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Question

The maximum and minimum values of absinx+b1a2cosx+c lie in the interval [|a|<1,b>0]

A
(bc,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
abcosxb1a2sinx=0
Now
Let
a=cosθ
Hence we get
b[cosθ.cosxsinθ.sinx]=0
Or
b(cos(θ+x))=0
Or
θ+x=π2,3π2
Hence
x1=π2θ and
x2=3π2θ.
f(x)=b[sin(θ+x)]+c
f(x1)
=b[sin(θ+π2θ)]+c
=b+c
Similarly
f(x2)=b+c.

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