Verified Solution • 2025–26 Session moderate

QJEEM19Maa3a0: JEE Main 2019 Mathematics - Class 11 JEEM19Maa3a0 Solution: Let S be the set of all R such that the equation, cos2x + sinx…

Let S be the set of all α\alpha ∈\in R such that the equation, cos2x + α\alpha sinx = 2 α\alpha – 7 has a solution. Then S is equal to :

Multiple Choice Options

  • A [2, 6]
  • B [3, 7]
  • C [1, 4]
  • D R

Step-by-Step Verified Solution

1 - 2sin2x + α\alpha sin x = 2 α\alpha - 7

⇒\Rightarrow 2sin2x - α\alpha sin x + (2 α\alpha - 8) = 0

sin⁡x=α±α2−8(2α−8)4\sin x = {{\alpha \pm \sqrt {{\alpha ^2} - 8(2\alpha - 8)} } \over 4}

⇒α±α2−16α+644=α±(α−8)4\Rightarrow {{\alpha \pm \sqrt {{\alpha ^2} - 16\alpha + 64} } \over 4} = {{\alpha \pm (\alpha - 8)} \over 4}

α±(α−8)4⇒2α−84,2{{\alpha \pm (\alpha - 8)} \over 4} \Rightarrow {{2\alpha - 8} \over 4},2

⇒α−42,2\Rightarrow {{\alpha - 4} \over 2},2 (rejected)

To exists solutions

−1≤α−42≤1⇒−2≤α−4≤2⇒2≤α≤6- 1 \le {{\alpha - 4} \over 2} \le 1 \Rightarrow - 2 \le \alpha - 4 \le 2 \Rightarrow 2 \le \alpha \le 6

∴\therefore α∈[2,6]\alpha \in \left[ {2,6} \right]

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