Difference between revisions of "1953 AHSME Problems/Problem 35"
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+ | ==Problem== | ||
If <math>f(x)=\frac{x(x-1)}{2}</math>, then <math>f(x+2)</math> equals: | If <math>f(x)=\frac{x(x-1)}{2}</math>, then <math>f(x+2)</math> equals: | ||
Line 5: | Line 6: | ||
\textbf{(C)}\ x(x+2)f(x) \qquad | \textbf{(C)}\ x(x+2)f(x) \qquad | ||
\textbf{(D)}\ \frac{xf(x)}{x+2}\ \textbf{(E)}\ \frac{(x+2)f(x+1)}{x} </math> | \textbf{(D)}\ \frac{xf(x)}{x+2}\ \textbf{(E)}\ \frac{(x+2)f(x+1)}{x} </math> | ||
+ | |||
+ | ==Solution== | ||
+ | First simplify <math>f(x+2)</math>: | ||
+ | <cmath>f(x+2)=\frac{(x+2)(x+1)}{2}.</cmath> | ||
+ | Then simplify the answers and see which one matches. | ||
+ | |||
+ | <math>f(x)+f(2)=\frac{x(x-1)}{2} + 1,</math> | ||
+ | |||
+ | <math>(x+2)f(x)=\frac{(x+2)x(x-1)}{2},</math> | ||
+ | |||
+ | <math>x(x+2)f(x)=\frac{(x+2)x^2(x-1)}{2},</math> | ||
+ | |||
+ | <math>\frac{xf(x)}{x+2}=\frac{x^2(x-1)}{2(x+2)},</math> | ||
+ | |||
+ | <math>\frac{(x+2)f(x+1)}{x}=\frac{(x+2)(x+1)}{2}.</math> | ||
+ | |||
+ | Since <math>f(x+2)=\frac{(x+2)f(x+1)}{x}=\frac{(x+2)(x+1)}{2}</math>, the answer is <math>\boxed{\textbf{(E)}\ \frac{(x+2)f(x+1)}{x}}.</math> | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AHSME 50p box|year=1953|num-b=34|num-a=36}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 20:41, 24 January 2020
Problem
If , then
equals:
Solution
First simplify :
Then simplify the answers and see which one matches.
Since , the answer is
See Also
1953 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 34 |
Followed by Problem 36 | |
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