Mathematics - Application of Derivative Question with Solution | TestHub

Given: $\mathrm{f}(\mathrm{t})=\frac{\sin \mathrm{t}}{\mathrm{t}}$ At $t=0$, we will check continuity of the function. $\begin{array}{l}\mathrm{LHL}=\mathrm{f}(0-\mathrm{h})\\ ={\lim }_{\mathrm{h}\to 0}\frac{\sin (0-\mathrm{h})}{(0-\mathrm{h})}={\lim }_{\mathrm{h}\to 0}\frac{-\sin \mathrm{h}}{-\mathrm{h}}=1\\ \mathrm{RHL}=\mathrm{f}(0+\mathrm{h})\\ {\lim }_{\mathrm{h}\to 0}\frac{\sin (0+\mathrm{h})}{(0+\mathrm{h})}\\ ={\lim }_{\mathrm{h}\to 0}\frac{\sin \mathrm{h}}{\mathrm{h}}=1\end{array}$ and $\mathrm{f}(0)=1$ $\mathrm{LHL}=\mathrm{RHL}=\mathrm{f}(0)$ So, the function is continuous at $\mathrm{t}=0$ Now, we check the function is maximum or minimum. $\begin{array}{l}{\mathrm{f}}^{\prime }(\mathrm{t})=\frac{1}{\mathrm{t}}\cos \mathrm{t}-\frac{1}{{\mathrm{t}}^2}\sin \mathrm{t}\\ \text{ and }{\mathrm{f}}^{\prime \prime }(\mathrm{t})=\frac{-1}{\mathrm{t}}\sin \mathrm{t}-\frac{1}{{\mathrm{t}}^2}\cos \mathrm{t}-\frac{1}{{\mathrm{t}}^2}\cos \mathrm{t}+\frac{2}{{\mathrm{t}}^3}\sin \mathrm{t}\\ =\frac{-\sin \mathrm{t}}{\mathrm{t}}-\frac{2\cos \mathrm{t}}{{\mathrm{t}}^2}+\frac{2\sin \mathrm{t}}{{\mathrm{t}}^3}\end{array}$ For maximum or minimum value of $\mathrm{f}(\mathrm{x})$, $\begin{array}{rl}& \text{ put }\\ & {\mathrm{f}}^{\prime }(\mathrm{x})=0\\ \Rightarrow & \frac{\cos \mathrm{t}}{\mathrm{t}}-\frac{\sin \mathrm{t}}{{\mathrm{t}}^2}=0\Rightarrow \frac{\tan \mathrm{t}}{\mathrm{t}}=1\end{array}$ Now ${\lim }_{t\to 0}{f}^{\prime \prime }(t)$ $\begin{array}{l}=-{\lim }_{t\to 0}\left(\frac{\sin t}{t}\right)-2{\lim }_{t\to 0}\left(\frac{t\cos t-\sin t}{t^3}\right)\\ =-1-2{\lim }_{t\to 0}\left(\frac{\cos t-t\sin t-\cos t}{3t^2}\right)\end{array}$ [using L'Hospital rule] $\begin{array}{l}=-1+\frac{2}{3}{\lim }_{t\to 0}\frac{\sin t}{t}\\ =-1+\frac{2}{3}\times 1=\frac{-1}{3}<0\end{array}$ So, function $\mathrm{f}(\mathrm{t})$ is maximum at $\mathrm{t}=0$