<p>Let \(a\) denotes the number of non-negative values of \(p\) for which the equation \(p2^x + 2^{-x} = 5\) possess a unique solution. If \(a, \alpha_1, \alpha_2, \ldots, \alpha_{20}, 6\) are in H.P. and \(a, \beta_1, \beta_2, \ldots, \beta_{20}, 6\) are in A.P., find \(\alpha_{18}\beta_3\).</p>
Step-by-Step Solution
Key Concept: For the equation p·2^x + 2^(-x) = 5 to have a unique solution, we must find when the function has exactly one critical point or intersection. By substituting t = 2^x (where t > 0), we transform this into a quadratic whose discriminant and parameter conditions determine uniqueness.
<p><strong>Step 1: Transform the equation using substitution.</strong></p><p>Let t = 2^x where t > 0. The equation becomes:</p><p>pt + 1/t = 5</p><p>Multiplying by t: pt² - 5t + 1 = 0</p><p><strong>Step 2: Determine conditions for unique solution in t > 0.</strong></p><p>For the original equation to have a unique solution in x, we need exactly one positive value of t.</p><p>Case 1: p = 0 → t = 1/5 (one positive root) ✓</p><p>Case 2: p ≠ 0 → pt² - 5t + 1 = 0</p><p>Discriminant: Δ = 25 - 4p</p><p>For unique positive root: either Δ = 0 (one repeated root that's positive) or one root positive, one negative.</p><p>When Δ = 0: p = 25/4, giving t = 5/(2p) = 2/5 > 0 ✓</p><p>When Δ > 0 with roots of opposite signs: product of roots = 1/p < 0, so p < 0 (not non-negative)</p><p>When Δ > 0 with both roots positive: need 1/p > 0 and -(-5)/p > 0, both satisfied when p > 0, giving infinitely many values.</p><p><strong>Step 3: Count non-negative values of p.</strong></p><p>Only p = 0 and p = 25/4 give unique solutions.</p><p>Therefore, a = 2</p><p><strong>Step 4: Find α₁₈ using H.P. condition.</strong></p><p>a, α₁, α₂, ..., α₂₀, 6 are in H.P. (22 terms total)</p><p>Their reciprocals form an A.P.: 1/a, 1/α₁, ..., 1/α₂₀, 1/6</p><p>This is an A.P. with 22 terms. First term = 1/2, last term = 1/6</p><p>Common difference: d = (1/6 - 1/2)/(22-1) = (-1/3)/21 = -1/63</p><p>The (20)th term of the reciprocal A.P. is: 1/α₁₈ = 1/2 + 19(-1/63) = 1/2 - 19/63 = (63-38)/126 = 25/126</p><p>Therefore, α₁₈ = 126/25</p><p><strong>Step 5: Find β₃ using A.P. condition.</strong></p><p>a, β₁, β₂, ..., β₂₀, 6 are in A.P. (22 terms total)</p><p>First term = 2, last term = 6</p><p>Common difference: D = (6-2)/(22-1) = 4/21</p><p>The 3rd term is: β₃ = 2 + 2(4/21) = 2 + 8/21 = (42+8)/21 = 50/21</p><p><strong>Step 6: Calculate α₁₈β₃.</strong></p><p>α₁₈β₃ = (126/25) × (50/21) = (126 × 50)/(25 × 21) = (6 × 50)/25 = 300/25 = 12</p><p><strong>∴ Answer: 12</strong></p>
Correct Answer: 12