Sets, Relations & Functions
Invertible Functions
Grade 11
Question:
<p>Let \(f : A \to B\) be a function defined as \(f(x) = \dfrac{x-1}{x-2}\), where \(A = R - \{2\}\) and \(B = R - \{1\}\). Then <i>f</i> is</p>
<p>invertible and \(f^{-1}(y) = \dfrac{3y-1}{y-1}\)</p>
<p>invertible and \(f^{-1}(y) = \dfrac{2y-1}{y-1}\)</p>
<p>invertible and \(f^{-1}(y) = \dfrac{2y+1}{y-1}\)</p>
<p>not invertible</p>
Step-by-Step Solution
Key Concept: A function is bijective if it's both injective (one-to-one) and surjective (onto). For f(x) = (x-1)/(x-2), check injectivity by solving f(x₁) = f(x₂) and verify that every element in B = ℝ - {1} has a preimage in A = ℝ - {2}.
<p><strong>Step 1: Check Injectivity (One-to-One)</strong></p><p>Assume f(x₁) = f(x₂):<br/>$$\frac{x_1-1}{x_1-2} = \frac{x_2-1}{x_2-2}$$<br/>Cross-multiply: $(x_1-1)(x_2-2) = (x_2-1)(x_1-2)$<br/>Expand: $x_1x_2 - 2x_1 - x_2 + 2 = x_1x_2 - 2x_2 - x_1 + 2$<br/>Simplify: $-2x_1 - x_2 = -2x_2 - x_1$<br/>Therefore: $x_1 = x_2$ ✓ <strong>f is injective</strong></p><p><strong>Step 2: Check Surjectivity (Onto)</strong></p><p>For any y ∈ B = ℝ - {1}, find x such that f(x) = y:<br/>$$y = \frac{x-1}{x-2}$$<br/>$$y(x-2) = x-1$$<br/>$$yx - 2y = x - 1$$<br/>$$x(y-1) = 2y - 1$$<br/>$$x = \frac{2y-1}{y-1}$$</p><p>Since y ≠ 1, we have y - 1 ≠ 0, so x is well-defined. Also, x ≠ 2 (verify: if x = 2, then 2(y-1) = 2y-1 gives -2 = -1, contradiction). Thus every y ∈ B has a preimage in A. ✓ <strong>f is surjective</strong></p><p><strong>Step 3: Conclusion</strong></p><p>Since f is both injective and surjective, <strong>f is bijective</strong>.</p><p>∴ Answer: B</p>
Correct Answer: B