Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

Let f and g be continuous functions on [0, a] such that f(x) = f(a – x) and g(x) + g(a – x) = 4, then $$\int\limits_0^a \, $$f(x) g(x) dx is equal to :

A

4$$\int\limits_0^a \, $$f(x)dx

B

$$-$$ 3$$\int\limits_0^a \, $$f(x)dx

C

$$\int\limits_0^a \, $$f(x)dx

D

2$$\int\limits_0^a \, $$f(x)dx

$${\rm I} = \int_0^a {f\left( x \right)g\left( x \right)dx} $$

$${\rm I} = \int_0^a {f\left( {a - x} \right)g\left( {a - x} \right)dx} $$

$${\rm I} = \int_0^a {f\left( x \right)\left( {4 - g\left( x \right.} \right)dx} $$

$${\rm I} = 4\int_0^a {f\left( x \right)dx - {\rm I}} $$

$$ \Rightarrow {\rm I} = 2\int_0^a {f\left( x \right)dx} $$

$${\rm I} = \int_0^a {f\left( {a - x} \right)g\left( {a - x} \right)dx} $$

$${\rm I} = \int_0^a {f\left( x \right)\left( {4 - g\left( x \right.} \right)dx} $$

$${\rm I} = 4\int_0^a {f\left( x \right)dx - {\rm I}} $$

$$ \Rightarrow {\rm I} = 2\int_0^a {f\left( x \right)dx} $$

2

MCQ (Single Correct Answer)

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

A

0

B

4

C

1

D

2

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

3

MCQ (Single Correct Answer)

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

MCQ (Single Correct Answer)

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.

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