The function f(x)=(sin x)/x is decreasing in the interval (a) (-π / 2…

We have,

$f(x)=\frac{\sin x}{x}$

$\Rightarrow f'(x)=\frac{x \cos x-\sin x}{x^2}$

$\Rightarrow f'(x)=\frac{g(x)}{x^2}$, where $g(x)=x \cos x-\sin x$

Now, $g'(x)=-x \sin x$

Consider the interval $(-\pi / 2,0)$

In this interval we observe that

$g'(x)<0$

$\Rightarrow g(x)$ is decreasing on $(-\pi / 2,0)$

$\Rightarrow g(x)>g(0)$ for all $x \in(-\pi / 2,0)$

$\Rightarrow g(x)>0$ for all $x \in(-\pi / 2,0)$

∴ $f'(x)=\frac{g(x)}{x^2}>0$ for all $x \in(-\pi / 2,0)$

⇒ f(x) is increasing on $(-\pi / 2,0)$

Consider now the interval $(0, \pi / 2)$

In this interval, we have

g'(x) < 0 for all $x \in(0 \pi / 2)$

⇒ g(x) is decreasing on $(0, \pi / 2)$

⇒ g(x) < g(0) for all $x \in(0, \pi / 2)$

⇒ g(x) < 0 for all $x \in(0, \pi / 2)$

⇒ f'(x) < 0 for $x \in(0, \pi / 2)$

⇒ f(x) is decreasing on $(0, \pi / 2)$

For all $x \in(0, \pi)$, we observe that

g'(x) < 0

⇒ g(x) is decreasing on $(0, \pi)$ 

⇒ g(x) < g(0) for all $x \in(0, \pi)$

⇒ g(x) < 0 for all $x \in(0, \pi)$ 

⇒ f'(x) < 0 for all $x \in(0, \pi)$

⇒ f(x) is decreasing on $(0, \pi)$