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Q.

Let a2+b2=α2+β2=2then the maximum value of S=1−a1−b+1−α1−β  is 

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a

4

b

8

c

2

d

6

answer is B.

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Detailed Solution

a=2  cosθ,b=2 sinθ

α=2cosϕ,β=2sinϕ

⇒s=2−2sinθ+π4+sinθ+π4+2cosθ−ι

=2+4+2=8

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