\(f(x)= \begin{cases}2 a-x & \text { in }-a < x < a \\ 3 x-2 a & \text { in } a \leq x\end{cases}\)
Then, which of the following is true ?

Given, \(\quad \mathrm{f}(\mathrm{x})=\left\{\begin{array}{llc}2 \mathrm{a}-\mathrm{x} & \text { in } & -a < x < a \\ 3 \mathrm{x}-2 \mathrm{a} & \text { in } & \mathrm{a} \leq \mathrm{x}\end{array}\right.\) At \(x=a\) \(\mathrm{LHL}=\lim _{\mathrm{x} \rightarrow \mathrm{a}^{-}} \mathrm{f}(\mathrm{x})\) \( =\lim _{x \rightarrow a} 2 a-x=a \) \( \begin{aligned} \mathrm{RHL} &=\lim _{\mathrm{x} \rightarrow \mathrm{a}^{+}} \mathrm{f}(\mathrm{x}) \\ &=\lim _{\mathrm{x} \rightarrow \mathrm{a}} 3 \mathrm{x}-2 \mathrm{a}=\mathrm{a} \end{aligned} \) and \(\quad \mathrm{f}(\mathrm{a})=3(\mathrm{a})-2 \mathrm{a}=\mathrm{a}\) \(\therefore \quad \mathrm{LHL}=\mathrm{RHL}=\mathrm{f}(\mathrm{a})\) Hence, it is continuous at \(\mathrm{x}=\mathrm{a}\) Again, at \(x=a\) \( \begin{aligned} \mathrm{LHD} &=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(\mathrm{a}-\mathrm{h})-\mathrm{f}(\mathrm{a})}{-\mathrm{h}} \\ &=\lim _{\mathrm{h} \rightarrow 0} \frac{2 \mathrm{a}-(\mathrm{a}-\mathrm{h})-\mathrm{a}}{-\mathrm{h}}=\frac{\mathrm{h}}{-\mathrm{h}}=-1 \end{aligned} \) \( \text { and } \begin{aligned} \mathrm{RHD}=& \lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(\mathrm{a}+\mathrm{h})-\mathrm{f}(\mathrm{a})}{h} \\ =& \lim _{\mathrm{h} \rightarrow 0} \frac{3(\mathrm{a}+\mathrm{h})-2 \mathrm{a}-\mathrm{a}}{\mathrm{h}}=3 \\ & \mathrm{LHD} \neq \mathrm{RHD} \end{aligned} \) Hence, it is not differentiable at \(\mathrm{x}=\mathrm{a}\). Alternate Solution Let \(y=f(x)= \begin{cases}2 a-x & \text { in }-a < x < a \\ 3 x-2 a & \text { in } \quad a \leq x\end{cases}\)It is clear from the graph that \(\mathrm{f}(\mathrm{x})\) is continuous everywhere it is also differentiable everywhere except a point \(\mathrm{A}\) at \(\mathrm{x}=\mathrm{a}\).