1

### JEE Main 2019 (Online) 11th January Morning Slot

Let f : R $\to$ R be defined by f(x) = ${x \over {1 + {x^2}}},x \in R$.   Then the range of f is :
A
$\left[ { - {1 \over 2},{1 \over 2}} \right]$
B
$R - \left[ { - {1 \over 2},{1 \over 2}} \right]$
C
($-$ 1, 1) $-$ {0}
D
R $-$ [$-$1, 1]

## Explanation

f(0) = 0 & f(x) is odd

Further, if x > 0 then

f(x) = $f(x) = {1 \over {x + {1 \over x}}} \in \left( {0,{1 \over 2}} \right]$

Hence,  $f(x) \in \left[ { - {1 \over 2},{1 \over 2}} \right]$
2

### JEE Main 2019 (Online) 11th January Morning Slot

Let fk(x) = ${1 \over k}\left( {{{\sin }^k}x + {{\cos }^k}x} \right)$ for k = 1, 2, 3, ... Then for all x $\in$ R, the value of f4(x) $-$ f6(x) is equal to
A
${1 \over 4}$
B
${5 \over {12}}$
C
${{ - 1} \over {12}}$
D
${1 \over {12}}$

## Explanation

f4(x) $-$ f6(x)

= ${1 \over 4}$ (sin4 x + cos4 x) $-$ ${1 \over 6}$ (sin6 x + cos6 x)

= ${1 \over 4}$ (1$-$ ${1 \over 2}$ sin2 2x) $-$ ${1 \over 6}$ (1 $-$ ${3 \over 4}$ sin2 2x) = ${1 \over 12}$
3

### JEE Main 2019 (Online) 11th January Evening Slot

Let a function f : (0, $\infty$) $\to$ (0, $\infty$) be defined by f(x) = $\left| {1 - {1 \over x}} \right|$. Then f is :
A
not injective but it is surjective
B
neiter injective nor surjective
C
injective only
D
both injective as well as surjective

## Explanation

$f\left( x \right) = \left| {1 - {1 \over x}} \right| = {{\left| {x - 1} \right|} \over x} = \left\{ {\matrix{ {{{1 - x} \over x}} & {0 < x \le 1} \cr {{{x - 1} \over x}} & {x \ge 1} \cr } } \right.$ $\Rightarrow$  f(x) is not injective

but range of function is $\left[ {0,\infty } \right)$

Remarks :  If co-domain is $\left[ {0,\infty } \right)$, then f(x) will be surjective.
4

### JEE Main 2019 (Online) 12th January Morning Slot

Let S = {1, 2, 3, … , 100}. The number of non-empty subsets A of S such that the product of elements in A is even is :
A
250 – 1
B
250 (250 $-$ 1)
C
2100 $-$ 1
D
250 + 1

## Explanation

S = {1,2,3, . . . .100}

= Total non empty subsets-subsets with product of element is odd

= 2100 $-$ 1 $-$ 1[(250 $-$ 1)]

= 2100 $-$ 250

= 250 (250 $-$ 1)