Answer:
Option D
Explanation:
Given $f(t)=\frac{\sin t}{t}$
At t=0, we will check continuity of the function
LHL=f(0-h)
$=\lim_{h \rightarrow 0}\frac{\sin(0-h)}{(0-h)}=\lim_{h \rightarrow 0}\frac{-\sin h}{-h}=1$
RHL=f(0+h)
$\lim_{h \rightarrow 0}\frac{\sin(0+h)}{(0+h)}=\lim_{h \rightarrow 0}\frac{\sin h}{h}=1$
and f(0)=1
LHL=RHL= f(0)
So, the function is continuous at t=0
Now, we check the function is maximum or minimum
$f '(t)=\frac{1}{t}\cos t-\frac{1}{t^{2}}\sin t$
and $f ''(t)=\frac{-1}{t}\sin t-\frac{1}{t^{2}}\cos t-\frac{1}{t^{2}}\cos t+\frac{2}{t^{3}}\sin t$
=$\frac{-\sin t}{t}-\frac{2\cos t}{t^{2}}+\frac{2\sin t}{t^{3}}$
For maximum or minimum value of f(x) , put f '(x)=0
$\Rightarrow\frac{\cos t}{t}-\frac{\sin t}{t^{2}}=0\Rightarrow\frac{\tan t}{t^{}}=1$
Now $\lim_{t \rightarrow 0}f ''(t)$
$=-\lim_{t \rightarrow 0}\left(\frac{\sin t}{t}\right)-2\lim_{t \rightarrow 0}\left(\frac{t\cos t-\sin t}{t^{3}}\right)$
$[\frac{0}{0} form]$
$= -1-2\lim_{t \rightarrow 0}\left(\frac{\cos t-t\sin t-\cos t}{3t^{2}}\right)$ [Using L' Hosptial rule ]
$= -1+\frac{2}{3}\lim_{t \rightarrow 0}\frac{\sin t}{t}$
$= -1+\frac{2}{3}\times1=\frac{-1}{3}<0$
So , function f(t) is maximum at t=0
