(a) Find the value of k for which the function f given as f ( x )={ (1- cos x)/(2 x^2), if x ≠ 0; k, if x =0 is continuous at x =0..

Here, $$\mathrm{f}(\mathrm{x})=\frac{1-\cos \mathrm{x}}{2 \mathrm{x}^{2}}=\frac{2 \sin ^{2} \dfrac{\mathrm{x}}{2}}{2 \mathrm{x}^{2}}=\left(\frac{\sin \dfrac{\mathrm{x}}{2}}{2 \dfrac{\mathrm{x}}{2}}\right)^{2} $$ $$\Rightarrow \lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 0} \frac{1}{4}\left(\frac{\sin \dfrac{\mathrm{x}}{2}}{\dfrac{\mathrm{x}}{2}}\right)=\frac{1}{4} $$ Given $\displaystyle \mathrm{x}=\mathrm{a} \cos \mathrm{t}$ and $\displaystyle \mathrm{y}=\mathrm{b} \sin \mathrm{t}$, we have $$\frac{d \mathrm{x}}{d \mathrm{t}}=-\mathrm{a} \sin \mathrm{t} \text { and } \frac{d \mathrm{y}}{d \mathrm{t}}=\mathrm{b} \cos \mathrm{t} $$ $$\therefore \frac{d \mathrm{y}}{d \mathrm{x}}=\frac{d \mathrm{y}}{d \mathrm{t}} \cdot \frac{d \mathrm{t}}{d \mathrm{x}}=\frac{\mathrm{b} \cos \mathrm{t}}{-\mathrm{a} \sin \mathrm{t}}=-\frac{\mathrm{b}}{\mathrm{a}} \cot \mathrm{t} $$ $$\frac{d^{2} \mathrm{y}}{d \mathrm{x}^{2}}=\frac{d}{d \mathrm{t}}\left(-\frac{\mathrm{b}}{\mathrm{a}} \cot \mathrm{t}\right) \cdot \frac{d \mathrm{t}}{d \mathrm{x}} $$ $$\begin{aligned} & =\frac{\mathrm{b}}{\mathrm{a}} \operatorname{cosec}^{2} \mathrm{t} \cdot \frac{1}{-\mathrm{a} \sin \mathrm{t}} \\ & =-\frac{\mathrm{b}}{\mathrm{a}^{2}} \cdot \frac{1}{\sin ^{3} \mathrm{t}} \text { or }-\frac{\mathrm{b}}{\mathrm{a}^{2}} \operatorname{cosec}^{3} \mathrm{t} \end{aligned} $$