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

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

Answers (1)

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

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