The points in the interval [0,2π] , where the function f(x)=cos 2x+2s…

The correct answer is Option (3) → B and C only

Since \( 2\sin x\cos x=\sin 2x \), the function is \( f(x)=\cos 2x+\sin 2x \).

    \( \displaystyle f'(x)=-2\sin 2x+2\cos 2x \)
    \( \displaystyle f''(x)=-4\cos 2x-4\sin 2x=-4f(x) \)

f′(x) = 0 gives tan 2x = 1. For x in [0, 2π], 2x lies in [0, 4π]:

    \( \displaystyle 2x=\frac{\pi}{4},\ \frac{5\pi}{4},\ \frac{9\pi}{4},\ \frac{13\pi}{4} \)

    \( \displaystyle x=\frac{\pi}{8},\ \frac{5\pi}{8},\ \frac{9\pi}{8},\ \frac{13\pi}{8} \)

Since f″(x) = −4f(x), a critical point is a local minimum when f is negative there:

    \( \displaystyle x=\frac{\pi}{8}:\ f=\cos\frac{\pi}{4}+\sin\frac{\pi}{4}=\sqrt{2}\ \text{(local maximum)} \)

    \( \displaystyle x=\frac{5\pi}{8}:\ f=\cos\frac{5\pi}{4}+\sin\frac{5\pi}{4}=-\sqrt{2}\ \text{(local minimum)} \)

    \( \displaystyle x=\frac{9\pi}{8}:\ f=\cos\frac{9\pi}{4}+\sin\frac{9\pi}{4}=\sqrt{2}\ \text{(local maximum)} \)

    \( \displaystyle x=\frac{13\pi}{8}:\ f=\cos\frac{13\pi}{4}+\sin\frac{13\pi}{4}=-\sqrt{2}\ \text{(local minimum)} \)

So the critical points are π/8, 5π/8, 9π/8 and 13π/8, and the local minima are at 5π/8 and 13π/8.

Checking the statements

(A) \( \pi/8 \) — Incorrect. f(π/8) = √2, a local maximum.

(B) \( 5\pi/8 \) — Correct. f(5π/8) = −√2, a local minimum.

(C) \( 13\pi/8 \) — Correct. f(13π/8) = −√2, a local minimum.

(D) \( 11\pi/8 \) — Incorrect. Not a critical point: \( \tan\frac{11\pi}{4}=-1\ne 1 \).

Shortcut: Write \( f(x)=\sqrt{2}\sin\left(2x+\frac{\pi}{4}\right) \). Its minimum −√2 occurs where \( 2x+\frac{\pi}{4}=\frac{3\pi}{2}+2k\pi \), that is \( x=\frac{5\pi}{8}+k\pi \), giving \( \frac{5\pi}{8} \) and \( \frac{13\pi}{8} \) in [0, 2π].