Prove that f (x) = sin x + √3 cos x has maximum value

Solution:

Given, $f(x)=\sin x+\sqrt{3} \cos x=2\left(\frac{1}{2} \sin x+\frac{\sqrt{3}}{2} \cos x\right)$

$=2\left(\cos \frac{\pi}{3} \sin x+\sin \frac{\pi}{3} \cos x\right)=2 \sin \left(x+\frac{\pi}{3}\right)$

$f^{\prime}(x)=2 \cos \left(x+\frac{\pi}{3}\right) ; f^{\prime \prime}(x)=-2 \sin \left(x+\frac{\pi}{3}\right)$

$f^{\prime \prime}(x)_{x=\frac{\pi}{6}}=-2 \sin \left(\frac{\pi}{6}+\frac{\pi}{3}\right)$

$=-2 \sin \frac{\pi}{2}=-2.1=-2<0$ (Maxima)

$=-2 \times \frac{\sqrt{3}}{2}=-\sqrt{3}<0$ (Maxima)

Maximum value of the function at $x=\frac{\pi}{6}$ is

$\sin \frac{\pi}{6}+\sqrt{3} \cos \frac{\pi}{6}=\frac{1}{2}+\sqrt{3} \cdot \frac{\sqrt{3}}{2}=2$

Given, $f(x)=\sin x+\sqrt{3} \cos x=2\left(\frac{1}{2} \sin x+\frac{\sqrt{3}}{2} \cos x\right)$

$=2\left(\cos \frac{\pi}{3} \sin x+\sin \frac{\pi}{3} \cos x\right)=2 \sin \left(x+\frac{\pi}{3}\right)$

$f^{\prime}(x)=2 \cos \left(x+\frac{\pi}{3}\right) ; f^{\prime \prime}(x)=-2 \sin \left(x+\frac{\pi}{3}\right)$

$f^{\prime \prime}(x)_{x=\frac{\pi}{6}}=-2 \sin \left(\frac{\pi}{6}+\frac{\pi}{3}\right)$

$=-2 \sin \frac{\pi}{2}=-2.1=-2<0$ (Maxima)

$=-2 \times \frac{\sqrt{3}}{2}=-\sqrt{3}<0$ (Maxima)

Maximum value of the function at $x=\frac{\pi}{6}$ is

$\sin \frac{\pi}{6}+\sqrt{3} \cos \frac{\pi}{6}=\frac{1}{2}+\sqrt{3} \cdot \frac{\sqrt{3}}{2}=2$

Therefore, the given function has maximum value at x = π/6 and the maximum value is 2.