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

JEE · Math · previous-year question

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

Answer

A. [2, 6]

Explanation

1 - 2sin2x + $$\alpha $$ sin x = 2$$\alpha $$ - 7 $$ \Rightarrow $$ 2sin2x - $$\alpha $$ sin x + (2$$\alpha $$ - 8) = 0 $$\sin x = {{\alpha \pm \sqrt {{\alpha ^2} - 8(2\alpha - 8)} } \over 4}$$ $$ \Rightarrow {{\alpha \pm \sqrt {{\alpha ^2} - 16\alpha + 64} } \over 4} = {{\alpha \pm (\alpha - 8)} \over 4}$$ $${{\alpha \pm (\alpha - 8)} \over 4} \Rightarrow {{2\alpha - 8} \over 4},2$$ $$ \Rightarrow {{\alpha - 4} \over 2},2$$ (rejected) To exists solutions $$ - 1 \le {{\alpha - 4} \over 2} \le 1 \Rightarrow - 2 \le \alpha - 4 \le 2 \Rightarrow 2 \le \alpha \le 6$$ $$ \therefore $$ $$\alpha \in \left[ {2,6} \right]$$

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