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

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Geomatics Engineering Or Surveying

Environmental Engineering

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Engineering Mathematics

General Aptitude

1

Let f be a polynomial function such that

f (3x) = f ' (x) . f '' (x), for all x $$ \in $$**R**. Then :

f (3x) = f ' (x) . f '' (x), for all x $$ \in $$

A

f (2) + f ' (2) = 28

B

f '' (2) $$-$$ f ' (2) = 0

C

f '' (2) $$-$$ f (2) = 4

D

f (2) $$-$$ f ' (2) + f '' (2) = 10

2

For each t $$ \in R$$, let [t] be the greatest integer less than or equal to t.

Then $$\mathop {\lim }\limits_{x \to {0^ + }} x\left( {\left[ {{1 \over x}} \right] + \left[ {{2 \over x}} \right] + ..... + \left[ {{{15} \over x}} \right]} \right)$$

Then $$\mathop {\lim }\limits_{x \to {0^ + }} x\left( {\left[ {{1 \over x}} \right] + \left[ {{2 \over x}} \right] + ..... + \left[ {{{15} \over x}} \right]} \right)$$

A

does not exist in R

B

is equal to 0

C

is equal to 15

D

is equal to 120

Given,

$$\mathop {lim}\limits_{x \to {0^ + }} \,\,x\,\left( {\left[ {{1 \over x}} \right] + \left[ {{2 \over x}} \right]} \right. + $$ $$\left. {\,.\,.\,.\,.\,.\, + \left[ {{{15} \over x}} \right]} \right)$$

as we know that

$${1 \over x} = \left[ {{1 \over x}} \right] + \left\{ {{1 \over x}} \right\}$$

$$ \Rightarrow \,\,\,\,\left[ {{1 \over x}} \right] = {1 \over x} - \left\{ {{1 \over x}} \right\}$$

$$ = \,\,\,\,\mathop {\lim }\limits_{x \to {0^ + }} \,\,x\left[ {{1 \over x} - \left\{ {{1 \over x}} \right\} + {2 \over 2} - \left\{ {{2 \over x}} \right\} + ........{{15} \over x} - \left\{ {{{15} \over x}} \right\}} \right]$$

$$ = \,\,\,\,\mathop {\lim }\limits_{x \to {0^ + }} \,\left[ {x.{1 \over x} + x.{2 \over x} + .....x.{{15} \over x}} \right]$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \,\left[ {x.\left\{ {{1 \over 2}} \right\} + .. + x.\left\{ {{{15} \over x}} \right\}} \right]$$

We know $$\left\{ {{1 \over x}} \right\}$$ is fractional part of $${1 \over x}.$$

So, the range of $$\,\left\{ {{1 \over x}} \right\}$$ is $$0 \le \left\{ {{1 \over x}} \right\} < 1$$

So, $$\mathop {lim}\limits_{x \to {0^ + }} \,\,x\,\left\{ {{1 \over x}} \right\} = 0.$$ (finite no) $$=0$$

Similarly $$\mathop {\lim }\limits_{x \to {0^ + }} x.\left\{ {{2 \over x}} \right\} = 0$$

$$ = \,\,\,\,\left( {1 + 2 + ... + 15} \right) - \left( {0 + 0...} \right)$$

$$ = \,\,\,\,{{15 \times 16} \over 2}$$

$$ = \,\,\,\,120$$

$$\mathop {lim}\limits_{x \to {0^ + }} \,\,x\,\left( {\left[ {{1 \over x}} \right] + \left[ {{2 \over x}} \right]} \right. + $$ $$\left. {\,.\,.\,.\,.\,.\, + \left[ {{{15} \over x}} \right]} \right)$$

as we know that

$${1 \over x} = \left[ {{1 \over x}} \right] + \left\{ {{1 \over x}} \right\}$$

$$ \Rightarrow \,\,\,\,\left[ {{1 \over x}} \right] = {1 \over x} - \left\{ {{1 \over x}} \right\}$$

$$ = \,\,\,\,\mathop {\lim }\limits_{x \to {0^ + }} \,\,x\left[ {{1 \over x} - \left\{ {{1 \over x}} \right\} + {2 \over 2} - \left\{ {{2 \over x}} \right\} + ........{{15} \over x} - \left\{ {{{15} \over x}} \right\}} \right]$$

$$ = \,\,\,\,\mathop {\lim }\limits_{x \to {0^ + }} \,\left[ {x.{1 \over x} + x.{2 \over x} + .....x.{{15} \over x}} \right]$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \,\left[ {x.\left\{ {{1 \over 2}} \right\} + .. + x.\left\{ {{{15} \over x}} \right\}} \right]$$

