Questions
The equation
a
4a+b+3=0
b
2a+b+1=0
c
b=0, a=–4/3
d
None of these
detailed solution
Correct option is B
Consider the function f(x)=x3+2ax2+bx Obviously f(x) is continuous in [0,1] and differentiable in ]0,1[.Also f(0)=0 if f(1)=1+2a+b=0 , then all conditions of Rolle’s theorem are satisfied∴ f'(c)=0 for atleast one c in ]0,1[Hence, f'(x)=3x2+4ax+b=0 at least once in ]0,1[i.e., the equation 3x2+4ax+b=0 has atleast one root in ]0,1[.So 2a+b+1=0 .Talk to our academic expert!
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