Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

A spherical iron ball $$10$$ cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of $$50$$ cm$$^3$$ /min. When the thickness of ice is $$5$$ cm, then the rate at which the thickness of ice decreases is

A

$${1 \over {36\pi }}$$ cm/min

B

$${1 \over {18\pi }}$$ cm/min

C

$${1 \over {54\pi }}$$ cm/min

D

$${5 \over {6\pi }}$$ cm/min

Given that

$${{dv} \over {dt}} = 50\,c{m^3}/\min $$

$$ \Rightarrow {d \over {dt}}\left( {{4 \over 3}\pi {r^3}} \right) = 50$$

$$ \Rightarrow 4\pi {r^2}{{dr} \over {dt}} = 50$$

$$ \Rightarrow {{dr} \over {dt}} = {{50} \over {4\pi {{\left( {15} \right)}^2}}} = {1 \over {18\pi }}\,\,cm/\min \,\,$$

(here $$r=10+5)$$

$${{dv} \over {dt}} = 50\,c{m^3}/\min $$

$$ \Rightarrow {d \over {dt}}\left( {{4 \over 3}\pi {r^3}} \right) = 50$$

$$ \Rightarrow 4\pi {r^2}{{dr} \over {dt}} = 50$$

$$ \Rightarrow {{dr} \over {dt}} = {{50} \over {4\pi {{\left( {15} \right)}^2}}} = {1 \over {18\pi }}\,\,cm/\min \,\,$$

(here $$r=10+5)$$

2

MCQ (Single Correct Answer)

The normal to the curve

$$x = a\left( {\cos \theta + \theta \sin \theta } \right),y = a\left( {\sin \theta - \theta \cos \theta } \right)$$ at any point

$$\theta\, '$$ is such that

$$x = a\left( {\cos \theta + \theta \sin \theta } \right),y = a\left( {\sin \theta - \theta \cos \theta } \right)$$ at any point

$$\theta\, '$$ is such that

A

it passes through the origin

B

it makes an angle $${\pi \over 2} + \theta $$ with the $$x$$-axis

C

it passes through $$\left( {a{\pi \over 2}, - a} \right)$$

D

it is at a constant distance from the origin

$$x = a\left( {\cos \theta + \theta \sin \theta } \right)$$

$$ \Rightarrow {{dx} \over {d\theta }} = a\left( { - \sin \theta + \sin \theta + \theta \cos \theta } \right)$$

$$ \Rightarrow {{dx} \over {d\theta }} = a\theta \cos \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$$y = a\left( {\sin \theta - \theta \cos \theta } \right)$$

$${{dy} \over {d\theta }} = a\left[ {\cos \theta - \cos \theta + \theta \sin \theta } \right]$$

$$ \Rightarrow {{dy} \over {d\theta }} = a\theta \sin \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

From equations $$(1)$$ and $$(2),$$ we get

$${{dy} \over {dx}} = \tan \theta \Rightarrow $$ Slope of normal $$ = - \cot \,\theta $$

Equation of normal at

$$'\theta '$$ is $$y - a\left( {\sin \theta - \theta \cos \theta } \right)$$

$$ = - \cot \theta \left( {x - a\left( {\cos \theta + \theta \sin \theta } \right)} \right.$$

$$ \Rightarrow y\sin \theta - a{\sin ^2}\theta + a\,\,\theta \cos \theta \sin \theta $$

$$ = - x\cos \theta + a{\cos ^2}\theta + a\,\theta \sin \theta \cos \theta $$

$$ \Rightarrow x\cos \theta + y\sin \theta = a$$

Clearly this is an equation of straight line -

which is at a constant distance $$'a'$$ from origin.

$$ \Rightarrow {{dx} \over {d\theta }} = a\left( { - \sin \theta + \sin \theta + \theta \cos \theta } \right)$$

$$ \Rightarrow {{dx} \over {d\theta }} = a\theta \cos \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$$y = a\left( {\sin \theta - \theta \cos \theta } \right)$$

$${{dy} \over {d\theta }} = a\left[ {\cos \theta - \cos \theta + \theta \sin \theta } \right]$$

$$ \Rightarrow {{dy} \over {d\theta }} = a\theta \sin \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

