Functional Equation Problem from SMO, 2013 - Senior Section

We cant this sum by assuming a,b,c as fixed quantity.

Let \( f(x) = ax^2 + bx + c \).

Then \( f(-1) = a - b + c \) ; \( f(0) = c \) ; \( f(1) = a + b + c\) ;

Try to do the rest of the sum ................................

Suppose \( |f(x)|\leq 1\) for all \(|x|\leq 1 \) . Then

\( |2ax + b| = | (x - \frac {1}{2} ) f(-1) - 2 f(0) x + (x+\frac {1}{2} f(1) |\)

\(\leq |x - \frac {1}{2}| + 2 |x| + |x + \frac {1}{2}|\)

\(\leq |x - \frac {1}{2} | + |x+\frac {1}{2}| + 2 \)

\(\leq 4 \)

Now I guess you have already got the answer but if not .............

From the last step we can conclude ,

\(|2 x^2 - 1|\leq 1 \) whenever \(|x|\leq 4\) and \(|2x| = 4 \)

is achieved at \(x = \pm 1\).

AMC - AIME Boot Camp... for brilliant students.

Use our exclusive one-on-one plus group class system to prepare for Math Olympiad

Learn More

Subscribe to Cheenta at Youtube