What is the x-coordinate of the maximum stationary point of the curve y = \sin(2x) - x in … — JAMB UTME

What is the $x$ -coordinate of the maximum stationary point of the curve $y = \sin(2x) - x$ in the interval $0 \le x \le \pi$ ?

The

x

-coordinate of the maximum stationary point is

\frac{\pi}{6}

Differentiating

y = \sin(2x) - x

gives

\frac{dy}{dx} = 2\cos(2x) - 1

. Setting this derivative to zero yields

\cos(2x) = \frac{1}{2}

, giving

x = \frac{\pi}{6}

and

x = \frac{5\pi}{6}

in the given interval. Checking the second derivative

\frac{d^2y}{dx^2} = -4\sin(2x)

x = \frac{\pi}{6}

gives

-2\sqrt{3} < 0

, which confirms that the maximum stationary point occurs at

\frac{\pi}{6}

Find the first derivative of

y = \sin(2x) - x

using the chain rule.

\frac{dy}{dx} = 2\cos(2x) - 1

Stationary points occur where the rate of change

\frac{dy}{dx} = 0

Set

\frac{dy}{dx} = 0

and solve for

x

within

0 \le x \le \pi

2\cos(2x) - 1 = 0 \implies \cos(2x) = \frac{1}{2} \implies 2x = \frac{\pi}{3}

2x = \frac{5\pi}{3} \implies x = \frac{\pi}{6}

x = \frac{5\pi}{6}

The trigonometric equation

\cos(\theta) = \frac{1}{2}

has solutions

\frac{\pi}{3}

and

\frac{5\pi}{3}

[0, 2\pi]

Evaluate the second derivative

\frac{d^2y}{dx^2}

to determine the nature of the stationary points.

\frac{d^2y}{dx^2} = -4\sin(2x)

. At

x = \frac{\pi}{6}

\frac{d^2y}{dx^2} = -4\sin\left(\frac{\pi}{3}\right) = -2\sqrt{3} < 0

A negative second derivative (

\frac{d^2y}{dx^2} < 0

) indicates a local maximum.

Determining maximum stationary points of trigonometric functions using the first and second derivative tests.

Soru