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

If $$\mathop {\lim }\limits_{n \to \infty } \,\,{{{1^a} + {2^a} + ...... + {n^a}} \over {{{(n + 1)}^{a - 1}}\left[ {\left( {na + 1} \right) + \left( {na + 2} \right) + ..... + \left( {na + n} \right)} \right]}} = {1 \over {60}}$$

for some positive real number a, then a is equal to :

for some positive real number a, then a is equal to :

A

7

B

8

C

$${{15} \over 2}$$

D

$${{17} \over 2}$$

$$\mathop {\lim }\limits_{n \to \infty } {{{1 \over {\left( {a + 1} \right)}}\,.\,{n^{a + 1}} + {a_1}{n^a} + {a_2}{n^{a - 1}} + .\,.\,.\,.} \over {{{\left( {n + 1} \right)}^{a - 1}}.\,{n^2}\left( {a + {{1 + {1 \over n}} \over 2}} \right)}} = {1 \over {60}}$$

$$ \Rightarrow $$ $$\mathop {\lim }\limits_{n \to \infty } {{{{\left( {{1 \over n}} \right)}^2} + {{\left( {{2 \over n}} \right)}^a} + ....... + {{\left( {{n \over n}} \right)}^a}} \over {{{\left( {n + 1} \right)}^{a - 1}}\left[ {{a^2}a + {{n\left( {n + 1} \right)} \over 2}} \right]}} = {1 \over {60}}$$

$$ = {{\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 1}^n {{{\left( {{r \over n}} \right)}^a}} } \over {{{\left( {1 + {1 \over n}} \right)}^{a - 1}}\left[ {a + {1 \over 2}\left( {1 + {1 \over n}} \right)} \right]}} = {1 \over {60}}$$

$$ = {{\int_0^1 {{x^a}dx} } \over {\left( {a + {1 \over 2}} \right)}} = {1 \over {60}} = {{{1 \over {a + 1}}} \over {a + {1 \over 2}}} = {1 \over {60}}$$

$$ \Rightarrow $$ $${{{1 \over {a + 1}}} \over {\left( {a + {1 \over 2}} \right)}} = {1 \over {60}} \Rightarrow \left( {a + 1} \right)\left( {2a + 1} \right) = 120$$

$$ \Rightarrow $$ 2a^{2} + 3a $$-$$ 119 $$=$$ 0

$$ \Rightarrow $$ 2a^{2} + 17a $$-$$ 14a $$-$$ 119 $$=$$ 0

$$ \Rightarrow $$ (a $$-$$ 7) (2a + 17) $$=$$ 0

$$ \Rightarrow $$ a $$=$$ 7, $$-$$ $${{17} \over 2}$$

$$ \Rightarrow $$ $$\mathop {\lim }\limits_{n \to \infty } {{{{\left( {{1 \over n}} \right)}^2} + {{\left( {{2 \over n}} \right)}^a} + ....... + {{\left( {{n \over n}} \right)}^a}} \over {{{\left( {n + 1} \right)}^{a - 1}}\left[ {{a^2}a + {{n\left( {n + 1} \right)} \over 2}} \right]}} = {1 \over {60}}$$

$$ = {{\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 1}^n {{{\left( {{r \over n}} \right)}^a}} } \over {{{\left( {1 + {1 \over n}} \right)}^{a - 1}}\left[ {a + {1 \over 2}\left( {1 + {1 \over n}} \right)} \right]}} = {1 \over {60}}$$

$$ = {{\int_0^1 {{x^a}dx} } \over {\left( {a + {1 \over 2}} \right)}} = {1 \over {60}} = {{{1 \over {a + 1}}} \over {a + {1 \over 2}}} = {1 \over {60}}$$

$$ \Rightarrow $$ $${{{1 \over {a + 1}}} \over {\left( {a + {1 \over 2}} \right)}} = {1 \over {60}} \Rightarrow \left( {a + 1} \right)\left( {2a + 1} \right) = 120$$

$$ \Rightarrow $$ 2a

$$ \Rightarrow $$ 2a

$$ \Rightarrow $$ (a $$-$$ 7) (2a + 17) $$=$$ 0

$$ \Rightarrow $$ a $$=$$ 7, $$-$$ $${{17} \over 2}$$

2

Let g(x) = cosx^{2}, f(x) = $$\sqrt x $$ and $$\alpha ,\beta \left( {\alpha < \beta } \right)$$ be the roots of the quadratic equation 18x^{2} - 9$$\pi $$x + $${\pi ^2}$$ = 0. Then the area (in sq. units) bounded by the curve

