If $x^{2}$ + 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

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$\frac{π}{2}$

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$π$

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$\frac{3 π}{2}$

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Solution

The correct option is C
$π$

$x^{2}$ + 5 = 2x - 4 cos (a + bx)
$x^{2}$ - 2x + 1 + 4 = - 4 cos (a + bx)
$(x - 1)^{2}$ + 4 (1 + cos (a + bx)) = 0
$∴$ x - 1 = 0, cos (a + bx) = -1
$∴$ minimum of a + bx = $π$

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