#### Please Solve R.D.Sharma class 12 Chapter 9 Differentiability Exercise 9.2 Question 3 Maths textbook Solution.

Answer: f  (1) = f  (2) $\Rightarrow$ L.H.S = R.H.S

Hint: Differentiate f(x) to find f (x). Put x = 1 and x = 2 in f (x). Then, we will check L.H.S and R.H.S are equal or not.

Given: $f\left ( x \right )=2x^{3}-9x^{2}+12x+9$

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(2 \mathrm{x}^{3}-9 \mathrm{x}^{2}+12 \mathrm{x}+9\right) \\ &=\frac{d}{d x}\left(2 \mathrm{x}^{3}\right)+\frac{d}{d x}\left(-9 \mathrm{x}^{2}\right)+\frac{d}{d x}(12 \mathrm{x})+\frac{d}{d x}(9) \end{aligned}

$\left[\because \frac{d}{d x}\left(a x^{3}+b x^{2}+c x+d\right)=\frac{d}{d x}\left(a x^{3}\right)+\frac{d}{d x}\left(b x^{2}\right)+\frac{d}{d x}(c x)+\frac{d}{d x}(d)\right]$

\begin{aligned} &=2 \frac{d}{d x}\left(x^{3}\right)-9 \frac{d}{d x}\left(x^{2}\right)+12 \frac{d}{d x}(x)+0 \\ &=2\left(3 x^{3-1}\right)-9\left(2 x^{2-1}\right)+12\left(1 x^{1-1}\right) \end{aligned}                                $\left [ \because \frac{d}{dx}\left ( constant \right )=0\left [ \because \frac{d}{dx}\left ( ax^{n}=a\frac{d}{dx}\left ( x^{n} \right ) \right ) \right ] \right ]$

$=6x^{2}-18x+12$                                                                    $\left [ \because \frac{d}{dx}\left ( x^{n} \right )=nx^{n-1} \right ]$

Now,$f'\left ( 1 \right )=6\left ( 1\right )^{2}-18\left ( 1 \right )+12$

$=6-18+12$

$=18-18$

$=0$

$f'\left ( 2 \right )=6\left ( 2 \right )^{2}-18\left ( 2 \right )+12$

$=24-36+12$

$=36-36$

$=0$

$\therefore f'\left ( 1 \right )=f'\left ( 2 \right )=0$                                                    [Equal]

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