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Question

The least value of a for which the equation 4sinx+11sinx=a has atleast one solution in the interval (0,π2) is

A
5
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B
7
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C
8
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D
9
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Solution

The correct option is D 9
a=4sinx+11sinxdadx=(4sin2x+1(1sinx)2)cosx
dadx=(3sinx2)(2sinx)sin2x(1sinx)2cosx
Critical points will occur, when sinx=23,sinx=0,sinx=1,cosx=0
But given that x(0,π2)
sinx=23,suppose x=x0(0,π2) is the only possible critical point.
As, sinx is increasing in the given interval.
So,f(x) changes sign from negative to positive as x crosses x0 from left to right OR sinx crosses 23 from left to right. So f(x) has local minima at sinx=23
Now, value of a at sinx=23
a=42/3+112/3
a=6+3=9

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