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

Irrigation

Geomatics Engineering Or Surveying

Environmental Engineering

Transportation Engineering

Engineering Mathematics

General Aptitude

1

Let P = {$$\theta $$ : sin$$\theta $$ $$-$$ cos$$\theta $$ = $$\sqrt 2 \,\cos \theta $$}

and Q = {$$\theta $$ : sin$$\theta $$ + cos$$\theta $$ = $$\sqrt 2 \,\sin \theta $$} be two sets. Then

and Q = {$$\theta $$ : sin$$\theta $$ + cos$$\theta $$ = $$\sqrt 2 \,\sin \theta $$} be two sets. Then

A

P $$ \subset $$ Q and Q $$-$$ P $$ \ne $$ $$\phi $$

B

Q $$ \not\subset $$ P

C

P $$ \not\subset $$ Q

D

P = Q

Given,

sin$$\theta $$ $$-$$ cos$$\theta $$ = $$\sqrt 2 $$cos$$\theta $$

$$ \Rightarrow $$ sin$$\theta $$ = $$\left( {\sqrt 2 + 1} \right)$$cos$$\theta $$

$$ \Rightarrow $$ sin$$\theta $$ = $${{2 - 1} \over {\sqrt 2 - 1}}$$cos$$\theta $$

$$ \Rightarrow $$ $$\left( {\sqrt 2 - 1} \right)$$sin$$\theta $$ = cos$$\theta $$ . . . (1)

This is for set P.

sin$$\theta $$ + cos$$\theta $$ = $${\sqrt 2 }$$sin$$\theta $$

$$ \Rightarrow $$ cos$$\theta $$ = $$\left( {\sqrt 2 - 1} \right)$$sin$$\theta $$ . . . .(2)

This is from set Q.

So, P = Q

sin$$\theta $$ $$-$$ cos$$\theta $$ = $$\sqrt 2 $$cos$$\theta $$

$$ \Rightarrow $$ sin$$\theta $$ = $$\left( {\sqrt 2 + 1} \right)$$cos$$\theta $$

$$ \Rightarrow $$ sin$$\theta $$ = $${{2 - 1} \over {\sqrt 2 - 1}}$$cos$$\theta $$

$$ \Rightarrow $$ $$\left( {\sqrt 2 - 1} \right)$$sin$$\theta $$ = cos$$\theta $$ . . . (1)

This is for set P.

sin$$\theta $$ + cos$$\theta $$ = $${\sqrt 2 }$$sin$$\theta $$

$$ \Rightarrow $$ cos$$\theta $$ = $$\left( {\sqrt 2 - 1} \right)$$sin$$\theta $$ . . . .(2)

This is from set Q.

So, P = Q

2

The function $$f:R \to \left[ { - {1 \over 2},{1 \over 2}} \right]$$ defined as

$$f\left( x \right) = {x \over {1 + {x^2}}}$$, is

$$f\left( x \right) = {x \over {1 + {x^2}}}$$, is

A

invertible

B

injective but not surjective.

C

surjective but not injective

D

neither injective nor surjective.

$$f\left( x \right) = {x \over {1 + {x^2}}}$$

$$ \therefore $$ $$f\left( {{1 \over x}} \right) = {{{1 \over x}} \over {1 + {1 \over {{x^2}}}}} = {x \over {1 + {x^2}}} = f\left( x \right)$$

$$ \therefore $$ f(x) is many-one function.

Now let y = f(x) = $${x \over {1 + {x^2}}}$$

$$ \Rightarrow $$ y + x^{2}y = x

$$ \Rightarrow $$ yx - x + y = 0

As x $$ \in $$ R

$$ \therefore $$ (-1)^{2} - 4(y)(y) $$ \ge $$ 0

$$ \Rightarrow $$ 1 - 4y^{2} $$ \ge $$ 0

$$ \Rightarrow $$ y $$ \in $$ $$\left[ { - {1 \over 2},{1 \over 2}} \right]$$

$$ \therefore $$ Range = Codomain = $$\left[ { - {1 \over 2},{1 \over 2}} \right]$$

So, f(x) is surjective.

$$ \therefore $$ f(x) is surjective but not injective

$$ \therefore $$ $$f\left( {{1 \over x}} \right) = {{{1 \over x}} \over {1 + {1 \over {{x^2}}}}} = {x \over {1 + {x^2}}} = f\left( x \right)$$

$$ \therefore $$ f(x) is many-one function.

