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Difficulty: MediumStationary Points, Maxima, and Minima

What is the x-coordinate of the maximum stationary point of the curve y=sin⁡x+cos⁡xy = \sin x + \cos x in the interval 0≤x≤π0 \le x \le \pi?

  1. π4\frac{\pi}{4}Answer
  2. B
    3π4\frac{3\pi}{4}
  3. C
    π2\frac{\pi}{2}
  4. D
    π\pi

Answer

The x-coordinate of the maximum stationary point is π4\frac{\pi}{4}.
To find stationary points, we differentiate y=sin⁡x+cos⁡xy = \sin x + \cos x to get dydx=cos⁡x−sin⁡x\frac{dy}{dx} = \cos x - \sin x. Setting dydx=0\frac{dy}{dx} = 0 gives sin⁡x=cos⁡x\sin x = \cos x, or tan⁡x=1\tan x = 1. Within 0≤x≤π0 \le x \le \pi, the solution is x=π4x = \frac{\pi}{4}. Evaluating the second derivative d2ydx2=−sin⁡x−cos⁡x\frac{d^2y}{dx^2} = -\sin x - \cos x at x=π4x = \frac{\pi}{4} yields −2<0-\sqrt{2} < 0, confirming a local maximum.

Step-by-Step Solution

1
Differentiate the function y=sin⁡x+cos⁡xy = \sin x + \cos x with respect to xx.
dydx=cos⁡x−sin⁡x\frac{dy}{dx} = \cos x - \sin x
Stationary points occur where the first derivative dydx=0\frac{dy}{dx} = 0.
2
Set the first derivative to zero and solve for xx in the interval 0≤x≤π0 \le x \le \pi.
\cos x - \sin x = 0 \implies \sin x = \cos x \implies \tan x = 1 \implies x = \frac{\pi}{4}
Dividing both sides by cos⁡x\cos x gives tan⁡x=1\tan x = 1, which has the solution x=π4x = \frac{\pi}{4} in the given domain.
3
Verify the nature of the stationary point using the second derivative test.
\frac{d^2y}{dx^2} = -\sin x - \cos x \implies \left.\frac{d^2y}{dx^2}\right|_{x=\frac{\pi}{4}} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2} < 0
A negative second derivative confirms that x=π4x = \frac{\pi}{4} is a local maximum point.

Key Concept

Finding stationary points and determining their nature using derivatives of trigonometric functions.
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Topics

Stationary Points, Maxima, and Minima

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