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Q.

If minimum value of (sin−1x)2+(cos−1x)2 is π2kthen the value of K is

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a

4

b

1

c

8

d

10

answer is C.

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Detailed Solution

(sin−1x)2+(cos−1x)2=(sin−1x)2+(π2−sin−1x)2=2(sin−1x)2−πsin−1x+π24 Its minimum value is 4.2.π24−π28=π28

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