Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

Let $$a,b \in R$$ be such that the function $$f$$ given by $$f\left( x \right) = In\left| x \right| + b{x^2} + ax,\,x \ne 0$$ has extreme values at $$x=-1$$ and $$x=2$$

**Statement-1 :** $$f$$ has local maximum at $$x=-1$$ and at $$x=2$$.

**Statement-2 :** $$a = {1 \over 2}$$ and $$b = {-1 \over 4}$$

A

Statement - 1 is false, Statement - 2 is true.

B

Statement - 1 is true , Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1.

C

Statement - 1 is true, Statement - 2 is true; Statement - 2 is **not** a correct explanation for Statement - 1.

D

Statement - 1 is true, Statement - 2 is false.

Given, $$f\left( x \right) = \ln \left| x \right| + b{x^2} + ax$$

$$\therefore$$ $$f'\left( x \right) = {1 \over x} + 2bx + a$$

At $$x=-1,$$ $$f'\left( x \right) = - 1 - 2b + a = 0$$

$$ \Rightarrow a - 2b = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

At $$x=2,$$ $$\,\,f'\left( x \right) = {1 \over 2} + 4b + a = 0$$

$$ \Rightarrow a + 4b = - {1 \over 2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

On solving $$(i)$$ and $$(ii)$$ we get $$a = {1 \over 2},b = - {1 \over 4}$$

Thus, $$f'\left( x \right) = {1 \over x} - {x \over 2} + {1 \over 2} = {{2 - {x^2} + x} \over {2x}}$$

$$ = {{ - {x^2} + x + 2} \over {2x}} = {{ - \left( {{x^2} - x - 2} \right)} \over {2x}} = {{ - \left( {x + 1} \right)\left( {x - 2} \right)} \over {2x}}$$

So maximum at $$x=-1,2$$

Hence both the statements are true and statement $$2$$ is a correct explanation for $$1.$$

$$\therefore$$ $$f'\left( x \right) = {1 \over x} + 2bx + a$$

At $$x=-1,$$ $$f'\left( x \right) = - 1 - 2b + a = 0$$

$$ \Rightarrow a - 2b = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

At $$x=2,$$ $$\,\,f'\left( x \right) = {1 \over 2} + 4b + a = 0$$

$$ \Rightarrow a + 4b = - {1 \over 2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

On solving $$(i)$$ and $$(ii)$$ we get $$a = {1 \over 2},b = - {1 \over 4}$$

Thus, $$f'\left( x \right) = {1 \over x} - {x \over 2} + {1 \over 2} = {{2 - {x^2} + x} \over {2x}}$$

$$ = {{ - {x^2} + x + 2} \over {2x}} = {{ - \left( {{x^2} - x - 2} \right)} \over {2x}} = {{ - \left( {x + 1} \right)\left( {x - 2} \right)} \over {2x}}$$

So maximum at $$x=-1,2$$

Hence both the statements are true and statement $$2$$ is a correct explanation for $$1.$$

2

MCQ (Single Correct Answer)

A spherical balloon is filled with $$4500\pi $$ cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of $$72\pi $$ cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases $$49$$ minutes after the leakage began is :

A

$${{9 \over 7}}$$

B

$${{7 \over 9}}$$

C

$${{2 \over 9}}$$

D

$${{9 \over 2}}$$

Volume of spherical balloon $$ = V = {4 \over 3}\pi {r^3}$$

$$ \Rightarrow 4500\pi = {{4\pi {r^3}} \over 3}$$

( as Given, volume $$ = 4500\pi {m^3}$$ )

Differentiating both the sides, $$w.r.t't'$$ we get,

$${{dV} \over {dt}} = 4\pi {r^2}\left( {{{dr} \over {dt}}} \right)$$

Now, it is given that $${{dV} \over {dt}} = 72\pi $$

$$\therefore$$ After $$49$$ min, Volume -

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {4500 - 49 \times 72} \right)\pi $$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {4500 - 3528} \right)\pi $$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 972\pi {m^3}$$

$$ \Rightarrow V = 972\,\,\pi {m^3}$$

$$\therefore$$ $$972\pi = {4 \over 3}\pi r{}^3$$

$$ \Rightarrow {r^3} = 3 \times 243 = 3 \times {3^5} = {3^6} = {\left( {{3^2}} \right)^3} \Rightarrow r = 9$$

Also, we have $${{dV} \over {dt}} = 72\pi $$

$$\therefore$$ $$72\pi = 4\pi \times 9 \times 9\left( {{{dr} \over {dt}}} \right) \Rightarrow {{dr} \over {dt}} = \left( {{2 \over 9}} \right)$$

$$ \Rightarrow 4500\pi = {{4\pi {r^3}} \over 3}$$

( as Given, volume $$ = 4500\pi {m^3}$$ )

