Practicing Success
The function $f(x)=\frac{\sin x}{x}$ is decreasing in the interval (a) $(-\pi / 2,0)$ |
(a), (c) (a), (d) (b), (c) (b), (d) |
(b), (c) |
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 $(0, \pi)$ |