Let y=f(x) and f:R®Rbe an odd function which is differentiable such that f′"(x)>0 and f(a,b)=sin8a+cos8b+2−4sin2acos2b. If f′′(f(a,b))=0thensin2a+sin2b=
A
1
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B
2
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C
3
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Solution
The correct option is A 1 f"(x)is an odd function f(a,b)=0Φ(sin4a−1)2+(cos4b−1)2+2(sin2a−cos2b)2=0 ⇒sin2α+sin2β=1