MCQ
The minimum value of $\frac{{\tan \,\left( {x\,\, + \,\,{\textstyle{\pi \over 6}}} \right)}}{{\tan \,x}}$ is
- A$0$
- B$1/2$
- C$1$
- ✓$3$
✓
Answer
Correct option: D.
$3$
d
$f(x)$ has a period equal to $\pi$ & can take values $(-\infty , \infty )$ $==> 3$ is the local minimum value.
$f(x)$ has a period equal to $\pi$ & can take values $(-\infty , \infty )$ $==> 3$ is the local minimum value.
$y = \frac{{2\,\,\sin \,\,\left( {x\,\, + \,\,{\textstyle{\pi \over 6}}} \right)\,\,\cos \,x}}{{2\,\,\sin \,x\,\,\cos \,\,\left( {x\,\, + \,\,{\textstyle{\pi \over 6}}} \right)}} = \frac{{\sin \,\,\left( {2\,x\,\, + \,\,{\textstyle{\pi \over 6}}} \right)\,\, + \,\,\sin \,\,{\textstyle{\pi \over 6}}}}{{\sin \,\,\left( {2\,x\,\, + \,\,{\textstyle{\pi \over 6}}} \right)\,\, - \,\,\sin \,\,{\textstyle{\pi \over 6}}}} $
$ = 1 + \frac{1}{{\sin \,\,\left( {2\,x\,\, + \,\,{\textstyle{\pi \over 6}}} \right)\,\, - \,\,\sin \,\,{\textstyle{\pi \over 6}}}}$
$y$ is minimum if $2 x + \frac{\pi }{6} = \frac{\pi }{2}$
$==> x = \frac{\pi }{6}$ $ ==> y_{min} = 1 + 2 = 3 $
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