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

Revision as of 16:37, 18 August 2011

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

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

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

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

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

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

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

@@ Line 5: / Line 5: @@
 == Solution ==
-<math>f(-x) = a(-x)^4 - b(-x)^2 - x + 5 = ax^4 - bx^2 - x + 5 = f(x) - 2x</math>. Thus <math>f(3) = f(-3)-2(-3) = 8 \Rightarrow \mathrm{(E)}</math>.
+<math>f(-x) = a(-x)^4 - b(-x)^2 - x + 5
+</math>f(-x) = ax^4 - bx^2 - x + 5
+<math>f(-x) = (ax^4 + bx^2 + x + 5) - 2x</math>
+<math>f(-x) =  f(x) - 2x</math>.
+Thus <math>f(3) = f(-3)-2(-3) = 8 \Rightarrow \mathrm{(E)}</math>.
 == See also ==