GATE CS 2014 Set 1 Q.57 (CS Q.47): Calculus

A function $f(x)$ is continuous in the interval $[0, 2]$ . It is known that $f(0) = f(2) = -1$ and $f(1) = 1$ . Which one of the following statements must be true?

There exists a $c$ in the interval $(0, 1)$ such that $f(c) = f(c + 1)$

For every $c$ in the interval $(0, 1)$ , $f(c) = f(2 - c)$

The maximum value of the function in the interval $(0, 2)$ is 1

There exists a $c$ in the interval $(0, 1)$ such that $f(c) = -f(2 - c)$

Official answer and solution

Official answer: A
From: Official GATE 2014 answer key

Solution

Define an auxiliary function $g(x) = f(x) - f(x+1)$ on $[0, 1]$ and apply the Intermediate Value Theorem.

Given that $f(x)$ is continuous on $[0, 2]$ , with:

f(0) = -1, \quad f(1) = 1, \quad f(2) = -1

Define an auxiliary function $g(x) = f(x) - f(x + 1)$ on the closed interval $[0, 1]$ .
Since $f$ is continuous on $[0, 2]$ , $g$ is continuous on $[0, 1]$ .

Evaluate $g(x)$ at the endpoints:

g(0) = f(0) - f(1) = -1 - 1 = -2 < 0

g(1) = f(1) - f(2) = 1 - (-1) = 2 > 0

Because $g(0) < 0$ and $g(1) > 0$ , by the Intermediate Value Theorem, there must exist at least one $c \in (0, 1)$ such that:

g(c) = 0

f(c) - f(c + 1) = 0 \implies f(c) = f(c + 1)

This proves Statement A must be true.

The other statements are not guaranteed:

Statement B assumes symmetry $f(c) = f(2 - c)$ for all $c$ , which need not hold.
Statement C claims the maximum is 1, but continuous functions can exceed 1 in $(0, 1)$ or $(1, 2)$ .
Statement D asserts $f(c) = -f(2 - c)$ , which is not universally true.

Final answer

Common trap

Assuming the function must be symmetric about $x = 1$ , which is only one specific realization.

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

Source

Paper: GATE 2014 Computer Science and Information Technology (CS) Session 1
Question: 57 of 65, printed as CS Q.47
Section: Computer Science and Information Technology

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