Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geotechnical Engineering

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Reinforced Cement Concrete

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General Aptitude

1

Let f be any continuous function on [0, 2] and twice differentiable on (0, 2). If f(0) = 0, f(1) = 1 and f(2) = 2, then

A

f''(x) = 0 for all x $$\in$$ (0, 2)

B

f''(x) = 0 for some x $$\in$$ (0, 2)

C

f'(x) = 0 for some x $$\in$$ [0, 2]

D

f''(x) > 0 for all x $$\in$$ (0, 2)

f(0) = 0, f(1) = 1 and f(2) = 2

Let h(x) = f(x) $$-$$ x has three roots

By Rolle's theorem h'(x) = f'(x) $$-$$ 1 has at least two roots

h''(x) = f''(x) = 0 has at least one roots

Let h(x) = f(x) $$-$$ x has three roots

By Rolle's theorem h'(x) = f'(x) $$-$$ 1 has at least two roots

h''(x) = f''(x) = 0 has at least one roots

2

If $$\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{{{\tan }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$$ and $$\beta = \mathop {\lim }\limits_{x \to 0 } {(\cos x)^{\cot x}}$$ are the roots of the equation, ax^{2} + bx $$-$$ 4 = 0, then the ordered pair (a, b) is :

A

(1, $$-$$3)

B

($$-$$1, 3)

C

($$-$$1, $$-$$3)

D

(1, 3)

$$\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{{{\tan }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}};{0 \over 0}$$ form

Using L Hospital rule

$$\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{3{{\tan }^2}x{{\sec }^2}x - {{\sec }^2}x} \over { - \sin \left( {x + {\pi \over 4}} \right)}}$$

$$\alpha$$ = $$-$$4

$$\beta = \mathop {\lim }\limits_{x \to 0} {(\cos x)^{\cot x}} = {e^{\mathop {\lim }\limits_{x \to 0} {{(\cos x - 1)} \over {\tan x}}}}$$

$$\beta = {e^{\mathop {\lim }\limits_{x \to 0} {{ - (1 - \cos x)} \over {{x^2}}}.{{{x^2}} \over {{{\left( {{{\tan x} \over x}} \right)}^x}}}}}$$

$$\beta = {e^{\mathop {\lim }\limits_{x \to 0} \left( {{{ - 1} \over 2}} \right).{x \over 1}}} = {e^0} \Rightarrow \beta = 1$$

$$\alpha$$ = $$-$$4; $$\beta$$ = 1

If ax^{2} + bx $$-$$ 4 = 0 are the roots then

16a $$-$$ 4b $$-$$ 4 = 0 & a + b $$-$$ 4 = 0

$$\Rightarrow$$ a = 1 & b = 3

Using L Hospital rule

$$\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{3{{\tan }^2}x{{\sec }^2}x - {{\sec }^2}x} \over { - \sin \left( {x + {\pi \over 4}} \right)}}$$

$$\alpha$$ = $$-$$4

$$\beta = \mathop {\lim }\limits_{x \to 0} {(\cos x)^{\cot x}} = {e^{\mathop {\lim }\limits_{x \to 0} {{(\cos x - 1)} \over {\tan x}}}}$$

$$\beta = {e^{\mathop {\lim }\limits_{x \to 0} {{ - (1 - \cos x)} \over {{x^2}}}.{{{x^2}} \over {{{\left( {{{\tan x} \over x}} \right)}^x}}}}}$$

$$\beta = {e^{\mathop {\lim }\limits_{x \to 0} \left( {{{ - 1} \over 2}} \right).{x \over 1}}} = {e^0} \Rightarrow \beta = 1$$

$$\alpha$$ = $$-$$4; $$\beta$$ = 1

If ax

16a $$-$$ 4b $$-$$ 4 = 0 & a + b $$-$$ 4 = 0

$$\Rightarrow$$ a = 1 & b = 3

3

$$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}\left( {\pi {{\cos }^4}x} \right)} \over {{x^4}}}$$ is equal to :

A

$${\pi ^2}$$

B

$$2{\pi ^2}$$

C

$$4{\pi ^2}$$

D

$$4\pi $$

$$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}\left( {\pi {{\cos }^4}x} \right)} \over {{x^4}}}$$

= $$\mathop {\lim }\limits_{x \to 0} {{1 - \cos \left( {2\pi {{\cos }^4}x} \right)} \over {2{x^4}}}$$

