1970 AHSME Problems/Problem 5

Problem

If $f(x)=\frac{x^4+x^2}{x+1}$ , then $f(i)$ , where $i=\sqrt{-1}$ , is equal to

$\text{(A) } 1+i\quad \text{(B) } 1\quad \text{(C) } -1\quad \text{(D) } 0\quad \text{(E) } -1-i$

$i^4 = 1$ and $i^2=-1$ , so the numerator is $0$ . As long as the denominator is not $0$ , which it isn't, the answer is $0 \Rightarrow$ $\fbox{D}$

1970 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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