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 the area of the region bounded by the curves, $$y = {x^2},y = {1 \over x}$$ and the lines y = 0 and x= t (t >1) is 1 sq. unit, then t is equal to :

A

$${e^{{3 \over 2}}}$$

B

$${4 \over 3}$$

C

$${3 \over 2}$$

D

$${e^{{2 \over 3}}}$$

Point of intersection of y = x

put y = $${1 \over x}$$ in y = x

$${1 \over x} = {x^2}$$

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

$$ \Rightarrow $$ $$\,\,\,$$ x = 1

$$\therefore\,\,\,$$ y = 1

$$\therefore\,\,\,$$ point B = (1, 1)

Area of region ABCDA

= $$\int\limits_0^1 {{x^2}} $$ dx + $$\int\limits_1^t {{1 \over x}} $$ dx

$$=$$ $$\left[ {{{{x^3}} \over 3}} \right]_0^1$$ + $$\left[ {\ell n\,x} \right]_1^t$$

= $${1 \over 3}$$ + $$\ell n\,t$$ $$-$$ $$\ell n$$ 1

= $${1 \over 3}$$ + $$\ell n\,t$$ [ as $$\ell n$$ 1 = 0]

given this Area = 1 sq unit.

$$\therefore\,\,\,$$ $${1 \over 3}$$ + $$\ell n\,t$$ = 1

$$ \Rightarrow $$ $$\ell n\,t$$ = $${2 \over 3}$$

$$ \Rightarrow $$ t = e$${^{{2 \over 3}}}$$

2

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

3

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

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

= $${4 \over 3}$$

4

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

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

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

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

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

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

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

= $${8 \over 3}$$

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Limits, Continuity and Differentiability *keyboard_arrow_right*

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Definite Integrals and Applications of Integrals *keyboard_arrow_right*

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Straight Lines and Pair of Straight Lines *keyboard_arrow_right*

Circle *keyboard_arrow_right*

Conic Sections *keyboard_arrow_right*