Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geotechnical Engineering

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Engineering Mathematics

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Geomatics Engineering Or Surveying

General Aptitude

1

The area (in sq. units) of the region described by

A= {(x, y) $$\left| {} \right.$$y$$ \ge $$ x^{2} $$-$$ 5x + 4, x + y $$ \ge $$ 1, y $$ \le $$ 0} is :

A= {(x, y) $$\left| {} \right.$$y$$ \ge $$ x

A

$${7 \over 2}$$

B

$${{19} \over 6}$$

C

$${{13} \over 6}$$

D

$${{17} \over 6}$$

Required Area

= A

= $$\left| {\int\limits_1^3 {\left( {1 - x} \right)} dx} \right| + \left| {\int\limits_3^4 {\left( {{x^2} - 5x + 4} \right)dx} } \right|$$

= $$\left| {\left[ {x - {{{x^2}} \over 2}} \right]_1^3} \right| + \left| {\left[ {{{{x^3}} \over 3} - {5 \over 2}{x^2} + 4x} \right]_3^4} \right|$$

= $$\left| {\left[ {\left( {3 - {9 \over 2}} \right) - \left( {1 - {1 \over 2}} \right)} \right]} \right| + \left| {\left[ {\left( {{{64} \over 3} - 40 + 16} \right) - \left( {9 - {{45} \over 2} + 12} \right)} \right]} \right|$$

= $$\left| {\left( {2 - 4} \right)} \right| + \left| {\left( {{{ - 8} \over 3} + {3 \over 2}} \right)} \right|$$

= 2 + $${7 \over 6}$$

= $${{19} \over 6}$$ sq. unit.

2

The value of the integral

$$\int\limits_4^{10} {{{\left[ {{x^2}} \right]dx} \over {\left[ {{x^2} - 28x + 196} \right] + \left[ {{x^2}} \right]}}} ,$$

where [x] denotes the greatest integer less than or equal to x, is :

$$\int\limits_4^{10} {{{\left[ {{x^2}} \right]dx} \over {\left[ {{x^2} - 28x + 196} \right] + \left[ {{x^2}} \right]}}} ,$$

where [x] denotes the greatest integer less than or equal to x, is :

A

6

B

3

C

7

D

$${1 \over 3}$$

Let I = $$\int\limits_4^{10} {{{\left[ {{x^2}} \right]\,dx} \over {\left[ {{x^2} - 28x + 196} \right] + \left[ {{x^2}} \right]}}} $$

= $$\int\limits_4^{10} {{{\left[ {{x^2}} \right]dx} \over {\left[ {{{\left( {x - 14} \right)}^2}} \right] + \left[ {{x^2}} \right]}}} \,\,\,.....(1)$$

Using,

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

I = $$\int\limits_4^{10} {{{{{\left( {x - 14} \right)}^2}} \over {\left[ {{x^2}} \right] + \left[ {{{\left( {x - 14} \right)}^2}} \right]}}} \,dx\,\,....(2)$$

Adding (1) and (2)

2I = $$\int\limits_4^{10} {{{\left[ {{{\left( {x - 14} \right)}^2}} \right] + \left[ {{x^2}} \right]} \over {\left[ {{x^2}} \right] + \left[ {{{\left( {x - 14} \right)}^2}} \right]}}} \,dx$$

$$ \Rightarrow $$$$\,\,\,$$ 2I = $$\int\limits_4^{10} {dx} = \left[ x \right]_4^{10}$$ = 6

= I = 3

= $$\int\limits_4^{10} {{{\left[ {{x^2}} \right]dx} \over {\left[ {{{\left( {x - 14} \right)}^2}} \right] + \left[ {{x^2}} \right]}}} \,\,\,.....(1)$$

Using,

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

I = $$\int\limits_4^{10} {{{{{\left( {x - 14} \right)}^2}} \over {\left[ {{x^2}} \right] + \left[ {{{\left( {x - 14} \right)}^2}} \right]}}} \,dx\,\,....(2)$$

Adding (1) and (2)

2I = $$\int\limits_4^{10} {{{\left[ {{{\left( {x - 14} \right)}^2}} \right] + \left[ {{x^2}} \right]} \over {\left[ {{x^2}} \right] + \left[ {{{\left( {x - 14} \right)}^2}} \right]}}} \,dx$$

$$ \Rightarrow $$$$\,\,\,$$ 2I = $$\int\limits_4^{10} {dx} = \left[ x \right]_4^{10}$$ = 6

= I = 3

3

For x $$ \in $$ **R**, x $$ \ne $$ 0, if y(x) is a differentiable function such that

x $$\int\limits_1^x y $$ (t) dt = (x + 1) $$\int\limits_1^x ty $$ (t) dt, then y (x) equals :

(where C is a constant.)

x $$\int\limits_1^x y $$ (t) dt = (x + 1) $$\int\limits_1^x ty $$ (t) dt, then y (x) equals :

(where C is a constant.)