y = (gof)(x) and the lines $$x = \alpha $$, $$x = \beta $$ and y = 0 is

y = (gof)(x) and the lines $$x = \alpha $$, $$x = \beta $$ and y = 0 is

A

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

B

$${1 \over 2}\left( {\sqrt 3 - 1} \right)$$

C

$${1 \over 2}\left( {\sqrt 3 + 1} \right)$$

D

$${1 \over 2}\left( {\sqrt 3 - \sqrt 2 } \right)$$

Given quadratic equation,

$$18{x^2} - 9\pi x + {\pi ^2} = 0$$

$$ \Rightarrow \,\,\,18{x^2} - 6\pi x - 3\pi x + {\pi ^2} = 0$$

$$ \Rightarrow \,\,\,\,6x\left( {3x - \pi } \right) - \pi \left( {3x - \pi } \right) = 0$$

$$ \Rightarrow \,\,\,\,\left( {3x - \pi } \right)\left( {6x - \pi } \right) = 0$$

$$\therefore$$ $$\,\,\,\,x = {\pi \over 3},{\pi \over 6}$$

as $$\,\,\,\, \propto < B$$

$$\therefore$$ $$\,\,\,\, \propto = {\pi \over 6}$$ $$\,\,\,\,$$ and $$\,\,\,\,$$ $$\beta = {\pi \over 3}$$

Given, $$g\left( x \right) = \cos {x^2}$$ and $$ + \left( x \right) = \sqrt x $$

$$y = \left( {gof} \right)x$$

$$ = \,\,\,\,\,g\left( {f\left( x \right)} \right)$$

$$ = \,\,\,\,\cos \left( {f{{\left( x \right)}^2}} \right)$$

$$ = \,\,\,\,\cos {\left( {\sqrt x} \right)^2}$$

$$ = \,\,\,\,\cos x$$

So, the required area in the curve is

Area $$ = \int\limits_{{\pi \over 6}}^{{\pi \over 3}} {\cos \,\,dx} $$

$$ = \left[ {\sin x} \right]_{{\pi \over 6}}^{{\pi \over 3}}$$

$$ = \sin {\pi \over 3} - \sin {\pi \over 6}$$

$$ = {{\sqrt 3} \over 2} - {1 \over 2}$$

$$ = {{\sqrt 3 - 1} \over 2}$$

$$18{x^2} - 9\pi x + {\pi ^2} = 0$$

$$ \Rightarrow \,\,\,18{x^2} - 6\pi x - 3\pi x + {\pi ^2} = 0$$

$$ \Rightarrow \,\,\,\,6x\left( {3x - \pi } \right) - \pi \left( {3x - \pi } \right) = 0$$

$$ \Rightarrow \,\,\,\,\left( {3x - \pi } \right)\left( {6x - \pi } \right) = 0$$

$$\therefore$$ $$\,\,\,\,x = {\pi \over 3},{\pi \over 6}$$

as $$\,\,\,\, \propto < B$$

$$\therefore$$ $$\,\,\,\, \propto = {\pi \over 6}$$ $$\,\,\,\,$$ and $$\,\,\,\,$$ $$\beta = {\pi \over 3}$$

Given, $$g\left( x \right) = \cos {x^2}$$ and $$ + \left( x \right) = \sqrt x $$

$$y = \left( {gof} \right)x$$

$$ = \,\,\,\,\,g\left( {f\left( x \right)} \right)$$

$$ = \,\,\,\,\cos \left( {f{{\left( x \right)}^2}} \right)$$

$$ = \,\,\,\,\cos {\left( {\sqrt x} \right)^2}$$

$$ = \,\,\,\,\cos x$$

So, the required area in the curve is

Area $$ = \int\limits_{{\pi \over 6}}^{{\pi \over 3}} {\cos \,\,dx} $$

$$ = \left[ {\sin x} \right]_{{\pi \over 6}}^{{\pi \over 3}}$$

$$ = \sin {\pi \over 3} - \sin {\pi \over 6}$$

$$ = {{\sqrt 3} \over 2} - {1 \over 2}$$

$$ = {{\sqrt 3 - 1} \over 2}$$

3

The value of $$\int\limits_{ - \pi /2}^{\pi /2} {{{{{\sin }^2}x} \over {1 + {2^x}}}} dx$$ is

