Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

If $${I_1} = \int\limits_0^1 {{2^{{x^2}}}dx,{I_2} = \int\limits_0^1 {{2^{{x^3}}}dx,\,{I_3} = \int\limits_1^2 {{2^{{x^2}}}dx} } } $$ and $${I_4} = \int\limits_1^2 {{2^{{x^3}}}dx} $$ then

A

$${I_2} > {I_1}$$

B

$${I_1} > {I_2}$$

C

$${I_3} = {I_4}$$

D

$${I_3} > {I_4}$$

$${I_1} = \int\limits_0^1 {{2^{{x^2}}}} dx,\,{I_2} = \int\limits_0^1 {{2^{{x^3}}}} dx,$$

$$ = {I_3} = \int\limits_0^1 {{2^{{x^2}}}} dx,\,$$

$${I_4} = \int\limits_0^1 {{2^{{x^3}}}} dx\,\,$$

$$\forall 0 < x < 1,\,{x^2} > {x^3}$$

$$ \Rightarrow \int\limits_0^1 {{2^{{x^2}}}} \,dx > \int\limits_0^1 {{2^{{x^3}}}} dx$$

and $$\int\limits_1^2 {{2^{{x^3}}}dx} > \int\limits_1^2 {{2^{{x^2}}}dx} $$

$$ \Rightarrow {I_1} > {I_2}$$ and $${I_4} > {I_3}$$

$$ = {I_3} = \int\limits_0^1 {{2^{{x^2}}}} dx,\,$$

$${I_4} = \int\limits_0^1 {{2^{{x^3}}}} dx\,\,$$

$$\forall 0 < x < 1,\,{x^2} > {x^3}$$

$$ \Rightarrow \int\limits_0^1 {{2^{{x^2}}}} \,dx > \int\limits_0^1 {{2^{{x^3}}}} dx$$

and $$\int\limits_1^2 {{2^{{x^3}}}dx} > \int\limits_1^2 {{2^{{x^2}}}dx} $$

$$ \Rightarrow {I_1} > {I_2}$$ and $${I_4} > {I_3}$$

2

MCQ (Single Correct Answer)

The area of the region bounded by the curves

$$y = \left| {x - 2} \right|,x = 1,x = 3$$ and the $$x$$-axis is

$$y = \left| {x - 2} \right|,x = 1,x = 3$$ and the $$x$$-axis is

A

$$4$$

B

$$2$$

C

$$3$$

D

$$1$$

The required area is shown by shaded region

$$A = \int\limits_1^3 {\left| {x - 2} \right|dx = 2\int\limits_2^3 {\left( {x - 2} \right)} } dx$$

$$ = 2\left[ {{{{x^2}} \over 2} - 2x} \right]_2^3 = 1$$

$$A = \int\limits_1^3 {\left| {x - 2} \right|dx = 2\int\limits_2^3 {\left( {x - 2} \right)} } dx$$

$$ = 2\left[ {{{{x^2}} \over 2} - 2x} \right]_2^3 = 1$$

3

MCQ (Single Correct Answer)

If $$f\left( x \right) = {{{e^x}} \over {1 + {e^x}}},{I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {xg\left\{ {x\left( {1 - x} \right)} \right\}dx} $$

and $${I_2} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {g\left\{ {x\left( {1 - x} \right)} \right\}dx} ,$$ then the value of $${{{I_2}} \over {{I_1}}}$$ is

and $${I_2} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {g\left\{ {x\left( {1 - x} \right)} \right\}dx} ,$$ then the value of $${{{I_2}} \over {{I_1}}}$$ is

A

$$1$$

B

$$-3$$

C

$$-1$$

D

$$2$$

$$f\left( x \right) = {{{e^x}} \over {1 + {e^x}}}$$

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

$$\therefore$$ $$f\left( x \right) + f\left( { - x} \right) = 1\forall x$$

Now $${I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {xg\left\{ {x\left( {1 - x} \right)} \right\}} dx$$

$$ = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {\left( {1 - x} \right)} g\left\{ {x\left( {1 - x} \right)} \right\}dx$$

$$\left[ {} \right.$$ using $$\int\limits_a^b {f\left( x \right)} dx\,a$$

$$ = \int\limits_a^b {f\left( {a + b - x} \right)dx} $$ $$\left. \, \right]$$

$$ = {I_2} - {I_1} \Rightarrow 2{I_1} = {I_2}$$

$$\therefore$$ $${{{I_2}} \over {{I_1}}} = 2$$

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

$$\therefore$$ $$f\left( x \right) + f\left( { - x} \right) = 1\forall x$$

Now $${I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {xg\left\{ {x\left( {1 - x} \right)} \right\}} dx$$

$$ = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {\left( {1 - x} \right)} g\left\{ {x\left( {1 - x} \right)} \right\}dx$$

$$\left[ {} \right.$$ using $$\int\limits_a^b {f\left( x \right)} dx\,a$$

$$ = \int\limits_a^b {f\left( {a + b - x} \right)dx} $$ $$\left. \, \right]$$

$$ = {I_2} - {I_1} \Rightarrow 2{I_1} = {I_2}$$

$$\therefore$$ $${{{I_2}} \over {{I_1}}} = 2$$

4

MCQ (Single Correct Answer)

If $$\int\limits_0^\pi {xf\left( {\sin x} \right)dx = A\int\limits_0^{\pi /2} {f\left( {\sin x} \right)dx,} } $$ then $$A$$ is

A

$$2\pi $$

B

$$\pi $$

C

$${\pi \over 4}$$

D

$$0$$

Let $$I = \int\limits_0^\pi {xf\left( {\sin x} \right)} dx$$

$$ = \int\limits_0^\pi {\left( {\pi - x} \right)} f\left( {\sin x} \right)dx$$

$$\therefore$$ $$2I = \pi \int\limits_2^\pi {f\left( {\sin x} \right)} dx$$

$$ = \pi .2\int\limits_0^{{\pi \over 2}} {f\left( {\sin x} \right)} dx$$

$$\therefore$$ $$I = \pi \int\limits_0^{{\pi \over 2}} {f\left( {\sin x} \right)} dx$$

$$ \Rightarrow A = \pi $$

$$ = \int\limits_0^\pi {\left( {\pi - x} \right)} f\left( {\sin x} \right)dx$$

$$\therefore$$ $$2I = \pi \int\limits_2^\pi {f\left( {\sin x} \right)} dx$$

$$ = \pi .2\int\limits_0^{{\pi \over 2}} {f\left( {\sin x} \right)} dx$$

$$\therefore$$ $$I = \pi \int\limits_0^{{\pi \over 2}} {f\left( {\sin x} \right)} dx$$

$$ \Rightarrow A = \pi $$

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