If $a{x^2} + 2hxy + b{y^2} = 1$, then $\frac{{{d^2}y}}{{d{x^2}}}$ equals

$\frac{{{h^2} - ab}}{{{{\left( {hx + by} \right)}^2}}}$

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

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

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

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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Hard

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Practice format and chapter quiz

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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1 Hard

1 of 30

Application +4 | -1

The value of derivative of

\left| {x - 1} \right| + \left| {x - 3} \right|

x = 2

, is

View answer and explanation

Correct option: B

Option B (0) is correct because at x=2, the function is constant, and the derivative of a constant is zero.

2 Hard

2 of 30

Analyzing +4 | -1

y = {\cos ^{ - 1}}\cos \left( {\left| x \right| - f\left( x \right)} \right)

, where

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {1,if\,\,x > 0}\\ { - 1,if\,\,x < 0}\\ {0,if\,\,x = 0} \end{array}} \right.

Then,

\left( {dy/dx} \right)x = \frac{{5\pi }}{4}

is equal to

View answer and explanation

Correct option: B

Option B (1) is correct because, for x > 0, y = cos⁻¹(cos(x-1)), and for x in the interval [1, π+1], y = x-1, so dy/dx = 1.

3 Hard

3 of 30

Application +4 | -1

f(x)=|\cos x|

, then

f^{\prime}(3 \pi / 4)

is equal to

View answer and explanation

Correct option: B

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

Option B is correct because it is the result of differentiating the function in the given neighborhood and evaluating at the specified point.

4 Hard

4 of 30

Analyzing +4 | -1

Let

f(x) = 15 - \left| {x - 10} \right|;\,\,x \in R

. Then, the set of all values of

x

, at which the function,

g(x) = f(f(x))

is not differentiable, is

View answer and explanation

Correct option: B

\left\{ {5,\,10,\,15} \right\}

Option B ({5, 10, 15}) is correct because these are the points where the absolute value function inside g(x) changes its behavior, leading to non-differentiability.

5 Hard

5 of 30

Analyzing +4 | -1

f(x)=\left\{\begin{array}{cc}\frac{x\left(3 e^{1 / x}+4\right)}{2-e^{1 / x}}, x \neq 0 \\ 0, \quad x=0\end{array}\right.

then

f(x)

View answer and explanation

Correct option: B

Continuous but not differentiable at

x=0

Option B (Continuous but not differentiable at

x=0

) is correct because the left-hand and right-hand limits of the function at x=0 are equal, but the left-hand and right-hand derivatives are not equal.

6 Hard

6 of 30

Analyzing +4 | -1

Let

f(x)=3 x^2+4 x g^{\prime}(1)+g^{\prime \prime}(2)

and,

g(x)=2 x^2+3 x f^{\prime}(2)+f^{\prime \prime}(3)

for all

x \in R

. Then,

View answer and explanation

Correct option: D

All the above

Option D (All the above) is correct because all the derived relationships between the derivatives of f(x) and g(x) are satisfied.

7 Hard

7 of 30

Analyzing +4 | -1

y^2=p(x)

is a polynomial of degree 3 , then

2 \frac{d}{d x}\left(y^3 \cdot \frac{d^2 y}{d x^2}\right)

is equal to

View answer and explanation

Correct option: C

p(x) \cdot p^{\prime \prime \prime}(x)

Option C (

p(x) \cdot p^{\'\''}(x)

) is correct because it is the result of differentiating the given expression and simplifying using the given condition.

8 Hard

8 of 30

Analyzing +4 | -1

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

and

\phi \left( x \right) - {2^x} + 2x\log 2.

f'\left( x \right) > \phi '\left( x \right),

then

View answer and explanation

Correct option: C

x > 0

Option C (

x > 0

) is correct because solving the inequality

f'(x) > \phi'(x)

leads to the condition

x > 0

9 Hard

9 of 30

Application +4 | -1

x=e^{\tan ^{-1}\left(\frac{y-x^{2}}{x^{2}}\right)}

, then

\frac{d y}{d x}

View answer and explanation

Correct option: A

x\left(2+\frac{y^{2}}{x^{4}}-\frac{2 y}{x^{2}}\right)+\frac{2 y}{x}

The correct answer is A (

x\left(2+\frac{y^{2}}{x^{4}}-\frac{2 y}{x^{2}}\right)+\frac{2 y}{x}

) because it is the derivative of y with respect to x after implicit differentiation and simplification.