A

$${C \over x}{e^{ - {1 \over x}}}$$

B

$${C \over {{x^2}}}{e^{ - {1 \over x}}}$$

C

$${C \over {{x^3}}}{e^{ - {1 \over x}}}$$

D

$$C{x^3}\,{1 \over {{e^x}}}$$

$$x\int\limits_1^x {y\left( t \right)dt} = x\int\limits_1^x {ty} \left( t \right)dt + \int\limits_1^x {ty\left( t \right)} \,\,dt$$

Differentiate w.r. to x.

$$\int\limits_1^x {y\left( t \right)dt + x\left[ {y\left( x \right) - y\left( 1 \right)} \right]} $$

$$ = \int\limits_1^x {ty\left( t \right)dt + x\left[ {xy\left( x \right) - y\left( 1 \right)} \right] + xy\left( x \right) - y\left( 1 \right)} $$

$$\int\limits_1^x {y\left( t \right)dt = \int\limits_1^x {ty\left( t \right)dt + {x^2}y\left( x \right) - y\left( 1 \right)} } $$

Diff. again w.r.t to x

$$y\left( x \right) - y\left( a \right) = xy\left( x \right) - y\left( a \right) + 2x\,y\left( x \right) + {x^2}{y^3}\left( x \right)$$

$$\left( {1 - 3x} \right)y\left( x \right) = {x^2}{y^3}\,\left( x \right)$$

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

$${{1dy} \over {ydx}} = {{1 - 3x} \over {{x^2}}} \Rightarrow \ln \,y = - {1 \over x} - 3\ln x$$

$$\ln \left( {y{x^3}} \right) = - {1 \over x}$$

$$y{x^3} = - {e^{ - 1/x}}$$

$$y = {{{e^{ - {1 \over x}}}} \over {{x^3}}}$$

or $$y = {{c{e^{ - {1 \over x}}}} \over {{x^3}}}$$

Differentiate w.r. to x.

$$\int\limits_1^x {y\left( t \right)dt + x\left[ {y\left( x \right) - y\left( 1 \right)} \right]} $$

$$ = \int\limits_1^x {ty\left( t \right)dt + x\left[ {xy\left( x \right) - y\left( 1 \right)} \right] + xy\left( x \right) - y\left( 1 \right)} $$

$$\int\limits_1^x {y\left( t \right)dt = \int\limits_1^x {ty\left( t \right)dt + {x^2}y\left( x \right) - y\left( 1 \right)} } $$

Diff. again w.r.t to x

$$y\left( x \right) - y\left( a \right) = xy\left( x \right) - y\left( a \right) + 2x\,y\left( x \right) + {x^2}{y^3}\left( x \right)$$

$$\left( {1 - 3x} \right)y\left( x \right) = {x^2}{y^3}\,\left( x \right)$$

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

$${{1dy} \over {ydx}} = {{1 - 3x} \over {{x^2}}} \Rightarrow \ln \,y = - {1 \over x} - 3\ln x$$

$$\ln \left( {y{x^3}} \right) = - {1 \over x}$$

$$y{x^3} = - {e^{ - 1/x}}$$

$$y = {{{e^{ - {1 \over x}}}} \over {{x^3}}}$$

or $$y = {{c{e^{ - {1 \over x}}}} \over {{x^3}}}$$

4

The integral $$\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {1 + \cos x}}} $$ is equal to

A

2

B

4

C

$$-$$ 1

D

$$-$$ 2

$$\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {1 + \cos x}}} $$

= $$\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {2{{\cos }^2}{x \over 2}}}} $$

= $${1 \over 2}\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{\sec }^2}} {x \over 2}\,dx$$

$${1 \over 2}\left[ {{{\tan {x \over 2}} \over {{1 \over 2}}}} \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}$$

= $$\left[ {\tan {x \over 2}} \right]_{{\pi \over 4}}^{{3 \over 4}}$$

= tan $${{3\pi } \over 8}$$ $$-$$ tan$${\pi \over 8}$$

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

= 2

= $$\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {2{{\cos }^2}{x \over 2}}}} $$

= $${1 \over 2}\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{\sec }^2}} {x \over 2}\,dx$$

$${1 \over 2}\left[ {{{\tan {x \over 2}} \over {{1 \over 2}}}} \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}$$

= $$\left[ {\tan {x \over 2}} \right]_{{\pi \over 4}}^{{3 \over 4}}$$

= tan $${{3\pi } \over 8}$$ $$-$$ tan$${\pi \over 8}$$

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

= 2

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