If the equation $\sin (x)+\cos (x)=k$ has a real solution for $x$, then the range of $k$ is

Solution

1. The given equation is $\sin (x)+\cos (x)=k$.
2. Using the harmonic addition formula, we can rewrite the expression as $$\begin{array}{rl}\sqrt{2}(\frac{1}{\sqrt{2}}\sin (x)+\frac{1}{\sqrt{2}}\cos (x))& =\sqrt{2}\sin (x+\frac{\pi }{4})\\ & =k\end{array}$$.
3. Since the range of the function $\sin (x+\frac{\pi }{4})$ is $[-1,1]$, the range of $\sqrt{2}\sin (x+\frac{\pi }{4})$ is $[-\sqrt{2},\sqrt{2}]$.
4. Hence the answer is (B).