Question

Let $f(x)=x\left(e^{x^2}-e^{-x^2}\right)-2 x-\int\left(e^{x^2}-e^{-x^2}\right) d x$. If $f(x)$ is decreasing in $\left(x_1, x_2\right)$ then $\left(x_1+x_2\right)$ equals

• A. 0
• B. $2 e$
• C. $\frac{ e ^2-1}{ e }$
• D. $\frac{3}{2} e$

Answer 1

Answer: (A)

$f(x)=x\left(e^{x^2}-e^{-x^2}\right)-2 x-\int\left(e^{x^2}-e^{-x^2}\right) d x$

$f ^{\prime}( x )=2 x ^2\left( e ^{ x ^2}+ e ^{- x ^2}\right)+\left( e ^{ x ^2}- e ^{- x ^2}\right)-2- e ^{ x ^2}+ e ^{- x ^2} $
$f ^{\prime}( x )=2 x ^2\left( e ^{ x ^2}- e ^{- x ^2}\right)-2<0$
$\therefore e ^{ x ^2}+ e ^{- x ^2}<\frac{1}{ x ^2}$
$e ^{ t }+ e ^{- t }<\frac{1}{ t } ; t >0$
decreasing in (-a, a)
$\therefore x _1+ x _2=0$