10 Hard

10 of 30

Analyzing +4 | -1

Let

f\left( {x + y} \right) = f\left( x \right)f\left( y \right)

and

f\left( x \right) = 1 + xg\left( x \right)G\left( x \right)

, where

\lim\limits_{x \to 0} g\left( x \right) = a

and

\lim\limits_{x \to 0} G\left( x \right) = b

. Then

f'\left( x \right)

is equal to

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the derivative

f'(x)

is equal to

ab f(x)

, which is not listed in the options.

11 Hard

11 of 30

Analyzing +4 | -1

The following functions are differentiable on

\left( { - 1,2} \right)

View answer and explanation

Correct option: C

\int \limits_x^{2x} \frac{{1 - t + {t^2}}}{{1 + t + {t^2}}}dt

Option C (

\int \limits_x^{2x} \frac{{1 - t + {t^2}}}{{1 + t + {t^2}}}dt

) is correct because the integrand is continuous on the interval (-1, 2), ensuring differentiability of the integral.

12 Hard

12 of 30

Application +4 | -1

f(x)=(1+x)^{n}

, then the value of

f(0)+f^{\prime}(0)+\frac{f^{\prime \prime}(0)}{2 !}+\cdots

+\frac{1}{n !} f^{n}(0)

View answer and explanation

Correct option: B

2^{n}

The correct answer is B (

2^{n}

) because the given expression is the binomial expansion of

(1+1)^n

13 Hard

13 of 30

Analyzing +4 | -1

Let

f(x)

and

g(x)

be two functions having finite non-zero

3 r d

order derivatives

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

and

g^{\prime \prime \prime}(x)

for all

x \in R

. If

f(x) g(x)=1

for all

x \in R

, then

\frac{f^{\prime \prime \prime}}{f^{\prime}}-\frac{g^{\prime \prime \prime}}{g^{\prime}}

is equal to

View answer and explanation

Correct option: B

3\left(\frac{f^{\prime \prime}}{f}-\frac{g^{\prime \prime}}{g}\right)

Option B (3(

\frac{f^{\prime\prime}}{f}-\frac{g^{\prime\prime}}{g}

)) is correct because after differentiating the equation f(x)g(x) = 1 three times and rearranging, the expression simplifies to 3(

\frac{f^{\prime\prime}}{f}-\frac{g^{\prime\prime}}{g}

14 Hard

14 of 30

Analyzing +4 | -1

The expression of

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

of the function

y = {a^{{x^a}{{^x}^{ \ldots \infty }}}}

, is

View answer and explanation

Correct option: C

\frac{{{y^2}\log y}}{{x\left( {1 - y\log x\log y} \right)}}

The correct answer is C (

\frac{{{y^2}\log y}}{{x\left( {1 - y\log x\log y} \right)}}

) because it is the result of logarithmic differentiation and solving for

\frac{dy}{dx}

15 Hard

15 of 30

Application +4 | -1

x\sqrt {1 + y} + y\sqrt {1 + x} = 0

, then

\frac{{d\mathrm{y}}}{{\mathrm{d}x}}

is equal to

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it is the derivative of the function obtained after simplifying the given equation.

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16 Hard

16 of 30

Analyzing +4 | -1

a{x^2} + 2hxy + b{y^2} = 1

, then

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

equals

View answer and explanation

Correct option: D

\frac{{{h^2} - ab}}{{{{\left( {hx + by} \right)}^3}}}

Option D is correct because the second derivative is found by implicit differentiation and simplification, resulting in

\frac{{{h^2} - ab}}{{{{\left( {hx + by} \right)}^3}}}

17 Hard

17 of 30

Analyzing +4 | -1

Function

f\left( x \right) = \left| {x - 1} \right| + \left| {x - 2} \right|,x \in R

View answer and explanation

Correct option: B

Except

x = 1

and

x = 2

differentiable everywhere in

R

Option B (Except x = 1 and x = 2 differentiable everywhere in R) is correct because the function is not differentiable at x=1 and x=2 due to the absolute value functions.

18 Hard

18 of 30

Application +4 | -1

f(x)=x^n

, then the value of

f(1)-\frac{f^{\prime}(1)}{1 !}+\frac{f^{\prime \prime}(1)}{2 !}-\frac{f^{\prime \prime \prime}(1)}{3 !}+\frac{f^{i v}(1)}{4 !} \ldots+\frac{(-1)^n f^n(1)}{n !}

, is

View answer and explanation

Correct option: D

Option D is correct because it correctly applies the binomial theorem to the given expression.

