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Question

If x2 + 5 = 2x - 4 cos (a + bx) where a, b ϵ (0, 5) is satisfied for alteast one real x, then the minimum value of a + bx is


A

0

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B

π2

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C

π

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D

3π2

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Solution

The correct option is C

π


x2 + 5 = 2x - 4 cos (a + bx)
x2 - 2x + 1 + 4 = - 4 cos (a + bx)
(x1)2 + 4 (1 + cos (a + bx)) = 0
x - 1 = 0, cos (a + bx) = -1
minimum of a + bx = π


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