Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

Hydrology

Transportation Engineering

Strength of Materials Or Solid Mechanics

Reinforced Cement Concrete

Steel Structures

Irrigation

Environmental Engineering

Engineering Mathematics

Structural Analysis

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

General Aptitude

1

Let a, b $$ \in $$ **R**, (a $$ \ne $$ 0). If the function *f* defined as

$$f\left( x \right) = \left\{ {\matrix{ {{{2{x^2}} \over a}\,\,,} & {0 \le x < 1} \cr {a\,\,\,,} & {1 \le x < \sqrt 2 } \cr {{{2{b^2} - 4b} \over {{x^3}}},} & {\sqrt 2 \le x < \infty } \cr } } \right.$$

is continuous in the interval [0, $$\infty $$), then an ordered pair ( a, b) is :

$$f\left( x \right) = \left\{ {\matrix{ {{{2{x^2}} \over a}\,\,,} & {0 \le x < 1} \cr {a\,\,\,,} & {1 \le x < \sqrt 2 } \cr {{{2{b^2} - 4b} \over {{x^3}}},} & {\sqrt 2 \le x < \infty } \cr } } \right.$$

is continuous in the interval [0, $$\infty $$), then an ordered pair ( a, b) is :

A

$$\left( {\sqrt 2 ,1 - \sqrt 3 } \right)$$

B

$$\left( { - \sqrt 2 ,1 + \sqrt 3 } \right)$$

C

$$\left( {\sqrt 2 , - 1 + \sqrt 3 } \right)$$

D

$$\left( { - \sqrt 2 ,1 - \sqrt 3 } \right)$$

f(x) is continuous at x = 1

$$ \therefore $$ $${{2{{\left( 1 \right)}^2}} \over a} = a$$

$$ \Rightarrow $$ a^{2} = 2

$$ \Rightarrow $$ a = $$ \pm $$ $$\sqrt 2 $$

Also f(x) is continuous at x = $$\sqrt 2 $$

$$ \therefore $$ a = $${{2{b^2} - 4b} \over {2\sqrt 2 }}$$

Now, when a = $$\sqrt 2 ,$$ then

4 = 2b^{2} $$-$$ 4b

$$ \Rightarrow $$ b^{2} $$-$$ 2b = 2

$$ \Rightarrow $$ b^{2} $$-$$ 2b $$-$$ 2 = 0

$$ \Rightarrow $$ b = $${{2 \mp \sqrt {4 + 4.2} } \over 2}$$

= 1 $$ \pm $$ $$\sqrt 3 $$

$$ \therefore $$ (a, b) = $$\left( {\sqrt 2 ,1 \pm \sqrt 3 } \right)$$

When a = $$-$$ $$\sqrt 2 $$, then

$$-$$ $$\sqrt 2 $$ = $${{2{b^2} - 4b} \over {2\sqrt 2 }}$$

$$ \Rightarrow $$ $$-$$ 4 = 2b^{2} $$-$$ 4b

$$ \Rightarrow $$ b^{2} $$-$$ 2b + 2 = 0

$$ \therefore $$ b = $${{2 \pm \sqrt {4 - 8} } \over 2}$$ = 1 $$ \pm $$ i

As b $$ \in $$ Real number so,

b = 1 $$ \pm $$ i is not accepted.

$$ \therefore $$ (a, b) = $$\left( {\sqrt 2 ,1 + \sqrt 3 } \right)$$ or $$\left( {\sqrt 2 ,1 - \sqrt 3 } \right)$$

$$ \therefore $$ $${{2{{\left( 1 \right)}^2}} \over a} = a$$

$$ \Rightarrow $$ a

$$ \Rightarrow $$ a = $$ \pm $$ $$\sqrt 2 $$

Also f(x) is continuous at x = $$\sqrt 2 $$

$$ \therefore $$ a = $${{2{b^2} - 4b} \over {2\sqrt 2 }}$$

Now, when a = $$\sqrt 2 ,$$ then

4 = 2b

$$ \Rightarrow $$ b

$$ \Rightarrow $$ b

$$ \Rightarrow $$ b = $${{2 \mp \sqrt {4 + 4.2} } \over 2}$$

= 1 $$ \pm $$ $$\sqrt 3 $$

$$ \therefore $$ (a, b) = $$\left( {\sqrt 2 ,1 \pm \sqrt 3 } \right)$$

When a = $$-$$ $$\sqrt 2 $$, then

$$-$$ $$\sqrt 2 $$ = $${{2{b^2} - 4b} \over {2\sqrt 2 }}$$

$$ \Rightarrow $$ $$-$$ 4 = 2b

$$ \Rightarrow $$ b

$$ \therefore $$ b = $${{2 \pm \sqrt {4 - 8} } \over 2}$$ = 1 $$ \pm $$ i

As b $$ \in $$ Real number so,

b = 1 $$ \pm $$ i is not accepted.

