If 0

Question

(B) Consider the function $f(x) = \frac{\sin x}{x}$ for $x \in (0, \pi/2)$.Taking the derivative,$f'(x) = \frac{x \cos x - \sin x}{x^2}$.Let $u(x) = x

Vedclass · Accepted Answer

$\frac{2}{\pi} < \frac{\sin x}{x}$

Vedclass · Answer

(B) Consider the function $f(x) = \frac{\sin x}{x}$ for $x \in (0, \pi/2)$.Taking the derivative,$f'(x) = \frac{x \cos x - \sin x}{x^2}$.Let $u(x) = x \cos x - \sin x$. Then $u'(x) = \cos x - x \sin x - \cos x = -x \sin x$.Since $x \in (0, \pi/2)$,$u'(x) Since $u(0) = 0$ and $u(x)$ is decreasing,$u(x) Thus,$f'(x) As $x 	o 0^+$,$f(x) 	o 1$,and at $x = \pi/2$,$f(\pi/2) = \frac{\sin(\pi/2)}{\pi/2} = \frac{2}{\pi}$.Since $f(x)$ is decreasing,for $0 Therefore,$\frac{2}{\pi} Hence,option $B$ is correct.

If $0 < x < \pi / 2$,then

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