Difference between revisions of "1953 AHSME Problems/Problem 35"

(Created page with "If <math>f(x)=\frac{x(x-1)}{2}</math>, then <math>f(x+2)</math> equals: <math>\textbf{(A)}\ f(x)+f(2) \qquad \textbf{(B)}\ (x+2)f(x) \qquad \textbf{(C)}\ x(x+2)f(x) \qquad \...")
 
Line 1: Line 1:
 +
==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==

Revision as of 21:19, 24 January 2020

Problem

If $f(x)=\frac{x(x-1)}{2}$, then $f(x+2)$ equals:

$\textbf{(A)}\ f(x)+f(2) \qquad \textbf{(B)}\ (x+2)f(x) \qquad \textbf{(C)}\ x(x+2)f(x) \qquad \textbf{(D)}\ \frac{xf(x)}{x+2}\\ \textbf{(E)}\ \frac{(x+2)f(x+1)}{x}$

Solution