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 $$f\left( x \right) = \left\{ {\matrix{
{ - 1} & { - 2 \le x < 0} \cr
{{x^2} - 1,} & {0 \le x \le 2} \cr
} } \right.$$ and

$$g(x) = \left| {f\left( x \right)} \right| + f\left( {\left| x \right|} \right).$$

Then, in the interval (–2, 2), g is :

$$g(x) = \left| {f\left( x \right)} \right| + f\left( {\left| x \right|} \right).$$

Then, in the interval (–2, 2), g is :

A

non continuous

B

differentiable at all points

C

not differentiable at two points

D

not differentiable at one point

$$\left| {f\left( x \right)} \right| = \left\{ {\matrix{
1 & , & { - 2 \le x < 0} \cr
{1 - {x^2}} & , & {0 \le x < 1} \cr
{{x^2} - 1} & , & {1 \le x \le 2} \cr
} } \right.$$

and $$f\left( {\left| x \right|} \right) = {x^2} - 1,x \in \left[ { - 2,2} \right]$$

Hence $$g(x) = \left\{ {\matrix{ {{x^2}} & , & {x \in \left[ { - 2,0} \right]} \cr 0 & , & {x \in \left[ {0,1} \right)} \cr {2\left( {{x^2} - 1} \right)} & , & {x \in \left[ {1,2} \right]} \cr } } \right.$$

It is not differentiable at x = 1

and $$f\left( {\left| x \right|} \right) = {x^2} - 1,x \in \left[ { - 2,2} \right]$$

Hence $$g(x) = \left\{ {\matrix{ {{x^2}} & , & {x \in \left[ { - 2,0} \right]} \cr 0 & , & {x \in \left[ {0,1} \right)} \cr {2\left( {{x^2} - 1} \right)} & , & {x \in \left[ {1,2} \right]} \cr } } \right.$$

It is not differentiable at x = 1

2

Let [x] denote the greatest integer less than or equal to x. Then $$\mathop {\lim }\limits_{x \to 0} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( {\left| x \right| - \sin \left( {x\left[ x \right]} \right)} \right)}^2}} \over {{x^2}}}$$

A

equals $$\pi $$ + 1

B

equals 0

C

does not exist

D

equals $$\pi $$

R.H.L. $$=$$ $$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( {\left| x \right| - \sin \left( {x\left[ x \right]} \right)} \right)}^2}} \over {{x^2}}}$$

(as x $$ \to $$ 0^{+} $$ \Rightarrow $$ [x] $$=$$ 0)

$$=$$ $$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {x^2}} \over {{x^2}}}$$

$$=$$ $$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right)} \over {\left( {\pi {{\sin }^2}x} \right)}} + 1 = \pi + 1$$

L.H.L. $$=$$ $$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( { - x + \sin x} \right)}^2}} \over {{x^2}}}$$

(as x $$ \to $$ 0^{$$-$$} $$ \Rightarrow $$ [x] $$=$$ $$-$$1)

$$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right)} \over {\pi {{\sin }^2}x}}\,.\,{{\pi {{\sin }^2}x} \over {{x^2}}} + {\left( { - 1 + {{\sin x} \over x}} \right)^2} \Rightarrow \pi $$

R.H.L. $$ \ne $$ L.H.L.

(as x $$ \to $$ 0

$$=$$ $$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {x^2}} \over {{x^2}}}$$

$$=$$ $$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right)} \over {\left( {\pi {{\sin }^2}x} \right)}} + 1 = \pi + 1$$

L.H.L. $$=$$ $$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( { - x + \sin x} \right)}^2}} \over {{x^2}}}$$

(as x $$ \to $$ 0

$$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right)} \over {\pi {{\sin }^2}x}}\,.\,{{\pi {{\sin }^2}x} \over {{x^2}}} + {\left( { - 1 + {{\sin x} \over x}} \right)^2} \Rightarrow \pi $$

R.H.L. $$ \ne $$ L.H.L.

3

Let K be the set of all real values of x where the function f(x) = sin |x| – |x| + 2(x – $$\pi $$) cos |x| is not differentiable. Then the set K is equal to :

A

{0, $$\pi $$}

B

$$\phi $$ (an empty set)

C

{ r }

D

{0}

f(x) = sin$$\left| x \right| - \left| x \right|$$ + 2(x $$-$$ $$\pi $$) cosx

$$ \because $$ sin$$\left| x \right|$$ $$-$$ $$\left| x \right|$$ is differentiable function at c = 0

$$ \therefore $$ k = $$\phi $$

$$ \because $$ sin$$\left| x \right|$$ $$-$$ $$\left| x \right|$$ is differentiable function at c = 0

$$ \therefore $$ k = $$\phi $$

4

$$\mathop {\lim }\limits_{x \to 0} {{x\cot \left( {4x} \right)} \over {{{\sin }^2}x{{\cot }^2}\left( {2x} \right)}}$$ is equal to :

A

0

B

4

C

1

D

2

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

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Trigonometric Functions & Equations *keyboard_arrow_right*

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Circle *keyboard_arrow_right*

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