3. If $f(x)=x^{2}+\sqrt{1-x^{2}}$ where $-1 \leq x \leq 1$, find the range of $f(x)$.
(A) $\frac{1}{2} \leq f(x) \leq 1$
(B) $1 \leq f(x) \leq \frac{5}{4}$
(C) $1 \leq f(x) \leq \frac{1+2 \sqrt{3}}{4}$
(D) $\frac{\sqrt{3}}{2} \leq f(x) \leq 1$
(E) $\frac{1}{2} \leq f(x) \leq \frac{\sqrt{3}}{2}$
3. Answer: (B)
Let $x^{2}+\sqrt{1-x^{2}}=M$. Then $\left(x^{2}-M\right)^{2}=1-x^{2}$, which leads to $x^{4}-(2 M-1) x^{2}+M^{2}-1=0$. For real $x$, we must have $(2 M-1)^{2}-4\left(M^{2}-1\right)>0$, which gives $M \leq \frac{5}{4}$.
Note that $f(-1)=f(0)=f(1)=1$. Assume that $f(x)<1$ for some $x \neq-1,0,1$. Then $x^{2}+\sqrt{1-x^{2}}<1$, or $\sqrt{1-x^{2}}<1-x^{2}$, which is not possible since $-1<x<1$. Thus $f(x) \geq 1$, and hence $1 \leq f(x) \leq \frac{5}{4}$.