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

If the function f defined as

$$f\left( x \right) = {1 \over x} - {{k - 1} \over {{e^{2x}} - 1}},x \ne 0,$$ is continuous at

x = 0, then the ordered pair (k, f(0)) is equal to :

$$f\left( x \right) = {1 \over x} - {{k - 1} \over {{e^{2x}} - 1}},x \ne 0,$$ is continuous at

x = 0, then the ordered pair (k, f(0)) is equal to :

A

(3, 2)

B

(3, 1)

C

(2, 1)

D

$$\left( {{1 \over 3},\,2} \right)$$

If the function is continuous at x = 0, then

$$\mathop {\lim }\limits_{x \to 0} $$ f(x) will exist and f(0) = $$\mathop {\lim }\limits_{x \to 0} $$ f(x)

Now, $$\mathop {\lim }\limits_{x \to 0} $$ f(x) = $$\mathop {\lim }\limits_{x \to 0} \left( {{1 \over x} - {{k - 1} \over {{e^{2x}} - 1}}} \right)$$

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

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

= $$\mathop {\lim }\limits_{x \to 0} \left[ {{{\left( {3 - k} \right)x + {{4{x^2}} \over {2!}} + {{8{x^3}} \over {3!}} + ...} \over {\left( {2{x^2} + {{4{x^3}} \over {2!}} + {{8{x^3}} \over {3!}} + ....} \right)}}} \right]$$

For the limit to exist, power of x in the numerator should be greater than or equal to the power of x in the denominator. Therefore, coefficient of x in numerator is equal to zero

$$ \Rightarrow $$ 3 $$-$$ k = 0

$$ \Rightarrow $$ k = 3

So the limit reduces to

$$\mathop {\lim }\limits_{x \to 0} {{\left( {{x^2}} \right)\left( {{4 \over {2!}} + {{8x} \over {3!}} + ...} \right)} \over {\left( {{x^2}} \right)\left( {2 + {{4x} \over {2!}} + {{8{x^2}} \over {3!}} + ...} \right)}}$$

= $$\mathop {\lim }\limits_{x \to 0} {{{4 \over {2!}} + {{8x} \over {3!}} + ...} \over {2 + {{4x} \over {2!}} + {{8{x^2}} \over {3!}} + ...}}$$ = 1

Hence, f(0) = 1

$$\mathop {\lim }\limits_{x \to 0} $$ f(x) will exist and f(0) = $$\mathop {\lim }\limits_{x \to 0} $$ f(x)

Now, $$\mathop {\lim }\limits_{x \to 0} $$ f(x) = $$\mathop {\lim }\limits_{x \to 0} \left( {{1 \over x} - {{k - 1} \over {{e^{2x}} - 1}}} \right)$$

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

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

= $$\mathop {\lim }\limits_{x \to 0} \left[ {{{\left( {3 - k} \right)x + {{4{x^2}} \over {2!}} + {{8{x^3}} \over {3!}} + ...} \over {\left( {2{x^2} + {{4{x^3}} \over {2!}} + {{8{x^3}} \over {3!}} + ....} \right)}}} \right]$$

For the limit to exist, power of x in the numerator should be greater than or equal to the power of x in the denominator. Therefore, coefficient of x in numerator is equal to zero

$$ \Rightarrow $$ 3 $$-$$ k = 0

$$ \Rightarrow $$ k = 3

So the limit reduces to

$$\mathop {\lim }\limits_{x \to 0} {{\left( {{x^2}} \right)\left( {{4 \over {2!}} + {{8x} \over {3!}} + ...} \right)} \over {\left( {{x^2}} \right)\left( {2 + {{4x} \over {2!}} + {{8{x^2}} \over {3!}} + ...} \right)}}$$

= $$\mathop {\lim }\limits_{x \to 0} {{{4 \over {2!}} + {{8x} \over {3!}} + ...} \over {2 + {{4x} \over {2!}} + {{8{x^2}} \over {3!}} + ...}}$$ = 1

Hence, f(0) = 1

2

Let f : R $$ \to $$ R be a function defined as

$$f(x) = \left\{ {\matrix{ 5 & ; & {x \le 1} \cr {a + bx} & ; & {1 < x < 3} \cr {b + 5x} & ; & {3 \le x < 5} \cr {30} & ; & {x \ge 5} \cr } } \right.$$

Then, f is

$$f(x) = \left\{ {\matrix{ 5 & ; & {x \le 1} \cr {a + bx} & ; & {1 < x < 3} \cr {b + 5x} & ; & {3 \le x < 5} \cr {30} & ; & {x \ge 5} \cr } } \right.$$

