1

### JEE Main 2019 (Online) 10th January Evening Slot

Let N be the set of natural numbers and two functions f and g be defined as f, g : N $\to$ N such that

f(n) = $\left\{ {\matrix{ {{{n + 1} \over 2};} & {if\,\,n\,\,is\,\,odd} \cr {{n \over 2};} & {if\,\,n\,\,is\,\,even} \cr } \,\,} \right.$;

and g(n) = n $-$($-$ 1)n.

Then fog is -
A
neither one-one nor onto
B
onto but not one-one
C
both one-one and onto
D
one-one but not onto

## Explanation

f(x) = $\left\{ {\matrix{ {{{n + 1} \over 2};} & {if\,\,n\,\,is\,\,odd} \cr {{n \over 2};} & {if\,\,n\,\,is\,\,even} \cr } \,\,} \right.$;

g(x) = n $-$ ($-$ 1)n $\left\{ {\matrix{ {n + 1;\,\,\,\,n\,\,is\,\,odd} \cr {n - 1;\,\,\,\,n\,\,is\,\,even} \cr } } \right.$

f(g(n)) = $\left\{ {\matrix{ {{n \over 2};\,\,\,\,n\,\,is\,\,even} \cr {{{n + 1} \over 2};\,\,\,\,n\,\,is\,\,odd} \cr } } \right.$

$\therefore$  many one but onto
2

### 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,  $f(x) \in \left[ { - {1 \over 2},{1 \over 2}} \right]$
3

### 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) = ${1 \over 12}$
4

### 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.