Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

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Reinforced Cement Concrete

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Engineering Mathematics

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General Aptitude

1

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

2

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.

3

Two sets A and B are as under :

A = {($$a$$, b) $$ \in $$ R $$ \times $$ R : |$$a$$ - 5| < 1 and |b - 5| < 1};

B = {($$a$$, b) $$ \in $$ R $$ \times $$ R : 4($$a$$ - 6)^{2} + 9(b - 5)^{2} $$ \le $$ 36 };

Then

A = {($$a$$, b) $$ \in $$ R $$ \times $$ R : |$$a$$ - 5| < 1 and |b - 5| < 1};

B = {($$a$$, b) $$ \in $$ R $$ \times $$ R : 4($$a$$ - 6)

Then

A

neither A $$ \subset $$ B nor B $$ \subset $$ A

B

B $$ \subset $$ A

C

A $$ \subset $$ B

D

A $$ \cap $$ B = $$\phi $$ ( an empty set )

Given,

$$4{\left( {a - 6} \right)^2} + 9{\left( {b - 5} \right)^2} \le 36$$

Let $$a - 6 = x$$ and $$b - 5 = y$$

$$\therefore\,\,\,\,$$ $$4{x^2} + 9{y^2} \le 36$$

$$ \Rightarrow \,\,\,\,{{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1$$

This is a equation of ellipse.

This ellipse will look like this,

According to set A,

$$\left| {a - 5} \right| < 1$$

as $$a - 6 = x$$ then $$a - 5 = x + 1$$

$$\therefore\,\,\,$$ $$\left| {x + 1} \right| < 1$$

$$ \Rightarrow \,\,\, - 1 < x + 1 < 1$$

$$ \Rightarrow \,\,\, - 2 < x < 0$$

$$\left| {b - 5} \right| < 1$$

as $$b - 5 = y$$

$$\therefore\,\,\,$$ $$\left| y \right| < 1$$

$$ \Rightarrow \,\,\,\, - 1 < y < 1$$

This will look like this,

By combining both those graphs it will look like this,

To check entire set A is inside of B or not put any paint of set A on the equation $${{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1$$ and if this inequality satisfies, then it is inside B.

By looking at the graph you can surely say (0, 1) or (0, $$-$$1) inside the graph. But to check point ($$-2$$, 1) or ($$-$$2, $$-$$1) is inside of the ellipse or not, put on the $${{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1.$$

Putting ($$-$$2, 1) on the inequality,

LHS $$ = {{\left( { - 2} \right){}^2} \over 9} + {{{1^2}} \over 4}$$

$$ = {4 \over 9} + {1 \over 4}$$

$$ = {{25} \over {36}} < 1$$

$$\therefore\,\,\,$$ Inequality holds.

So, $$\left( { - 2,1} \right)$$ is inside the ellipse.

Similarly by checking we can see $$\left( { - 2, - 1} \right)$$ is also inside the ellipse.

Hence, we can say entire set A is inside of the set B.

$$\therefore\,\,\,\,$$ A $$ \subset $$ B

$$4{\left( {a - 6} \right)^2} + 9{\left( {b - 5} \right)^2} \le 36$$

Let $$a - 6 = x$$ and $$b - 5 = y$$

$$\therefore\,\,\,\,$$ $$4{x^2} + 9{y^2} \le 36$$

$$ \Rightarrow \,\,\,\,{{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1$$

This is a equation of ellipse.

This ellipse will look like this,

According to set A,

$$\left| {a - 5} \right| < 1$$

as $$a - 6 = x$$ then $$a - 5 = x + 1$$

$$\therefore\,\,\,$$ $$\left| {x + 1} \right| < 1$$

$$ \Rightarrow \,\,\, - 1 < x + 1 < 1$$

$$ \Rightarrow \,\,\, - 2 < x < 0$$

$$\left| {b - 5} \right| < 1$$

as $$b - 5 = y$$

$$\therefore\,\,\,$$ $$\left| y \right| < 1$$

$$ \Rightarrow \,\,\,\, - 1 < y < 1$$

This will look like this,

By combining both those graphs it will look like this,

To check entire set A is inside of B or not put any paint of set A on the equation $${{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1$$ and if this inequality satisfies, then it is inside B.

By looking at the graph you can surely say (0, 1) or (0, $$-$$1) inside the graph. But to check point ($$-2$$, 1) or ($$-$$2, $$-$$1) is inside of the ellipse or not, put on the $${{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1.$$

Putting ($$-$$2, 1) on the inequality,

LHS $$ = {{\left( { - 2} \right){}^2} \over 9} + {{{1^2}} \over 4}$$

$$ = {4 \over 9} + {1 \over 4}$$

$$ = {{25} \over {36}} < 1$$

$$\therefore\,\,\,$$ Inequality holds.

So, $$\left( { - 2,1} \right)$$ is inside the ellipse.

Similarly by checking we can see $$\left( { - 2, - 1} \right)$$ is also inside the ellipse.

Hence, we can say entire set A is inside of the set B.

$$\therefore\,\,\,\,$$ A $$ \subset $$ B

4

Consider the following two binary relations on the set A = {a, b, c} :

R_{1} = {(c, a), (b, b), (a, c), (c, c), (b, c), (a, a)} and

R_{2} = {(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c)}.

Then :

R

R

Then :

A

both R_{1} and R_{2} are not symmetric.

B

R_{1} is not symmetric but it is transitive.

C

R_{2} is symmetric but it is not transitive.

D

both R_{1} and R_{2} are transitive.

Here both R_{1} and R_{2} are symmetric as for any (x, y) $$ \in $$ R_{1}, we have (y, x) $$ \in $$ R_{1} and similarly for any (x, y) $$ \in $$ R_{2}, we have (y, x) $$ \in $$ R_{2}

In R_{1}, (b, c) $$ \in $$ R_{1}, (c, a) $$ \in $$ R_{1} but (b,a) $$ \notin $$ R_{1}

Similarly in R_{2}, (b, a) $$ \in $$ R_{2}, (a, c) $$ \in $$ R_{2} but (b, c) $$ \notin $$ R_{2}

$$ \therefore $$ R_{1} and R_{2} are not transitive.

In R

Similarly in R

$$ \therefore $$ R

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