Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

Consider :

f $$\left( x \right) = {\tan ^{ - 1}}\left( {\sqrt {{{1 + \sin x} \over {1 - \sin x}}} } \right),x \in \left( {0,{\pi \over 2}} \right).$$

f $$\left( x \right) = {\tan ^{ - 1}}\left( {\sqrt {{{1 + \sin x} \over {1 - \sin x}}} } \right),x \in \left( {0,{\pi \over 2}} \right).$$

A normal to $$y = $$ f$$\left( x \right)$$ at $$x = {\pi \over 6}$$ also passes through the point:

A

$$\left( {{\pi \over 6},0} \right)$$

B

$$\left( {{\pi \over 4},0} \right)$$

C

$$(0,0)$$

D

$$\left( {0,{{2\pi } \over 3}} \right)$$

$$f\left( x \right) = {\tan ^{ - 1}}\left( {\sqrt {{{1 + \sin \,x} \over {1 - \sin x}}} } \right)$$

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

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

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

$$ \Rightarrow y = {\pi \over 4} + {x \over 2}$$

$$ \Rightarrow {{dy} \over {dx}} = {1 \over 2}$$

Slope of normal $$ = {{ - 1} \over {\left( {{{dy} \over {dx}}} \right)}} = - 2$$

At $$\left( {{\pi \over 6},{\pi \over 4} + {\pi \over {12}}} \right)$$

$$y - \left( {{\pi \over 4} + {\pi \over {12}}} \right) = - 2\left( {x - {\pi \over 6}} \right)$$

$$y - {{4\pi } \over {12}} = - 2x + {{2\pi } \over 6}$$

$$y - {\pi \over 3} = - 2x + {\pi \over 3}$$

$$y = - 2x + {{2\pi } \over 3}$$

This equation is satisfied only by the point

$$\left( {0,{{2\pi } \over 3}} \right)$$

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

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

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

$$ \Rightarrow y = {\pi \over 4} + {x \over 2}$$

$$ \Rightarrow {{dy} \over {dx}} = {1 \over 2}$$

Slope of normal $$ = {{ - 1} \over {\left( {{{dy} \over {dx}}} \right)}} = - 2$$

At $$\left( {{\pi \over 6},{\pi \over 4} + {\pi \over {12}}} \right)$$

$$y - \left( {{\pi \over 4} + {\pi \over {12}}} \right) = - 2\left( {x - {\pi \over 6}} \right)$$

$$y - {{4\pi } \over {12}} = - 2x + {{2\pi } \over 6}$$

$$y - {\pi \over 3} = - 2x + {\pi \over 3}$$

$$y = - 2x + {{2\pi } \over 3}$$

This equation is satisfied only by the point

$$\left( {0,{{2\pi } \over 3}} \right)$$

2

MCQ (Single Correct Answer)

A wire of length $$2$$ units is cut into two parts which are bent respectively to form a square of side $$=x$$ units and a circle of radius $$=r$$ units. If the sum of the areas of the square and the circle so formed is minimum, then:

A

$$x=2r$$

B

$$2x=r$$

C

$$2x = \left( {\pi + 4} \right)r$$

D

$$\left( {4 - \pi } \right)x = \pi \,\, r$$

$$4x + 2\pi r = 2$$ $$\,\,\,$$ $$ \Rightarrow 2x + \pi r = 1$$

$$S = {x^2} + \pi {r^2}$$

$$S = {\left( {{{1 - \pi r} \over 2}} \right)^2} + \pi {r^2}$$

$${{dS} \over {dr}} = 2\left( {{{1 - \pi r} \over 2}} \right)\left( {{{ - \pi } \over 2}} \right) + 2\pi r$$

$$ \Rightarrow {{ - \pi } \over 2} + {{{\pi ^2}r} \over 2} + 2\pi r = 0$$

$$ \Rightarrow r = {1 \over {\pi + 4}}$$

$$ \Rightarrow x = {2 \over {\pi + 4}}\,$$

$$ \Rightarrow x = 2r$$

$$S = {x^2} + \pi {r^2}$$

$$S = {\left( {{{1 - \pi r} \over 2}} \right)^2} + \pi {r^2}$$

$${{dS} \over {dr}} = 2\left( {{{1 - \pi r} \over 2}} \right)\left( {{{ - \pi } \over 2}} \right) + 2\pi r$$

$$ \Rightarrow {{ - \pi } \over 2} + {{{\pi ^2}r} \over 2} + 2\pi r = 0$$

$$ \Rightarrow r = {1 \over {\pi + 4}}$$

$$ \Rightarrow x = {2 \over {\pi + 4}}\,$$

$$ \Rightarrow x = 2r$$

3

MCQ (Single Correct Answer)

Let $$f(x)$$ be a polynomial of degree four having extreme values

at $$x=1$$ and $$x=2$$. If $$\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3$$, then f$$(2)$$ is equal to :

at $$x=1$$ and $$x=2$$. If $$\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3$$, then f$$(2)$$ is equal to :

A

$$0$$

B

$$4$$

C

$$-8$$

D

$$-4$$

$$\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3 \Rightarrow \mathop {Lim}\limits_{x \to 0} {{f\left( x \right)} \over {{x^2}}} = 2$$