$$ \therefore $$ (a, b) = $$\left( {\sqrt 2 ,1 + \sqrt 3 } \right)$$ or $$\left( {\sqrt 2 ,1 - \sqrt 3 } \right)$$

2

$$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\cot x - \cos x} \over {{{\left( {\pi - 2x} \right)}^3}}}$$ equals

A

$${1 \over {16}}$$

B

$${1 \over 8}$$

C

$${1 \over {4}}$$

D

$${1 \over {24}}$$

$$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\cot x - \cos x} \over {{{\left( {\pi - 2x} \right)}^3}}}$$

= $$\mathop {\lim }\limits_{x \to {\pi \over 2}} {1 \over 8}.{{\cos x\left( {1 - \sin x} \right)} \over {\sin x{{\left( {{\pi \over 2} - x} \right)}^3}}}$$

Put $${{\pi \over 2} - x}$$ = t

$$ \Rightarrow $$ x $$ \to $$ $${{\pi \over 2}}$$

t $$ \to $$ 0

= $$\mathop {\lim }\limits_{t \to 0} {1 \over 8}.{{\cos \left( {{\pi \over 2} - t} \right)\left( {1 - \sin \left( {{\pi \over 2} - t} \right)} \right)} \over {\sin \left( {{\pi \over 2} - t} \right){{\left( {{\pi \over 2} - {\pi \over 2} + t} \right)}^3}}}$$

= $$\mathop {{1 \over 8}\lim }\limits_{t \to 0} {{\sin t\left( {1 - \cos t} \right)} \over {\cos t{{\left( t \right)}^3}}}$$

= $$\mathop {{1 \over 8}\lim }\limits_{t \to 0} {{\sin t\left( {2{{\sin }^2}{t \over 2}} \right)} \over {\cos t{{\left( t \right)}^3}}}$$

= $$\mathop {{1 \over 4}\lim }\limits_{t \to 0} {{\sin t\left( {{{\sin }^2}{t \over 2}} \right)} \over {\cos t{{\left( t \right)}^3}}}$$

= $$\mathop {\lim }\limits_{t \to 0} \left( {{{\sin t} \over t}} \right){\left( {{{\sin {t \over 2}} \over {{t \over 2}}}} \right)^2}{1 \over {\cos t}}{1 \over 4}$$

= $${1 \over 4} \times {1 \over 4}$$

= $${1 \over {16}}$$

= $$\mathop {\lim }\limits_{x \to {\pi \over 2}} {1 \over 8}.{{\cos x\left( {1 - \sin x} \right)} \over {\sin x{{\left( {{\pi \over 2} - x} \right)}^3}}}$$

Put $${{\pi \over 2} - x}$$ = t

$$ \Rightarrow $$ x $$ \to $$ $${{\pi \over 2}}$$

t $$ \to $$ 0

= $$\mathop {\lim }\limits_{t \to 0} {1 \over 8}.{{\cos \left( {{\pi \over 2} - t} \right)\left( {1 - \sin \left( {{\pi \over 2} - t} \right)} \right)} \over {\sin \left( {{\pi \over 2} - t} \right){{\left( {{\pi \over 2} - {\pi \over 2} + t} \right)}^3}}}$$

= $$\mathop {{1 \over 8}\lim }\limits_{t \to 0} {{\sin t\left( {1 - \cos t} \right)} \over {\cos t{{\left( t \right)}^3}}}$$

= $$\mathop {{1 \over 8}\lim }\limits_{t \to 0} {{\sin t\left( {2{{\sin }^2}{t \over 2}} \right)} \over {\cos t{{\left( t \right)}^3}}}$$

= $$\mathop {{1 \over 4}\lim }\limits_{t \to 0} {{\sin t\left( {{{\sin }^2}{t \over 2}} \right)} \over {\cos t{{\left( t \right)}^3}}}$$

= $$\mathop {\lim }\limits_{t \to 0} \left( {{{\sin t} \over t}} \right){\left( {{{\sin {t \over 2}} \over {{t \over 2}}}} \right)^2}{1 \over {\cos t}}{1 \over 4}$$

= $${1 \over 4} \times {1 \over 4}$$

= $${1 \over {16}}$$

3

If for $$x \in \left( {0,{1 \over 4}} \right)$$, the derivatives of

$${\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)$$ is $$\sqrt x .g\left( x \right)$$, then $$g\left( x \right)$$ equals

$${\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)$$ is $$\sqrt x .g\left( x \right)$$, then $$g\left( x \right)$$ equals