We know $$\left\{ {{1 \over x}} \right\}$$ is fractional part of $${1 \over x}.$$

So, the range of $$\,\left\{ {{1 \over x}} \right\}$$ is $$0 \le \left\{ {{1 \over x}} \right\} < 1$$

So, $$\mathop {lim}\limits_{x \to {0^ + }} \,\,x\,\left\{ {{1 \over x}} \right\} = 0.$$ (finite no) $$=0$$

Similarly $$\mathop {\lim }\limits_{x \to {0^ + }} x.\left\{ {{2 \over x}} \right\} = 0$$

$$ = \,\,\,\,\left( {1 + 2 + ... + 15} \right) - \left( {0 + 0...} \right)$$

$$ = \,\,\,\,{{15 \times 16} \over 2}$$

$$ = \,\,\,\,120$$

3

If $$f\left( x \right) = \left| {\matrix{
{\cos x} & x & 1 \cr
{2\sin x} & {{x^2}} & {2x} \cr
{\tan x} & x & 1 \cr
} } \right|,$$ then $$\mathop {\lim }\limits_{x \to 0} {{f'\left( x \right)} \over x}$$

A

does not exist.

B

exists and is equal to 2.

C

existsand is equal to 0.

D

exists and is equal to $$-$$ 2.

Given,

$$f\left( x \right) = \left| {\matrix{ {\cos x} & x & 1 \cr {2\sin x} & {{x^2}} & {2x} \cr {\tan x} & x & 1 \cr } } \right|$$

= cosx(x^{2} - 2x^{2}) - x(2 sinx - 2x tanx) + (2x sinx - x^{2} tanx)

= x^{2} (tanx - cosx)

$$ \therefore $$ $${f^{'}}(x)$$ = 2x (tanx - cosx) + x^{2}(sec^{2}x + sinx)

$$ \therefore $$ $$\mathop {\lim }\limits_{x \to 0} {{f'\left( x \right)} \over x}$$

= $$\mathop {\lim }\limits_{x \to o} {{2x(\tan x - \cos x) + {x^2}({{\sec }^2}x + \sin x)} \over x}$$

= $$\mathop {\lim }\limits_{x \to o} \,\,2(\tan x - \cos x) + x({\sec ^2}x + \sin x)$$

= 2 (0-1) + 0

= -2

$$f\left( x \right) = \left| {\matrix{ {\cos x} & x & 1 \cr {2\sin x} & {{x^2}} & {2x} \cr {\tan x} & x & 1 \cr } } \right|$$

= cosx(x

= x

$$ \therefore $$ $${f^{'}}(x)$$ = 2x (tanx - cosx) + x

$$ \therefore $$ $$\mathop {\lim }\limits_{x \to 0} {{f'\left( x \right)} \over x}$$

= $$\mathop {\lim }\limits_{x \to o} {{2x(\tan x - \cos x) + {x^2}({{\sec }^2}x + \sin x)} \over x}$$

= $$\mathop {\lim }\limits_{x \to o} \,\,2(\tan x - \cos x) + x({\sec ^2}x + \sin x)$$

= 2 (0-1) + 0

= -2

4

Let S = {($$\lambda $$, $$\mu $$) $$ \in $$ **R** $$ \times $$ **R** : f(t) = (|$$\lambda $$| e^{|t|} $$-$$ $$\mu $$). sin (2|t|), **t** $$ \in $$ **R**, is a differentiable function}. Then S is a subset of :

A

B

[0, $$\infty $$) $$ \times $$ **R**

C

D

($$-$$ $$\infty $$, 0) $$ \times $$ **R**

S = {($$\lambda $$, $$\mu $$) $$ \in $$ **R** $$ \times $$ **R** : f(t) = $$\left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right)$$ sin $$\left( {2\left| t \right|} \right),$$ t $$ \in $$ **R**

f(t) = $$\left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right)\sin \left( {2\left| t \right|} \right)$$

