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

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

Answers (1)

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}
 

 

Posted by

infoexpert26

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