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 $x = - 1$

If $x=\phi(t), y=\psi(t)$, then $\frac{d^2 y}{d x^2}$ is equal to

$\frac{\phi^{\prime \prime}}{\psi^{\prime \prime}}$

$\frac{\psi^{\prime \prime}}{\phi^{\prime \prime}}$

If $f(x)=\cos ^{-1}\left(4 x^3-3 x\right), x \in[-1,1]$ then

$f'(x) = \left\{ {\begin{array}{*{20}{l}} {\frac{3}{{\sqrt {1 - {x^2}} }},{\rm{ }}\,\,\,\,\,\,{\rm{if }}|x| < \frac{1}{2}}\\ {\frac{{ - 3}}{{\sqrt {1 - {x^2}} }},{\rm{ }}\,\,\,\,\,\,\,\,{\rm{if }} - 1 < x < \frac{{ - 1}}{2},\frac{1}{2} < x < 1} \end{array

$f^{\prime}(x)=\frac{-3}{\sqrt{1-x^2}}$ for all $x \in[-1,1]$

$f^{\prime}(x)=\frac{-3}{\sqrt{1-x^2}}$ for all $x \in(-1,1)$

$f^{\prime}(x)=\left\{\begin{array}{l}\frac{-3}{\sqrt{1-x^2}}, \text { if }|x|<\frac{1}{2} \\ \frac{3}{\sqrt{1-x^2}}, \text { if } \frac{1}{2}<|x|<1\end{array}\right.$

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}} $

If $f(x)=|x|^{|\sin x|}$, then $f^{\prime}\left(-\frac{\pi}{4}\right)$ equals

$\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{\sqrt{2}}{2} \ln \frac{\pi}{4}-\frac{2 \sqrt{2}}{\pi}\right)$

$\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{\sqrt{2}}{2} \ln \frac{\pi}{4}+\frac{2 \sqrt{2}}{\pi}\right)$

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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Easy

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331 Easy

331 of 500

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.

332 Easy

332 of 500

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.

333 Easy

333 of 500

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.

334 Easy

334 of 500

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.

335 Easy

335 of 500

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.

336 Easy

336 of 500

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.

337 Easy

337 of 500

Application +4 | -1

y = {e^{ax}}\sin \mathrm{b}x,

then

\frac{{{d^2}y}}{{d{x^2}}} - 2a\frac{{dy}}{{dx}} + {a^2}y

is equal to

View answer and explanation

Correct option: C

- {b^2}y

Option C (

- {b^2}y

) is correct because after differentiating the given function twice and substituting back into the given expression, it simplifies to

- {b^2}y

338 Easy

338 of 500

Application +4 | -1

Let

f(x) = x + 3\log (x - 2)

and

g(x) = x + 5\log (x - 1),

then the set of

x

satisfying the inequality

f'(x) < g'(x)

View answer and explanation

Correct option: D

(7/2,\infty )

Option D (

(7/2,\infty )

) is correct because it represents the interval where f'(x) < g'(x), considering the domain restrictions imposed by the logarithmic terms.

339 Easy

339 of 500

Application +4 | -1

Let

f

and

g

be differentiable functions satisfying

g^{\prime}(a)=2, g(a)=b

and

fog=I

(identity function). Then,

f^{\prime}(b)

is equal to

View answer and explanation

Correct option: C

\frac{1}{2}

Option C is correct because it correctly applies the chain rule and the properties of inverse functions to find the derivative.

340 Easy

340 of 500

Application +4 | -1

Let

f(x)=x+\cot x

and

f

is the inverse of

g

, then

g^{\prime}(x)

is equal to

View answer and explanation

Correct option: C

-\frac{1}{(x-g(x))^2}

Option C (

-\frac{{1}}{{(x-g(x))^2}}

) is correct because using the relationship between a function and its inverse, and differentiating implicitly, we arrive at the given expression.

341 Easy

341 of 500

Application +4 | -1

It is given that

f^{\prime}(a)

exists, then

\lim\limits _{x \rightarrow a} \frac{x f(a)-a f(x)}{x-a}

is equal to

View answer and explanation

Correct option: A

f(a)-a f^{\prime}(a)

Option A (

f(a)-a f^{\prime}(a)

) is correct because it is the result of evaluating the limit using the definition of the derivative.

