Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

The function $$f\left( x \right) = {\tan ^{ - 1}}\left( {\sin x + \cos x} \right)$$ is an incresing function in

A

$$\left( {0,{\pi \over 2}} \right)$$

B

$$\left( { - {\pi \over 2},{\pi \over 2}} \right)$$

C

$$\left( { {\pi \over 4},{\pi \over 2}} \right)$$

D

$$\left( { - {\pi \over 2},{\pi \over 4}} \right)$$

Given $$f\left( x \right) = {\tan ^{ - 1}}\left( {\sin x + \cos x} \right)$$

$$f'\left( x \right) = {1 \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}.\left( {\cos x - \sin x} \right)$$

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

$$ = {{\left( {\cos {\pi \over 4}.\cos x - \sin {\pi \over 4}.\sin x} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}$$

$$\therefore$$ $$f'\left( x \right) = {{\sqrt 2 \cos \left( {x + {\pi \over 4}} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}$$

if $$f'\left( x \right) > O$$ then $$f\left( x \right)$$ is increasing function.

Hence $$f(x)$$ is increasing, if $$ - {\pi \over 2} < x + {\pi \over 4} < {\pi \over 2}$$

$$ \Rightarrow - {{3\pi } \over 4} < x < {\pi \over 4}$$

Hence, $$f(x)$$ is increasing when $$n \in \left( { - {\pi \over 2},{\pi \over 4}} \right)$$

$$f'\left( x \right) = {1 \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}.\left( {\cos x - \sin x} \right)$$

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

$$ = {{\left( {\cos {\pi \over 4}.\cos x - \sin {\pi \over 4}.\sin x} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}$$

$$\therefore$$ $$f'\left( x \right) = {{\sqrt 2 \cos \left( {x + {\pi \over 4}} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}$$

if $$f'\left( x \right) > O$$ then $$f\left( x \right)$$ is increasing function.

Hence $$f(x)$$ is increasing, if $$ - {\pi \over 2} < x + {\pi \over 4} < {\pi \over 2}$$

$$ \Rightarrow - {{3\pi } \over 4} < x < {\pi \over 4}$$

Hence, $$f(x)$$ is increasing when $$n \in \left( { - {\pi \over 2},{\pi \over 4}} \right)$$

2

MCQ (Single Correct Answer)

A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length $$x$$. The maximum area enclosed by the park is

A

$${3 \over 2}{x^2}$$

B

$$\sqrt {{{{x^3}} \over 8}} $$

C

$${1 \over 2}{x^2}$$

D

$$\pi {x^2}$$

Area $$ = {1 \over 2}{x^2}\,\sin \,\theta $$

Maximum value of $$\sin \theta $$ is $$1$$ at $$\theta = {\pi \over 2}$$

$${A_{\max }} = {1 \over 2}{x^2}$$

Maximum value of $$\sin \theta $$ is $$1$$ at $$\theta = {\pi \over 2}$$

$${A_{\max }} = {1 \over 2}{x^2}$$

3

MCQ (Single Correct Answer)

The function $$f\left( x \right) = {x \over 2} + {2 \over x}$$ has a local minimum at

A

$$x=2$$

B

$$x=-2$$

C

$$x=0$$

D

$$x=1$$

$$f\left( x \right) = {x \over 2} + {2 \over x} \Rightarrow f'\left( x \right) = {1 \over 2} - {2 \over {{x^2}}} = 0$$

$$ \Rightarrow {x^2} = 4$$ or $$x=2,-2;$$ $$\,\,\,\,\,f''\left( x \right) = {4 \over {{x^3}}}$$

$$f''{\left. {\left( x \right)} \right]_{x = 2}} = + ve \Rightarrow f\left( x \right)$$

has local min at $$x=2.$$

$$ \Rightarrow {x^2} = 4$$ or $$x=2,-2;$$ $$\,\,\,\,\,f''\left( x \right) = {4 \over {{x^3}}}$$

$$f''{\left. {\left( x \right)} \right]_{x = 2}} = + ve \Rightarrow f\left( x \right)$$

has local min at $$x=2.$$

4

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 $$\alpha .$$

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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