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Value of $$c$$ using Rolle's theorem for  $$f\left ( x \right )=\begin{cases} x^{2}+1 &, \text{ when } 0\le x \le 1 \\ 3-x &, \text{ when } 1 < x\le 2\end{cases}$$ is:


A
1
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B
0
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C
1
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D
does not exist
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Solution

The correct option is D does not exist
$$f(x)=\left\{\begin{matrix}x^2+1
 & 0\leq x\leq 1\\ 3-x
 & 1<x\leq 2
\end{matrix}\right.$$

$$f^1(x)=\left\{\begin{matrix}
2x & 0\leq x\leq 1\\ -1
 & 1<x\leq 2
\end{matrix}\right.$$

$$f^1(1^-)\neq f^1(1^+)$$

$$L\cdot H\cdot S\neq R\cdot H\cdot S$$

So $$f'(x)$$ is not continuous.

$$\therefore$$ Rolle's theorem can't be applied.

Mathematics

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