19 Hard

19 of 30

Analyzing +4 | -1

y=\left|x-x^2\right|

, then

\frac{d y}{d x}

x=1

View answer and explanation

Correct option: C

Does not exist

Option C (Does not exist) is correct because the left-hand derivative and right-hand derivative at x=1 are not equal, indicating a non-differentiable point.

20 Hard

20 of 30

Analyzing +4 | -1

y = {\log ^n}x,

where

{\log ^n}

means

log\,\,log\,\,log \ldots

(repeated

n

times), then

x\log x\,\,{\log ^2}x\,\,{\log ^3}x \ldots {\log ^{n - 1}}x\,\,{\log ^n}x\,\,\frac{{dy}}{{dx}}

is equal to

View answer and explanation

Correct option: B

{\log ^n}x

The correct answer is B (

{\log ^n}x

) because it is obtained by repeatedly applying the chain rule to the logarithmic function.

21 Hard

21 of 30

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

22 Hard

22 of 30

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.

23 Hard

23 of 30

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.

24 Hard

24 of 30

Analyzing +4 | -1

Let

S=\left\{(\lambda, \mu) \in R \times R: f(t)=\left(|\lambda| e^{|t|}-\mu\right)\right.

\sin (2|t|), t \in R

, is a differentiable function

\}

. Then

S

is a subest of?

View answer and explanation

Correct option: A

R \times[0, \infty)

Option A is correct because the differentiability condition at t=0 implies that |λ| = μ, which means μ must be non-negative, hence S is a subset of R x [0, ∞).

25 Hard

25 of 30

Application +4 | -1

The value of the derivative of

\left| {x - 1} \right| + \left| {x - 3} \right|

x = 2

View answer and explanation

Correct option: C

Option C (0) is correct because at x=2, the function is constant, and the derivative of a constant is zero.

26 Hard

26 of 30

Analyzing +4 | -1

f: R \in R

is defined by

f(x)=\frac{x^2-a x+1}{x^2+a x+1}, 0 < a < 2

, then which of the following is true:

View answer and explanation

Correct option: A

(2+a)^2 f^{\prime \prime}(1)+(2-a)^2 f^{\prime \prime}(-1)=0

Option A is correct because

(2+a)^2 f^{\prime \prime}(1)+(2-a)^2 f^{\prime \prime}(-1)=0

is derived from calculating the first and second derivatives of the given function and substituting x=1 and x=-1.

27 Hard

27 of 30

Analyzing +4 | -1

f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3)

for all

x \in R

, then which of the following is false?

View answer and explanation

Correct option: D

f(1)+f(3)=f(0)

Option D,

f(1)+f(3)=f(0)

, is false because the correct relationship between the function values is

f(0)+f(2)=f(1)

28 Hard

28 of 30

Analyzing +4 | -1

y = \left( {\frac{{ax + b}}{{cx + d}}} \right),

then

2\frac{{dy}}{{dx}}.\frac{{{d^3}y}}{{d{x^3}}}

is equal to

View answer and explanation

Correct option: C

3{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2}

Option C (

3{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2}

) is correct because after differentiating the given function three times, the expression

2\frac{{dy}}{{dx}}.\frac{{{d^3}y}}{{d{x^3}}}

simplifies to

3{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^2}

29 Hard

29 of 30

Analyzing +4 | -1

Let

f\left( x \right)

be defined on

R

such that

f\left( 1 \right) = 2,f\left( 2 \right) = 8

and

f\left( {u + \upsilon } \right) = f\left( u \right) + ku\upsilon - 2{\upsilon ^2}

for all

u,\upsilon \in R

(

k

is a fixed constant). Then,

View answer and explanation

Correct option: A

f'\left( x \right) = 8x

Option A (

f'(x) = 8x

) is correct because by substituting

u = x

and

\upsilon = h

into the given functional equation and taking the limit as

h

approaches 0, we find the derivative.

30 Hard

30 of 30

Analyzing +4 | -1

f\left( x \right) = \left| {\begin{array}{cccccccccccccccccccc} {{x^3}}&{{x^2}}&{3{x^2}}\\ 1&{ - 6}&4\\ P&{{P^2}}&{{P^3}} \end{array}} \right|,

here

p

is a constant, then

\frac{{{d^3}f\left( x \right)}}{{d{x^3}}}

View answer and explanation

Correct option: D

A constant

Option D (A constant) is correct because after taking the third derivative of the determinant function, the result is a constant value.

Revision notes and key takeaways

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

Key Takeaways

✓ This set contains 30 single-correct-answer MCQs on DIFFERENTIATION.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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