Relations & Functions
Composite functions
Grade 12
Question:
<p>Let \(f(x) = 2^{10} \cdot x + 1\) and \(g(x) = 3^{10} \cdot x - 1\). If \((f \circ g)(x) = x\), then \(x\) is equal to</p>
<p>\(\dfrac{3^{10}-1}{3^{10}-2^{-10}}\)</p>
<p>\(\dfrac{2^{10}-1}{2^{10}-3^{-10}}\)</p>
<p>\(\dfrac{1-3^{-10}}{2^{10}-3^{-10}}\)</p>
<p>\(\dfrac{1-2^{-10}}{3^{10}-2^{-10}}\)</p>
Step-by-Step Solution
Key Concept: For (f ∘ g)(x) = x to hold, we need f(g(x)) = x, which means g must be the left inverse of f. This occurs only at specific values where the composition equation is satisfied.
<p><strong>Step 1:</strong> Write out the composition f(g(x)).</p><p>f(g(x)) = f(3^{10}·x - 1) = 2^{10}(3^{10}·x - 1) + 1</p><p><strong>Step 2:</strong> Expand and simplify.</p><p>f(g(x)) = 2^{10}·3^{10}·x - 2^{10} + 1 = (2·3)^{10}·x - 2^{10} + 1 = 6^{10}·x - (2^{10} - 1)</p><p><strong>Step 3:</strong> Set f(g(x)) = x and solve for x.</p><p>6^{10}·x - (2^{10} - 1) = x</p><p>6^{10}·x - x = 2^{10} - 1</p><p>x(6^{10} - 1) = 2^{10} - 1</p><p><strong>Step 4:</strong> Solve for x.</p><p>x = (2^{10} - 1)/(6^{10} - 1)</p><p>∴ Answer: D</p>
Correct Answer: D