Try this beautiful** Functional Equation** Problem from SMO, Singapore Mathematics Olympiad, 2013.

## Problem – Functional Equation (SMO Test)

Let M be a positive integer .It is known that whenever \(|ax^2 + bx +c|\leq 1\) for all

\(|x|\leq 1\) then \(|2ax + b |\leq M \) for all \(|x|\leq 1\). Find the smallest possible value of M.

- 4
- 5
- 6
- 10

**Key Concepts**

**Functional Equation**

**Function**

## Check the Answer

But try the problem first…

**Answer: 4**

**Singapore Mathematics Olympiad **

**Challenges and Thrills – Pre – College Mathematics**

## Try with Hints

First Hint ………………….

*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 …………………………..**

Second Hint ……………………..

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

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

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

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

\(\leq 4 \)

Final Step ………………………….

** 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 \(x = \pm 1\). **