Question

$f\left(x\right)=\sin x$ has a local minima at x = in the interval $[0,2\pi ]$

• A. $\frac{\pi }{2}$
• B. $\frac{3\pi }{2}$
• C. $\frac{2\pi }{3}$
• D. $\frac{\pi }{3}$

Answer 1

The correct option is B $\frac{3\pi }{2}$

If f(x) has a local minima at x = a, then
f'(a) = 0 and f''(a) > 0

Given, f(x) = sin x
f'(x) = cos x
f''(x) = –sin x.

When f'(x) = 0, cos x = 0.
$\Rightarrow x=\frac{\pi }{2}$ or $x=\frac{3\pi }{2}$
(The values in the interval $[0,2\pi ]$)

When $x=\frac{\pi }{2}$, f''(x) = $-\sin \frac{\pi }{2}$, which is –1

When $x=\frac{3\pi }{2}$, f''(x) = $-\sin \frac{3\pi }{2}$, which is –(–1) = 1

Hence, when $x=\frac{3\pi }{2}$, f'(x) = 0 and f''(x) > 0. Thus, f(x) has a local minima at $x=\frac{3\pi }{2}$.