GATE CS 2014 Set 1 Q47 — Calculus: Limits, Continuity, Differentiability

To determine which statement must be true, let us define a new function $g(y)$ on the interval $[0, 1]$ as:

Since $f(x)$ is continuous on $[0, 2]$, $g(y)$ is also continuous on $[0, 1]$. Let us evaluate $g(y)$ at the endpoints of its domain:

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• At $y = 0$: $g(0) = f(0) - f(1) = -1 - 1 = -2 < 0$
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• At $y = 1$: $g(1) = f(1) - f(2) = 1 - (-1) = 2 > 0$
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Since $g(0) < 0$ and $g(1) > 0$, by the Intermediate Value Theorem (IVT), there must exist some $y ext{ in } (0, 1)$ such that:

This confirms that Option A is always true.

The other options are not necessarily true because we can easily construct counterexamples (e.g., asymmetric functions, functions with a maximum value greater than $1$, or functions that remain positive over most of the interval preventing $f(y) = -f(2-y)$).