Question

At $t=0$, the function $f\left(t\right)=\frac{sint}{t}$ has

• A. a minimum
• B. a discontinuity
• C. a point of inflexion
• D. a maximum

Answer 1

Answer: (D)

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