Question

If $f\left(x\right)=\frac{\tan (x+\frac{\pi }{6})}{\tan x}$ attains local minimum at $x=a\pi$ in the interval $(\frac{\pi }{3},\pi )$ and the local minimum value is $b,$ then the value of $a+b$ is

• A. $0$
• B. $\frac{7}{2}$
• C. $1$
• D. $3$

Answer 1

The correct option is C $1$

$f\left(x\right)=\frac{\tan (x+\frac{\pi }{6})}{\tan x}=\frac{2\sin (x+\frac{\pi }{6})\cos x}{2\sin x\cos (x+\frac{\pi }{6})}=\frac{\sin (2x+\frac{\pi }{6})+\sin \frac{\pi }{6}}{\sin (2x+\frac{\pi }{6})-\sin \frac{\pi }{6}}=1+\frac{1}{\sin (2x+\frac{\pi }{6})-\sin \frac{\pi }{6}}$

$f\left(x\right)$ attains local minimum if $2x+\frac{\pi }{6}=\frac{3\pi }{2}$ or, $x=\frac{2\pi }{3}$
and $f\left(\frac{2\pi }{3}\right)=1-\frac{2}{3}=\frac{1}{3}$
$\therefore a+b=1$