Question

If $e^{sinx}-e^{-sinx}=a$ has alteast one real solution, then

• A. $|a|\in (\frac{e^2-1}{e},\infty )$
• B. $|a|\in [0,\frac{e^2-1}{e}]$
• C. $|a|\in (e,\infty )$
• D. $|a|\in [0,e]$

Answer 1

The correct option is B

$|a|\in [0,\frac{e^2-1}{e}]$

$e^{sinx}-e^{-sinx}=a..............\left(1\right)$
Let $e^{sinx}=t\Rightarrow t\in [\frac{1}{e},e]$
$\therefore \left(1\right)\Rightarrow t-\frac{1}{t}-a=t^2-at-1=0$
Discriminant $=a^2+4>0\forall$ a $\in$ R.
Also product of roots =-1
i.e. both the roots can't belong to $[\frac{1}{e},e]$
Let it have only one root in $[\frac{1}{e},e]$, then
$(e^2-ae-1)(\frac{1}{e^2}-\frac{a}{e}-1)\le 0=\left[\right(e^2-1)-ae]\left[\right(e^2-1)+ae]\ge 0\Rightarrow |a|\le \frac{e^2-1}{e}$