342 Easy

342 of 500

Application +4 | -1

g

is the inverse of a function

f

and

f^{\prime}(x)=\frac{1}{1+x^n}

, then

g^{\prime}(x)

is equal to

View answer and explanation

Correct option: B

1+\{g(x)\}^n

Option B (

1+\{g(x)\}^n

) is correct because it directly follows from the inverse function differentiation rule.

343 Easy

343 of 500

Analyzing +4 | -1

(\sin y)^{\sin (\pi x / 2)}-\frac{\sqrt{3}}{2} \operatorname{Sec}^{-1}(2 x)+2^x \tan (\log (x+2))=0

then

d y / d x

x=-1

View answer and explanation

Correct option: C

\frac{3}{\pi \sqrt{\pi^2-3}}

Option C (

\frac{{3}}{{\pi \sqrt{{\pi^2-3}}}}

) is correct because by differentiating the given equation implicitly and substituting x = -1 and solving for dy/dx, we obtain this result.

344 Easy

344 of 500

Application +4 | -1

f\left( x \right) = \left| {{x^2} - 5x + 6} \right|

, then

f'\left( x \right)

equals

View answer and explanation

Correct option: B

5 - 2x

for

2 < x < 3

Option B (

5 - 2x

for

2 < x < 3

) is correct because it represents the derivative of the absolute value function in the interval where the expression inside the absolute value is negative.

345 Easy

345 of 500

Application +4 | -1

x=\phi(t), y=\psi(t)

, then

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

is equal to

View answer and explanation

Correct option: B

\frac{\phi^{\prime} \psi^{\prime \prime}-\psi^{\prime} \phi^{\prime \prime}}{\left(\phi^{\prime}\right)^3}

Option B is correct because it is the result of applying the chain rule and quotient rule to find the second derivative of a parametric function.

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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

346 Easy

346 of 500

Application +4 | -1

Let

f(x)=(a x+b) \cos x+(c x+d) \sin x

and

f^{\prime}(x)=x \cos x

for all

x.

Then,

View answer and explanation

Correct option: A

a=0, b=c=1, d=0

Option A (

a=0, b=c=1, d=0

) is correct because these values satisfy the given condition

f'(x)=x \cos x

when substituted into the derivative of

f(x)

347 Easy

347 of 500

Application +4 | -1

y = x - {x^2}

, then the derivatives of

{y^2}

w.r.t.

{x^2}

View answer and explanation

Correct option: B

2{x^2} - 3x + 1

Option B (

2{x^2} - 3x + 1

) is correct because it represents the derivative of

y^2

with respect to

x^2

, calculated using the chain rule and the given relationship

y = x - x^2

348 Easy

348 of 500

Application +4 | -1

The functions

u=e^x \sin x, v=e^x \cos x

satisfy the equation

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because all the given relations are satisfied by the functions u and v after differentiation and substitution.

349 Easy

349 of 500

Analyzing +4 | -1

f(x)=\sin (|x|)

is not differentiable at

View answer and explanation

Correct option: C

x=0

Option C (

x=0

) is correct because the function

f(x) = \sin(|x|)

has a sharp corner at

x=0

, making it non-differentiable at that point.

350 Easy

350 of 500

Analyzing +4 | -1

y = \frac{x}{{a + \frac{x}{{b + \frac{x}{{a + \frac{x}{{b + .....\infty }}}}}}}},

then

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

equals

View answer and explanation

Correct option: D

\frac{b}{{ab + 2ay}}

Option D (

\frac{b}{{ab + 2ay}}

) is correct because it is the derivative of y with respect to x after substituting the continued fraction.

