Question

If ${\ell }^2+m^2=1$ then the maximum value of $\ell +m$ is

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

Answer 1

Answer: (B)

$\sqrt{2}$

$l^2+m^2=1\Rightarrow m^{2}1-l^2\Rightarrow m=\sqrt{1-l^2}$

Let

$y=l+m=l+\sqrt{1-l^2}\frac{dy}{dl}=1+\frac{1}{2\sqrt{1-l^2}}\times -2l\frac{dy}{dl}=0\Rightarrow 1-\frac{2l}{2\sqrt{1-l^2}}=01=\frac{l}{\sqrt{1-l^2}}\Rightarrow 1-l^2=l^2\Rightarrow x^2=1l=\frac{1}{\sqrt{2}}$

$\therefore$ maximum value of $l+m=l+\sqrt{1-l^2}=\frac{1}{\sqrt{2}}+\sqrt{1-\frac{1}{(\sqrt{2}{)}^2}}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$