Question

Shortest distance from origin to the curve $y=\frac{e^x+e^{-x}}{2}$

• A. $1$
• B. $\sqrt{2}$
• C. $3\sqrt{2}$
• D. $4\sqrt{2}$

Answer 1

The correct option is A $1$

Let (x, y) is the point on the curve $y=\frac{e^x+e^{-x}}{2}$ nearer to origin
$\therefore o{p}^2=x^2+y^2=x^2+{\left(\frac{e^x+e^{-x}}{2}\right)}^2$
$Letf\left(x\right)=x^2+\frac{1}{4}(e^{2x}+e^{-2x}+2)$
$\Rightarrow f^{\prime }\left(x\right)=2x+\frac{1}{2}(e^{2x}-e^{-2x})$
For maximum (or) minimum, we have $f^{\prime }\left(x\right)=0\Rightarrow e^{-2x}-e^{2x}=4x$
$\Rightarrow x=0also{f}^{\prime \prime }\left(0\right)>0$
$\therefore$ at x = 0 shortest distance exist i.e ‘1’