Question

lf $I^2+m^2=1$, then the $max$ values of $I+m$ is

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

Answer 1

The correct option is B $\sqrt{2}$

$f\left(I\right)=I+\sqrt{1-I^2}$
$f^{\prime }\left(I\right)=1-\frac{I}{\sqrt{1-I^2}}$
$0=\sqrt{1-I^2}-I$
$I=\pm \frac{1}{\sqrt{2}}$ $m=\pm \frac{1}{\sqrt{2}}$
$f\left(x\right)$ is max when $I=\frac{1}{\sqrt{2}}$
$I+m=\sqrt{2}$