f(x) = 2 sin x | cos x |, x ∈ [0, 6π]
f(x) = \(\begin{cases}2\,sin\,x\,cos\,x,&x\,\in[0,\frac{\pi}{2}]\cup[\frac{3\pi}{2},\frac{5\pi}{2}]\cup[\frac{7\pi}{2},\frac{9\pi}{2}]\cup[\frac{11\pi}{2},6\pi]\\-2\,sin\,x\,cos\,x,&x\,\in[\frac{\pi}{2},\frac{3\pi}{2}]\cup[\frac{5\pi}{2},\frac{7\pi}{2}]\cup[\frac{9\pi}{2},\frac{11\pi}{2}]\end{cases}\)
⇒ f(x) = \(\begin{cases}sin\,2x,&x\,\in[0,\frac{\pi}{2}]\cup[\frac{3\pi}{2},\frac{5\pi}{2}]\cup[\frac{7\pi}{2},\frac{9\pi}{2}]\cup[\frac{11\pi}{2},6\pi]\\-sin\,2x,&x\,\in[\frac{\pi}{2},\frac{3\pi}{2}]\cup[\frac{5\pi}{2},\frac{7\pi}{2}]\cup[\frac{9\pi}{2},\frac{11\pi}{2}]\end{cases}\)
By differentiating f(x) w.r.t. x, we obtain,
f'(x) = \(\begin{cases}2\,cos\,2x,&x\,\in[0,\frac{\pi}{2}]\cup[\frac{3\pi}{2},\frac{5\pi}{2}]\cup[\frac{7\pi}{2},\frac{9\pi}{2}]\cup[\frac{11\pi}{2},6\pi]\\-2\,cos\,2x,&x\,\in[\frac{\pi}{2},\frac{3\pi}{2}]\cup[\frac{5\pi}{2},\frac{7\pi}{2}]\cup[\frac{9\pi}{2},\frac{11\pi}{2}]\end{cases}\)
For critical points, we have to find these x which gives f'(x) = 0
Now, f'(x) = 0 gives cos 2x = 0 = cos(2n+1)π/2
⇒ 2x = (2n+1)π/2
⇒ x = (2n+1)π/4, n ∈ 0, 1, 2, .... , 12.
Hence, number of critical points are 13.
Alternative :→ f(x) = 2 sin x | cos x |
⇒ f(x) = | sin 2x |
Then f'(x) = | 2 cos 2x |
Thus, f'(x) = 0 gives | cos 2x | = 0
⇒ cos 2x = 0 (∵ |x| = 0 ⇒ x = 0)
⇒ cos 2x = 0 = cos(2n+1)π/2
⇒ 2x = (2n+1)π/2
⇒ x = (2n+1)π/4
Since, x ∈ [0, 6π]
∴ x = (2n+1)π/4, n = 0, 1, 2, .... , 12 are critical points of function f(x).
Hence, number of critical points = 13.