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Let $a^{2} + b^{2} = α^{2} + β^{2} = 2$ then the maximum value of $S = (1 - a) (1 - b) + (1 - α) (1 - β) i s$

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answer is B.

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Detailed Solution

$a = \sqrt{2} \cos θ, b = \sqrt{2} \sin θ$

$α = \sqrt{2} \cos ϕ, β = \sqrt{2} \sin ϕ$

$\Rightarrow s = 2 - 2 [\sin (θ + \frac{π}{4}) + \sin (θ + \frac{π}{4})] + 2 [\cos (θ - ι)]$

$= 2 + 4 + 2 = 8$

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