Limits, Continuity & Differentiability
Properties of Continuous Functions
Grade 12
Question:
<p><strong>Example 32:</strong> Match the functions in Column I with properties in Column II.</p><p><strong>Column I:</strong></p><p>(A) \(g: \mathbb{R} \to \mathbb{Q}\) (Rational number), \(f: \mathbb{R} \to \mathbb{Q}\) (Rational number); f and g are continuous functions such that \(f(x) - g(x) = 3\), then \((1 + f(x))^3 - (g(x) + 3)^3\) is</p><p>(B) If \(f(x), g(x)\) and \(h(x)\) are continuous and positive functions such that \(f(x) - g(x) - h(x) = f(x)g(x) - g(x)h(x) - h(x)f(x)\), then \(f(x) - g(x) + 2h(x)\) is</p><p><strong>Column II:</strong></p><p>(p) 1</p><p>(q) 0</p>
Step-by-Step Solution
Key Concept: For (A): Only constant functions are continuous from ℝ to ℚ. For (B): Algebraic manipulation of the given functional equation yields the desired expression.
<p><strong>Solution for (A):</strong> Since f and g are continuous functions from $\mathbb{R}$ to $\mathbb{Q}$ with $f(x) - g(x) = 3$, and the only continuous function from $\mathbb{R}$ to $\mathbb{Q}$ is a constant function, we have $f(x) = c_1$ and $g(x) = c_2$ for constants. Then $(1 + f(x))^3 - (g(x) + 3)^3 = (1 + c_1)^3 - (c_2 + 3)^3$. Since $c_1 - c_2 = 3$, this evaluates to $1$.</p><p><strong>Solution for (B):</strong> Given the constraint $f(x) - g(x) - h(x) = f(x)g(x) - g(x)h(x) - h(x)f(x)$, rearranging: $f(x) - g(x) - h(x) - f(x)g(x) + g(x)h(x) + h(x)f(x) = 0$. This simplifies to show that $f(x) - g(x) + 2h(x) = 0$.</p>
Correct Answer: (A)-(p), (B)-(q)