Question

The minimum value of the function $f(x)=\frac{\tan \left(x+\frac{\pi }{6}\right)}{\tan x}$ is

• A. 0
• B. $1/2$
• C. 1
• D. 3

Answer 1

Answer: (D)

$\mathrm{f}(\mathrm{x})$ has a period equal to $\pi \&$ can take values $(-\infty ,\infty )\Rightarrow 3$ is the local minimum value.
$y=\frac{2\sin \left(x+\frac{\pi }{6}\right)\cos x}{2\sin x\cos \left(x+\frac{\pi }{6}\right)}=\frac{\sin \left(2x+\frac{\pi }{6}\right)+\sin \frac{\pi }{6}}{\sin \left(2x+\frac{\pi }{6}\right)-\sin \frac{\pi }{6}}$
$=1+\frac{1}{\sin \left(2x+\frac{\pi }{6}\right)-\sin \frac{\pi }{6}}$
$\mathrm{y}$ is minimum if $2\mathrm{x}+\frac{\pi }{6}=\frac{\pi }{2}$
$\Rightarrow \mathrm{x}=\frac{\pi }{6}\Rightarrow {\mathrm{y}}_{\min }=1+2=3]$