<p>78. If the roots of the quadratic equation \((4p - p^2 - 5)x^2 - (2p-1)x + 3p = 0\) lie on either side of unity, then the number of integral values of \(p\) is</p>
Step-by-Step Solution
Key Concept: For roots to lie on either side of unity, the quadratic must change sign at x=1 (i.e., f(1) and leading coefficient must have opposite signs), which means f(1) < 0 when coefficient of x² > 0, or f(1) > 0 when coefficient of x² < 0. This is a sign-change condition, not a discriminant condition.
<p><strong>Step 1: Set up the condition</strong></p><p>For roots to lie on either side of unity (one root < 1, one root > 1), we need:</p><p>f(1) · (leading coefficient) < 0</p><p><strong>Step 2: Calculate f(1)</strong></p><p>f(1) = (4p - p² - 5)(1)² - (2p-1)(1) + 3p</p><p>= 4p - p² - 5 - 2p + 1 + 3p</p><p>= -p² + 5p - 4</p><p>= -(p² - 5p + 4)</p><p>= -(p-1)(p-4)</p><p><strong>Step 3: Identify leading coefficient constraint</strong></p><p>For a valid quadratic: 4p - p² - 5 ≠ 0</p><p>⟹ -p² + 4p - 5 ≠ 0</p><p>⟹ p² - 4p + 5 ≠ 0</p><p>Since discriminant = 16 - 20 = -4 < 0, this is always true.</p><p><strong>Step 4: Apply the sign condition</strong></p><p>Need: [-(p-1)(p-4)] · [4p - p² - 5] < 0</p><p>⟹ (p-1)(p-4) · (4p - p² - 5) > 0</p><p>⟹ (p-1)(p-4) · [-(p² - 4p + 5)] > 0</p><p>⟹ -(p-1)(p-4)(p² - 4p + 5) > 0</p><p>Since p² - 4p + 5 > 0 always:</p><p>⟹ -(p-1)(p-4) > 0</p><p>⟹ (p-1)(p-4) < 0</p><p>⟹ 1 < p < 4</p><p><strong>Step 5: Count integral values</strong></p><p>Integral values in (1, 4): p ∈ {2, 3}</p><p>∴ Answer: <strong>2</strong></p>
Correct Answer: 2