GATE CS 2014 Set 2 Q5 — Sets, Relations and Functions

A polynomial $f(x)$ of degree 3 with roots at $x=1$, $x=2$, and $x=3$ can be uniquely represented in the form:

where $a$ is a non-zero real constant ($a \neq 0$).

Let us evaluate $f(x)$ at $x=0$ and $x=4$:

• For $x=0$: $$f(0) = a(0-1)(0-2)(0-3) = a(-1)(-2)(-3) = -6a$$
• For $x=4$: $$f(4) = a(4-1)(4-2)(4-3) = a(3)(2)(1) = 6a$$

Now, let us analyze the options by computing the product $f(0)f(4)$:

Since $a \neq 0$, we know that $a^2 > 0$. Therefore, $-36a^2 < 0$, which means:

This confirms that Option A must always be true.

Checking other options:

• Option B: $f(0)f(4) > 0$ is false because $-36a^2$ is strictly negative.
• Options C and D: $f(0) + f(4) = -6a + 6a = 0$. Thus, neither $f(0) + f(4) > 0$ nor $f(0) + f(4) < 0$ is true.