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 f(x) = $$\left\{ {\matrix{
{{{\left( {x - 1} \right)}^{{1 \over {2 - x}}}},} & {x > 1,x \ne 2} \cr
{k\,\,\,\,\,\,\,\,\,\,\,\,\,\,} & {,x = 2} \cr
} } \right.$$

Thevaue of k for which f s continuous at x = 2 is :

Thevaue of k for which f s continuous at x = 2 is :

A

1

B

e

C

e^{-1}

D

e^{-2}

Since f(x) is continuous at x = 2.

$$\therefore\,\,\,$$$$\mathop {\lim }\limits_{x \to 2} $$ f(x) = f(2)

$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 2} $$ $${\left( {x - 1} \right)^{{1 \over {2 - x}}}}$$ = k ($${{1^\infty }}$$ form)

$$ \therefore $$ $${{e^l}}$$ = k

Where $$l$$ = $$\mathop {\lim }\limits_{x \to 2} $$(x $$-$$ $$l$$ $$-$$ 1) $$ \times $$ $${1 \over {2 - x}}$$ = $$\mathop {\lim }\limits_{x \to 2} {{x - 2} \over {2 - x}}$$

= $$\mathop {\lim }\limits_{x \to 2} \left( {{{x - 2} \over {x - 2}}} \right)$$

$$ \Rightarrow $$ k = e^{$$-$$1}

$$\therefore\,\,\,$$$$\mathop {\lim }\limits_{x \to 2} $$ f(x) = f(2)

$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 2} $$ $${\left( {x - 1} \right)^{{1 \over {2 - x}}}}$$ = k ($${{1^\infty }}$$ form)

$$ \therefore $$ $${{e^l}}$$ = k

Where $$l$$ = $$\mathop {\lim }\limits_{x \to 2} $$(x $$-$$ $$l$$ $$-$$ 1) $$ \times $$ $${1 \over {2 - x}}$$ = $$\mathop {\lim }\limits_{x \to 2} {{x - 2} \over {2 - x}}$$

= $$\mathop {\lim }\limits_{x \to 2} \left( {{{x - 2} \over {x - 2}}} \right)$$

$$ \Rightarrow $$ k = e

2

If f(x) = sin^{-1} $$\left( {{{2 \times {3^x}} \over {1 + {9^x}}}} \right),$$ then f'$$\left( { - {1 \over 2}} \right)$$ equals :

A

$$ - \sqrt 3 {\log _e}\sqrt 3 $$

B

$$ \sqrt 3 {\log _e}\sqrt 3 $$

C

$$ - \sqrt 3 {\log _e}\, 3 $$

D

$$ \sqrt 3 {\log _e}\, 3 $$

Since f(x) = sin$$\left( {{{2 \times {3^x}} \over {1 + {9^x}}}} \right)$$

Suppose 3^{x} = tan t

$$ \Rightarrow $$ f(x) = sin^{$$-$$1} $$\left( {{{2\tan t} \over {1 + {{\tan }^2}t}}} \right)$$ = sin^{$$-$$1} (sin2t) = 2t

= 2tan^{$$-$$1} (3x)

So, f'(x) = $${2 \over {1 + {{\left( {{3^x}} \right)}^2}}} \times {3^x}.{\log _e}3$$

$$ \therefore $$ f '$$\left( { - {1 \over 2}} \right)$$ = $${2 \over {1 + {{\left( {{3^{ - {1 \over 2}}}} \right)}^2}}} \times {3^{ - {1 \over 2}}}$$ . log_{e} 3

= $${1 \over 2} \times \sqrt 3 \times {\log _e}3$$ = $$\sqrt 3 \times {\log _e}\sqrt 3 $$

Suppose 3

$$ \Rightarrow $$ f(x) = sin

= 2tan

So, f'(x) = $${2 \over {1 + {{\left( {{3^x}} \right)}^2}}} \times {3^x}.{\log _e}3$$

$$ \therefore $$ f '$$\left( { - {1 \over 2}} \right)$$ = $${2 \over {1 + {{\left( {{3^{ - {1 \over 2}}}} \right)}^2}}} \times {3^{ - {1 \over 2}}}$$ . log

= $${1 \over 2} \times \sqrt 3 \times {\log _e}3$$ = $$\sqrt 3 \times {\log _e}\sqrt 3 $$

3

Let f(x) be a polynomial of degree $$4$$ having extreme values at $$x = 1$$ and $$x = 2.$$

If $$\mathop {lim}\limits_{x \to 0} \left( {{{f\left( x \right)} \over {{x^2}}} + 1} \right) = 3$$ then f($$-$$1) is equal to :

If $$\mathop {lim}\limits_{x \to 0} \left( {{{f\left( x \right)} \over {{x^2}}} + 1} \right) = 3$$ then f($$-$$1) is equal to :

A

$${9 \over 2}$$

B

$${5 \over 2}$$

C

$${3 \over 2}$$

D

$${1 \over 2}$$

$$ \because $$ f(x) has extremum values at x = 1, and x = 2

$$ \because $$ f'(1) = 0 and f'(2) = 0

As, f(x) is a polynomial of degree 4.

