Question

The shortest distance of (0, 0) from the curve $y=\frac{e^x+e^{-x}}{2}$ is

• A. 1/2
• B. 1/3
• C. 2
• D. none of these

Answer 1

The correct option is D none of these

$point(0,0),curvey=\frac{e^x+e^{-x}}{2}le{t}^{\prime }stakeapoint(x_1,\frac{e^{x_1}+e^{{-x}_1}}{2})oncurve.sodistancet=\sqrt{{x_1}^2+{\left(\frac{e^{x_1}+e^{{-x}_1}}{2}\right)}^2}\left(betweenpointoncurveandorigin\right)t=\sqrt{{x_1}^2+\left(\frac{e^{{2x}_1}+e^{{-2x}_1}+2}{4}\right)}forttobeminimum{t}^{\prime }=0\frac{d\left(t\right)}{dt}=\frac{-1}{2\sqrt{t}}(2x_1+\frac{2e^{{2x}_1}}{4}-\frac{2e^{{2x}_1}}{4})=0itcanonlybe0for{x}_1=0soatx=0t=1$