Question

If $f\left(x\right) = \frac{x}{sin\,x}$ and $g\left(x\right) = \frac{x}{tan\,x}, 0 < x \le 1$, then in the interval,

• A. both $f(x)$ and $g(x)$ are increasing functions
• B. both $f(x)$ and $g(x)$ are decreasing functions
• C. $f(x)$ is an increasing function
• D. $g(x)$ is an increasing function

Answer 1

Answer: (C)

$f'\left(x\right) = \frac{sin\,x - x\,cos\,x}{sin^{2}\,x}$,
$g'\left(x\right) = \frac{tan\,x - x\,sec^{2}\,x}{tan^{2}\,x}$
Now, $\frac{d}{dx}\left(sin\,x - x\,cos\,x\right) = x\,sin\,x > 0$ for $0 < x \le 1$
$\therefore sin\,x - x\,cos\,x$ is an increasing function.
Also, at $x = 0$, $sin\, x - x\, cos\, x = 0$
$\Rightarrow 0 < x \le 1$, $sin\, x-x \,cos \, x > 0$
$\therefore f\left(x\right) > 0$ for $0 < x \le 1$
$\therefore f\left(x\right)$ is increasing on the interval $(0, 1]$.
Again, $\frac{d}{dx}\left(tan\,x - x\,sec^{2}\,x\right) = -x \cdot 2\,sec^{2}\,x\cdot tan\,x < 0$
for $0 < x \le 1$
$\Rightarrow tan\,x - x\,sec^{2}\,x$ is an decreasing function.
Also at $x = 0$, $tan\,x - x\,sec^{2}\,x = 0$
$\Rightarrow 0 < x \le 1, tan\,x - x\,sec^{2}\,x < 0$
$g\left(x \right) < 0$ for $0 < x \le 1$.
$\therefore g\left(x\right)$ is decreasing in $(0,1]$.