The maximum and minimum values of absinx+b√1−a2cosx+c lie in the interval [|a|<1,b>0]
A
(b−c,b+c)
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B
[−b+c,b+c]
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C
[b,b+c]
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D
[−c,c]
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Solution
The correct option is C[−b+c,b+c] Differentiating with respect to x, we get abcosx−b√1−a2sinx=0 Now Let a=cosθ Hence we get b[cosθ.cosx−sinθ.sinx]=0 Or b(cos(θ+x))=0 Or θ+x=π2,3π2 Hence x1=π2−θ and x2=3π2−θ. f(x)=b[sin(θ+x)]+c f(x1) =b[sin(θ+π2−θ)]+c =b+c Similarly f(x2)=−b+c.