1

### JEE Main 2018 (Online) 15th April Evening Slot

Let f : A $\to$ B be a function defined as f(x) = ${{x - 1} \over {x - 2}},$ Where A = R $-$ {2} and B = R $-$ {1}. Then   f   is :
A
invertible and ${f^{ - 1}}(y) =$ ${{3y - 1} \over {y - 1}}$
B
invertible and ${f^{ - 1}}\left( y \right) = {{2y - 1} \over {y - 1}}$
C
invertible and ${f^{ - 1}}\left( y \right) = {{2y + 1} \over {y - 1}}$
D
not invertible

## Explanation

Assume,

y = f(x)

$\Rightarrow$ y = ${{x - 1} \over {x - 2}}$

$\Rightarrow$ yx - 2y = x - 1

$\Rightarrow$ (y - 1)x = 2y - 1

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

As on the given domain the function is invertible and its inverse can be computed as shown above.
2

### JEE Main 2018 (Online) 16th April Morning Slot

Let N denote the set of all natural numbers. Define two binary relations on N as R = {(x, y) $\in$ N $\times$ N : 2x + y = 10} and R2 = {(x, y) $\in$ N $\times$ N : x + 2y = 10}. Then :
A
Range of R1 is {2, 4, 8).
B
Range of R2 is {1, 2, 3, 4}.
C
Both R1 and R2 are symmetric relations.
D
Both R1 and R2 are transitive relations.

## Explanation

For R1; 2x + y = 10 and x, y $\in$ N possible values for x and y are :

x = 1, y = 8    i.e.   (1, 8);

x = 2, y = 6    i.e    (2, 6);

x = 3, y = 4    i.e    (3, 4);

x = 4, y = 2    i.e    (4, 2)

$\therefore\,\,\,$ R1 = { (1, 8), (2, 6), (3, 4), (4, 2) }

$\therefore\,\,\,$ Range of R1 is {2, 4, 6, 8}

R1 is not symmetric.

R1 is not transitive also as

(3, 4), (4, 2) $\in$ R , but (3, 2) $\notin$ R1

For R2 : x + 2y = 10 and x, y $\in$ N

Possible values of x, and y are :

x = 8, y= 1    i.e    (8, 1)

x = 6, y = 2    i.e    (6, 2)

x = 4, y = 3    i.e    (4, 3) and

x = 2, y = 4    i.e    (2, 4)

$\therefore\,\,\,$ R2 = {(8, 1) (6, 2) (4, 3) (2, 4)}

$\therefore\,\,\,$ Range of R2 = $\left\{ {1,2,3,4} \right\}$

R2 is not symmetric and transitive
3

### 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) = ${{{1 \over x}} \over {{1 \over x} - 1}}$

= ${1 \over {1 - x}} = {f_3}\left( x \right)$
4

### 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)