Differentiating both the sides, $$w.r.t't'$$ we get,

$${{dV} \over {dt}} = 4\pi {r^2}\left( {{{dr} \over {dt}}} \right)$$

Now, it is given that $${{dV} \over {dt}} = 72\pi $$

$$\therefore$$ After $$49$$ min, Volume -

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {4500 - 49 \times 72} \right)\pi $$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {4500 - 3528} \right)\pi $$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 972\pi {m^3}$$

$$ \Rightarrow V = 972\,\,\pi {m^3}$$

$$\therefore$$ $$972\pi = {4 \over 3}\pi r{}^3$$

$$ \Rightarrow {r^3} = 3 \times 243 = 3 \times {3^5} = {3^6} = {\left( {{3^2}} \right)^3} \Rightarrow r = 9$$

Also, we have $${{dV} \over {dt}} = 72\pi $$

$$\therefore$$ $$72\pi = 4\pi \times 9 \times 9\left( {{{dr} \over {dt}}} \right) \Rightarrow {{dr} \over {dt}} = \left( {{2 \over 9}} \right)$$

3

MCQ (Single Correct Answer)

The shortest distance between line $$y-x=1$$ and curve $$x = {y^2}$$ is

A

$${{3\sqrt 2 } \over 8}$$

B

$${8 \over {3\sqrt 2 }}$$

C

$${4 \over {\sqrt 3 }}$$

D

$${{\sqrt 3 } \over 4}$$

Shortest distance between two curve occurred along -

the common normal

Slope of normal to $${y^2} = x$$ at point

$$P\left( {{t^2},t} \right)$$ is $$-2t$$ and

slope of line $$y - x = 1$$ is $$1.$$

As they are perpendicular to each other

$$\therefore$$ $$\left( { - 2t} \right) = - 1 \Rightarrow t = {1 \over 2}$$

$$\therefore$$ $$P\left( {{1 \over 4},{1 \over 2}} \right)$$

and shortest distance $$ = \left| {{{{1 \over 2} - {1 \over 4} - 1} \over {\sqrt 2 }}} \right|$$

So shortest distance between them is $${{3\sqrt 2 } \over 8}$$

the common normal

Slope of normal to $${y^2} = x$$ at point

$$P\left( {{t^2},t} \right)$$ is $$-2t$$ and

slope of line $$y - x = 1$$ is $$1.$$

As they are perpendicular to each other

$$\therefore$$ $$\left( { - 2t} \right) = - 1 \Rightarrow t = {1 \over 2}$$

$$\therefore$$ $$P\left( {{1 \over 4},{1 \over 2}} \right)$$

and shortest distance $$ = \left| {{{{1 \over 2} - {1 \over 4} - 1} \over {\sqrt 2 }}} \right|$$

So shortest distance between them is $${{3\sqrt 2 } \over 8}$$

4

MCQ (Single Correct Answer)

For $$x \in \left( {0,{{5\pi } \over 2}} \right),$$ define $$f\left( x \right) = \int\limits_0^x {\sqrt t \sin t\,dt.} $$ Then $$f$$ has

A

local minimum at $$\pi $$ and $$2\pi $$

B

local minimum at $$\pi $$ and local maximum at $$2\pi $$

C

local maximum at $$\pi $$ and local minimum at $$2\pi $$

D

local maximum at $$\pi $$ and $$2\pi $$

$$f'\left( x \right) = \sqrt x \sin x$$

At local maxima or minima, $$f'\left( x \right) = 0$$

$$ \Rightarrow x = 0$$ or $$sin$$ $$x=0$$

$$ \Rightarrow x = 2\pi ,\,\,\pi \in \left( {0,{{5\pi } \over 2}} \right)$$

$$f''\left( x \right) = \sqrt x \cos \,x + {1 \over {2\sqrt x }}\sin \,x$$

$$ = {1 \over {2\sqrt x }}\left( {2x\,\cos \,x + \sin \,x} \right)$$

At $$x = \pi ,$$ $$f''\left( x \right) < 0$$

Hence, local maxima at $$x = \pi $$

At $$x = 2\pi ,\,\,\,f''\left( x \right) > 0$$

Hence local minima at $$x = 2\pi $$

At local maxima or minima, $$f'\left( x \right) = 0$$

$$ \Rightarrow x = 0$$ or $$sin$$ $$x=0$$

$$ \Rightarrow x = 2\pi ,\,\,\pi \in \left( {0,{{5\pi } \over 2}} \right)$$

$$f''\left( x \right) = \sqrt x \cos \,x + {1 \over {2\sqrt x }}\sin \,x$$

$$ = {1 \over {2\sqrt x }}\left( {2x\,\cos \,x + \sin \,x} \right)$$

At $$x = \pi ,$$ $$f''\left( x \right) < 0$$

Hence, local maxima at $$x = \pi $$

At $$x = 2\pi ,\,\,\,f''\left( x \right) > 0$$

Hence local minima at $$x = 2\pi $$

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Complex Numbers

Quadratic Equation and Inequalities

Permutations and Combinations

Mathematical Induction and Binomial Theorem

Sequences and Series

Matrices and Determinants

Vector Algebra and 3D Geometry

Probability

Statistics

Mathematical Reasoning

Trigonometric Functions & Equations

Properties of Triangle

Inverse Trigonometric Functions

Straight Lines and Pair of Straight Lines

Circle

Conic Sections

Functions

Limits, Continuity and Differentiability

Differentiation

Application of Derivatives

Indefinite Integrals

Definite Integrals and Applications of Integrals

Differential Equations