= $$\left\{ {\matrix{ {\left( {\left| \lambda \right|{e^t} - \mu } \right)\sin 2t,} & {t > 0} \cr {\left( {\left| \lambda \right|{e^{ - t}} - \mu } \right)\left( { - \sin 2t} \right),} & {t < 0} \cr } } \right.$$

f'(t) = $$\left\{ {\matrix{ {\left( {\left| \lambda \right|{e^t}} \right)\sin 2t + \left( {\left| \lambda \right|{e^t} - \mu } \right)\left( {2\cos 2t} \right),\,\,t > 0} \cr {\left| \lambda \right|{e^{ - t}}\sin 2t + \left( {\left| \lambda \right|{e^{ - t}} - \mu } \right)\left( {-2\cos 2t} \right),t < 0} \cr } } \right.$$

As, f(t) is differentiable

$$ \therefore $$**LHD** = **RHD** at t = 0

$$ \Rightarrow $$ $$\left| \lambda \right|$$ . sin2(0) + $$\left( {\left| \lambda \right|{e^0} - \mu } \right)$$2cos(0)

= $$\left| \lambda \right|{e^{ - 0}}\,$$ . sin 2(0) $$-$$ 2 cos (0) $$\left( {\left| \lambda \right|{e^{ - 0}} - \mu } \right)$$

$$ \Rightarrow $$$$\,\,\,$$ 0 + $$\left( {\left| \lambda \right| - \mu } \right)$$2 = 0 $$-$$ 2$$\left( {\left| \lambda \right| - \mu } \right)$$

$$ \Rightarrow $$ 4$$\left( {\left| \lambda \right| - \mu } \right)$$ = 0

$$ \Rightarrow $$ $$\left| \mu \right|$$ = $$\mu $$

So, S $$ \equiv $$ ($$\lambda $$, $$\mu $$) = {$$\lambda $$ $$ \in $$**R** & $$\mu $$ $$ \in $$ [0, $$\infty $$)}

Therefore set S is subset of R $$ \times $$ [0, $$\infty $$)

f(t) = $$\left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right)\sin \left( {2\left| t \right|} \right)$$

= $$\left\{ {\matrix{ {\left( {\left| \lambda \right|{e^t} - \mu } \right)\sin 2t,} & {t > 0} \cr {\left( {\left| \lambda \right|{e^{ - t}} - \mu } \right)\left( { - \sin 2t} \right),} & {t < 0} \cr } } \right.$$

f'(t) = $$\left\{ {\matrix{ {\left( {\left| \lambda \right|{e^t}} \right)\sin 2t + \left( {\left| \lambda \right|{e^t} - \mu } \right)\left( {2\cos 2t} \right),\,\,t > 0} \cr {\left| \lambda \right|{e^{ - t}}\sin 2t + \left( {\left| \lambda \right|{e^{ - t}} - \mu } \right)\left( {-2\cos 2t} \right),t < 0} \cr } } \right.$$

As, f(t) is differentiable

$$ \therefore $$

$$ \Rightarrow $$ $$\left| \lambda \right|$$ . sin2(0) + $$\left( {\left| \lambda \right|{e^0} - \mu } \right)$$2cos(0)

= $$\left| \lambda \right|{e^{ - 0}}\,$$ . sin 2(0) $$-$$ 2 cos (0) $$\left( {\left| \lambda \right|{e^{ - 0}} - \mu } \right)$$

$$ \Rightarrow $$$$\,\,\,$$ 0 + $$\left( {\left| \lambda \right| - \mu } \right)$$2 = 0 $$-$$ 2$$\left( {\left| \lambda \right| - \mu } \right)$$

$$ \Rightarrow $$ 4$$\left( {\left| \lambda \right| - \mu } \right)$$ = 0

$$ \Rightarrow $$ $$\left| \mu \right|$$ = $$\mu $$

So, S $$ \equiv $$ ($$\lambda $$, $$\mu $$) = {$$\lambda $$ $$ \in $$

Therefore set S is subset of R $$ \times $$ [0, $$\infty $$)

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