A

$${{{3x\sqrt x } \over {1 - 9{x^3}}}}$$

B

$${{{3x} \over {1 - 9{x^3}}}}$$

C

$${{3 \over {1 + 9{x^3}}}}$$

D

$${{9 \over {1 + 9{x^3}}}}$$

Let y = $${\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)$$

= $${\tan ^{ - 1}}\left[ {{{2.\left( {3{x^{{3 \over 2}}}} \right)} \over {1 - {{\left( {3{x^{{3 \over 2}}}} \right)}^2}}}} \right]$$

= 2$${\tan ^{ - 1}}\left( {3{x^{{3 \over 2}}}} \right)$$

$$ \therefore $$ $${{dy} \over {dx}} = 2.{1 \over {1 + {{\left( {3{x^{{3 \over 2}}}} \right)}^2}}}.3 \times {3 \over 2}{\left( x \right)^{{1 \over 2}}}$$

= $${9 \over {1 + 9{x^3}}}.\sqrt x $$

$$ \therefore $$ g(x) = $${9 \over {1 + 9{x^3}}}$$

= $${\tan ^{ - 1}}\left[ {{{2.\left( {3{x^{{3 \over 2}}}} \right)} \over {1 - {{\left( {3{x^{{3 \over 2}}}} \right)}^2}}}} \right]$$

= 2$${\tan ^{ - 1}}\left( {3{x^{{3 \over 2}}}} \right)$$

$$ \therefore $$ $${{dy} \over {dx}} = 2.{1 \over {1 + {{\left( {3{x^{{3 \over 2}}}} \right)}^2}}}.3 \times {3 \over 2}{\left( x \right)^{{1 \over 2}}}$$

= $${9 \over {1 + 9{x^3}}}.\sqrt x $$

$$ \therefore $$ g(x) = $${9 \over {1 + 9{x^3}}}$$

4

$$\mathop {\lim }\limits_{x \to 3} $$ $${{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}$$ is equal to :

A

$$\sqrt 3 $$

B

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

C

$${{\sqrt 3 } \over 2}$$

D

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

Given,

$$\mathop {\lim }\limits_{x \to 3} $$ $${{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}$$

Here if you put x = 3 in $${{\sqrt {3x} - 3} \over {\sqrt {2x - 4 - \sqrt 2 } }}$$

you will get $${0 \over 0}$$ form.

So, we can apply L' Hospital rule

$$\therefore\,\,\,$$ $$\mathop {\lim }\limits_{x \to 3} {{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}$$

= $$\mathop {\lim }\limits_{x \to 3} $$ $${{\sqrt 3 .{1 \over {2\sqrt x }}} \over {{2 \over {2\sqrt {2x - 4} }}}}$$ (applying L' Hospital rule)

= $${{\sqrt 3 .{1 \over {2\sqrt 3 }}} \over {{1 \over {\sqrt 6 - 4}}}}$$

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

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

$$\mathop {\lim }\limits_{x \to 3} $$ $${{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}$$

Here if you put x = 3 in $${{\sqrt {3x} - 3} \over {\sqrt {2x - 4 - \sqrt 2 } }}$$

you will get $${0 \over 0}$$ form.

So, we can apply L' Hospital rule

$$\therefore\,\,\,$$ $$\mathop {\lim }\limits_{x \to 3} {{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}$$

= $$\mathop {\lim }\limits_{x \to 3} $$ $${{\sqrt 3 .{1 \over {2\sqrt x }}} \over {{2 \over {2\sqrt {2x - 4} }}}}$$ (applying L' Hospital rule)

= $${{\sqrt 3 .{1 \over {2\sqrt 3 }}} \over {{1 \over {\sqrt 6 - 4}}}}$$

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

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

Number in Brackets after Paper Name Indicates No of Questions

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Trigonometric Functions & Equations *keyboard_arrow_right*

Properties of Triangle *keyboard_arrow_right*

Inverse Trigonometric Functions *keyboard_arrow_right*

Complex Numbers *keyboard_arrow_right*

Quadratic Equation and Inequalities *keyboard_arrow_right*

Permutations and Combinations *keyboard_arrow_right*

Mathematical Induction and Binomial Theorem *keyboard_arrow_right*

Sequences and Series *keyboard_arrow_right*

Matrices and Determinants *keyboard_arrow_right*

Vector Algebra and 3D Geometry *keyboard_arrow_right*

Probability *keyboard_arrow_right*

Statistics *keyboard_arrow_right*

Mathematical Reasoning *keyboard_arrow_right*

Functions *keyboard_arrow_right*

Limits, Continuity and Differentiability *keyboard_arrow_right*

Differentiation *keyboard_arrow_right*

Application of Derivatives *keyboard_arrow_right*

Indefinite Integrals *keyboard_arrow_right*

Definite Integrals and Applications of Integrals *keyboard_arrow_right*

Differential Equations *keyboard_arrow_right*

Straight Lines and Pair of Straight Lines *keyboard_arrow_right*

Circle *keyboard_arrow_right*

Conic Sections *keyboard_arrow_right*