So, $$f(x)$$ contains terms in $$x{}^2,{x^3}$$ and $${x^4}$$

Let $$f\left( x \right) = {a_1}{x^2} + {a_2}{x^3} + {a_3}{x^4}$$

Since $$\mathop {\lim }\limits_{x \to 0} {{f\left( x \right)} \over {{x^2}}} = 2 \Rightarrow {a_1} = 2$$

Hence, $$f\left( x \right) = 2{x^2} + {a_2}{x^3} + {a_3}{x^4}$$

$$f'\left( x \right) = 4x + 3{a_2}{x^2} + 4{a_3}{x^3}$$

As given: $$f'\left( 1 \right) = 0$$ and $$f'\left( 2 \right) = 0$$

Hence, $$4 + 3{a_2} + 4{a_3} = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

and $$8 + 12{a_2} + 32{a_3} = 0\,\,\,\,\,...\left( 1 \right)$$

By $$4x\left( {eq1} \right) - eq\left( 2 \right),$$ we get

$$16 + 12{a_2} + 16{a_3} - \left( {8 + 12{a_2} + 32{a_3}} \right) = 0$$

$$ \Rightarrow 8 - 16{a_3} = 0 \Rightarrow {a_3} = 1/2$$

and by eqn. $$\left( 1 \right),4 + 3{a_2} + 4/2 = 0 \Rightarrow {a_2} = - 2$$

$$ \Rightarrow f\left( x \right) = 2{x^2} - 2{x^3} + {1 \over 2}{x^4}$$

$$f\left( 2 \right) = 2 \times 4 - 2 \times 8 + {1 \over 2} \times 16 = 0$$

So, $$f(x)$$ contains terms in $$x{}^2,{x^3}$$ and $${x^4}$$

Let $$f\left( x \right) = {a_1}{x^2} + {a_2}{x^3} + {a_3}{x^4}$$

Since $$\mathop {\lim }\limits_{x \to 0} {{f\left( x \right)} \over {{x^2}}} = 2 \Rightarrow {a_1} = 2$$

Hence, $$f\left( x \right) = 2{x^2} + {a_2}{x^3} + {a_3}{x^4}$$

$$f'\left( x \right) = 4x + 3{a_2}{x^2} + 4{a_3}{x^3}$$

As given: $$f'\left( 1 \right) = 0$$ and $$f'\left( 2 \right) = 0$$

Hence, $$4 + 3{a_2} + 4{a_3} = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

and $$8 + 12{a_2} + 32{a_3} = 0\,\,\,\,\,...\left( 1 \right)$$

By $$4x\left( {eq1} \right) - eq\left( 2 \right),$$ we get

$$16 + 12{a_2} + 16{a_3} - \left( {8 + 12{a_2} + 32{a_3}} \right) = 0$$

$$ \Rightarrow 8 - 16{a_3} = 0 \Rightarrow {a_3} = 1/2$$

and by eqn. $$\left( 1 \right),4 + 3{a_2} + 4/2 = 0 \Rightarrow {a_2} = - 2$$

$$ \Rightarrow f\left( x \right) = 2{x^2} - 2{x^3} + {1 \over 2}{x^4}$$

$$f\left( 2 \right) = 2 \times 4 - 2 \times 8 + {1 \over 2} \times 16 = 0$$

4

MCQ (Single Correct Answer)

If $$f$$ and $$g$$ are differentiable functions in $$\left[ {0,1} \right]$$ satisfying

$$f\left( 0 \right) = 2 = g\left( 1 \right),g\left( 0 \right) = 0$$ and $$f\left( 1 \right) = 6,$$ then for some $$c \in \left] {0,1} \right[$$

$$f\left( 0 \right) = 2 = g\left( 1 \right),g\left( 0 \right) = 0$$ and $$f\left( 1 \right) = 6,$$ then for some $$c \in \left] {0,1} \right[$$

A

$$f'\left( c \right) = g'\left( c \right)$$

B

$$f'\left( c \right) = 2g'\left( c \right)$$

C

$$2f'\left( c \right) = g'\left( c \right)$$

D

$$2f'\left( c \right) = 3g'\left( c \right)$$

Since, $$f$$ and $$g$$ both are continuous function on $$\left[ {0,1} \right]$$

and differentiable on $$\left( {0,1} \right)$$ then $$\exists c \in \left( {0,1} \right)$$ such that

$$f'\left( c \right) = {{f\left( 1 \right) - f\left( 0 \right)} \over 1} = {{6 - 2} \over 1} = 4$$

and $$g'\left( c \right) = {{g\left( 1 \right) - g\left( 0 \right)} \over 1} = {{2 - 0} \over 1} = 2$$

Thus, we get $$f'\left( c \right) = 2g'\left( c \right)$$

and differentiable on $$\left( {0,1} \right)$$ then $$\exists c \in \left( {0,1} \right)$$ such that

$$f'\left( c \right) = {{f\left( 1 \right) - f\left( 0 \right)} \over 1} = {{6 - 2} \over 1} = 4$$

and $$g'\left( c \right) = {{g\left( 1 \right) - g\left( 0 \right)} \over 1} = {{2 - 0} \over 1} = 2$$

Thus, we get $$f'\left( c \right) = 2g'\left( c \right)$$

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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

Limits, Continuity and Differentiability

Differentiation

Application of Derivatives

Indefinite Integrals

Definite Integrals and Applications of Integrals

Differential Equations