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

Let M and m be respectively the absolute maximum and the absolute minimum values of the function, f(x) = 2x^{3} $$-$$ 9x^{2} + 12x + 5 in the interval [0, 3]. Then M $$-$$m is equal to :

A

5

B

9

C

4

D

1

Here, f(x) = 2x^{3} $$-$$ 9x^{2} + 12x + 5

$$ \Rightarrow $$ f'(x) = 6x^{2} $$-$$ 18x + 12 = 0

For maxima or minima put f'(x) = 0

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

$$ \Rightarrow $$ x = 1 or x = 2

Now, f''(x) = 12x $$-$$ 18

$$ \Rightarrow $$ f''(1) = 12(1) $$-$$ 18 = $$-$$ 6 < 0

Hence, f(x) has maxima at x = 1

$$ \therefore $$ maximum value = M = f(1) = 2 $$-$$ 9 + 12 + 5 = 10

And, f''(2) = 12(2) $$-$$ 18 = 6 > 0.

Hence, f(x) has minima at x = 2.

$$ \therefore $$ minimum value = m = f(2) = 2(8) $$-$$ 9(4) + 12(2) + 5 = 9

$$ \therefore $$ M $$-$$ m = 10 $$-$$ 9 = 1

$$ \Rightarrow $$ f'(x) = 6x

For maxima or minima put f'(x) = 0

$$ \Rightarrow $$ x

$$ \Rightarrow $$ x = 1 or x = 2

Now, f''(x) = 12x $$-$$ 18

$$ \Rightarrow $$ f''(1) = 12(1) $$-$$ 18 = $$-$$ 6 < 0

Hence, f(x) has maxima at x = 1

$$ \therefore $$ maximum value = M = f(1) = 2 $$-$$ 9 + 12 + 5 = 10

And, f''(2) = 12(2) $$-$$ 18 = 6 > 0.

Hence, f(x) has minima at x = 2.

$$ \therefore $$ minimum value = m = f(2) = 2(8) $$-$$ 9(4) + 12(2) + 5 = 9

$$ \therefore $$ M $$-$$ m = 10 $$-$$ 9 = 1

2

If $${{dy} \over {dx}} + {3 \over {{{\cos }^2}x}}y = {1 \over {{{\cos }^2}x}},\,\,x \in \left( {{{ - \pi } \over 3},{\pi \over 3}} \right)$$ and $$y\left( {{\pi \over 4}} \right) = {4 \over 3},$$ then $$y\left( { - {\pi \over 4}} \right)$$ equals -

A

$${1 \over 3} + {e^6}$$

B

$${1 \over 3}$$

C

$${1 \over 3}$$ + e^{3}

D

$$-$$ $${4 \over 3}$$

$${{dy} \over {dx}} + 3{\sec ^2}x.y = {\sec ^2}x$$

I.F. = $${e^{3\int {{{\sec }^2}xdx} }} = {e^{3\tan x}}$$

or $$y.e{}^{3\tan x} = \int {{{\sec }^2}x.{e^{3\tan x}}} $$

or $$y.{e^{3\tan x}} = {1 \over 3}{e^{3\tan x}} + C$$

Given

$$y\left( {{\pi \over 4}} \right) = {4 \over 3}$$

$$ \therefore $$ $${4 \over 3}.{e^3} = {1 \over 3}{e^3} + C$$

$$ \therefore $$ C = e^{3}

I.F. = $${e^{3\int {{{\sec }^2}xdx} }} = {e^{3\tan x}}$$

or $$y.e{}^{3\tan x} = \int {{{\sec }^2}x.{e^{3\tan x}}} $$

or $$y.{e^{3\tan x}} = {1 \over 3}{e^{3\tan x}} + C$$

Given

$$y\left( {{\pi \over 4}} \right) = {4 \over 3}$$

$$ \therefore $$ $${4 \over 3}.{e^3} = {1 \over 3}{e^3} + C$$

$$ \therefore $$ C = e

3

The shortest distance between the point $$\left( {{3 \over 2},0} \right)$$ and the curve y = $$\sqrt x $$, (x > 0), is -

A

$${{\sqrt 3 } \over 2}$$

B

$${5 \over 4}$$

C

$${3 \over 2}$$

D

$${{\sqrt 5 } \over 2}$$

Let points $$\left( {{3 \over 2},0} \right),\left( {{t^2},t} \right),t > 0$$

Distance = $$\sqrt {{t^2} + {{\left( {{t^2} - {3 \over 2}} \right)}^2}} $$

= $$\sqrt {{t^4} - 2{t^2} + {9 \over 4}} = \sqrt {{{\left( {{t^2} - 1} \right)}^2} + {5 \over 4}} $$

So minimum distance is $$\sqrt {{5 \over 4}} = {{\sqrt 5 } \over 2}$$

Distance = $$\sqrt {{t^2} + {{\left( {{t^2} - {3 \over 2}} \right)}^2}} $$

= $$\sqrt {{t^4} - 2{t^2} + {9 \over 4}} = \sqrt {{{\left( {{t^2} - 1} \right)}^2} + {5 \over 4}} $$

So minimum distance is $$\sqrt {{5 \over 4}} = {{\sqrt 5 } \over 2}$$

4

The tangent to the curve, y = xe^{x2} passing through the point (1, e) also passes through the point

A

$$\left( {{4 \over 3},2e} \right)$$

B

(3, 6e)

C

(2, 3e)

D

$$\left( {{5 \over 3},2e} \right)$$

y = xe^{x2}

$${\left. {{{dy} \over {dx}}} \right|_{(1,e)}}{\left. { = \left( {e.e{x^2}.2x + {e^{{x^2}}}} \right)} \right|_{(1,e)}}$$

$$ = 2 \cdot e + e = 3e$$

T : y $$-$$ e = 3e (x $$-$$ 1)

y = 3ex $$-$$ 3e + e

y = $$\left( {3e} \right)x - 2e$$

$$\left( {{4 \over 3},2e} \right)$$ lies on it

$${\left. {{{dy} \over {dx}}} \right|_{(1,e)}}{\left. { = \left( {e.e{x^2}.2x + {e^{{x^2}}}} \right)} \right|_{(1,e)}}$$

$$ = 2 \cdot e + e = 3e$$

T : y $$-$$ e = 3e (x $$-$$ 1)

y = 3ex $$-$$ 3e + e

y = $$\left( {3e} \right)x - 2e$$

$$\left( {{4 \over 3},2e} \right)$$ lies on it

Number in Brackets after Paper Name Indicates No of Questions

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Trigonometric Functions & Equations *keyboard_arrow_right*

Properties of Triangle *keyboard_arrow_right*

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Complex Numbers *keyboard_arrow_right*

Quadratic Equation and Inequalities *keyboard_arrow_right*

Permutations and Combinations *keyboard_arrow_right*

Mathematical Induction and Binomial Theorem *keyboard_arrow_right*

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Vector Algebra and 3D Geometry *keyboard_arrow_right*

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Limits, Continuity and Differentiability *keyboard_arrow_right*

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Application of Derivatives *keyboard_arrow_right*

Indefinite Integrals *keyboard_arrow_right*

Definite Integrals and Applications of Integrals *keyboard_arrow_right*

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