#### Please Solve R.D.Sharma class 12 Chapter 9 Differentiability Exercise 9.2 Question 4 Maths textbook Solution.

Answer:$\Phi =9$

Hint: Differentiate f(x) to find f (x). Put x = 5 in f (x). To find the value of $\Phi$, equate R.H.S of f ` (5) from given and calculate.

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

Solution:

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

$\Rightarrow \frac{d}{d x}\{\mathrm{f}(\mathrm{x})\}=\frac{d}{d x}\left(\Phi \mathrm{x}^{2}+7 \mathrm{x}-4\right) \; \; \; \; \; \; \; \; \; \quad\left[\because \frac{d}{d x}\left(a x^{2}+b x+c\right)=\frac{d}{d x}\left(a x^{2}\right)+\frac{d}{d x}(b x)+\frac{d}{d x}(c)\right]$

$\begin{array}{ll} =\frac{d}{d x}\left(\Phi \mathrm{x}^{2}\right)+\frac{d}{d x}(7 \mathrm{x})+\frac{d}{d x}(-4) & {\left[\because \frac{d}{d x}(\text { constant })=0\right]} \; \; \; \\ =\Phi \frac{d}{d x}\left(\mathrm{x}^{2}\right)+7 \frac{d}{d x}(\mathrm{x})+0 & {\left[\because \frac{d}{d x}\left(a x^{n}\right)=a \frac{d}{d x}\left(x^{n}\right)\right]} \\ =\Phi\left(2 \mathrm{x}^{2-1}\right)+7\left(1 \mathrm{x}^{1-1}\right)+0 & {\left[\because \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right]} \end{array}$

$=2\Phi x+7$

$\therefore f'\left ( x \right )=2\Phi\; x+7$

We Know that

$f'\left ( 5 \right )=97$

$2\: \Phi \left ( 5 \right )+7=97$

$10\: \Phi =97-7$

$10\: \Phi =90$

$\Phi =9$