A) \[\alpha =-6,\beta =\frac{1}{2}\]
B) \[\alpha =-6,\beta =-\frac{1}{2}\]
C) \[\alpha =2,\beta =-\frac{1}{2}\]
D) \[\alpha =2,\beta =\frac{1}{2}\]
Correct Answer: C
Solution :
Here \[f'(x)=\frac{\alpha }{x}+2\beta x+1;\]\[f'(-1)=0\]is an extreme point)\[-\alpha -2\beta +1=0\] ?(1) Also, f′ (2) = 0 gives\[\frac{\alpha }{2}+4\beta +1=0\]?(2) Solving (1) and (2) gives \[\alpha =2,\beta =-\frac{1}{2}\]
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