#### Please Solve RD Sharma Class 12 Chapter Function Exercise 2.4 Question 9 Maths Textbook Solution.

$f$is invertible with the function.

$f^{-1}\left ( x \right )=\frac{\sqrt{x-6}-1}{3}$

Given:

$f:R_{+}\rightarrow \left [ 5,\infty \right ]$ given by $f\left ( x \right )=9x^{2}+6x-5$

Hint:

Bijection function should fulfil the injection and surjection condition.

Solution:

Let  $x,y$ be two elements of domain$\left ( R^{+} \right )$, such that$f\left ( x \right )=f\left ( y \right )$.

$\! \! \! \! \! \! \! \! 9x^{2}+6x-5=9y^{2}+6y-5\\\\9x^{2}+6x=9y^{2}+6y\\\\x=y\left ( as,x,y\: \epsilon \: R^{+} \right )$

$f$is one-one.

Subjectivity of $f$:

Let $y$ is in the co-domain$\left ( Q \right )$ such that $f\left ( x \right )=y$

$\! \! \! \! \! \! \! \! \! 9x^{2}+6x-5=y\\\\9x^{2}+6x=y+5$

$9x^{2}+6x+1=y+6$                                                                                 [Adding $1$ on both sides]
$\left ( 3x+1 \right )^{2}=6+y\\\\3x+1=\sqrt{y+6}\\\\3x=\sqrt{y+6}-1\\\\x=\frac{\sqrt{y+6}1}{3}\: \epsilon \: R^{+}\left ( domain \right )$

$f$ is onto.

So, $f$ is a bijection and hence, it is invertible.

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

$\! \! \! \! \! \! \! \! x=f\left ( y \right )\\\\x=9y^{2}+6y-5\\\\x+5=9y^{2}+6y$

$x+6=\left ( 3y+1 \right )^{2}$                                                                                 [Adding  on both sides]

$\! \! \! \! \! \! \! \! \! 3y+1=\sqrt{x+6}\\\\3y=\sqrt{x+6}-1\\\\y=\frac{\sqrt{x+6}-1}{3}$

$f^{-1}\left ( x \right )=\frac{\sqrt{x+6}-1}{3}$                                                                                                         [From (i)]