Question

The maximum value of the function $\sin x+\cos x$ is

• A. 1
• B. $\sqrt{2}$
• C. 2
• D. None of these

Answer 1

Answer: (B)

Let $f(x)=\sin x+\cos x\Rightarrow f^{\prime }(x)=\cos x-\sin x$
and $f^{\prime \prime }(x)=-\sin x-\cos x=-(\sin x+\cos x)$
For maxima or minima put $f^{\prime }(x)=0$.
$\therefore \cos x-\sin x=0$
$\Rightarrow \sin x=\cos x$
$\Rightarrow \frac{\sin x}{\cos x}=1$
$\Rightarrow \tan x=1$
$\Rightarrow x=\frac{\pi }{4},\frac{5\pi }{4},\dots$
Now, $f^{\prime \prime }(x)$ will be negative when $(\sin x+\cos x)$ is positive i.e., when $\sin x$ and $\cos x$ are both positive. Also, we know that $\sin x$ and $\cos x$ both are positive in the first quadrant.
Then, $f^{\prime \prime }(x)$ will be negative when $x\in \left(0,\frac{\pi }{2}\right)$.
Thus, we consider $x=\frac{\pi }{4}$.
$f^{\prime \prime }\left(\frac{\pi }{4}\right)=-\left(\sin \frac{\pi }{4}+\cos \frac{\pi }{4}\right)$
$=-\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=-\frac{2}{\sqrt{2}}=-\sqrt{2}<0$
By second derivative test, $f$ will be maximum at $x=\frac{\pi }{4}$ and the maximum value of $f$ is
$f\left(\frac{\pi }{4}\right)=\sin \frac{\pi }{4}+\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$