Theory of EquationseasyFree

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)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Theory of Equations · medium
If one of the roots of ax² + ax + a + 1 = 0 is less than 1 and the other is greater than…
Theory of Equations · medium
If a, b, c, d are non-zero real numbers such that c and d are the roots of x² + ax + b =…
Theory of Equations · medium
If α, β, γ be the roots of the equation x³ + ax + a = 0 (where a ∈ R , a ≠ 0 ) satisfying…
Theory of Equations · medium
If α + (1)/(α) and β + (1)/(β) are the zeroes of f(x) = x² - 5x - a where α, β ∈ (0, ∞) ,…

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath