Joint Entrance Examination

Graduate Aptitude Test in Engineering

Strength of Materials Or Solid Mechanics

Structural Analysis

Construction Material and Management

Reinforced Cement Concrete

Steel Structures

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

Hydrology

Irrigation

Geomatics Engineering Or Surveying

Environmental Engineering

Transportation Engineering

Engineering Mathematics

General Aptitude

1

If $$f(x) = \int\limits_0^x {t\left( {\sin x - \sin t} \right)dt\,\,\,} $$ then :

A

f'''(x) + f''(x) = sinx

B

f'''(x) + f''(x) $$-$$ f'(x) = cosx

C

f'''(x) + f'(x) = cosx $$-$$ 2x sinx

D

f'''(x) $$-$$ f''(x) = cosx $$-$$ 2x sinx

f(x) = $$\int_0^x {t(\sin x - \sin t).dt} $$

= sin x$$\int_0^x {t.dt - \int_0^x {t\sin t.dt} } $$

= $${{{x^2}} \over 2}$$ sin x +$$\left[ {t\cos t} \right]_0^x$$ + sin x

$$ \Rightarrow $$f(x) = $${{{x^2}} \over 2}$$ sinx + xcosx + sinx

f'(x) = $${{{x^2}} \over 2}$$ cosx + 2cos x

f''(x) = x cos x $$-$$ $${{{x^2}} \over 2}$$ sin x $$-$$ 2sin x

f'''(x) = cos x $$-$$ 2x sin x $$-$$ $${{{x^2}} \over 2}$$ cos x $$-$$ 2cos x

$$\therefore\,\,\,$$ f'''(x) + f'(x) = cos x $$-$$ 2x sin x

= sin x$$\int_0^x {t.dt - \int_0^x {t\sin t.dt} } $$

= $${{{x^2}} \over 2}$$ sin x +$$\left[ {t\cos t} \right]_0^x$$ + sin x

$$ \Rightarrow $$f(x) = $${{{x^2}} \over 2}$$ sinx + xcosx + sinx

f'(x) = $${{{x^2}} \over 2}$$ cosx + 2cos x

f''(x) = x cos x $$-$$ $${{{x^2}} \over 2}$$ sin x $$-$$ 2sin x

f'''(x) = cos x $$-$$ 2x sin x $$-$$ $${{{x^2}} \over 2}$$ cos x $$-$$ 2cos x

$$\therefore\,\,\,$$ f'''(x) + f'(x) = cos x $$-$$ 2x sin x

2

The value of $$\int\limits_0^\pi {{{\left| {\cos x} \right|}^3}} \,dx$$ is :

A

$$4 \over 3$$

B

$$-$$ $$4 \over 3$$

C

0

D

$$2 \over 3$$

$$\int\limits_0^\pi {{{\left| {\cos x} \right|}^3}} \,dx$$

The period of $$\left| {\cos x} \right|$$ = $${\pi \over 2}$$

$$ \therefore $$ I = 2 $$\int\limits_0^{{\pi \over 2}} {{{\left| {\cos x} \right|}^3}} \,dx$$

as in the range 0 to $${\pi \over 2}$$ $$\left| {\cos x} \right|$$ is positive.

So, $$\left| {\cos x} \right|$$ = $$cosx$$

$$ \therefore $$ I = 2 $$\int\limits_0^{{\pi \over 2}} {{{\cos }^3}x\,dx} $$

= 2$$\int\limits_0^{{\pi \over 2}} {\left( {{{\cos 3x + 3\cos x} \over 4}} \right)} dx$$

I = $${1 \over 2}\left[ {{{\sin 3x} \over 3} + 3\sin x} \right]_0^{{\pi \over 2}}$$

I = $${1 \over 2}\left[ {{1 \over 3}\left( {{{3\pi } \over 2}} \right) + 3.\sin {\pi \over 2}} \right]$$

I = $${1 \over 2}\left[ { - {1 \over 3} + 3} \right]$$

= $${1 \over 2}\left( {{8 \over 3}} \right)$$

= $${4 \over 3}$$

The period of $$\left| {\cos x} \right|$$ = $${\pi \over 2}$$

$$ \therefore $$ I = 2 $$\int\limits_0^{{\pi \over 2}} {{{\left| {\cos x} \right|}^3}} \,dx$$

as in the range 0 to $${\pi \over 2}$$ $$\left| {\cos x} \right|$$ is positive.

So, $$\left| {\cos x} \right|$$ = $$cosx$$

$$ \therefore $$ I = 2 $$\int\limits_0^{{\pi \over 2}} {{{\cos }^3}x\,dx} $$

= 2$$\int\limits_0^{{\pi \over 2}} {\left( {{{\cos 3x + 3\cos x} \over 4}} \right)} dx$$

I = $${1 \over 2}\left[ {{{\sin 3x} \over 3} + 3\sin x} \right]_0^{{\pi \over 2}}$$

I = $${1 \over 2}\left[ {{1 \over 3}\left( {{{3\pi } \over 2}} \right) + 3.\sin {\pi \over 2}} \right]$$

I = $${1 \over 2}\left[ { - {1 \over 3} + 3} \right]$$

= $${1 \over 2}\left( {{8 \over 3}} \right)$$

= $${4 \over 3}$$

3

The area (in sq. units) bounded by the parabolae y = x^{2} – 1, the tangent at the point (2, 3) to it and the y-axis is :

A

$$56\over3$$

B

$$32\over3$$

C

$$8\over3$$

D

$$14\over3$$

Equation of tangent at (2, 3) on the parabola y = x

$${{y + 3} \over 2} = 2x - 1$$

$$ \Rightarrow $$ y + 3 = 4x $$-$$ 2

$$ \Rightarrow $$ y = 4x $$-$$ 5

When x = 0 then for the tangent y = $$-$$ 5

$$ \therefore $$ Tangent cuts x y axis at (0, $$-$$ 5) point.

$$ \therefore $$ Area of the bounded region is

= $$\int\limits_{ - 5}^3 {{{y + 5} \over 4}} \,\,\,dy - \int\limits_{ - 1}^3 {\sqrt {y + 1} } \,\,\,dy$$

= $${1 \over 4}\left[ {{{{y^2}} \over 2} + 5y} \right]_{ - 5}^3 - \left[ {{2 \over 3} \times {{\left( {y + 1} \right)}^{{3 \over 2}}}} \right]_{ - 1}^3$$

$${1 \over 4}\left[ {\left( {{9 \over 2} + 15} \right) - \left( {{{25} \over 2} - 25} \right)} \right] - {2 \over 3}{\left( 4 \right)^{{3 \over 2}}}$$

= $${1 \over 4}\left[ {{{93} \over 2} + {{25} \over 2}} \right] - {2 \over 3} \times 8$$

= $${1 \over 4} \times {{64} \over 2} - {{16} \over 3}$$

= $$8 - {{16} \over 3}$$

= $${8 \over 3}$$

4

The area of the region

A = {(x, y) : 0 $$ \le $$ y $$ \le $$x |x| + 1 and $$-$$1 $$ \le $$ x $$ \le $$1} in sq. units, is :

A = {(x, y) : 0 $$ \le $$ y $$ \le $$x |x| + 1 and $$-$$1 $$ \le $$ x $$ \le $$1} in sq. units, is :

A

$${2 \over 3}$$

B

2

C

$${4 \over 3}$$

D

$${1 \over 3}$$

Required area

$$ = \int\limits_{ - 1}^1 {\left( {x\left| x \right| + 1} \right)} dx$$

$$ = 0 + \left( x \right)_{ - 1}^1 = 2$$

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