Q.Find the value of the following: At what points in the interval $[0, 2\pi]$, does the function $\sin 2x$ attain its maximum value?

Concept: Critical Points Analysis — we find where the derivative is zero or undefined, then check endpoints and evaluate.

1. Let $f(x) = \sin 2x$ on $[0, 2\pi]$.

   $f'(x) = 2\cos 2x$. Set $f'(x) = 0$:

   $\cos 2x = 0 \implies 2x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$

   So $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.

2. Evaluate $f$ at these critical points and the endpoints $0, 2\pi$:

   $f(0) = 0$, $f(2\pi) = 0$,

   $f(\pi/4) = \sin(\pi/2) = 1$,

   $f(3\pi/4) = \sin(3\pi/2) = -1$,

   $f(5\pi/4) = \sin(5\pi/2) = 1$, …

We need to find where $\sin 2x$ reaches its highest value in the closed interval $[0, 2\pi]$. The sine function itself peaks at $1$, so we’re really asking: for which $x$ in $[0, 2\pi]$ does $\sin 2x = 1$? But we must also consider that the maximum could occur at the boundaries of the interval, so a full critical-point analysis is the reliable method.

The key idea: for a continuous function on a closed interval, the maximum occurs either at critical points (where derivative is zero or undefined) or at the endpoints. Since $\sin 2x$ is differentiable everywhere, we only need to find where its derivative vanishes and then compare function values.

1. Find the derivative and critical points. Let $f(x) = \sin 2x$. Then $f'(x) = 2\cos 2x$. Set $f'(x) = 0$:

In the interval $[0, 2\pi]$, $2x$ ranges from $0$ to $4\pi$. The cosine function is zero at odd multiples of $\frac{\pi}{2}$:

Solving for $x$:

These four points are the critical points inside $(0, 2\pi)$.

2. Evaluate $f(x)$ at critical points and endpoints. Endpoints: $x = 0$ and $x = 2\pi$.

Critical points:

$f\left(\frac{7\pi}{4}\right) = \sin\left(2 \cdot \frac{7\pi}{4}\right) = \sin\frac{7\pi}{2} = \sin\left(3\pi + \frac{\pi}{2}\right) = -1.$ …

Method: Locating Where a Periodic Function Attains Its Maximum on a Closed Interval

This is the closed-interval candidates test specialised to a trigonometric function whose argument is scaled (like $\sin 2x$ instead of $\sin x$), where the key extra care is in correctly solving the trig equation over the full given range.

Steps

Step 1: Differentiate and set the derivative equal to zero.

For $f(x)=\sin(kx)$, $f'(x)=k\cos(kx)=0 \implies \cos(kx)=0$.

Step 2: Solve for the argument $kx$ over its full extended range, not just one period.

Because the interval is given in terms of $x$ (here $[0,2\pi]$), the argument $kx$ sweeps through a wider range (here $[0,4\pi]$ for $k=2$) — list every solution of $\cos(kx)=0$ in that whole extended range before dividing back by $k$ to recover $x$.

Step 3: Evaluate $f$ at every resulting critical point plus the two endpoints. …

Common Mistakes

Mistake 1: Confusing the period of $\sin 2x$ with that of $\sin x$.

Why it's wrong: $\sin x$ completes one cycle over $[0,2\pi]$ and peaks once, but $\sin 2x$ completes two full cycles over the same interval and therefore peaks twice — assuming a single peak at $x=\frac{\pi}{2}$ (borrowed from $\sin x$'s behaviour) misses the second, genuine peak. Correct approach: always substitute the scaled argument $2x$ into the standard $\sin$-peak condition, rather than reusing the unscaled peak location.

Mistake 2: Stopping after finding the first critical point instead of listing every solution across the full interval. …