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Difference between revisions of "1995 AHSME Problems/Problem 14"

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<math> \mathrm{(A) \ -5 } \qquad \mathrm{(B) \ -2 } \qquad \mathrm{(C) \ 1 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 8 }  </math>
 
<math> \mathrm{(A) \ -5 } \qquad \mathrm{(B) \ -2 } \qquad \mathrm{(C) \ 1 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 8 }  </math>
  
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== Solution ==
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== Solution 1==
 
<math>f(-x) = a(-x)^4 - b(-x)^2 - x + 5</math>  
 
<math>f(-x) = a(-x)^4 - b(-x)^2 - x + 5</math>  
  
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Thus <math>f(3) = f(-3)-2(-3) = 8 \Rightarrow \mathrm{(E)}</math>.
 
Thus <math>f(3) = f(-3)-2(-3) = 8 \Rightarrow \mathrm{(E)}</math>.
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==Solution 2==
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If <math>f(-3) = 2</math>, then <math>81a - 9b + -3 + 5 = 2</math>.  Simplifying, we get <math>81a - 9b = 0</math>.
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Getting an expression for <math>f(3)</math>, we find <math>f(3) = 81a - 9b + 3 + 5</math>.  Since the first two terms sum up to zero, we get <math>f(3) = 8</math>, which is answer <math>\mathrm{(E)}</math>
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==Solution 3==
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Substituting <math>x = -3</math>, we get
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<cmath>f(-3) = 81a - 9b - 3 + 5 = 81a - 9b + 2.</cmath>
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But <math>f(-3) = 2</math>, so <math>81a - 9b + 2 = 2</math>, which means <math>81a - 9b = 0</math>. Then
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<cmath>f(3) = 81a - 9b + 3 + 5 = 0 + 3 + 5 = \boxed{8}.</cmath>
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 17:05, 6 April 2016

Problem

If $f(x) = ax^4 - bx^2 + x + 5$ and $f( - 3) = 2$, then $f(3) =$

$\mathrm{(A) \ -5 } \qquad \mathrm{(B) \ -2 } \qquad \mathrm{(C) \ 1 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 8 }$

Solution 1

$f(-x) = a(-x)^4 - b(-x)^2 - x + 5$

$f(-x) = ax^4 - bx^2 - x + 5$

$f(-x) = (ax^4 - bx^2 + x + 5) - 2x$

$f(-x) = f(x) - 2x$.

Thus $f(3) = f(-3)-2(-3) = 8 \Rightarrow \mathrm{(E)}$.

Solution 2

If $f(-3) = 2$, then $81a - 9b + -3 + 5 = 2$. Simplifying, we get $81a - 9b = 0$.

Getting an expression for $f(3)$, we find $f(3) = 81a - 9b + 3 + 5$. Since the first two terms sum up to zero, we get $f(3) = 8$, which is answer $\mathrm{(E)}$


Solution 3

Substituting $x = -3$, we get \[f(-3) = 81a - 9b - 3 + 5 = 81a - 9b + 2.\] But $f(-3) = 2$, so $81a - 9b + 2 = 2$, which means $81a - 9b = 0$. Then \[f(3) = 81a - 9b + 3 + 5 = 0 + 3 + 5 = \boxed{8}.\]

See also

1995 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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
All AHSME Problems and Solutions

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