#### Need solution for RD Sharma maths class 12 chapter 9 Differentiability exercise Fill in the blanks question  7

Answer: $\mathrm{a}=\frac{1}{2}$

Hint:

Use differentiability condition

f(x) is said to be differentiable at x=b

Given:

$f(x)= \begin{cases}a x^{2}+3 & x>1 \\ x+\frac{5}{2} & x \leq 1\end{cases}$  is differentiable at x=1

Solution:

we have $\mathrm{f}(\mathrm{x})= \begin{cases}\mathrm{ax}^{2}+3 & \mathrm{x}>1 \\ \mathrm{x}+\frac{5}{2} & \mathrm{x} \leq 1\end{cases}$

\begin{aligned} \text { L.H.D } &=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(1-\mathrm{h})-\mathrm{f}(1)}{-\mathrm{h}} \\\\ &=\lim _{\mathrm{h} \rightarrow 0} \frac{(1-\mathrm{h})+5 / 2-(1+5 / 2)}{-\mathrm{h}} \\\\ &=\lim _{\mathrm{h} \rightarrow 0} \frac{(1-\mathrm{h}+5 / 2-1-5 / 2}{-\mathrm{h}} \end{aligned}

$=\lim _{\mathrm{h} \rightarrow 0} \frac{-\mathrm{h}}{-\mathrm{h}}=+1$

\begin{aligned} \mathrm{R} \cdot \mathrm{H} \cdot \mathrm{D}=& \lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(1+\mathrm{h})-\mathrm{f}(1)}{\mathrm{h}} \\ &=\lim _{\mathrm{n} \rightarrow 0} \frac{\mathrm{a}(1+\mathrm{h})^{2}+3-\left(\mathrm{a}(1)^{2}+3\right)}{\mathrm{h}} \end{aligned}

\begin{aligned} &=\lim _{\mathrm{n} \rightarrow 0} \frac{\mathrm{a}\left(1+\mathrm{h}^{2}+2 \mathrm{~h}\right)+3-\mathrm{a}-3}{\mathrm{~h}} \\\\ &=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{a}+\mathrm{ah}^{2}+2 \mathrm{ah}+3-\mathrm{a}-3}{\mathrm{~h}} \\\\ &=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{h}(\mathrm{ah}+2 \mathrm{a})}{\mathrm{h}} \end{aligned}

\begin{aligned} &\lim _{\mathrm{h} \rightarrow 0} \mathrm{ah}+2 \mathrm{a} \\ &\mathrm{a} \times 0+2 \mathrm{a}=2 \mathrm{a} \end{aligned}

Since $f(x)$ is differentiable at $x=1$

So,$LHD= RHD$

\begin{aligned} &\Rightarrow 1=2 \mathrm{a} \\ &\Rightarrow\mathrm{a}=\frac{1}{2} \end{aligned}