Question

Let S be the set of all \(\alpha \in \mathbb{R}\) such that the equation, \(\cos 2x + \alpha \sin x = 2\alpha - 7\) has a solution. Then S is equal to :

• A. \([2, 6]\)
• B. \([3, 7]\)
• C. \([1, 4]\)
• D. \(\mathbb{R}\)

Hint:

Write everything in terms of \(\sin x = s\) with \(s\in[-1,1]\), solve for \(\alpha\) as a function of s, and find its range.

Answer 1

The Correct Option is A

Let \(s=\sin x\in[-1,1]\). Then \(\cos 2x = 1-2s^2\).
The equation becomes \(1 - 2s^2 + \alpha s = 2\alpha - 7\), i.e. \(\alpha(2-s) = 8 - 2s^2 = 2(2-s)(2+s)\).
Since \(s\ne 2\) for \(s\in[-1,1]\), \(\alpha = 2(2+s) = 4+2s\).
As s varies over \([-1,1]\), \(\alpha\) takes all values in \([2, 6]\).
So \(S=[2,6]\).
Answer: (1) \([2, 6]\)