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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Find whether the function is differentiable at x = 1 and x = 2
$\text{f(x)}=\begin{cases}\text{x} & \text{x}\leq1\\2-\text{x} & 1\leq\text{x}\leq2\\-2+3\text{x}&\text{x}>2\end{cases}$
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Answer

$\text{f(x)}=\begin{cases}\text{x} & \text{x}\leq1\\2-\text{x} & 1\leq\text{x}\leq2\\-2+3\text{x}&\text{x}>2\end{cases}$
$\text{f(x)}=\begin{cases}\text{x} & \text{x}\leq1\\-1 & 1\leq\text{x}\leq2\\3-2\text{x}&\text{x}>2\end{cases}$
Now,
LHL $=\lim_\limits{\text{x}\rightarrow1^{-}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow1^{-}}1=1$
RHL $=\lim_\limits{\text{x}\rightarrow1^{+}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow1^{+}}-1=-1$
Since, at $\text{x}=1,\text{LHL}\neq\text{RHL}$
Hence, f(x) is not differentiable at x = 1
Again,
LHL $=\lim_\limits{\text{x}\rightarrow2^{-}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow2^{-}}-1=-1$
RHL $=\lim_\limits{\text{x}\rightarrow2^{+}}\text{f}'(\text{x})=\lim_\limits{\text{x}\rightarrow2^{+}}3-2\text{x}=3-4=-1$
Since, at x = 2, LHL = RHL
Hence, f(x) is differentiable at x = 2.

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