Question

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

• A. $a\in R$
  
• B. $a\in [\frac{-e}{4},\frac{-1}{4}]$
  
• C. $a\in \left[\frac{-1-e^2}{4e}\right],\infty$
  
• D. None

Answer 1

The correct option is D

None

Let $e^{|sinx|}=tt\in [1,e]\therefore t+\frac{1}{t}+4a=0\Rightarrow t^2+4at+1=0.\therefore$ This quadratic equation has 2 distinct roots in [1,e] as x$\in [0,2\pi ]\Rightarrow 16a^2-4>0,1+4a+1\ge 0,e^2+4at+1\ge 0,1<-2a<e|a|>\frac{1}{2},a\ge \frac{-1}{2},a\ge \frac{-1-e^2}{4e},\frac{-e}{2}<a<\frac{-1}{2}\therefore$ No value satisfy the above inequalities simultaneously.