From equations $$(1)$$ and $$(2),$$ we get

$${{dy} \over {dx}} = \tan \theta \Rightarrow $$ Slope of normal $$ = - \cot \,\theta $$

Equation of normal at

$$'\theta '$$ is $$y - a\left( {\sin \theta - \theta \cos \theta } \right)$$

$$ = - \cot \theta \left( {x - a\left( {\cos \theta + \theta \sin \theta } \right)} \right.$$

$$ \Rightarrow y\sin \theta - a{\sin ^2}\theta + a\,\,\theta \cos \theta \sin \theta $$

$$ = - x\cos \theta + a{\cos ^2}\theta + a\,\theta \sin \theta \cos \theta $$

$$ \Rightarrow x\cos \theta + y\sin \theta = a$$

Clearly this is an equation of straight line -

which is at a constant distance $$'a'$$ from origin.

3

MCQ (Single Correct Answer)

Area of the greatest rectangle that can be inscribed in the

ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$

ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$

A

$$2ab$$

B

$$ab$$

C

$$\sqrt {ab} $$

D

$${a \over b}$$

Area of rectangle $$ABCD$$ $$ = 2a\,\cos \,\theta $$

$$\left( {2b\,\sin \,\theta } \right) = 2ab\,\sin \,2\theta $$

$$ \Rightarrow $$ Area of greatest rectangle is equal to $$2ab$$

When $$\sin \,2\theta = 1.$$

$$\left( {2b\,\sin \,\theta } \right) = 2ab\,\sin \,2\theta $$

$$ \Rightarrow $$ Area of greatest rectangle is equal to $$2ab$$

When $$\sin \,2\theta = 1.$$

4

MCQ (Single Correct Answer)

The normal to the curve x = a(1 + cos $$\theta $$), $$y = a\sin \theta $$ at $$'\theta '$$ always passes through the fixed point

A

$$(a, a)$$

B

$$(0, a)$$

C

$$(0, 0)$$

D

$$(a, 0)$$

$${{dx} \over {d\theta }} = - a\sin \theta $$ and $${{dy} \over {d\theta }} = a\cos \theta $$

$$\therefore$$ $${{dy} \over {dx}} = - \cot \theta .$$

$$\therefore$$ The slope of the normal at $$\theta $$ = $$ - {1 \over { - \cot \theta }}$$$$= \tan \theta $$

$$\therefore$$ The equation of the normal at $$\theta $$ is

$$y - a\sin \theta = \tan \theta \left( {x - a - a\cos \theta } \right)$$

$$ \Rightarrow $$ $$y - a\sin \theta =$$ $${{\sin \theta } \over {\cos \theta }}$$ $$\left( {x - a - a\cos \theta } \right)$$

$$ \Rightarrow y\cos \theta - a\sin \theta \cos \theta $$

$$\,\,\,\,\,\,\,\,\,\,\,$$ $$ = x\sin \theta - a\sin \theta - a\sin \theta \cos \theta $$

$$ \Rightarrow x\sin \theta - y\cos \theta = a\sin \theta $$

$$ \Rightarrow y = \left( {x - a} \right)\tan \theta $$

which always passes through $$(a, 0)$$

$$\therefore$$ $${{dy} \over {dx}} = - \cot \theta .$$

$$\therefore$$ The slope of the normal at $$\theta $$ = $$ - {1 \over { - \cot \theta }}$$$$= \tan \theta $$

$$\therefore$$ The equation of the normal at $$\theta $$ is

$$y - a\sin \theta = \tan \theta \left( {x - a - a\cos \theta } \right)$$

$$ \Rightarrow $$ $$y - a\sin \theta =$$ $${{\sin \theta } \over {\cos \theta }}$$ $$\left( {x - a - a\cos \theta } \right)$$

$$ \Rightarrow y\cos \theta - a\sin \theta \cos \theta $$

$$\,\,\,\,\,\,\,\,\,\,\,$$ $$ = x\sin \theta - a\sin \theta - a\sin \theta \cos \theta $$

$$ \Rightarrow x\sin \theta - y\cos \theta = a\sin \theta $$

$$ \Rightarrow y = \left( {x - a} \right)\tan \theta $$

which always passes through $$(a, 0)$$

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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