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

If $$x = \sqrt {{2^{\cos e{c^{ - 1}}}}} $$ and $$y = \sqrt {{2^{se{c^{ - 1}}t}}} \,\,\left( {\left| t \right| \ge 1} \right),$$ then $${{dy} \over {dx}}$$ is equal to :

A

$${y \over x}$$

B

$${x \over y}$$

C

$$-$$ $${y \over x}$$

D

$$-$$ $${x \over y}$$

x = $$\sqrt {{2^{\cos e{c^{ - 1}}t}}} $$

$$\therefore\,\,\,\,$$ $${{dx} \over {dt}}$$ = $${1 \over {2\sqrt {{2^{\cos e{c^{ - 1}}t}}} }}$$ $$ \times $$ ($${2^{\cos e{c^{ - 1}}t}}\,.\,\log 2$$) $$ \times $$ $${{ - 1} \over {t\sqrt {{t^2} - 1} }}$$

$${{dy} \over {dt}}$$ = $${1 \over {2\sqrt {{2^{{{\sec }^{ - 1}}t}}} }}$$ $$ \times $$ $$\left( {{2^{{{\sec }^{ - 1}}t}}\log 2} \right)$$ $$ \times $$ $${1 \over {t\sqrt {{t^2} - 1} }}$$

$$\therefore\,\,\,\,$$ $${{dy} \over {dx}}$$

= $${{{{dy} \over {dt}}} \over {{{dx} \over {dt}}}}$$

= $${{ - \sqrt {{2^{\cos e{c^{ - 1}}t}}} } \over {\sqrt {{2^{\cos e{c^{ - 1}}t}}} }}$$ $$ \times $$ $${{{2^{{{\sec }^{ - 1}}t}}} \over {{2^{\cos e{c^{ - 1}}t}}}}$$

= $$ - \sqrt {{{{2^{{{\sec }^{ - 1}}t}}} \over {{2^{\cos e{c^{ - 1}}t}}}}} $$

= $$-$$ $${y \over x}$$

$$\therefore\,\,\,\,$$ $${{dx} \over {dt}}$$ = $${1 \over {2\sqrt {{2^{\cos e{c^{ - 1}}t}}} }}$$ $$ \times $$ ($${2^{\cos e{c^{ - 1}}t}}\,.\,\log 2$$) $$ \times $$ $${{ - 1} \over {t\sqrt {{t^2} - 1} }}$$

$${{dy} \over {dt}}$$ = $${1 \over {2\sqrt {{2^{{{\sec }^{ - 1}}t}}} }}$$ $$ \times $$ $$\left( {{2^{{{\sec }^{ - 1}}t}}\log 2} \right)$$ $$ \times $$ $${1 \over {t\sqrt {{t^2} - 1} }}$$

$$\therefore\,\,\,\,$$ $${{dy} \over {dx}}$$

= $${{{{dy} \over {dt}}} \over {{{dx} \over {dt}}}}$$

= $${{ - \sqrt {{2^{\cos e{c^{ - 1}}t}}} } \over {\sqrt {{2^{\cos e{c^{ - 1}}t}}} }}$$ $$ \times $$ $${{{2^{{{\sec }^{ - 1}}t}}} \over {{2^{\cos e{c^{ - 1}}t}}}}$$

= $$ - \sqrt {{{{2^{{{\sec }^{ - 1}}t}}} \over {{2^{\cos e{c^{ - 1}}t}}}}} $$

= $$-$$ $${y \over x}$$

2

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

3

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.

4

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

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.

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

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

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

Number in Brackets after Paper Name Indicates No of Questions

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Straight Lines and Pair of Straight Lines *keyboard_arrow_right*

Circle *keyboard_arrow_right*

Conic Sections *keyboard_arrow_right*

Complex Numbers *keyboard_arrow_right*

Quadratic Equation and Inequalities *keyboard_arrow_right*

Permutations and Combinations *keyboard_arrow_right*

Mathematical Induction and Binomial Theorem *keyboard_arrow_right*

Sequences and Series *keyboard_arrow_right*

Matrices and Determinants *keyboard_arrow_right*

Vector Algebra and 3D Geometry *keyboard_arrow_right*

Probability *keyboard_arrow_right*

Statistics *keyboard_arrow_right*

Mathematical Reasoning *keyboard_arrow_right*

Trigonometric Functions & Equations *keyboard_arrow_right*

Properties of Triangle *keyboard_arrow_right*

Inverse Trigonometric Functions *keyboard_arrow_right*

Functions *keyboard_arrow_right*

Limits, Continuity and Differentiability *keyboard_arrow_right*

Differentiation *keyboard_arrow_right*

Application of Derivatives *keyboard_arrow_right*

Indefinite Integrals *keyboard_arrow_right*

Definite Integrals and Applications of Integrals *keyboard_arrow_right*

Differential Equations *keyboard_arrow_right*