If 4x−22+x+5+||b−1|−3|=|siny|, where x,y,b∈R has a real solution, then the maximum possible value of b is
A
8
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B
−2
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C
0
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D
4
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Solution
The correct option is D4 4x−22+x+5+||b−1|−3|=|siny|⇒22x−4⋅2x+4+1+||b−1|−3|=|siny|⇒(2x−2)2+1+||b−1|−3|=|siny|
We know that |siny|∈[0,1]
Also, (2x−2)2+1+||b−1|−3|≥1
So equality is possible only when (2x−2)2=0and||b−1|−3|=0⇒siny=±1∴y=(2n+1)π2andx=1andb=−2,4