Then, f is

A

continuous if a = 0 and b = 5

B

continuous if a = –5 and b = 10

C

continuous if a = 5 and b = 5

D

not continuous for any values of a and b

Checking

if f(x) is continuous at x = 1 :

f(1^{$$-$$}) = 5

f(1) = 5

f(1^{+}) = a + b

if f(x) is continuous at x = 1,

then

f(1^{$$-$$}) = f(1) = f(1^{+})

$$ \Rightarrow $$ 5 = 5 = a + b

$$ \therefore $$ a + b = 5 . . . . . . . . (1)

checking if f(x) is continuous at x = 3 :

f(3^{$$-$$}) = a + 3b

f(3) = b + 15

f(3^{+}) = b + 15

if f(x) = is continuous at x = 3

then,

f(3^{$$-$$}) = f(3) = f(3^{+})

$$ \Rightarrow $$ a + 3b = b + 15 = b + 15

$$ \Rightarrow $$ a + 2b = 15 . . . . . (2)

checking if f(x) is continuous at x = 5 :

f(5^{$$-$$}) = b + 25

f(5) = 30

f(5^{+}) = 30

if f(x) is continuous at x = 5 then,

f(5^{$$-$$}) = f(5) = f(5^{+})

$$ \Rightarrow $$ b + 25 = 30 = 30

$$ \Rightarrow $$ b = 5

By putting this value in equation (2), we get,

a + 2(5) = 15

$$ \Rightarrow $$ a = 5

when a = 5 and b = 5 then equation (1)

a + b = 5 does not satisfy.

$$ \therefore $$ f is not continuous for any value of a and b.

if f(x) is continuous at x = 1 :

f(1

f(1) = 5

f(1

if f(x) is continuous at x = 1,

then

f(1

$$ \Rightarrow $$ 5 = 5 = a + b

$$ \therefore $$ a + b = 5 . . . . . . . . (1)

checking if f(x) is continuous at x = 3 :

f(3

f(3) = b + 15

f(3

if f(x) = is continuous at x = 3

then,

f(3

$$ \Rightarrow $$ a + 3b = b + 15 = b + 15

$$ \Rightarrow $$ a + 2b = 15 . . . . . (2)

checking if f(x) is continuous at x = 5 :

f(5

f(5) = 30

f(5

if f(x) is continuous at x = 5 then,

f(5

$$ \Rightarrow $$ b + 25 = 30 = 30

$$ \Rightarrow $$ b = 5

By putting this value in equation (2), we get,

a + 2(5) = 15

$$ \Rightarrow $$ a = 5

when a = 5 and b = 5 then equation (1)

a + b = 5 does not satisfy.

$$ \therefore $$ f is not continuous for any value of a and b.

3

$$\mathop {\lim }\limits_{y \to 0} {{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$$

A

exists and equals $${1 \over {2\sqrt 2 }}$$

B

exists and equals $${1 \over {4\sqrt 2 }}$$

C

exists and equals $${1 \over {2\sqrt 2 (1 + \sqrt {2)} }}$$

D

does not exists

$$\mathop {\lim }\limits_{y \to 0} {{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$$

If you put y = 0 at $${{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$$ it is in $${0 \over 0}$$ form. So we can use L' Hospital's Rule.

= $$\mathop {\lim }\limits_{y \to 0} {{{1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times \left( {{1 \over {2\sqrt {1 + {y^4}} }}} \right) \times 4{y^3}} \over {4{y^3}}}$$

= $$\mathop {\lim }\limits_{y \to 0} {1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times {1 \over {2\sqrt {1 + {y^4}} }}$$

= $${1 \over {4\sqrt 2 }}$$

If you put y = 0 at $${{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$$ it is in $${0 \over 0}$$ form. So we can use L' Hospital's Rule.

= $$\mathop {\lim }\limits_{y \to 0} {{{1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times \left( {{1 \over {2\sqrt {1 + {y^4}} }}} \right) \times 4{y^3}} \over {4{y^3}}}$$

= $$\mathop {\lim }\limits_{y \to 0} {1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times {1 \over {2\sqrt {1 + {y^4}} }}$$

= $${1 \over {4\sqrt 2 }}$$

4

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

Then $$\mathop {\lim }\limits_{x \to {0^ - }} \,\,{{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]} \over {\left| x \right|}}$$ is equal to :

Then $$\mathop {\lim }\limits_{x \to {0^ - }} \,\,{{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]} \over {\left| x \right|}}$$ is equal to :

A

$$-$$ sin 1

B

1

C

sin 1

D

0

$$\mathop {\lim }\limits_{x \to {0^ - }} {{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]} \over {\left| x \right|}}$$

$$ = \mathop {\lim }\limits_{h \to 0} {{\left( {0 - h} \right)\left( {\left[ {0 - h} \right] + \left| {0 - h} \right|} \right)\sin \left[ {0 - h} \right]} \over {\left| {0 - h} \right|}}$$

$$ = \mathop {\lim }\limits_{h \to 0} {{ - h\left( { - 1 + h} \right)\sin \left( { - 1} \right)} \over h}$$

$$ = \mathop {\lim }\limits_{h \to 0} \left( { - 1 + h} \right)\sin \left( 1 \right) = - \sin 1$$

$$ = \mathop {\lim }\limits_{h \to 0} {{\left( {0 - h} \right)\left( {\left[ {0 - h} \right] + \left| {0 - h} \right|} \right)\sin \left[ {0 - h} \right]} \over {\left| {0 - h} \right|}}$$

$$ = \mathop {\lim }\limits_{h \to 0} {{ - h\left( { - 1 + h} \right)\sin \left( { - 1} \right)} \over h}$$

$$ = \mathop {\lim }\limits_{h \to 0} \left( { - 1 + h} \right)\sin \left( 1 \right) = - \sin 1$$

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