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Question

If x2+5=2x4cos(a+bx) where a,b(0,5) is satisfied for alteast one real x, then the minimum value of a+bx is


A

π

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B

0

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C

π2

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D

3π2

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Solution

The correct option is A

π


x2+5=2x4cos(a+bx)
x2+52x=2x4cos(a+bx)2x
x22x+5=4cos(a+bx)
x22x+1+4=4cos(a+bx)
x22x+1+4+4cos(a+bx)=4cos(a+bx)+4cos(a+bx)
(x1)2+4(1+cos(a+bx))=0
x1=0 & cos(a+bx)+1=0
x=1 & cos(a+bx)=1
Minimum value of a+bx=π (cosπ=1)


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