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Please Solve RD Sharma Class 12 Chapter Function Exercise 2.4 Question 14 Maths Textbook Solution.

Answers (1)

Answer:

$f^{-1}\left ( y \right )g\left ( y \right )=\frac{\sqrt{25y+5 4}-3}{5}$

Given:

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

We need to prove that $f^{-1}\left ( y \right )=\frac{\sqrt{54+25y}-3}{5}$

Solution:

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

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

Dividing by $5$ ,

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

$= x^{2}+2x\frac{3}{5}+\frac{9}{25}-\frac{9}{25}-\frac{9}{5}$

$= x^{2}+\left ( 2x\times \frac{3}{5} \right )+\left ( \frac{3}{5} \right )^{2}-\frac{9}{25}-\frac{9}{5}$ $\left [ \because \left ( a+b \right )^{2}=a^{2}+2ab+b^{2} \right ]$

$= \left ( x+\frac{3}{5} \right )^{2}-\frac{9-45}{25}$

$y= \left ( x+\frac{3}{5} \right )^{2}-\frac{54}{25}$

$\sqrt{y+\frac{54}{25}}=\left ( x+\frac{3}{5} \right )$

$\sqrt{\frac{25y+54}{25}}= x+\frac{3}{5}$

$x=\frac{\sqrt{25y+54}}{5}-\frac{3}{5}$

$x=\frac{\sqrt{25y+54}-3}{5}$

Let $g\left ( y \right )=\frac{\sqrt{25y+54}-3}{5}$

Now,

$fog\left ( y \right )=f\left ( g\left ( y \right ) \right )$

$=f\left (\frac{\sqrt{25y+54}-3}{5} \right )$

$=5\left (\frac{\sqrt{25y+54}-3}{5} \right )^{2}+6\left (\frac{\sqrt{25y+54}-3}{5} \right )-9$

$=5\left (\frac{25y+54+9-6\sqrt{25y+54-3}}{25} \right )+\left (\frac{6\sqrt{25y+54}-18}{5} \right )-9$

$=\frac{25y+63-6\sqrt{25y+54}}{5} +\frac{6\sqrt{25y+54}-18}{5} -9$

$=\frac{25y+63-18-45}{5}$

$=\frac{25y}{5}$

$=5y$

Identity function,

Also, $gof\left ( x \right )=g\left ( f\left ( x \right ) \right )$

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

$= \frac{\sqrt{5\left ( 5x^{2}+6x-9 \right )+54}-3}{5}$

$= \frac{\sqrt{25x^{2}+30x-45+54}-3}{5}$

$= \frac{\sqrt{25x^{2}+30x+9}-3}{5}$

$=\frac{\sqrt{\left ( 5x+3 \right )^{2}}-3}{5}$

$=\frac{5x+3-3}{5}$

$= x$

So, $f$ is invertible, an identity function.

$f\left ( y \right )=g\left ( y \right )=\frac{\sqrt{25y+54}-3}{5}$

Posted by

infoexpert27

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