Question

The number of turning points of the curve $f(x) = 2\cos x - \sin 2x$ in the interval $[-\pi, \pi]$ is

• A. 4
• B. 3
• C. 1
• D. 2

Hint:

Turning points: $f'(x)=0$. Solve the trigonometric equation within the given interval.

Answer 1

The Correct Option is B

The function given is \( f(x) = 2\cos x - \sin 2x \). To find the number of turning points in the interval \([- \pi, \pi]\), we begin by finding its derivative.

The first derivative is:

\[ f'(x) = \frac{d}{dx}(2\cos x - \sin 2x) = -2\sin x - 2\cos 2x \]

Using the double angle identity \(\cos 2x = 2\cos^2 x - 1\), we rewrite the equation:

\[ f'(x) = -2\sin x - 2(2\cos^2 x - 1) = -2\sin x - 4\cos^2 x + 2 \]

Simplifying:

\[ f'(x) = -2(\sin x + 2\cos^2 x - 1) \]

Setting \(f'(x) = 0\) for critical points,

\[ \sin x + 2\cos^2 x - 1 = 0 \]

Using \( \cos^2 x = 1 - \sin^2 x \), substitute:

\[ \sin x + 2(1 - \sin^2 x) - 1 = 0 \]

\[ \sin x + 2 - 2\sin^2 x - 1 = 0 \]

\[ -2\sin^2 x + \sin x + 1 = 0 \]

Let \( u = \sin x \), then:

\[ -2u^2 + u + 1 = 0 \]

Using the quadratic formula, \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = -2\), \(b = 1\), \(c = 1\),

\[ u = \frac{-1 \pm \sqrt{1 + 8}}{-4} = \frac{-1 \pm 3}{-4} \]

Therefore,

\[ u = \frac{2}{-4} = -\frac{1}{2}, \quad u = \frac{-4}{-4} = 1 \]

Turning Points are at \( \sin x = 1, \sin x = -\frac{1}{2} \).
For \(\sin x = 1\), \( x = \frac{\pi}{2} \).
For \(\sin x = -\frac{1}{2}\), \[ x = -\frac{\pi}{6}, -\frac{5\pi}{6} \] within \([- \pi, \pi]\).

Thus, there are 3 turning points: \(x = \frac{\pi}{2}, -\frac{\pi}{6}, -\frac{5\pi}{6}\).

The correct answer is 3.