Consider the equation ∫x0(t2−8t+13)dt=xsin(ax). If x takes the values for which the equation has a solution, then the number of values of aϵ[0,100] are,
A
2
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B
1
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C
5
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D
3
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Solution
The correct option is D3 After integrating above equation,
x33−4x2+13x=xsin(ax)
⇒x2−12x+39=3sin(ax), since x=0 is not in domain
⇒(x−6)2+3=3sin(ax)
Now the maximum possible value of RHS is 3 and the minimum possible value of LHS is 3