Question

If $e^{|\sin x|}+e^{-|\sin x|}+4a=0$ will have exactly four different solutions in $[0,2\pi ]$, then

• A. $a\in R$
• B. $a\in [-\frac{e}{4},-\frac{1}{4}]$
• C. $a\in [\frac{-1-e^2}{4e},\infty ]$
• D. no real value of $a$ exists

Answer 1

The correct option is D no real value of $a$ exists

Given : $e^{|\sin x|}+e^{-|\sin x|}+4a=0$
Let $t=e^{|\sin x|}\Rightarrow t\in [1,e]$
Now,
$t+\frac{1}{t}+4a=0\Rightarrow t^2+4at+1=0$

For the given equation to have four different solutions, this quadratic expression should have two distinct roots in $[1,e]$.
So,required conditions are
$\left(1\right)D>0\Rightarrow 16a^2-4>0\Rightarrow |a|>\frac{1}{2}$
$\left(2\right)f\left(1\right)=1+4a+1\ge 0\Rightarrow a\ge -\frac{1}{2}$
$\left(3\right)f\left(e\right)=e^2+4ae+1\ge 0\Rightarrow a\ge \frac{-1-e^2}{4e}$
$\left(4\right)1<\frac{b}{2a}<e\Rightarrow 1<-2a<e\Rightarrow -\frac{e}{2}<a<-\frac{1}{2}$
Clearly , there is no value of $a$ satisfying the above inequalities simultaneously.