GATE CS 2014 Set 2 Q.15 (CS Q.5): Calculus

A non-zero polynomial $f(x)$ of degree 3 has roots at $x = 1, x = 2$ and $x = 3$ . Which one of the following must be TRUE?

$f(0) f(4) < 0$

$f(0) f(4) > 0$

$f(0) + f(4) > 0$

$f(0) + f(4) < 0$

Official answer and solution

Solution

Express the cubic polynomial in terms of its three real roots and evaluate the sign of the product f(0) * f(4).

A polynomial of degree 3 with roots at $x = 1, x = 2, x = 3$ can be factored uniquely as:
$f(x) = c(x - 1)(x - 2)(x - 3)$
for some non-zero real constant $c \ne 0$ .
Evaluate at $x = 0$ :
$f(0) = c(0 - 1)(0 - 2)(0 - 3) = c(-1)(-2)(-3) = -6c.$
Evaluate at $x = 4$ :
$f(4) = c(4 - 1)(4 - 2)(4 - 3) = c(3)(2)(1) = 6c.$
Evaluate $f(0) f(4)$ :
$f(0) f(4) = (-6c)(6c) = -36c^2.$
Because $c$ is non-zero and real, $c^2 > 0$ , which guarantees:
$f(0) f(4) = -36c^2 < 0.$
This condition holds regardless of whether $c$ is positive or negative.

Hence, $f(0) f(4) < 0$ must be TRUE (option A).

Final answer

Common trap

Assuming c must be positive, which would lead to guessing about f(0) + f(4).

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

GATE questions are the work of the organising institutes. This site is independent and is not affiliated with them.