A

$${\pi \over 4}$$

B

$${\pi \over 8}$$

C

$${\pi \over 2}$$

D

$${4\pi }$$

As we know,

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

Let $$I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\,\,{{{{\sin }^2}x} \over {1 + {2^x}}}} \,dx$$

$$ \Rightarrow \,\,\,\,I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\,\,{{{{\sin }^2}\left( { - {\pi \over 2} + {\pi \over 2} - x} \right)} \over {1 + {2^{ - {\pi \over 2} + {\pi \over 2} - x}}}}} $$

$$ \Rightarrow\,\,\,\, I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\sin }^2}x} \over {1 + {2^{ - x}}}}} dx$$

$$ \Rightarrow \,\,\,\,I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{2^x}{{\sin }^2}x} \over {1 + {2^x}}}} \,\,dx$$

$$\therefore\,\,\,$$ $$2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{{{{\sin }^2}x} \over {1 + {2^x}}} + {{{2^x}{{\sin }^2}x} \over {1 + {2^x}}}} \right)} \,\,dx$$

$$ \Rightarrow \,\,\,\,2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^2}x\left( {{{1 + {2^x}} \over {1 + {2^x}}}} \right)} \,\,dx$$

$$ \Rightarrow \,\,\,\,2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx$$

$$ \Rightarrow \,\,\,\,2I = 2\int\limits_0^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx\,\,\,$$ [ as sin^{2}
x is an even function ]

$$ \Rightarrow \,\,\,\,I = \int\limits_0^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx\,\,\,\,$$

[ as $$\,\,\,\,$$ $$\int\limits_0^a {f\left( a \right)\,\,dx = \int\limits_0^a {f\left( {a - x} \right)\,dx]} } $$

So, $$\,\,\,$$ $$I = \int\limits_0^{{\pi \over 2}} {{{\sin }^2}\left( {{b \over 2} - x} \right)} \,\,dx\,\,\,\,$$

$$\,\,\,I = \int\limits_0^{{\pi \over 2}} {{{\cos }^2}x} \,\,dx\,\,\,\,$$

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

$$ \Rightarrow \,\,\,\,2I = \int\limits_0^{{\pi \over 2}} \, \,\,dx\,\,\,\,$$

$$ \Rightarrow \,\,\,\,2I = {\left[ x \right]_0}^{{\pi \over 2}}$$

$$ \Rightarrow \,\,\,\,2I = {\pi \over 2}$$

$$ \Rightarrow \,\,\,\,I = {\pi \over 4}$$

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

Let $$I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\,\,{{{{\sin }^2}x} \over {1 + {2^x}}}} \,dx$$

$$ \Rightarrow \,\,\,\,I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\,\,{{{{\sin }^2}\left( { - {\pi \over 2} + {\pi \over 2} - x} \right)} \over {1 + {2^{ - {\pi \over 2} + {\pi \over 2} - x}}}}} $$

$$ \Rightarrow\,\,\,\, I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{{\sin }^2}x} \over {1 + {2^{ - x}}}}} dx$$

$$ \Rightarrow \,\,\,\,I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{{2^x}{{\sin }^2}x} \over {1 + {2^x}}}} \,\,dx$$

$$\therefore\,\,\,$$ $$2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{{{{\sin }^2}x} \over {1 + {2^x}}} + {{{2^x}{{\sin }^2}x} \over {1 + {2^x}}}} \right)} \,\,dx$$

$$ \Rightarrow \,\,\,\,2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^2}x\left( {{{1 + {2^x}} \over {1 + {2^x}}}} \right)} \,\,dx$$

$$ \Rightarrow \,\,\,\,2I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx$$

$$ \Rightarrow \,\,\,\,2I = 2\int\limits_0^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx\,\,\,$$ [ as sin

$$ \Rightarrow \,\,\,\,I = \int\limits_0^{{\pi \over 2}} {{{\sin }^2}x} \,\,dx\,\,\,\,$$

[ as $$\,\,\,\,$$ $$\int\limits_0^a {f\left( a \right)\,\,dx = \int\limits_0^a {f\left( {a - x} \right)\,dx]} } $$

