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Theory of Equations: Values Quadratic Equation Real Solution

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

Question
The values of α\alpha and β\beta for which the quadratic equation x2+2x+2+e2α−cos⁡β=0x^2 + 2x + 2 + e^{2\alpha} - \cos\beta = 0 has a real solution is
Aα,β∈R\alpha, \beta \in \mathbb{R}
Bα∈(0,1)\alpha \in (0,1), β∈(π2,2π)\beta \in \left(\dfrac{\pi}{2}, 2\pi\right)
Cα∈(0,∞)\alpha \in (0,\infty), β∈(π2,2π)\beta \in \left(\dfrac{\pi}{2}, 2\pi\right)
DNo real value of α,β\alpha, \beta possiblecorrect
Solution
Step 1: For x2+2x+(2+e2α−cos⁡β)=0x^2 + 2x + (2 + e^{2\alpha} - \cos\beta) = 0 to have real solutions, the discriminant must be non-negative:
D=4−4(2+e2α−cos⁡β)≥0  ⟹  cos⁡β≥1+e2αD = 4 - 4(2 + e^{2\alpha} - \cos\beta) \geq 0 \implies \cos\beta \geq 1 + e^{2\alpha}
 Step 2: Since e2α>0e^{2\alpha} > 0 for all real α\alpha: 1+e2α>11 + e^{2\alpha} > 1. Step 3: But cos⁡β≤1\cos\beta \leq 1 always. Therefore cos⁡β≥1+e2α>1\cos\beta \geq 1 + e^{2\alpha} > 1 is impossible. No real values of α\alpha and β\beta satisfy the condition. Correct answer: (4)
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