Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

If $$p$$ and $$q$$ are positive real numbers such that $${p^2} + {q^2} = 1$$, then the maximum value of $$(p+q)$$ is

A

$${1 \over 2}$$

B

$${1 \over {\sqrt 2 }}$$

C

$${\sqrt 2 }$$

D

$$2$$

Given that $${p^2} + {q^2} = 1$$

$$\therefore$$ $$p = \cos \theta $$ and $$q = \sin \theta $$

Then $$p+q$$ $$ = \cos \theta + \sin \theta $$

We know that

$$ - \sqrt {{a^2} + {b^2}} \le a\cos \theta + b\sin \theta \le \sqrt {{a^2} + {b^2}} $$

$$\therefore$$ $$ - \sqrt 2 \le \cos \theta + \sin \theta \le \sqrt 2 $$

Hence max. value of $$p + q$$ is $$\sqrt 2 $$

$$\therefore$$ $$p = \cos \theta $$ and $$q = \sin \theta $$

Then $$p+q$$ $$ = \cos \theta + \sin \theta $$

We know that

$$ - \sqrt {{a^2} + {b^2}} \le a\cos \theta + b\sin \theta \le \sqrt {{a^2} + {b^2}} $$

$$\therefore$$ $$ - \sqrt 2 \le \cos \theta + \sin \theta \le \sqrt 2 $$

Hence max. value of $$p + q$$ is $$\sqrt 2 $$

2

MCQ (Single Correct Answer)

A value of $$c$$ for which conclusion of Mean Value Theorem holds for the function $$f\left( x \right) = {\log _e}x$$ on the interval $$\left[ {1,3} \right]$$ is

A

$${\log _3}e$$

B

$${\log _e}3$$

C

$$2\,\,{\log _3}e$$

D

$${1 \over 2}{\log _3}e$$

Using Lagrange's Mean Value Theorem

Let $$f(x)$$ be a function defined on $$\left[ {a,b} \right]$$

then, $$f'\left( c \right) = {{f\left( b \right) - f\left( a \right)} \over {b - a}}\,\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)$$

$$c\,\, \in \left[ {a,b} \right]$$

$$\therefore$$ Given $$f\left( x \right) = {\log _e}x$$

$$\therefore$$ $$f'\left( x \right) = {1 \over x}$$

$$\therefore$$ equation $$(i)$$ become $${1 \over c} = {{f\left( 3 \right) - f\left( 1 \right)} \over {3 - 1}}$$

$$ \Rightarrow {1 \over c} = {{{{\log }_e}3 - {{\log }_e}1} \over 2} = {{{{\log }_e}3} \over 2}$$

$$ \Rightarrow c = {2 \over {{{\log }_e}3}} \Rightarrow c = 2\,{\log _3}e$$

Let $$f(x)$$ be a function defined on $$\left[ {a,b} \right]$$

then, $$f'\left( c \right) = {{f\left( b \right) - f\left( a \right)} \over {b - a}}\,\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)$$

$$c\,\, \in \left[ {a,b} \right]$$

$$\therefore$$ Given $$f\left( x \right) = {\log _e}x$$

$$\therefore$$ $$f'\left( x \right) = {1 \over x}$$

$$\therefore$$ equation $$(i)$$ become $${1 \over c} = {{f\left( 3 \right) - f\left( 1 \right)} \over {3 - 1}}$$

$$ \Rightarrow {1 \over c} = {{{{\log }_e}3 - {{\log }_e}1} \over 2} = {{{{\log }_e}3} \over 2}$$

$$ \Rightarrow c = {2 \over {{{\log }_e}3}} \Rightarrow c = 2\,{\log _3}e$$

3

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

4

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 $$\theta = {\pi \over 2}$$

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

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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