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Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

If the equation $${a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ........... + {a_1}x = 0$$

$${a_1} \ne 0,n \ge 2,$$ has a positive root $$x = \alpha $$, then the equation

$$n{a_n}{x^{n - 1}} + \left( {n - 1} \right){a_{n - 1}}{x^{n - 2}} + ........... + {a_1} = 0$$ has a positive root, which is

$${a_1} \ne 0,n \ge 2,$$ has a positive root $$x = \alpha $$, then the equation

$$n{a_n}{x^{n - 1}} + \left( {n - 1} \right){a_{n - 1}}{x^{n - 2}} + ........... + {a_1} = 0$$ has a positive root, which is

A

greater than $$\alpha $$

B

smaller than $$\alpha $$

C

greater than or equal to smaller than $$\alpha $$

D

equal to smaller than $$\alpha $$

Let $$f\left( x \right) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ........... + {a_1}x = 0$$

The other given equation,

$$na{}_n{x^{n - 1}} + \left( {n - 1} \right){a_{n - 1}}{x^{n - 2}} + .... + {a_1} = 0 = f'\left( x \right)$$

Given $${a_1} \ne 0 \Rightarrow f\left( 0 \right) = 0$$

Again $$f(x)$$ has root $$\alpha ,\,\,\, \Rightarrow f\left( \alpha \right) = 0$$

$$\therefore$$ $$\,\,\,\,\,f\left( 0 \right) = f\left( \alpha \right)$$

$$\therefore$$ By Roll's theorem $$f'(x)=0$$ has root be-

tween $$\left( {0,\alpha } \right)$$

Hence $$f'(x)$$ has a positive root smaller than $$\alpha .$$

The other given equation,

$$na{}_n{x^{n - 1}} + \left( {n - 1} \right){a_{n - 1}}{x^{n - 2}} + .... + {a_1} = 0 = f'\left( x \right)$$

Given $${a_1} \ne 0 \Rightarrow f\left( 0 \right) = 0$$

Again $$f(x)$$ has root $$\alpha ,\,\,\, \Rightarrow f\left( \alpha \right) = 0$$

$$\therefore$$ $$\,\,\,\,\,f\left( 0 \right) = f\left( \alpha \right)$$

$$\therefore$$ By Roll's theorem $$f'(x)=0$$ has root be-

tween $$\left( {0,\alpha } \right)$$

Hence $$f'(x)$$ has a positive root smaller than $$\alpha .$$

2

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 $$r=10+5)$$

3

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.

4

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

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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

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Properties of Triangle

Inverse Trigonometric Functions

Straight Lines and Pair of Straight Lines

Circle

Conic Sections

Functions

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Differentiation

Application of Derivatives

Indefinite Integrals

Definite Integrals and Applications of Integrals

Differential Equations