Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiability5 Marks
Question
Write the number of points where f(x) = |x| + |x − 1| is continuous but not differentiable.
✓
Answer
Given:
f(x) = |x| + |x - 1|
$\Rightarrow\text{f(x)}=\begin{cases}-\text{x}-(\text{x}-1),&\text{x}<0\\\text{x}-(\text{x}-1),&0\leq\text{x}<1\\\text{x}+(\text{x}-1),&\text{x}\geq1\end{cases}$
$\Rightarrow\text{f(x)}=\begin{cases}-2\text{x}+1,&\text{x}<0\\1,&0\leq\text{x}<12\\\text{x}-1,&\text{x}\geq1\end{cases}$
When x < 0, we have:
f(x) = -2x + 1 which, being a polynomial function is continuous and differentiable.
When $0\leq\text{x}<1,$ we have:
f(x) = 1 which, being a polynimial function is continuous and differentiable on (0, 1).
When $\text{x}\leq1,$ we have:
f(x) = 2x - 1 which, being a polynimial function is continuous and differentiable on x > 2.
Thus, the possible points of non differentiability of f(x) are 0 and 1.
Now,
(LHL at x = 0)
$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$\lim\limits_{\text{x}\rightarrow0^{-}}\frac{2\text{x}+1-1}{\text{x}-0}\ [\because\text{f(x)}=-2\text{x}+1,\text{x}<0]$
$\lim\limits_{\text{x}\rightarrow0}\frac{-2\text{x}}{\text{x}}=-2$
(RHL at x = 0)
$=\lim\limits_{\text{x}\rightarrow0^{+}}\text{f(x)}-\text{f}(0)\text{x}-0$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{1-1}{\text{x}-1}$
$=0\ [\because\text{f(x)}=1,0\leq\text{x}<1]$
Thus, (LHL at x = 1) $\neq$ (RHL at x = 1)
Hence f(x) is not differentiable at x = 1.
Therefore, 0, 1 are the points where f(x) is continuous but not differentiable.
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