Question

at what point in the interval [0,2π], does the function sin 2x attain its maximum value?​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) \: = \: sin2x \: \: \: where \: x \: \in \: [0, \: 2\pi]$

Now, Differentiate both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} f(x) \: = \dfrac{d}{dx} \: sin2x \: \: \:$

$\rm :\longmapsto\:f'(x) \: = cos2x \: \dfrac{d}{dx} \: 2x \: \: \:$

$\rm :\longmapsto\:f'(x) \: = cos2x \: \times 2$

$\rm :\longmapsto\:f'(x) \: =2 \: cos2x \:$

For maxima or minima,

$\rm :\longmapsto\:f'(x) \: = \: 0 \:$

$\rm :\longmapsto\:2cos2x = 0$

$\rm :\longmapsto\:cos2x = 0$

$\rm :\longmapsto\:cos2x = cos\dfrac{\pi}{2}$

$\rm :\implies\:2x = 2n\pi \: \pm \: \dfrac{\pi}{2} \: \: where \: n \: is \: integer$

$\rm :\implies\:x = n\pi \: \pm \: \dfrac{\pi}{4} \: \: where \: n \: is \: integer$

As

$\rm :\longmapsto\:x \: \in \: [0,\: 2\pi]$

So,

x can attains the value

$\rm :\longmapsto\:x = 0, \: \dfrac{\pi}{4}, \: \dfrac{3\pi}{4}, \: \dfrac{5\pi}{4}, \: \dfrac{7\pi}{4}, \: 2\pi$

Now, we have to find out the value of f(x), on these critical points.

As,

$\rm :\longmapsto\:f(x) = sin2x$

So,

$\red{\rm :\longmapsto\:f(0) = sin0 = 0}$

$\red{\rm :\longmapsto\:f\bigg(\dfrac{\pi}{4} \bigg) = sin2\bigg(\dfrac{\pi}{4} \bigg) = sin\bigg(\dfrac{\pi}{2} \bigg) = 1}$

$\red{\rm :\longmapsto\:f\bigg(\dfrac{3\pi}{4} \bigg) = sin2\bigg(\dfrac{3\pi}{4} \bigg) = sin\bigg(\dfrac{3\pi}{2} \bigg) } \\ \red{ \rm \: = \: sin\bigg(\pi + \dfrac{\pi}{2} \bigg) = - sin\bigg(\dfrac{\pi}{2} \bigg) = - 1}$

$\red{\rm :\longmapsto\:f\bigg(\dfrac{5\pi}{4} \bigg) = sin2\bigg(\dfrac{5\pi}{4} \bigg) = sin\bigg(\dfrac{5\pi}{2} \bigg) } \\ \red{ \rm \: = \: sin\bigg(2\pi + \dfrac{\pi}{2} \bigg) = sin\bigg(\dfrac{\pi}{2} \bigg) = 1}$

$\red{\rm :\longmapsto\:f\bigg(\dfrac{7\pi}{4} \bigg) = sin2\bigg(\dfrac{7\pi}{4} \bigg) = sin\bigg(\dfrac{7\pi}{2} \bigg) } \\ \red{ \rm \: = \: sin\bigg(3\pi + \dfrac{\pi}{2} \bigg) = - sin\bigg(\dfrac{\pi}{2} \bigg) = - 1}$

$\red{\rm :\longmapsto\:f(2\pi) = sin2\pi = 0}$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf f(x) = sin2x \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 0 \\ \\ \sf \dfrac{\pi}{4} & \sf 1 \\ \\ \sf \dfrac{3\pi}{4} & \sf - 1\\ \\ \sf \dfrac{5\pi}{4} & \sf 1\\ \\ \sf \dfrac{7\pi}{4} & \sf - 1 \\ \\ \sf 2\pi & \sf 0\end{array}} \\ \end{gathered}$

So,

$\boxed{ \bf{ \: Maximum \: value \: of \: f(x) = 1 \: at \: x \: = \: \bigg(\dfrac{\pi}{4} \bigg), \: \bigg(\dfrac{5\pi}{4} \bigg)}}$