1

### JEE Main 2019 (Online) 9th January Morning Slot

For $x \in R - \left\{ {0,1} \right\}$, Let f1(x) = $1\over x$, f2 (x) = 1 – x

and f3 (x) = $1 \over {1 - x}$ be three given

functions. If a function, J(x) satisfies

(f2 o J o f1) (x) = f3 (x) then J(x) is equal to :
A
f1 (x)
B
$1 \over x$ f3 (x)
C
f2 (x)
D
f3 (x)

## Explanation

Given,

f1(x) = ${1 \over x}$

f2(x) = 1 $-$ x

f3(x) = ${1 \over {1 - x}}$

(f2 $\cdot$ J $\cdot$ f1) (x) = f3(x)

$\Rightarrow$   f2 {J(f1(x))} = f3(x)

$\Rightarrow$   f2{J (${1 \over x}$)} = ${1 \over {1 - x}}$

$\Rightarrow$   1 $-$ J(${1 \over x}$) = ${1 \over {1 - x}}$

$\Rightarrow$  J(${{1 \over x}}$) = 1 $-$ ${{1 \over {1 - x}}}$

$\Rightarrow$   J (${{1 \over { x}}}$) = ${{ - x} \over {1 - x}}$ = ${x \over {x - 1}}$

Put x inplace of ${1 \over x}$

$\therefore$   J(x) = ${{{1 \over x}} \over {{1 \over x} - 1}}$

= ${1 \over {1 - x}} = {f_3}\left( x \right)$
2

### JEE Main 2019 (Online) 9th January Evening Slot

Let A = {x$\in$R : x is not a positive integer}.

Define a function $f$ : A $\to$  R   as  $f(x)$ ${{2x} \over {x - 1}}$,

then $f$ is :
A
not injective
B
neither injective nor surjective
C
surjective but not injective
D
injective but not surjective

## Explanation

f(x) = ${{2x} \over {x - 1}}$

f(x) = 2 + ${2 \over {x - 1}}$

f'(x) = $-$ ${2 \over {{{\left( {x - 1} \right)}^2}}}$ < 0 $\forall$ x $\in$ R

Hence f(x) is strictly decreasing

So, f(x) is one-one

Range : Let y = ${{2x} \over {x - 1}}$

xy $-$ y = 2x

$\Rightarrow$  x(y $-$ 2) = y

$\Rightarrow$  x = ${y \over {y - 2}}$

given that x $\in$ R : x is not a +ve integer

$\therefore$  ${y \over {y - 2}} \ne$ N    (N $\to$ Natural number)

$\Rightarrow$  y $\ne$ Ny $-$ 2N

$\Rightarrow$  y $\ne$ ${{2N} \over {N - 1}}$

So range $\notin$ R (in to function)
3

### 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
4

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