So, $$\,\,\,$$ $$I = \int\limits_0^{{\pi \over 2}} {{{\sin }^2}\left( {{b \over 2} - x} \right)} \,\,dx\,\,\,\,$$

$$\,\,\,I = \int\limits_0^{{\pi \over 2}} {{{\cos }^2}x} \,\,dx\,\,\,\,$$

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

$$ \Rightarrow \,\,\,\,2I = \int\limits_0^{{\pi \over 2}} \, \,\,dx\,\,\,\,$$

$$ \Rightarrow \,\,\,\,2I = {\left[ x \right]_0}^{{\pi \over 2}}$$

$$ \Rightarrow \,\,\,\,2I = {\pi \over 2}$$

$$ \Rightarrow \,\,\,\,I = {\pi \over 4}$$

4

The area (in sq. units) of the region

{x $$ \in $$**R** : x $$ \ge $$ 0, y $$ \ge $$ 0, y $$ \ge $$ x $$-$$ 2 *and* y $$ \le $$ $$\sqrt x $$}, is :

{x $$ \in $$

A

$${{13} \over 3}$$

B

$${{8} \over 3}$$

C

$${{10} \over 3}$$

D

$${{5} \over 3}$$

y = $$\sqrt x $$

y = x $$-$$ 2

$$\therefore\,\,\,$$ $$\sqrt x $$ = x $$-$$ 2

$$ \Rightarrow $$$$\,\,\,$$ x = x^{2} $$-$$ 4x + 4

x^{2} $$-$$ 5x + 4 = 0

x^{2} $$-$$ 4x $$-$$ x + 4 = 0

$$ \Rightarrow $$$$\,\,\,$$ x(x $$-$$ 4) $$-$$ (x $$-$$ 4) = 0

$$ \Rightarrow $$$$\,\,\,$$ (x $$-$$ 4) (x $$-$$ 1) = 0

$$\therefore\,\,\,$$ x = 4, 1

and y = 2, $$-$$ 1

$$\therefore\,\,\,$$ Their point of intersection (4, 2) and (1, $$-$$ 1)

Required area is shown in the shaded figure.

$$\therefore\,\,\,$$ Required area

= $$\int\limits_0^2 {\sqrt x \,dx + \int\limits_2^4 {\left( {\sqrt x - x + 2} \right)\,dx} } $$

= $$\int\limits_0^4 {\sqrt x \,dx + \int\limits_2^4 {\left( {2 - x} \right)\,dx} } $$

= $$\left[ {{2 \over 3}{x^{{3 \over 2}}}} \right]_0^4 + \left[ {2x - {{{x^2}} \over 2}} \right]_2^4$$

= $${2 \over 3}\left( 8 \right)$$ + 2$$\left( {4 - 2} \right)$$ $$-$$ $${1 \over 2}$$ (16 $$-$$ 4)

= $${{16} \over 3}$$ + 4 $$-$$ 6

= $${{10} \over 3}$$

y = x $$-$$ 2

$$\therefore\,\,\,$$ $$\sqrt x $$ = x $$-$$ 2

$$ \Rightarrow $$$$\,\,\,$$ x = x

x

x

$$ \Rightarrow $$$$\,\,\,$$ x(x $$-$$ 4) $$-$$ (x $$-$$ 4) = 0

$$ \Rightarrow $$$$\,\,\,$$ (x $$-$$ 4) (x $$-$$ 1) = 0

$$\therefore\,\,\,$$ x = 4, 1

and y = 2, $$-$$ 1

$$\therefore\,\,\,$$ Their point of intersection (4, 2) and (1, $$-$$ 1)

Required area is shown in the shaded figure.

$$\therefore\,\,\,$$ Required area

= $$\int\limits_0^2 {\sqrt x \,dx + \int\limits_2^4 {\left( {\sqrt x - x + 2} \right)\,dx} } $$

= $$\int\limits_0^4 {\sqrt x \,dx + \int\limits_2^4 {\left( {2 - x} \right)\,dx} } $$

= $$\left[ {{2 \over 3}{x^{{3 \over 2}}}} \right]_0^4 + \left[ {2x - {{{x^2}} \over 2}} \right]_2^4$$

= $${2 \over 3}\left( 8 \right)$$ + 2$$\left( {4 - 2} \right)$$ $$-$$ $${1 \over 2}$$ (16 $$-$$ 4)

= $${{16} \over 3}$$ + 4 $$-$$ 6

= $${{10} \over 3}$$

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