Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geotechnical Engineering

Transportation Engineering

Irrigation

Engineering Mathematics

Construction Material and Management

Fluid Mechanics and Hydraulic Machines

Hydrology

Environmental Engineering

Engineering Mechanics

Structural Analysis

Reinforced Cement Concrete

Steel Structures

Geomatics Engineering Or Surveying

General Aptitude

1

The number of real roots of the equation

$${e^{4x}} + 2{e^{3x}} - {e^x} - 6 = 0$$ is :

$${e^{4x}} + 2{e^{3x}} - {e^x} - 6 = 0$$ is :

A

2

B

4

C

1

D

0

Let $${e^x} = t > 0$$

$$f(t) = {t^4} + 2{t^3} - t - 6 = 0$$

$$f'(t) = 4{t^3} + 6{t^2} - 1$$

$$f''(t) = 12{t^2} + 12t > 0$$

$$f(0) = - 6,f(1) = - 4,f(2) = 24$$

$$\Rightarrow$$ Number of real roots = 1

$$f(t) = {t^4} + 2{t^3} - t - 6 = 0$$

$$f'(t) = 4{t^3} + 6{t^2} - 1$$

$$f''(t) = 12{t^2} + 12t > 0$$

$$f(0) = - 6,f(1) = - 4,f(2) = 24$$

$$\Rightarrow$$ Number of real roots = 1

2

A box open from top is made from a rectangular sheet of dimension a $$\times$$ b by cutting squares each of side x from each of the four corners and folding up the flaps. If the volume of the box is maximum, then x is equal to :

A

$${{a + b - \sqrt {{a^2} + {b^2} - ab} } \over {12}}$$

B

$${{a + b - \sqrt {{a^2} + {b^2} + ab} } \over 6}$$

C

$${{a + b - \sqrt {{a^2} + {b^2} - ab} } \over 6}$$

D

$${{a + b + \sqrt {{a^2} + {b^2} + ab} } \over 6}$$

V = l . b . h = (a $$-$$ 2x)(b $$-$$ 2x) x

$$\Rightarrow$$ V(x) = (2x $$-$$ a)(2x $$-$$ b) x

$$\Rightarrow$$ V(x) = 4x

$$ \Rightarrow {d \over {dx}}v(x) = 12{x^2} - 4(a + b)x + ab$$

$${d \over {dx}}(v(x)) = 0 \Rightarrow 12{x^2} - 4(a + b)x + ab = 0 < _\beta ^\alpha $$

$$ \Rightarrow x = {{4(a + b) \pm \sqrt {16{{(a + b)}^2} - 48ab} } \over {2(12)}}$$

$$ = {{(a + b) \pm \sqrt {{a^2} + {b^2} - ab} } \over 6}$$

Let $$x = \alpha = {{(a + b) + \sqrt {{a^2} + {b^2} - ab} } \over 6}$$

$$\beta = {{(a + b) - \sqrt {{a^2} + {b^2} - ab} } \over 6}$$

Now, $$12(x - \alpha )(x - \beta ) = 0$$

$$\therefore$$ x = $$\beta$$ $$ = {{a + b - \sqrt {{a^2} + {b^2} - ab} } \over b}$$

3

A wire of length 20 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in meters) of the hexagon, so that the combined area of the square and the hexagon is minimum, is :

A

$${5 \over {2 + \sqrt 3 }}$$

B

$${{10} \over {2 + 3\sqrt 3 }}$$

C

$${5 \over {3 + \sqrt 3 }}$$

D

$${{10} \over {3 + 2\sqrt 3 }}$$

Let the wire is cut into two pieces of length x and 20 $$-$$ x.

Area of square = $${\left( {{x \over 4}} \right)^2}$$

Area of regular hexagon = $$6 \times {{\sqrt 3 } \over 4}{\left( {{{20 - x} \over 6}} \right)^2}$$

Total area = $$A(x) = {{{x^2}} \over {16}} + {{3\sqrt 3 } \over 2}{{{{(20 - x)}^2}} \over {36}}$$

$$A'(x) = {{2x} \over {16}} + {{3\sqrt 3 \times 2} \over {2 \times 36}}(20 - x)( - 1)$$

A'(x) = 0 at $$x = {{40\sqrt 3 } \over {3 + 2\sqrt 3 }}$$

Length of side of regular Hexagon $$ = {1 \over 6}(20 - x)$$

$$ = {1 \over 6}\left( {20 - {{4.\sqrt 3 } \over {3 + 2\sqrt 3 }}} \right)$$

$$ = {{10} \over {2 + 2\sqrt 3 }}$$

Area of square = $${\left( {{x \over 4}} \right)^2}$$

Area of regular hexagon = $$6 \times {{\sqrt 3 } \over 4}{\left( {{{20 - x} \over 6}} \right)^2}$$

Total area = $$A(x) = {{{x^2}} \over {16}} + {{3\sqrt 3 } \over 2}{{{{(20 - x)}^2}} \over {36}}$$

$$A'(x) = {{2x} \over {16}} + {{3\sqrt 3 \times 2} \over {2 \times 36}}(20 - x)( - 1)$$

A'(x) = 0 at $$x = {{40\sqrt 3 } \over {3 + 2\sqrt 3 }}$$

Length of side of regular Hexagon $$ = {1 \over 6}(20 - x)$$

$$ = {1 \over 6}\left( {20 - {{4.\sqrt 3 } \over {3 + 2\sqrt 3 }}} \right)$$

$$ = {{10} \over {2 + 2\sqrt 3 }}$$

4

Let 'a' be a real number such that the function f(x) = ax^{2} + 6x $$-$$ 15, x $$\in$$ R is increasing in $$\left( { - \infty ,{3 \over 4}} \right)$$ and decreasing in $$\left( {{3 \over 4},\infty } \right)$$. Then the function g(x) = ax^{2} $$-$$ 6x + 15, x$$\in$$R has a :

A

local maximum at x = $$-$$ $${{3 \over 4}}$$

B

local minimum at x = $$-$$$${{3 \over 4}}$$

C

local maximum at x = $${{3 \over 4}}$$

D

local minimum at x = $${{3 \over 4}}$$

$${{ - B} \over {2A}} = {3 \over 4}$$

$$ \Rightarrow {{ - (6)} \over {2a}} = {3 \over 4}$$

$$ \Rightarrow a = {{ - 6 \times 4} \over 6} \Rightarrow a = - 4$$

$$\therefore$$ $$g(x) = 4{x^2} - 6x + 15$$

Local max. at $$x = {{ - B} \over {2A}} = - {{( - 6)} \over {2( - 4)}}$$

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

$$ \Rightarrow {{ - (6)} \over {2a}} = {3 \over 4}$$

$$ \Rightarrow a = {{ - 6 \times 4} \over 6} \Rightarrow a = - 4$$

$$\therefore$$ $$g(x) = 4{x^2} - 6x + 15$$

Local max. at $$x = {{ - B} \over {2A}} = - {{( - 6)} \over {2( - 4)}}$$

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

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