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Mathematics Question on Definite Integral

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

A

a minimum

B

a discontinuity

C

a point of inflexion

D

a maximum

Answer

a maximum

Explanation

Solution

Given : f(t)=sin⁡ttf(t)=\frac{\sin t}{t}
At t=0t=0, we will check continuity of the function.
LHL=f(0−h)L H L=f(0-h)
=lim⁡h→0sin⁡(0−h)(0−h)=lim⁡h→0−sin⁡h−h=1=\displaystyle\lim _{h \rightarrow 0} \frac{\sin (0-h)}{(0-h)}=\lim _{h \rightarrow 0} \frac{-\sin h}{-h}=1
RHL=f(0+h)R H L=f(0+h)
=lim⁡h→0sin⁡(0+h)(0+h)=\displaystyle\lim _{h \rightarrow 0} \frac{\sin (0+h)}{(0+h)}
=lim⁡h→0sin⁡hh=1=\displaystyle\lim _{h \rightarrow 0} \frac{\sin h}{h}=1
and f(0)=1f(0)=1
LHL=RHL=f(0)L H L=R H L=f(0)
So, the function is continuous at t=0t=0
Now, we check the function is maximum or minimum.
f′(t)=1tcos⁡t−1t2sin⁡tf^{\prime}(t)=\frac{1}{t} \cos t-\frac{1}{t^{2}} \sin t
and f′′(t)=−1t2sin⁡t−1tcos⁡t−1t2cos⁡t−2t3sin⁡tf^{\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
=−sin⁡tt−2cos⁡tt2+2sin⁡tt3=\frac{-\sin t}{t}-\frac{2 \cos t}{t^{2}}+\frac{2 \sin t}{t^{3}}
For maximum or minimum value of f(x)f(x),put
f′(x)=0f^{\prime}(x)=0
⇒cos⁡tt−sin⁡tt2=0\Rightarrow \frac{\cos t}{t}-\frac{\sin t}{t^{2}}=0
⇒tan⁡tt=1\Rightarrow \frac{\tan t}{t}=1
Now lim⁡t→0f′′(t)\displaystyle\lim _{t \rightarrow 0} f^{\prime \prime}(t)
=−lim⁡t→0(sin⁡tt)−2lim⁡t→0(tcos⁡t−sin⁡tt3)=-\displaystyle\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)
[00\left[\frac{0}{0}\right. form ]]
=−1−2lim⁡t→0(cos⁡t−tsin⁡t−cos⁡t3t2)=-1-2 \displaystyle\lim _{t \rightarrow 0}\left(\frac{\cos t-t \sin t-\cos t}{3 t^{2}}\right)
[using L' Hospital rule]
=−+23lim⁡t→0sin⁡tt=-+\frac{2}{3} \displaystyle\lim _{t \rightarrow 0} \frac{\sin t}{t}
=−1+23×1=−13<0=-1+\frac{2}{3} \times 1=\frac{-1}{3} < 0
So, function f(t)f(t) is maximum at t=0t=0