Question

Maximum value of $f\left(x\right)=\sin x+\cos x$ is

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

Answer 1

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

$y=f\left(x\right)=\sin x+\cos x$
$\frac{dy}{dx}=\cos x-sinx$

For $\frac{dy}{dx}=0,\sin x=\cos x,\tan x=1$
$\therefore x={45}^{\circ }$

$y=\sin {45}^{\circ }+\cos {45}^{\circ }$
$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$