Question
Thu May 03, 2012 By: Student Learner

The equation kcosx - 3sinx = k + 1 is solvable only if k belongs to the interval: (a) [4,+infinity) (b) [4,4] (c) (-infinity,4] (d) None of these

Expert Reply
Thu May 03, 2012
k(cos x) - 3(sin x) = k + 1
k(cos x) - (k + 1) = 3(sin x)
[k(cos x) - (k + 1)]² = [3(sin x)]²
k²cos²x - 2k(k + 1)(cos x) + (k + 1)² = 9sin²x
k²cos²x - (2k² + 2k)(cos x) + (k + 1)² = 9(1 - cos²x)
(k² + 9)cos²x - (2k² + 2k)(cos x) + (k² + 2k - 8) = 0

Here we have a quadratic equation in terms of (cos x).
This equation is only solvable when the determinant b² - 4ac ? 0, otherwise the solution will be complex.

b² - 4ac ? 0
[-(2k² + 2k)]² - 4(k² + 9)(k² + 2k - 8) ? 0
(4k? + 8k³ + 4k²) - (4k? + 8k³ + 4k² + 72k - 288) ? 0
-72k + 288 ? 0
72k ? 288
k ? 4

i.e. -? < k ? 4

So, the answer is Option C: (-? , 4] .
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