Difference between revisions of "1970 AHSME Problems/Problem 5"

Latest revision as of 01:06, 12 March 2017

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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 == Solution ==
-<math>\fbox{D}</math>
+<math>i^4 = 1</math> and <math>i^2=-1</math>, so the numerator is <math>0</math>. As long as the denominator is not <math>0</math>, which it isn't, the answer is <math>0 \Rightarrow</math> <math>\fbox{D}</math>
 == See also ==