If x2-5-2 x-4 cos a-b x where a- b -0-5 is satisfied for alteast one real x- then the minimum value of a-b x is

The correct option is D 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) = ...

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Standard XIII

Mathematics

Converting to Perfect Square

If x2+5=2 x-4...

Question

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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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

The correct option is C π x2 + 5 = 2x - 4 cos (a + bx) x2 - 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

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

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 = $π$