Question

The absolute maximum and minimum values of the function \[ f(x)=\sin x+\sqrt3\cos x,\quad x\in[0,\pi] \] are:

• A. Minimum value \(=\dfrac1{\sqrt3}\), maximum value \(=2\)
• B. Minimum value \(=-\sqrt3\), maximum value \(=2\)
• C. Minimum value \(=\sqrt3\), maximum value \(=2\)
• D. Minimum value \(=-\dfrac1{\sqrt3}\), maximum value \(=2\)

Hint:

Convert expressions of the form \[ a\sin x+b\cos x \] into \[ R\sin(x+\alpha) \] to easily determine extrema.

Answer 1

The Correct Option is B

Concept: An expression of the form \[ a\sin x+b\cos x \] can be rewritten in the form: \[ R\sin(x+\alpha) \] where \[ R=\sqrt{a^2+b^2} \] This transformation helps determine maximum and minimum values easily.

Step 1: Rewrite the function. Given: \[ f(x)=\sin x+\sqrt3\cos x \] Compare with: \[ R\sin(x+\alpha) \] We know: \[ R=\sqrt{1^2+(\sqrt3)^2} \] \[ =\sqrt{1+3} \] \[ =2 \] Thus, \[ f(x)=2\sin(x+\alpha) \] Now, \[ \cos\alpha=\frac12, \qquad \sin\alpha=\frac{\sqrt3}{2} \] Therefore, \[ \alpha=\frac{\pi}{3} \] Hence, \[ f(x)=2\sin\left(x+\frac{\pi}{3}\right) \]

Step 2: Determine maximum value. Since \[ -1\le\sin\theta\le1 \] we get: \[ -2\le2\sin\left(x+\frac{\pi}{3}\right)\le2 \] Thus the maximum possible value is: \[ 2 \]

Step 3: Determine minimum value on the interval. For \[ x\in[0,\pi] \] the angle \[ x+\frac{\pi}{3} \] lies in: \[ \left[\frac{\pi}{3},\frac{4\pi}{3}\right] \] The minimum sine value in this interval is: \[ -\frac{\sqrt3}{2} \] Therefore, \[ f_{\min}=2\left(-\frac{\sqrt3}{2}\right) \] \[ =-\sqrt3 \] Thus, \[ \boxed{\text{Minimum value}=-\sqrt3,\ \text{Maximum value}=2} \]