(a) If f ( x )= {x^2, if x ≥ 1; x, if x <1}, then show that f is not differentiable at x =1.

Here $\displaystyle \mathrm{RHD}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(1+\mathrm{h})-\mathrm{f}(1)}{\mathrm{h}}=2$ $\displaystyle \mathrm{LHD}=\lim _{\mathrm{h} \rightarrow 0}\left[\frac{\mathrm{f}(1-\mathrm{h})-\mathrm{f}(1)}{-\mathrm{h}}\right]=1$ Since RHD $\displaystyle \neq \mathrm{LHD}$ $$\begin{aligned} & \lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 0}\left(\frac{\sin ^{2} \lambda \mathrm{x}}{\mathrm{x}^{2}}\right)=\lim _{\mathrm{x} \rightarrow 0}\left[\frac{\sin ^{2} \lambda \mathrm{x}}{(\lambda \mathrm{x})^{2}} \cdot \lambda^{2}\right]=\lambda^{2} \\ & \text { Since } \mathrm{f}(\mathrm{x}) \text { is continuous at } \mathrm{x}=0 \\ & \quad \lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=\mathrm{f}(0) \\ & \Rightarrow \lambda^{2}=1 \Rightarrow \lambda= \pm 1 \end{aligned} $$