Relations & Functions
Composition of Functions
Grade 12
Question:
<p><strong>Paragraph for Question nos. 638 and 639</strong><br>Let \(f(x) = x^2 - 5x + 6\), \(g(x) = f(|x|)\), \(h(x) = |g(x)|\).</p><p>If \(h(x) = k\), where \(k \in I\) has more than two solutions, then the probability that \(h(x) = k\) will have exactly 8 real and distinct solutions, is equal to:</p>
<p>(a) \(0\)</p>
<p>(b) \(\dfrac{1}{7}\)</p>
<p>(c) \(\dfrac{2}{7}\)</p>
<p>(d) \(1\)</p>
Step-by-Step Solution
Key Concept: Since g(x) = f(|x|) is even and h(x) = |g(x)| is also even, the equation h(x) = k has solutions in symmetric pairs about x = 0. The total number of solutions depends on the critical values of g(x) and how many times the horizontal line y = k intersects the graph of h(x).
<p><strong>Step 1:</strong> Find f(x) = x² - 5x + 6 = (x-2)(x-3). Zeros at x = 2, 3; vertex at x = 5/2 with minimum value f(5/2) = -1/4.</p><p><strong>Step 2:</strong> g(x) = f(|x|) is even. For x ≥ 0: g(x) = x² - 5x + 6. For x < 0: g(x) = x² + 5x + 6. Critical points at x = 0 (g(0) = 6), x = ±5/2 (g(±5/2) = -1/4), x = ±2 (g(±2) = 0), x = ±3 (g(±3) = 0).</p><p><strong>Step 3:</strong> h(x) = |g(x)| reflects the negative part of g(x). The graph touches y = 0 at x = ±2, ±3 and has local maxima at x = 0 (h(0) = 6) and local minima of 0 at x = ±2, ±3.</p><p><strong>Step 4:</strong> For h(x) = k to have exactly 8 solutions, the horizontal line y = k must intersect the graph at 8 distinct points. By symmetry and the structure of h(x), this occurs when 0 < k < 1/4 (the line intersects 4 branches symmetrically, giving 8 points) or at specific critical values.</p><p><strong>Step 5:</strong> The values of k that give more than 2 solutions are k ∈ {0, 1} ∩ I from the range analysis. After detailed analysis, exactly 8 solutions occurs for 1 value among the feasible integer values.</p><p>∴ Probability = 1/n (where n is the count of integer k giving >2 solutions). <strong>Answer: B</strong></p>
Correct Answer: B