Now let y = f(x) = $${x \over {1 + {x^2}}}$$

$$ \Rightarrow $$ y + x

$$ \Rightarrow $$ yx - x + y = 0

As x $$ \in $$ R

$$ \therefore $$ (-1)

$$ \Rightarrow $$ 1 - 4y

$$ \Rightarrow $$ y $$ \in $$ $$\left[ { - {1 \over 2},{1 \over 2}} \right]$$

$$ \therefore $$ Range = Codomain = $$\left[ { - {1 \over 2},{1 \over 2}} \right]$$

So, f(x) is surjective.

$$ \therefore $$ f(x) is surjective but not injective

3

Let f(x) = 2^{10}.x + 1 and g(x)=3^{10}.x $$-$$ 1. If (fog) (x) = x, then x is equal to :

A

$${{{3^{10}} - 1} \over {{3^{10}} - {2^{ - 10}}}}$$

B

$${{{2^{10}} - 1} \over {{2^{10}} - {3^{ - 10}}}}$$

C

$${{1 - {3^{ - 10}}} \over {{2^{10}} - {3^{ - 10}}}}$$

D

$${{1 - {2^{ - 10}}} \over {{3^{10}} - {2^{ - 10}}}}$$

(fog) (x) = x

$$ \Rightarrow $$$$\,\,\,$$ f (g(x)) = x

$$ \Rightarrow $$$$\,\,\,$$ f (3^{10}. x $$-$$ 1) = x [ as g(x) = 3^{10}. x $$-$$ 1]

$$ \Rightarrow $$$$\,\,\,$$ 2^{10} . (3^{10} . x $$-$$ 1) + 1 = x

$$ \Rightarrow $$$$\,\,\,$$ 3^{10} . x $$-$$ 1 + 2^{$$-$$10} = x . 2^{$$-$$10} [dividing by 2^{10}]

$$ \Rightarrow $$$$\,\,\,$$3^{10} . x $$-$$ 2^{$$-$$10} . x = 1 $$-$$ 2^{$$-$$10}

$$ \Rightarrow $$$$\,\,\,$$ x (3^{10} $$-$$ 2^{$$-$$ 10}) = 1$$-$$ 2^{$$-$$10}

$$ \Rightarrow $$ $$\,\,\,$$ x = $${{1 - {2^{ - 10}}} \over {{3^{10}} - {2^{ - 10}}}}$$

$$ \Rightarrow $$$$\,\,\,$$ f (g(x)) = x

$$ \Rightarrow $$$$\,\,\,$$ f (3

$$ \Rightarrow $$$$\,\,\,$$ 2

$$ \Rightarrow $$$$\,\,\,$$ 3

$$ \Rightarrow $$$$\,\,\,$$3

$$ \Rightarrow $$$$\,\,\,$$ x (3

$$ \Rightarrow $$ $$\,\,\,$$ x = $${{1 - {2^{ - 10}}} \over {{3^{10}} - {2^{ - 10}}}}$$

4

The function f : **N** $$ \to $$ **N** defined by f (x) = x $$-$$ 5 $$\left[ {{x \over 5}} \right],$$ Where **N** is the set of natural numbers and [x] denotes the greatest integer less than or equal to x, is :

A

one-one and onto

B

one-one but not onto.

C

onto but not one-one.

D

neither one-one nor onto.

f(1) = 1 - 5$$\left[ {{1 \over 5}} \right]$$ = 1

f(6) = 6 - 5$$\left[ {{6 \over 5}} \right]$$ = 1

So, this function is many to one.

f(10) = 10 - 5$$\left[ {{10 \over 5}} \right]$$ = 0 which is not present in the set of natural numbers.

So this function is neither one-one nor onto.

f(6) = 6 - 5$$\left[ {{6 \over 5}} \right]$$ = 1

So, this function is many to one.

f(10) = 10 - 5$$\left[ {{10 \over 5}} \right]$$ = 0 which is not present in the set of natural numbers.

So this function is neither one-one nor onto.

Number in Brackets after Paper Name Indicates No of Questions

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