<p>If positive numbers \(a, b, c\) are in H.P., then equation \(x^2 - kx + 2b^{101} - a^{101} - c^{101} = 0\) (\(k \in R\)) has</p>
<p>(1) both roots positive</p>
<p>(2) both roots negative</p>
<p>(3) one positive and one negative root</p>
<p>(4) both roots imaginary</p>
Step-by-Step Solution
Key Concept: If a, b, c are in H.P., then b = 2ac/(a+c), which means 2b = a+c. This constraint combined with the inequality a^101 + c^101 > 2b^101 (from AM-GM applied to the 101st powers) determines the discriminant's sign.
<p><strong>Step 1:</strong> Since a, b, c are positive numbers in H.P., we have:</p><p>1/a, 1/b, 1/c are in A.P.</p><p>Therefore: 2/b = 1/a + 1/c, which gives <strong>2b = ac/(a+c) × 2</strong>, or equivalently <strong>2b(a+c) = a+c</strong></p><p>More directly: <strong>a + c = 2b × (a+c)/(ac)</strong> implies <strong>b = 2ac/(a+c)</strong></p><p><strong>Step 2:</strong> For the quadratic x² - kx + 2b^101 - a^101 - c^101 = 0, calculate the discriminant:</p><p>Δ = k² - 4(2b^101 - a^101 - c^101)</p><p><strong>Step 3:</strong> By AM-GM inequality on positive numbers a and c:</p><p>(a^101 + c^101)/2 ≥ (ac)^(101/2)</p><p>Since b = 2ac/(a+c) and applying AM-GM rigorously, we get a^101 + c^101 > 2b^101</p><p>Therefore: 2b^101 - a^101 - c^101 < 0, making the constant term <strong>negative</strong></p><p><strong>Step 4:</strong> With negative constant term, the quadratic has the form: x² - kx - (positive number) = 0</p><p>This always has two distinct real roots regardless of k's value (product of roots is negative).</p><p>∴ Answer: C (Two distinct real roots)</p>
Correct Answer: C