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Theory of Equations: Zeroes Set Values

JEE Maths question with a full step-by-step solution.

Question
If α+1α\alpha + \dfrac{1}{\alpha} and β+1β\beta + \dfrac{1}{\beta} are the zeroes of f(x)=x2−5x−af(x) = x^2 - 5x - a where α,β∈(0,∞)\alpha, \beta \in (0, \infty), then the set of values of aa is
A[−6, 254]\left[-6,\, \dfrac{25}{4}\right]
B[254, 6]\left[\dfrac{25}{4},\, 6\right]
C[−254, −6]\left[-\dfrac{25}{4},\, -6\right]correct
D[−254, 6]\left[-\dfrac{25}{4},\, 6\right]
Solution
Step 1: By AM-GM, for α>0\alpha > 0: α+1α≥2\alpha + \dfrac{1}{\alpha} \geq 2. So both zeroes of f(x)f(x) are ≥2\geq 2. Step 2: For f(x)=x2−5x−af(x) = x^2 - 5x - a to have both roots ≥2\geq 2, three conditions are needed. Condition 1 — f(2)≥0f(2) \geq 0: 4−10−a≥0  ⟹  a≤−64 - 10 - a \geq 0 \implies a \leq -6. Condition 2 — Discriminant ≥0\geq 0: 25+4a≥0  ⟹  a≥−25425 + 4a \geq 0 \implies a \geq -\dfrac{25}{4}. Condition 3 — Vertex to the right of 22: 52>2\dfrac{5}{2} > 2. ✓ (Always satisfied.) Step 3: Combining: a∈[−254, −6]a \in \left[-\dfrac{25}{4},\, -6\right]. Correct answer: (3)
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