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Question

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 B 7π2
x2+4+3sin(ax+b)2x=0(x1)2+3=3sin(ax+b)
We know that,
(x1)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])

Hence, a+b=7π2

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