= $$\mathop {\lim }\limits_{x \to 0} {{1 - \cos \left( {2\pi - 2\pi {{\cos }^4}x} \right)} \over {{{\left[ {2\pi (1 - {{\cos }^4}x)} \right]}^2}}}4{\pi ^2}.{{{{\sin }^4}x} \over {2{x^4}}}{\left( {1 + {{\cos }^2}x} \right)^2}$$

$$ = {1 \over 2}.4{\pi ^2}.{1 \over 2}{(2)^2} = 4{\pi ^2}$$

= $$\mathop {\lim }\limits_{x \to 0} {{1 - \cos \left( {2\pi {{\cos }^4}x} \right)} \over {2{x^4}}}$$

= $$\mathop {\lim }\limits_{x \to 0} {{1 - \cos \left( {2\pi - 2\pi {{\cos }^4}x} \right)} \over {{{\left[ {2\pi (1 - {{\cos }^4}x)} \right]}^2}}}4{\pi ^2}.{{{{\sin }^4}x} \over {2{x^4}}}{\left( {1 + {{\cos }^2}x} \right)^2}$$

$$ = {1 \over 2}.4{\pi ^2}.{1 \over 2}{(2)^2} = 4{\pi ^2}$$

4

If the function

$$f(x) = \left\{ {\matrix{ {{1 \over x}{{\log }_e}\left( {{{1 + {x \over a}} \over {1 - {x \over b}}}} \right)} & , & {x < 0} \cr k & , & {x = 0} \cr {{{{{\cos }^2}x - {{\sin }^2}x - 1} \over {\sqrt {{x^2} + 1} - 1}}} & , & {x > 0} \cr } } \right.$$ is continuous

at x = 0, then $${1 \over a} + {1 \over b} + {4 \over k}$$ is equal to :

$$f(x) = \left\{ {\matrix{ {{1 \over x}{{\log }_e}\left( {{{1 + {x \over a}} \over {1 - {x \over b}}}} \right)} & , & {x < 0} \cr k & , & {x = 0} \cr {{{{{\cos }^2}x - {{\sin }^2}x - 1} \over {\sqrt {{x^2} + 1} - 1}}} & , & {x > 0} \cr } } \right.$$ is continuous

at x = 0, then $${1 \over a} + {1 \over b} + {4 \over k}$$ is equal to :

A

$$-$$5

B

5

C

$$-$$4

D

4

If f(x) is continuous at x = 0, RHL = LHL = f(0)

$$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^2}x - {{\sin }^2}x - 1} \over {\sqrt {{x^2} + 1} - 1}}.{{\sqrt {{x^2} + 1} + 1} \over {\sqrt {{x^2} + 1} + 1}}$$ (Rationalisation)

$$\mathop {\lim }\limits_{x \to {0^ + }} - {{2{{\sin }^2}x} \over {{x^2}}}.\left( {\sqrt {{x^2} + 1} + 1} \right) = - 4$$

$$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} {1 \over x}\ln \left( {{{1 + {x \over a}} \over {1 - {x \over b}}}} \right)$$

$$\mathop {\lim }\limits_{x \to {0^ - }} {{\ln \left( {1{x \over a}} \right)} \over {\left( {{x \over a}} \right).\,a}} + {{\ln \left( {1 - {x \over b}} \right)} \over {\left( { - {x \over b}} \right)\,.\,b}}$$$$ = {1 \over a} + {1 \over b}$$

So, $${1 \over a} + {1 \over b} = - 4 = k$$

$$ \Rightarrow {1 \over a} + {1 \over b} + {4 \over k} = - 4 - 1 = - 5$$

$$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} {{{{\cos }^2}x - {{\sin }^2}x - 1} \over {\sqrt {{x^2} + 1} - 1}}.{{\sqrt {{x^2} + 1} + 1} \over {\sqrt {{x^2} + 1} + 1}}$$ (Rationalisation)

$$\mathop {\lim }\limits_{x \to {0^ + }} - {{2{{\sin }^2}x} \over {{x^2}}}.\left( {\sqrt {{x^2} + 1} + 1} \right) = - 4$$

$$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} {1 \over x}\ln \left( {{{1 + {x \over a}} \over {1 - {x \over b}}}} \right)$$

$$\mathop {\lim }\limits_{x \to {0^ - }} {{\ln \left( {1{x \over a}} \right)} \over {\left( {{x \over a}} \right).\,a}} + {{\ln \left( {1 - {x \over b}} \right)} \over {\left( { - {x \over b}} \right)\,.\,b}}$$$$ = {1 \over a} + {1 \over b}$$

So, $${1 \over a} + {1 \over b} = - 4 = k$$

$$ \Rightarrow {1 \over a} + {1 \over b} + {4 \over k} = - 4 - 1 = - 5$$

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