Question:

<p><strong>Step 1: Understand what 'always false' means.</strong> We need to find a statement that cannot possibly be true for the given function and domain specified.</p><p><strong>Step 2: Analyze Option (A) - 'h(x) is one-one'.</strong> Without explicit definition of h(x), this could be true or false. It's not necessarily always false.</p><p><strong>Step 3: Analyze Option (B) - 'f(x) is one-one if x > 10'.</strong> For any function f(x), restricting the domain to x > 10 can potentially make it one-one. This is not always false.</p><p><strong>Step 4: Analyze Option (C) - 'g(x) is many-one if x ∈ (0, π/2) ∪ (π/2, 3)'.</strong> The domain is explicitly split into two disjoint intervals: (0, π/2) and (π/2, 3). For g(x) to be many-one on this domain, at least two different x-values from this domain must map to the same y-value. However, on a continuous domain consisting of two separate intervals, if g is continuous and strictly monotonic (as is typical for standard functions), g(x) cannot be many-one on the union of these disjoint intervals. Most standard functions like sin(x), cos(x), etc., are one-one on such restricted disjoint domains.</p><p><strong>Step 5: Analyze Option (D) - 'The values of k for which f(x) = k has exactly one solution is k = 2 or k = -2'.</strong> This is a specific claim about solutions that could be true for particular functions.</p><p><strong>Step 6: Conclusion.</strong> Option (C) is always false because a function restricted to disjoint intervals (0, π/2) ∪ (π/2, 3) cannot be many-one if it's a typical monotonic function. Standard continuous functions cannot map multiple points from these separated intervals to the same value, making the function one-one on this domain, not many-one.</p><p><strong>∴ Answer:</strong> C</p>