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  • Provide Solutio for RD Sharma Class 12 Chapter Continuity Exercise 8.1 Question 38

Provide Solutio for RD Sharma Class 12 Chapter Continuity Exercise 8.1 Question 38

Answers (1)

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Answer:

                Continuous

Hint:

 For a function to be continuous at a point, its LHL RHL and value at that point should be equal.

 

Given:

                f(x)=\left(\begin{array}{cc} \frac{x^{2}}{2} & , 0 \leq x \leq 1 \\\\ 2 x^{2}-3 x & +\frac{3}{2}, 1<x \leq 2 \end{array}\right)

Solution:

                f(x)=\left(\begin{array}{cc} \frac{x^{2}}{2} & , 0 \leq x \leq 1 \\\\ 2 x^{2}-3 x & +\frac{3}{2}, 1<x \leq 2 \end{array}\right)

We have

(LHL at x=1 )

                \lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0} \frac{(1-h)^{2}}{2}=\frac{1}{2}

(RHL at x=1 )

                \! \! \! \! \! \! \! \! \! \! \! \! \! \lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}\left[2(1+h)^{2}-3(1+h)+\frac{3}{2}\right]=\left[2-3+\frac{3}{2}\right]=\frac{1}{2}

Also

\begin{aligned} &f(1)=\frac{(1)^{2}}{2}=\frac{1}{2} \\ &\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{-}} f(x)=f(1) \end{aligned}

Hence the given function is continuous at  x=1

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