#### Please Solve R.D.Sharma class 12 Chapter 9 Differentiability Exercise 9.2 Question 2 Maths textbook Solution.

Answer: $f'\left ( 5 \right )=2\: f'\left ( \frac{7}{2} \right )\Rightarrow L.H.S=R.H.S$

Hint: First we will f (x) then put x = 5 and x = $\frac{7}{2}$  in f (x). Then, we will check L.H.S and R.H.S are equal or not.

Given: $f'\left ( x \right )=x^{2}-4x+7$

Solution:

Differentiating f(x) w.r.t x then,

\begin{aligned} &\Rightarrow \frac{d}{d x}\{\mathrm{f}(\mathrm{x})\}=\frac{d}{d x}\left(\mathrm{x}^{2}-4 \mathrm{x}+7\right) \\ &\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{d}{d x}\left(\mathrm{x}^{2}\right)+\frac{d}{d x}(-4 \mathrm{x})+\frac{d}{d x}(7) \\ &{\left[\because \frac{d}{d x}\left(a x^{2}+b x+c\right)=\frac{d}{d x}\left(a x^{2}\right)+\frac{d}{d x}(b x)+\frac{d}{d x}(c)\right]} \end{aligned}

$\Rightarrow f'\left ( x \right )=2x^{2-1}-4\left ( 1.x^{1-1} \right )+0$                                                                            $\left [ \because \frac{d}{dx}\left ( x^{n} \right ) nx^{n-1}\right ]$

$\Rightarrow f'\left ( x \right )=2x-4$                                                                                                    $\left [ \because \frac{d}{dx} \left ( constant \right )=0\right ]$

Now, put x = 5 and x = $\frac{7}{2}$ in f `(x).

$f'\left ( 5 \right )=2\left ( 5 \right ) and x=\frac{7}{2}in f'\left ( x \right )$

$f'\left ( 5 \right )=2\left ( 5 \right )-4$

$=10-4$

$=6$

$f'\left ( \frac{7}{2} \right )=2\frac{7}{2}-4$

$=7-4$

$=3$

L.H.S

$f'(5)=6$

R.H.S

$2\: f'\left ( \frac{7}{2} \right )=2\times 3=6$

$\therefore$ L.H.S=R.H.S