Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

The integral $$\int\limits_{\pi /6}^{\pi /4} {{{dx} \over {\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}} $$ equals :

A

$${\pi \over {40}}$$

B

$${1 \over {20}}{\tan ^{ - 1}}\left( {{1 \over {9\sqrt 3 }}} \right)$$

C

$${1 \over {10}}\left( {{\pi \over 4} - {{\tan }^{ - 1}}\left( {{1 \over {9\sqrt 3 }}} \right)} \right)$$

D

$${1 \over 5}\left( {{\pi \over 4}{{-\tan }^{ - 1}}\left( {{1 \over {3\sqrt 3 }}} \right)} \right)$$

I $$=$$ $$\int\limits_{\pi /6}^{\pi /4} {{{dx} \over {\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}} $$

$${\rm I} = {1 \over 2}\int\limits_{\pi /6}^{\pi /4} {{{{{\tan }^4}x{{\sec }^2}xdx} \over {\left( {1 + {{\tan }^{10}}x} \right)}}} $$ Put tan^{5}x $$=$$ t

$${\rm I} = {1 \over {10}}\int\limits_{{{\left( {{1 \over {\sqrt 3 }}} \right)}^5}}^1 {{{dt} \over {1 + {t^2}}}} = {1 \over {10}}\left( {{\pi \over 4} - {{\tan }^{ - 1}}{1 \over {9\sqrt 3 }}} \right)$$

$${\rm I} = {1 \over 2}\int\limits_{\pi /6}^{\pi /4} {{{{{\tan }^4}x{{\sec }^2}xdx} \over {\left( {1 + {{\tan }^{10}}x} \right)}}} $$ Put tan

$${\rm I} = {1 \over {10}}\int\limits_{{{\left( {{1 \over {\sqrt 3 }}} \right)}^5}}^1 {{{dt} \over {1 + {t^2}}}} = {1 \over {10}}\left( {{\pi \over 4} - {{\tan }^{ - 1}}{1 \over {9\sqrt 3 }}} \right)$$

2

MCQ (Single Correct Answer)

The area (in sq. units) in the first quadrant bounded by the parabola, y = x^{2} + 1, the tangent to it at the point (2, 5) and the coordinate axes is :

A

$${8 \over 3}$$

B

$${{14} \over 3}$$

C

$${{187} \over {24}}$$

D

$${{37} \over {24}}$$

Area $$ = \int\limits_0^2 {\left( {{x^2} + 1} \right)dx - {1 \over 2}} \left( {{5 \over 4}} \right)\left( 5 \right) = {{37} \over {24}}$$

3

MCQ (Single Correct Answer)

The area (in sq. units) of the region bounded by the curve x^{2} = 4y and the straight line x = 4y â€“ 2 is :

A

$${3 \over 4}$$

B

$${5 \over 4}$$

C

$${7 \over 8}$$

D

$${9 \over 8}$$

x = 4y $$-$$ 2 & x

$$ \Rightarrow $$ x

x = 2, $$-$$ 1

So, $$\int\limits_{ - 1}^2 {\left( {{{x + 2} \over 4} - {{{x^2}} \over 4}} \right)\,dx = {9 \over 8}} $$

4

MCQ (Single Correct Answer)

If $$\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}} $$ dx = A(x)$${\left( {\sqrt {1 - {x^2}} } \right)^m}$$ + C, for a suitable chosen integer m and a function A(x), where C is a constant of integration, then (A(x))^{m} equals :

A

$${1 \over {27{x^6}}}$$

B

$${{ - 1} \over {27{x^9}}}$$

C

$${1 \over {9{x^4}}}$$

D

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

$$\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}} $$ dx = A(x)$${\left( {\sqrt {1 - {x^2}} } \right)^m}$$ + C

$$\int {{{\left| x \right|\sqrt {{1 \over {{x^2}}} - 1} } \over {{x^4}}}} \,dx,$$

Put $${1 \over {{x^2}}} - 1 = t \Rightarrow {{dt} \over {dx}} = {{ - 2} \over {{x^3}}}$$

**Case-I** $$x \ge 0$$

$$ - {1 \over 2}\int {\sqrt t \,dt\, \Rightarrow {{{t^{3/2}}} \over 3}} + C$$

$$ \Rightarrow - {1 \over 3}{\left( {{1 \over {{x^2}}} - 1} \right)^{3/2}}$$

$$ \Rightarrow {{{{\left( {\sqrt {1 - {x^2}} } \right)}^3}} \over { - 3{x^2}}} + C$$

$$A(x) = - {1 \over {3{x^3}}}\,\,$$ and $$m = 3$$

$${(A(x))^m} = {\left( { - {1 \over {3{x^3}}}} \right)^3} = - {1 \over {27{x^9}}}$$

**Case-II** $$x \le 0$$

We get $${{{{\left( {\sqrt {1 - {x^2}} } \right)}^3}} \over { - 3{x^3}}} + C$$

$$A(x) = {1 \over { - 3{x^3}}},\,\,\,m = 3$$

$${(A(x))^m} = {{ - 1} \over {27{x^9}}}$$

$$\int {{{\left| x \right|\sqrt {{1 \over {{x^2}}} - 1} } \over {{x^4}}}} \,dx,$$

Put $${1 \over {{x^2}}} - 1 = t \Rightarrow {{dt} \over {dx}} = {{ - 2} \over {{x^3}}}$$

$$ - {1 \over 2}\int {\sqrt t \,dt\, \Rightarrow {{{t^{3/2}}} \over 3}} + C$$

$$ \Rightarrow - {1 \over 3}{\left( {{1 \over {{x^2}}} - 1} \right)^{3/2}}$$

$$ \Rightarrow {{{{\left( {\sqrt {1 - {x^2}} } \right)}^3}} \over { - 3{x^2}}} + C$$

$$A(x) = - {1 \over {3{x^3}}}\,\,$$ and $$m = 3$$

$${(A(x))^m} = {\left( { - {1 \over {3{x^3}}}} \right)^3} = - {1 \over {27{x^9}}}$$

We get $${{{{\left( {\sqrt {1 - {x^2}} } \right)}^3}} \over { - 3{x^3}}} + C$$

$$A(x) = {1 \over { - 3{x^3}}},\,\,\,m = 3$$

$${(A(x))^m} = {{ - 1} \over {27{x^9}}}$$

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Complex Numbers

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