Please Solve RD Sharma Class 12 Chapter Function Exercise 2.4 Question 19 Maths Textbook Solution.

$s=\left \{ 0,1 \right \}$

Given:

$f\left ( x \right )=\left ( x+1 \right )^{2}-1,x\geq -1$

Hint:

Bijection function should fulfil the injection and surjection condition.

Solution:

Injectivity: let $x$ and $y\: \epsilon \: \left [ -1,\infty \right )$, such that

$f\left ( x \right )=f\left ( y \right )$

$\left ( x+1 \right )^{2}-1=\left ( y+1 \right )^{2}-1$

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

$x+1 = y+1$

$x = y$

So, $f$is an injection.

Surjectivity: let$y\: \epsilon \: \left [ -1,\infty \right )$ , then

$f\left ( x \right )=y$

$\left ( x+1 \right )^{2}-1=y$

$x+1 =\sqrt{y+1}$

$x =\sqrt{y+1}-1$

$x =\sqrt{y+1}-1$  is real for all $y\geq -1$

$\Rightarrow f$ is a surjection.

$f$ is a bijection. Hence, $f$ is invertible.

Let

$f^{-1}\left ( x \right )= y$

$f\left ( y \right )= x$

$\left ( y+1 \right )^{2}-1= x$

$y\pm \sqrt{x+1}-1= f^{-1}\left ( x \right )$

$f^{-1}\left ( x \right )=\pm \sqrt{x+1}-1$

$f\left ( x \right )=f^{-1}\left ( x \right )$

$\left (x +1 \right )^{2}-1\pm \sqrt{x+1}-1$

$\left (x +1 \right )^{2}\pm \sqrt{x+1}$

Squaring on both sides

$\left ( x+1 \right )^{4}=x+1$

$\left ( x+1 \right )\left [ \left ( x+1 \right ) ^{3}-1\right ]=0$

$\left ( x+1 \right )=0\: or\: \left ( x+1 \right ) ^{3}-1=0$

$x=-1\: or\: \left ( x+1 \right ) ^{3}=1$

$\left ( x+1 \right ) ^{3}=1$

$x+1 =1$

$x=0$

$x=-1 \: or\: 0$ or

$s=\left \{ 0,1 \right \}$