1

### JEE Main 2016 (Online) 10th April Morning Slot

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
A
P $\subset$ Q and Q $-$ P $\ne$ $\phi$
B
Q $\not\subset$ P
C
P $\not\subset$ Q
D
P = Q

## Explanation

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
2

### JEE Main 2017 (Offline)

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
A
invertible
B
injective but not surjective.
C
surjective but not injective
D
neither injective nor surjective.

## Explanation

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

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

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

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

$\Rightarrow$ y + x2y = x

$\Rightarrow$ yx - x + y = 0

As x $\in$ R

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

$\Rightarrow$ 1 - 4y2 $\ge$ 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
3

### 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 = ${{1 - {2^{ - 10}}} \over {{3^{10}} - {2^{ - 10}}}}$
4

### 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.