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

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== Problem ==
 
== Problem ==
  
If <math>f(x)=\frac{x^4+x^2}{x+1}</math>, then <math>f(I)</math>, where <math>I=\sqrt{-1}</math>, is equal to
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If <math>f(x)=\frac{x^4+x^2}{x+1}</math>, then <math>f(i)</math>, where <math>i=\sqrt{-1}</math>, is equal to
  
 
<math>\text{(A) } 1+i\quad
 
<math>\text{(A) } 1+i\quad
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== Solution ==
 
== Solution ==
<math>\fbox{D}</math>
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<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 ==
 
== See also ==
{{AHSME box|year=1970|num-b=4|num-a=6}}   
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{{AHSME 35p box|year=1970|num-b=4|num-a=6}}   
  
 
[[Category: Introductory Algebra Problems]]
 
[[Category: Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 03:06, 12 March 2017

Problem

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

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

Solution

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

See also

1970 AHSC (ProblemsAnswer KeyResources)
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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