1
MCQ (Single Correct Answer)

JEE Main 2017 (Online) 8th April Morning Slot

Let f(x) = 210.x + 1 and g(x)=310.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
$${{1 - {2^{ - 10}}} \over {{3^{10}} - {2^{ - 10}}}}$$

Explanation

(fog) (x)   =   x

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

$$ \Rightarrow $$$$\,\,\,$$ f (310. x $$-$$ 1)   =   x    [ as    g(x) = 310. x $$-$$ 1]

$$ \Rightarrow $$$$\,\,\,$$ 210 . (310 . x $$-$$ 1) + 1   =   x

$$ \Rightarrow $$$$\,\,\,$$ 310 . x $$-$$ 1 + 2$$-$$10   =   x . 2$$-$$10   [dividing by 210]

$$ \Rightarrow $$$$\,\,\,$$310 . x $$-$$ 2$$-$$10 . x   =   1 $$-$$ 2$$-$$10

$$ \Rightarrow $$$$\,\,\,$$ x (310 $$-$$ 2$$-$$ 10)   =   1$$-$$ 2$$-$$10

$$ \Rightarrow $$ $$\,\,\,$$ x = $${{1 - {2^{ - 10}}} \over {{3^{10}} - {2^{ - 10}}}}$$
2
MCQ (Single Correct Answer)

JEE Main 2017 (Online) 9th April Morning Slot

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.

Explanation

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.
3
MCQ (Single Correct Answer)

JEE Main 2018 (Offline)

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
neither A $$ \subset $$ B nor B $$ \subset $$ A
B
B $$ \subset $$ A
C
A $$ \subset $$ B
D
A $$ \cap $$ B = $$\phi $$ ( an empty set )

Explanation

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
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 15th April Morning Slot

Consider the following two binary relations on the set A = {a, b, c} :
R1 = {(c, a), (b, b), (a, c), (c, c), (b, c), (a, a)} and
R2 = {(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c)}.
Then :
A
both R1 and R2 are not symmetric.
B
R1 is not symmetric but it is transitive.
C
R2 is symmetric but it is not transitive.
D
both R1 and R2 are transitive.

Explanation

Here both R1 and R2 are symmetric as for any (x, y) $$ \in $$ R1, we have (y, x) $$ \in $$ R1 and similarly for any (x, y) $$ \in $$ R2, we have (y, x) $$ \in $$ R2

In R1, (b, c) $$ \in $$ R1, (c, a) $$ \in $$ R1 but (b,a) $$ \notin $$ R1

Similarly in R2, (b, a) $$ \in $$ R2, (a, c) $$ \in $$ R2 but (b, c) $$ \notin $$ R2

$$ \therefore $$ R1 and R2 are not transitive.

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