If x2+4+3sin(ax+b)−2x=0 has atleast one real solution, where a,b∈[0,2π], then the value of a+b can be
A
5π2
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B
7π2
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C
9π2
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D
13π4
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Solution
The correct option is B7π2 x2+4+3sin(ax+b)−2x=0⇒(x−1)2+3=−3sin(ax+b) We know that, (x−1)2+3∈[3,∞)−3sin(ax+b)∈[−3,3] So, the solution exists, when x=1 and sin(a+b)=−1 ⇒a+b=3π2,7π2(∵a,b∈[0,2π])