Question

Find the condition of $f\left( t \right)$ at $t = 0$ where, $f(t) = \dfrac{{\sin t}}{t}$
a) A minimum
b) A discontinuity
c) A point of inflexion
d) A maximum

Answer 1

Hint: In this question you need to check the continuity of a function by finding out $(LHL)$ and $\left( {RHL} \right)$, to check maximum and minimum, differentiate the function twice. Is the resultant value being greater than zero then the given function is minimum else it is maximum?

Complete step-by-step answer:
In this question it is given that,
$f(t) = \dfrac{{\sin t}}{t}$
Now at $t = 0$ we will check continuity of a function $f\left( t \right)$ by finding out the left hand limit $(LHL)$ and right hand limit$\left( {RHL} \right)$.
First, let us find out $(LHL)$ whose formula when $t = 0$ is,
$LHL = f\left( {0 - h} \right)$
Now replace ‘$t$’ by $\left( {0 - h} \right)$ in equation (1) with $\mathop {\lim }\limits_{h \to 0} $, we get,
$ \Rightarrow LHL = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin (0 - h)}}{{(0 - h)}}$
On simplification,
$ \Rightarrow LHL = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin ( - h)}}{{( - h)}}$
We know that, $\sin ( - h) = - \sin (h)$
So above equation becomes,
$ \Rightarrow LHL = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - \sinh }}{{ - h}}$
On simplification we get,
$ \Rightarrow LHL = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sinh }}{h}$
We know that, $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$
Thus,
$ \Rightarrow LHL = 1$
Now, for $\left( {RHL} \right)$ we use below formula at $t = 0$
$ \Rightarrow RHL = f(0 + h)$
Using equation (1) replace ‘t’ by $\left( {0 + h} \right)$ and by adding $\mathop {\lim }\limits_{h \to 0} $ we get,
$ \Rightarrow RHL = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin (0 + h)}}{h}$
On simplification we get,
$ \Rightarrow RHL = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sinh }}{h}$
From equation (2) we get
$ \Rightarrow RHL = 1$
We will now find out $f\left( 0 \right)$, using equation (2)
$ \Rightarrow f(0) = 1$
Thus,
$ \Rightarrow LHL = RHL = f(0)$
If the above condition is satisfied, then we can conclude that the given function is continuous at $t = 0$.
Now, we will check minimum or maximum at a function,
Differentiating $f\left( t \right)$ using product rule,
$\dfrac{d}{{dx}}(u,v) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$
We get,
$ \Rightarrow f'(t) = \dfrac{1}{t}\cos t - \dfrac{1}{{{t^2}}}\sin t$
Now, again differentiate (3) with respect to ‘$t$’ using product rule. We get,
$ \Rightarrow f''(t) = - \dfrac{1}{t}\sin t - \dfrac{1}{{{t^2}}}\cos t - - \dfrac{1}{{{t^2}}}\cos t + \dfrac{2}{{{t^3}}}\sin t$
On simplification we get,
 $ \Rightarrow f''(t) = - \dfrac{{\sin t}}{t} - \dfrac{{2\cos t}}{{{t^2}}} + \dfrac{{2\sin t}}{{{t^3}}}$
$ \Rightarrow f''(t) = - \dfrac{{\sin t}}{t} - 2\left( {\dfrac{{t\cos t - \sin t}}{{{t^3}}}} \right)$
For maximum or minimum value of $f\left( t \right)$, put $f'\left( t \right) = 0$.
$ \Rightarrow \dfrac{{\cos t}}{t} - \dfrac{{\sin t}}{{{t^2}}} = 0$
On simplification we get,
$ \Rightarrow \dfrac{{t\cos t - \sin t}}{{{t^2}}} = 0$
$ \Rightarrow t\cos t - \sin t = 0$
$ \Rightarrow t\cos t = \sin t$
Dividing $\left( {t\cos t} \right)$ on both sides,
$ \Rightarrow 1 = \dfrac{{\sin t}}{{t\cos t}}$
We know that, $\tan t = \dfrac{{\sin t}}{{\cos t}}$
Thus equation becomes,
$ \Rightarrow \dfrac{{\tan t}}{t} = 1$
Now, we will find out $\mathop {\lim }\limits_{t \to 0} f''(t)$,
$ \Rightarrow \mathop {\lim }\limits_{t \to 0} f''(t) = \left[ { - \mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{\sin t}}{t}} \right) - 2\mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{t\cos t - \sin t}}{{{t^3}}}} \right)} \right]$
From equation (2) the above equation can be simplified.
$ \Rightarrow \mathop {\lim }\limits_{t \to 0} f''(t) = - 1 - 2\mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{t\cos t - \sin t}}{{{t^3}}}} \right)$
When we substitute in place of ‘$t$’ as ‘0’ we get $\left( {\dfrac{0}{0}} \right)$ form. Thus, we need to apply L’Hospital rule i.e., differentiating both numerator and denominator as a separate function we get,
$ \Rightarrow \mathop {\lim }\limits_{t \to 0} f''(t) = - 1 - 2\mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{\cos t - t\sin t - \cos t}}{{3{t^2}}}} \right)$
On simplification we get,
$ \Rightarrow \mathop {\lim }\limits_{t \to 0} f''(t) = - 1 - 2\mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{ - t\sin t}}{{3{t^2}}}} \right)$
$ \Rightarrow \mathop {\lim }\limits_{t \to 0} f''(t) = - 1 + \dfrac{2}{3}\mathop {\lim }\limits_{t \to 0} \left( {\dfrac{{\sin t}}{t}} \right)$
Using equation (2) we get,
$ \Rightarrow \mathop {\lim }\limits_{t \to 0} f''(t) = - 1 + \dfrac{2}{3} \times 1$
$ \Rightarrow \mathop {\lim }\limits_{t \to 0} f''(t) = - \dfrac{1}{3} < 0$
Since $ - \dfrac{1}{3}$ is less than zero, we can conclude that the given function $f\left( t \right)$ is maximum at $t = 0$.
Hence, option (d) is the correct answer.

Note: In this type of question you should know how to differentiate the given function and condition to check maximum and minimum.