6. If $x$ is a real number and $-1 \leq x \leq 1$, find the maximum value of $x+\sqrt{1-x^{2}}$.
(1 mark)若 $x$ 是實數, 且 $-1 \leq x \leq 1$, 求 $x+\sqrt{1-x^{2}}$ 的最大值。
6. $\sqrt{2}$
6. Let $x=\cos \theta$ where $0 \leq \theta \leq \pi$. Then $x+\sqrt{1-x^{2}}=\cos \theta+\sin \theta=\sqrt{2} \cos \left(\theta-\frac{\pi}{4}\right)$. Since the maximum value of the cosine function is 1 , the maximum value of $x+\sqrt{1-x^{2}}$ is $\sqrt{2}$, and equality is attained when $\theta=\frac{\pi}{4}$, i.e. when $x=\frac{\sqrt{2}}{2}$.
Remark. A solution using calculus obviously exists.