Question

 The difference between the greatest and the least values of the function, \[f\left( x \right)=\sin 2x-x\] on $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$.
A. $\pi $
B. 0
C. $\dfrac{\sqrt{3}}{2}+\dfrac{\pi }{3}$
D. $-\dfrac{\sqrt{3}}{2}+\dfrac{2\pi }{3}$

Answer 1

Hint: We will find the greatest and least value of function using the method of differentiation. The differentiation of $f\left( x \right)$ is denoted as $f'\left( x \right)$.
The find the value of x for which $f'\left( x \right)=0$, then the value of $f\left( x \right)$ at x is either greatest value or least value.

Complete step-by-step answer:

Now, we can see that $f\left( x \right)$ is given as $\sin 2x-x$ on interval $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$.
The differentiation of \[f\left( x \right)=\sin 2x-x\] with respect to x is $f'\left( x \right)$.
\[f\left( x \right)=\sin 2x-x\] ……..(1)
And we know that the differentiation of $\sin 2x=2\cos 2x$.
Now, differentiating equation (1) with respect to x,
$\Rightarrow f'\left( x \right)=2\cos 2x-1..........\left( 2 \right)$
And to find the value of x for which $f\left( x \right)$ is greatest or least, we will have to equal the equation (2) to zero i.e.
$\begin{align}
  & f'\left( x \right)=2\cos 2x-1=0 \\
 & \Rightarrow 2\cos 2x=1 \\
 & \Rightarrow \cos 2x=\dfrac{1}{2} \\
 & \Rightarrow 2x={{\cos }^{-1}}\dfrac{1}{2}..............\left( 3 \right) \\
\end{align}$
And we know that ${{\cos }^{-1}}\dfrac{1}{2}=\dfrac{\pi }{3},-\dfrac{\pi }{3}$ (as $x\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$, so there are only two value of ${{\cos }^{-1}}\dfrac{1}{2}$ which come in this interval).
Now,
$\begin{align}
  & \Rightarrow 2x=\dfrac{\pi }{3}\ \And \ 2x=-\dfrac{\pi }{3} \\
 & \Rightarrow x=\dfrac{\pi }{6}\ and-\dfrac{\pi }{6} \\
\end{align}$
Now, to find the greatest and least value of $f\left( x \right)$. First we will have to find the value of $f\left( x \right)$ at $-\dfrac{\pi }{2},\dfrac{\pi }{2},\dfrac{\pi }{6},-\dfrac{\pi }{6}$ (as it is close interval so we will have to check at boundary also).
$\begin{align}
  & f\left( x \right)=\sin 2x-x \\
 & \Rightarrow f\left( -\dfrac{\pi }{2} \right)=0-\left( -\dfrac{\pi }{2} \right)=\dfrac{\pi }{2},\ \left( \sin \left( -\pi \right)=-\sin \pi =0 \right) \\
 & f\left( \dfrac{\pi }{2} \right)=0-\dfrac{\pi }{2}=-\dfrac{\pi }{2}, \\
 & f\left( -\dfrac{\pi }{6} \right)=\sin 2\left( -\dfrac{\pi }{6} \right)-\left( -\dfrac{\pi }{6} \right) \\
 & =\sin \left( -\dfrac{\pi }{3} \right)+\dfrac{\pi }{6} \\
 & =-\sin \dfrac{\pi }{3}+\dfrac{\pi }{6} \\
 & f\left( -\dfrac{\pi }{6} \right)=-\dfrac{\sqrt{3}}{2}+\dfrac{\pi }{6}...............\left( 4 \right) \\
\end{align}$
$\begin{align}
  & f\left( \dfrac{\pi }{6} \right)=\sin 2\left( \dfrac{\pi }{6} \right)-\dfrac{\pi }{6} \\
 & =\sin \dfrac{\pi }{3}-\dfrac{\pi }{6} \\
 & f\left( \dfrac{\pi }{6} \right)=\dfrac{\sqrt{3}}{2}-\dfrac{\pi }{6}............\left( 5 \right) \\
\end{align}$
We can see that,
 \[\begin{align}
  & \dfrac{\sqrt{3}}{2}-\dfrac{\pi }{6}\simeq 0.34\ and\ -\dfrac{\sqrt{3}}{2}+\dfrac{\pi }{6}\simeq -0.34 \\
 & \dfrac{\pi }{2}\simeq 1.57 \\
 & -\dfrac{\pi }{2}\simeq -1.57 \\
\end{align}\]
From above, clearly we can see that the greatest value of $f\left( x \right)$ on $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ is $\dfrac{\pi }{2}$ and the least value of $f\left( x \right)$ is $-\dfrac{\pi }{2}$.
So, the difference between the greatest and least value of $f\left( x \right)=\dfrac{\pi }{2}-\left( -\dfrac{\pi }{2} \right)$
$\begin{align}
  & =\dfrac{\pi }{2}+\dfrac{\pi }{2} \\
 & =\pi \\
\end{align}$

Note: In this question, students can make mistake such as they don’t check the value of $f\left( x \right)$ at boundary and can find the answer as $\dfrac{\sqrt{3}}{2}-\dfrac{\pi }{6}$ is the greatest and $-\dfrac{\sqrt{3}}{2}-\dfrac{\pi }{6}$ is the least value and difference between them as,
 $\begin{align}
  & \dfrac{\sqrt{3}}{2}-\dfrac{\pi }{6}-\left( -\dfrac{\sqrt{3}}{2}+\dfrac{\pi }{6} \right) \\
 & =\dfrac{\sqrt{3}}{2}-\dfrac{\pi }{6}+\dfrac{\sqrt{3}}{2}-\dfrac{\pi }{6} \\
 & =\sqrt{3}-\dfrac{\pi }{3} \\
\end{align}$
But it is incorrect as the interval is the close interval. So, we have to check the value of $f\left( x \right)$ at boundary.