Question

If $e^{|\sin x|}+e^{-|\sin x|}+4a=0$ will have exactly $4$ different solutions in $[0,\pi ]$, then

• A. $a\in (-\frac{e}{2},-\frac{1}{2})$
• B. $a\in (\frac{1}{2},\infty )$
• C. $a\in (-\infty ,-\frac{1}{2})$
• D. $a\in \varphi$

Answer 1

The correct option is D $a\in \varphi$

Given : $e^{|\sin x|}+e^{-|\sin x|}+4a=0$
Let $t=e^{|\sin x|}\Rightarrow t\in [1,e]$
Now,
$t+\frac{1}{t}+4a=0\Rightarrow t^2+4at+1=0\cdots \left(1\right)$
For every $t\in (1,e)$, there is two possible solutions of $x\in [0,\pi ]$
For the equation to have $4$ solutions, equation $\left(1\right)$ has two distinct real roots in between $(1,e)$, so

$\left(1\right)D>0\Rightarrow 16a^2-4>0\Rightarrow a^2>\frac{1}{4}\Rightarrow a\in (-\infty ,-\frac{1}{2})\cup (\frac{1}{2},\infty )\left(2\right)f\left(1\right)>0\Rightarrow 1+4a+1>0\Rightarrow a>-\frac{1}{2}\left(3\right)f\left(e\right)>0\Rightarrow e^2+4ae+1>0\Rightarrow a>-\frac{e^2+1}{4e}\left(4\right)1<-\frac{b}{2a}<e\Rightarrow 1<-2a<e\Rightarrow -\frac{e}{2}<a<-\frac{1}{2}$
From above conditions, we get
$a\in \varphi$

Checking the boundary points,
When roots are $t=1$ and $t=e$, so
$e^{|\sin x|}=1\Rightarrow x=0,\pi e^{|\sin x|}=e\Rightarrow x=\frac{\pi }{2}$
We get $3$ solutions.

Hence, there is no value of $a$ for which the given equation has $4$ different solution.