Question
If $$a \cdot {3^{\tan x}} + a \cdot {3^{ - \tan x}} - 2 = 0$$ has real solutions, $$x \ne \frac{\pi }{2},0 \leqslant x \leqslant \pi ,$$ then the set of possible values of the parameter $$a$$ is
A.
$$\left[ { - 1,1} \right]$$
B.
$$\left[ { - 1,0} \right)$$
C.
$$\left( {0,1} \right]$$
D.
$$\left( {0, + \infty } \right)$$
Answer :
$$\left( {0,1} \right]$$
Solution :
$$\eqalign{
& {\text{Let }}{3^{\tan x}} = y.\,{\text{Then }}ay + \frac{a}{y} - 2 = 0\,\,{\text{or, }}a{y^2} - 2y + a = 0. \cr
& D \geqslant 0 \cr} $$
$$ \Rightarrow \,\,4 - 4{a^2} \geqslant 0.$$ Also roots are positive as $$y = {3^{\tan x}} > 0.$$
∴ sum of the root $$ = \frac{2}{a} > 0$$
$$ \Rightarrow \,\,a > 0.$$