Question

If $a\cdot 3^{tanx}+a\cdot 3^{-tanx}-2=0$ has real solution, $x\ne \frac{\pi }{2},0\le x\le \pi ,$ then the set of possible values of the parameter $a$ are

• A. $[-1,1]$
• B. $[-1,0)$
• C. $(0,1]$
• D. $(0,+\infty )$

Answer 1

The correct option is C $(0,1]$

Let $y=3^{\tan x}$
$\Rightarrow ay+\frac{a}{y}-2=0$
$\Rightarrow a{y}^2-2y+a=0$
Since, $0\le x\le \pi \Rightarrow -\infty <\tan x<\infty$
$\Rightarrow 0<3^{\tan x}<\infty$
$\Rightarrow 00$
Discriminant $\ge 0$
$\Rightarrow 8-8a^2\ge 0$
$\Rightarrow a^2-1⩽0$
$\Rightarrow a\in [-1,1]$
So, the intersection gives the solution set $a\in (0,1]$.
Hence, option 'C' is correct.