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  • Provide solution for rd sharma maths class 12 chapter 9 Differentiability exercise Multiple choice question, question 10

Provide solution for rd sharma maths class 12 chapter 9 Differentiability exercise Multiple choice question, question 10

Answers (1)

 (b)

HINTS: Understand the  definition of continuity and  differentiability

f(x)=x^{2}+\frac{x^{2}}{1+x^{2}}+\frac{x^{2}}{\left(1+x^{2}\right)^{2}}+\ldots

GIVEN:  then at x=0

SOLUTION:

f(x)=\sum_{n=0}^{\infty} \frac{x^{2}}{\left(1+x^{2}\right)^{n}}

Now f(x) is G.P with \mathrm{r}=\frac{1}{1+x^{2}}

So,

\begin{aligned} &f(x)=\frac{x^{2}\left(\frac{1}{1+x^{2}}\right)^{n}-1}{\frac{1}{1+x^{2}}-1}\left[a=x^{2} \text { and } s=\frac{a^{n-1}}{n-1}\right] \\ \end{aligned}

\begin{aligned} &f(x)=\frac{\frac{x^{2}-\left(1+x^{2}\right)^{n}}{\left(1+x^{2}\right)^{n}}}{\frac{1-1-x^{2}}{\left(1+x^{2}\right)}} \\ \end{aligned}

\begin{aligned} &f(x)=-1\left(1+x^{2}\right)^{-n+1}+x^{-2}\left(1+x^{2}\right) \\ &x=0 \\ \end{aligned}

\begin{aligned} &\lim _{x \rightarrow 0} f(x)=\lim _{h \rightarrow 0}-\left(1+h^{2}\right)^{-n-1}+h^{-2}\left(1+h^{2}\right)=\infty \end{aligned}

As LHL \neq RHL

Therefore, f(x)  is discontinues at x=0

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