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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Discuss the applicability of Rolle’s theorem on the function given by.
$\text{f(x)}=\begin{cases}\text{x}^2+1,&\text{if }0\leq\text{x}\leq1\\3-\text{x},&\text{if }1\leq\text{x}\leq2\end{cases}$
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Answer

Consider, $\text{f(x)}=\begin{cases}\text{x}^2+1,&\text{if }0\leq\text{x}\leq1\\3-\text{x},&\text{if }1\leq\text{x}\leq2\end{cases}$
We know that, polynomial function is everywhere continuous and differentiability.
So, f(x) is continuous and differentiable at all points except possibly at x = 1.
Now, check the differentiability at x = 1,
At x = 1 $\text{L.D.H}=\lim\limits_{\text{x}\rightarrow1^-}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}^2+1)-(1+1)}{\text{x}-1}$ $[\because\ \text{f(x)}=\text{x}^2+1,\forall\ 0\leq\text{x}\leq1]$
$=\lim\limits_{\text{x}\rightarrow1}\frac{\text{x}^2-1}{\text{x}-1}=\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}+1)(\text{x}-1)}{\text{x}-1}$
and $\text{R.D.H}=\lim\limits_{\text{x}\rightarrow1^+}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}=\lim\limits_{\text{x}\rightarrow1}\frac{(3-\text{x})\text{f}(1+1)}{(\text{x}-1)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{3-\text{x}-2}{\text{x}-1}=\lim\limits_{\text{x}\rightarrow1}\frac{-(\text{x}-1)}{\text{x}-1}=-1$
$\therefore$ L.H.D ≠ R.H.D
So, f(x) is not differentiable at x = 1.
Hence, polle’s theorem is not applicable on the interval [0, 2]

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