Suppose f(x) = Ax^{4} + Bx^{3} + cx^{2} + Dx + E

$$ \because $$ $$\mathop {\lim }\limits_{x \to 0} $$ $$\left( {{{f\left( x \right)} \over {{x^2}}} + 1} \right)$$ = 3

$$ \Rightarrow $$$$\,\,\,$$$$\mathop {\lim }\limits_{x \to 0} \left( {{{A{x^4} + B{x^3} + C{x^2} + Dx + E} \over {{x^2}}} + 1} \right)$$ = 3

$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 0} $$ $$\left( {A{x^2} + Bx + C + {D \over x} + {E \over {{x^2}}} + 1} \right)$$ = 3

As limit has finite value, so D = 0 and E = 0

Now A(0)^{2} + B(0) + C + 0 + 0 + 1 = 3

$$ \Rightarrow $$ c + 1 = 3 $$ \Rightarrow $$ c = 2

f'(x) = 4Ax^{3} + 3Bx^{2} + 2Cx + D

f'(1) = 0 $$ \Rightarrow $$ 4A(1) + 3B(1) + 2C(1) + D = 0

$$ \Rightarrow $$$$\,\,\,$$ 4A + 3B = $$-$$ 4 . . . .(1)

f'(2) = 0 $$ \Rightarrow $$ 4A(8) + 3B(4) + 2C(2) + D = 0

$$ \Rightarrow $$ 8A + 3B = $$-$$ 2 . . . . .(2)

From equations (1) and (2), we get

A = $${1 \over 2}$$ and B = $$-$$ 2

So, f(x) = $${{{x_4}} \over 2} - 2{x^3} + 2x{}^2$$

Therefore, f($$-$$ 1) = $${{{{( - 1)}^4}} \over 2} - 2{( - 1)^3} + 2{( - 1)^2}$$

= $${1 \over 2} + 2 + 2 = {9 \over 2}$$

Hence f($$-$$1) = $${9 \over 2}$$

$$ \because $$ f'(1) = 0 and f'(2) = 0

As, f(x) is a polynomial of degree 4.

Suppose f(x) = Ax

$$ \because $$ $$\mathop {\lim }\limits_{x \to 0} $$ $$\left( {{{f\left( x \right)} \over {{x^2}}} + 1} \right)$$ = 3

$$ \Rightarrow $$$$\,\,\,$$$$\mathop {\lim }\limits_{x \to 0} \left( {{{A{x^4} + B{x^3} + C{x^2} + Dx + E} \over {{x^2}}} + 1} \right)$$ = 3

$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 0} $$ $$\left( {A{x^2} + Bx + C + {D \over x} + {E \over {{x^2}}} + 1} \right)$$ = 3

As limit has finite value, so D = 0 and E = 0

Now A(0)

$$ \Rightarrow $$ c + 1 = 3 $$ \Rightarrow $$ c = 2

f'(x) = 4Ax

f'(1) = 0 $$ \Rightarrow $$ 4A(1) + 3B(1) + 2C(1) + D = 0

$$ \Rightarrow $$$$\,\,\,$$ 4A + 3B = $$-$$ 4 . . . .(1)

f'(2) = 0 $$ \Rightarrow $$ 4A(8) + 3B(4) + 2C(2) + D = 0

$$ \Rightarrow $$ 8A + 3B = $$-$$ 2 . . . . .(2)

From equations (1) and (2), we get

A = $${1 \over 2}$$ and B = $$-$$ 2

So, f(x) = $${{{x_4}} \over 2} - 2{x^3} + 2x{}^2$$

Therefore, f($$-$$ 1) = $${{{{( - 1)}^4}} \over 2} - 2{( - 1)^3} + 2{( - 1)^2}$$

= $${1 \over 2} + 2 + 2 = {9 \over 2}$$

Hence f($$-$$1) = $${9 \over 2}$$

4

$$\mathop {\lim }\limits_{x \to 0} \,\,{{{{\left( {27 + x} \right)}^{{1 \over 3}}} - 3} \over {9 - {{\left( {27 + x} \right)}^{{2 \over 3}}}}}$$ equals.

A

$${1 \over 3}$$

B

$$-$$ $${1 \over 3}$$

C

$$-$$ $${1 \over 6}$$

D

$${1 \over 6}$$

Given,

$$\mathop {lim}\limits_{x \to 0} \,{{{{\left( {27 + x} \right)}^{{1 \over 3}}} - 3} \over {9 - {{\left( {27 + x} \right)}^{{2 \over 3}}}}}$$

= $$\mathop {\lim }\limits_{x \to 0} \,{{3\left[ {{{\left( {1 + {x \over {27}}} \right)}^{{1 \over 3}}} - 1} \right]} \over {9\left[ {1 - {{\left( {1 + {x \over {27}}} \right)}^{{2 \over 3}}}} \right]}}$$

= $$\mathop {\lim }\limits_{x \to 0} \,\,{{\left( {1 + {x \over {3 \times 27}}} \right) - 1} \over {3\left[ {1 - \left( {1 + {{2x} \over {3 \times 27}}} \right)} \right]}}$$

= $$\mathop {\lim }\limits_{x \to 0} \,\,{{{x \over {81}}} \over {3\left( { - {{2x} \over {81}}} \right)}} = - {1 \over 6}$$

$$\mathop {lim}\limits_{x \to 0} \,{{{{\left( {27 + x} \right)}^{{1 \over 3}}} - 3} \over {9 - {{\left( {27 + x} \right)}^{{2 \over 3}}}}}$$

= $$\mathop {\lim }\limits_{x \to 0} \,{{3\left[ {{{\left( {1 + {x \over {27}}} \right)}^{{1 \over 3}}} - 1} \right]} \over {9\left[ {1 - {{\left( {1 + {x \over {27}}} \right)}^{{2 \over 3}}}} \right]}}$$

= $$\mathop {\lim }\limits_{x \to 0} \,\,{{\left( {1 + {x \over {3 \times 27}}} \right) - 1} \over {3\left[ {1 - \left( {1 + {{2x} \over {3 \times 27}}} \right)} \right]}}$$

= $$\mathop {\lim }\limits_{x \to 0} \,\,{{{x \over {81}}} \over {3\left( { - {{2x} \over {81}}} \right)}} = - {1 \over 6}$$

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Circle *keyboard_arrow_right*

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