351 Easy

351 of 500

Analyzing +4 | -1

f(x)=\cos ^{-1}\left(4 x^3-3 x\right), x \in[-1,1]

then

View answer and explanation

Correct option: D

f'(x) = \left\{ {\begin{array}{llllllllllllllllllll} {\frac{3}{{\sqrt {1 - {x^2}} }},\mathrm{ }\,\,\,\,\,\,\mathrm{if }|x| < \frac{1}{2}}\\ {\frac{{ - 3}}{{\sqrt {1 - {x^2}} }},\mathrm{ }\,\,\,\,\,\,\,\,\mathrm{if } - 1 < x < \frac{{ - 1}}{2},\frac{1}{2} < x < 1} \end{array}

Option D is correct because the derivative of arccos(4x^3-3x) is piecewise, being 3/sqrt(1-x²) for |x|<1/2 and -3/sqrt(1-x²) for 1/2<|x|<1 due to the properties of arccos.

352 Easy

352 of 500

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 differentiating the given equation twice and solving for

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

yields this result.

353 Easy

353 of 500

Analyzing +4 | -1

\sqrt {1 - {x^2}} + \sqrt {1 - {y^2}} = a\left( {x - y} \right)

, then

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

equals

View answer and explanation

Correct option: B

\sqrt {\frac{{1 - {y^2}}}{{1 - {x^2}}}}

Option B (

\sqrt {\frac{{1 - {y^2}}}{{1 - {x^2}}}}

) is correct because after substituting

x = \sin A

and

y = \sin B

, and differentiating implicitly, the expression simplifies to this result.

354 Easy

354 of 500

Application +4 | -1

f(x)=|x|^{|\sin x|}

, then

f^{\prime}\left(-\frac{\pi}{4}\right)

equals

View answer and explanation

Correct option: A

\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{\sqrt{2}}{2} \ln \frac{4}{\pi}-\frac{2 \sqrt{2}}{\pi}\right)

Option A is correct because it is the result of differentiating the given function and evaluating at the specified point, considering the absolute value.

355 Easy

355 of 500

Application +4 | -1

x=\frac{t+1}{t}, y=\frac{t-1}{t}

then

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

t=1

View answer and explanation

Correct option: A

Option A (0) is correct because the second derivative is zero, as the first derivative is a constant.

356 Easy

356 of 500

Application +4 | -1

y=\cos ^{-1}\left(\frac{3 \cos x+4 \sin x}{5}\right)

, then

\frac{d y}{d x}=

View answer and explanation

Correct option: B

Option B (1) is correct because the expression inside the inverse cosine simplifies to a cosine function, allowing for direct differentiation.

357 Easy

357 of 500

Understanding +4 | -1

Use the following table and the fact that

f(x)

is invertible and differentiable everywhere to find

f^{-1}(3)

\begin{array}{ccc}x & f(x) & f^{\prime}(x) \\ 3 & 1 & 7 \\ 6 & 2 & 10 \\ 9 & 3 & 5\end{array}

View answer and explanation

Correct option: B

\frac{1}{5}

Option B (

\frac{{1}}{{5}}

) is correct because it uses the inverse function derivative formula and the given table values.

358 Easy

358 of 500

Application +4 | -1

Let

f\left( x \right)

be a polynomial function of the second degree. If

f\left( 1 \right) = f\left( { - 1} \right)

and

{a_1},{a_2},{a_3}

are in AP, then

f'\left( {{a_1}} \right),f'\left( {{a_2}} \right),f'\left( {{a_3}} \right)

are in

View answer and explanation

Correct option: A

Option A (AP) is correct because the derivative of a quadratic function is a linear function, and the values of a linear function at points in arithmetic progression are also in arithmetic progression.

359 Easy

359 of 500

Application +4 | -1

y = {\tan ^{ - 1}}\sqrt {\frac{{1 - \sin x}}{{1 + \sin x}},}

then the value of

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

x = \frac{\pi }{6}

View answer and explanation

Correct option: A

- \frac{1}{2}

Option A is correct because simplifying the expression and then differentiating yields -1/2, which is independent of x.

360 Easy

360 of 500

Analyzing +4 | -1

f(x)=2 \operatorname{Sin}^{-1} \sqrt{1-x}+\operatorname{Sin}^{-1}(2 \sqrt{x(1-x)}

for

0 < x < 1/2

then

f'(x)

is equal to

View answer and explanation

Correct option: B

Option B is correct because the derivative of the given function simplifies to 0 for

0 < x < 1/2

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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