of

\sin ^{-1}\left(\frac{1-x^{2 n}}{1+x^{2 n}}\right)

with respect to

x^{2 n}

View answer and explanation

Correct option: A

-\frac{1}{x^n+x^{3 n}}

Option A (

-\frac{1}{x^n+x^{3 n}}

) is correct because it is the result of differentiating

\sin ^{-1}(\frac{1-x^{2 n}}{1+x^{2 n}})

with respect to

x^{2n}

using the chain rule and simplification.

619 Easy

619 of 932

Application +4 | -1

y = {e^x}.{e^{{x^2}}}.{e^{{x^3}}}{._.}...{e^{{x^n}}}...,

for

0 < x < 1

, then

\frac{{dy}}{{dx}}

x = \frac{1}{2}

View answer and explanation

Correct option: B

4e

Option B (

4e

) is correct because after simplifying the expression and differentiating, the derivative evaluated at

x = \frac{1}{2}

4e

620 Hard

620 of 932

Application +4 | -1

f(x)=(1-x) n, n \in N

then

f(0)+\frac{f^{\prime}(0)}{1 !}+\frac{f^{\prime \prime}(0)}{2 !}+\ldots .+\frac{f^n(0)}{n !}=

View answer and explanation

Correct option: B

Option B (0) is correct because the given expression is the binomial expansion of (1-1)^n which equals 0.

621 Hard

621 of 932

Application +4 | -1

Let

f(x)=\left\{\begin{array}{cc}a x, & x<2 \\ a x^2+b x+3, & x \geq 2\end{array}\right.

f

is differentiable for all

x

, then the value of (

a

, b) is equal to

View answer and explanation

Correct option: D

\left(\frac{3}{4},-\frac{9}{4}\right)

Option D (

\left(\frac{3}{4},-\frac{9}{4}\right)

) is correct because these values of a and b ensure that the function is both continuous and differentiable at x=2.

622 Medium

622 of 932

Application +4 | -1

f^{\prime}(x)=\frac{\cos (\log x)}{x}, f^{\prime}(1)-0

and

y=f\left(\frac{2 x+3}{3-2 x}\right)

, then

\frac{d y}{d x}=\ldots \ldots

View answer and explanation

Correct option: C

\frac{12}{(3-2 x)^2} \sin \left(\log \left(\frac{2 x+3}{3-2 x}\right)\right)

Option C is correct because it involves finding the derivative using the chain rule and the given derivative of f(x).

623 Easy

623 of 932

Application +4 | -1

\tan ^{-1} y-y+x=0

then

\frac{d^2 y}{d x^2}

is equal to

View answer and explanation

Correct option: A

\frac{-2\left(1+y^2\right)}{y^5}

Option A (

\frac{{-2(1+y^2)}}{{y^5}}

) is correct because it is the result of differentiating the given equation twice with respect to x.

624 Easy

624 of 932

Application +4 | -1

{2^x} + {2^y} = {2^{x + y}}

, then the value of

\frac{{dy}}{{dx}}

x = y = 1

, is

View answer and explanation

Correct option: B

- 1

Option B (-1) is correct because it is the value of dy/dx at x=y=1, obtained by implicit differentiation of the given equation.

625 Easy

625 of 932

Analyzing +4 | -1

The function

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {1 - 2x + 3{x^2} - 4{x^3} + \ldots \mathrm{to}\,\,\,\,\infty ,x \ne - 1}\\ {1,x = - 1} \end{array}} \right.

View answer and explanation

Correct option: B

Neither continuous nor derivable at

x = - 1

Option B (Neither continuous nor derivable at

x = - 1

) is correct because the function is not defined at

x = -1

and the limit as

x

approaches

-1

is infinity, indicating a discontinuity.

626 Easy

626 of 932

Application +4 | -1

y = \left( {1 + x} \right)\left( {1 + {x^2}} \right)\left( {1 + {x^4}} \right) \ldots (1 + {x^2}^n)

, then the value of

\frac{{dy}}{{dx}}

x = 0

View answer and explanation

Correct option: C

Option C (1) is correct because when evaluating the derivative at x=0, the result is 1.

627 Easy

627 of 932

Application +4 | -1

y=\sin ^{-1}\left(\frac{\sin \alpha \sin x}{1-\cos \alpha \sin x}\right)

, then

y^{\prime}(0)

is equal to

View answer and explanation

Correct option: D

\sin \alpha

Option D (

\sin \alpha

) is correct because it is the result of differentiating the given function and evaluating at x=0.

628 Easy

628 of 932

Application +4 | -1

\sqrt{r}-a e^{8 \cos \alpha}

, then

\frac{d^2 r}{d \theta^2}-4 r \cot ^2 \alpha=\ldots \ldots . \quad(a, \alpha \in \mathrm{R})

View answer and explanation

Correct option: D

Option D (0) is correct because after finding the first and second derivatives of r with respect to theta, the expression simplifies to zero.

629 Easy

629 of 932

Analyzing +4 | -1

f(x)=\left\{\begin{array}{cc}x e^{\left(-\frac{1}{|x|}+\frac{1}{x}\right)}, & x \neq 0 \\ 0, & x=0\end{array}\right.

then

f(x)

View answer and explanation

Correct option: C

continuous for all

x

but not differentiable at

x=0

Option C (continuous for all

x

but not differentiable at

x=0

) is correct because the function is continuous at

x=0

after evaluating the limits, but the left-hand derivative and right-hand derivative at

x=0

are not equal.

630 Easy

630 of 932

Analyzing +4 | -1

In the function

f

given by

f\left( x \right) = {x^3} - 3\left( {a - 2} \right){x^2} + 3ax + 7,

for some

a \in R

is increasing in

\left( {0,\,1} \right]

and decreasing in

\left[ {1,\,5} \right),

then a root of the equation,

\frac{{f\left( x \right) - 14}}{{{{\left( {x - 1} \right)}^2}}} = 0\,\,\,\left( {x \ne 1} \right)

View answer and explanation

Correct option: C

Option C (7) is correct because by finding the value of 'a' using the given increasing/decreasing conditions, we can determine the function f(x) and then solve for the root of the given equation.

Revision notes and key takeaways

Review DIFFERENTIATION in your textbook or class notes, then use this set to check your understanding.

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