A) \[-\,3,\text{ }2\]
B) \[2,\,\,-4\]
C) 1, 6
D) none of these
Correct Answer: D
Solution :
[d] \[f(x)=a{{x}^{2}}+b{{x}^{2}}+11x-6\] Satisfies condition of Roll's theorem in [1, 3]. Therefore, \[f(1)=f(3)\] or \[a+b+11-6=27a+9b+33-6\] or \[13a+4b=-11\] ...(1) And \[f'(x)=3a{{x}^{2}}+2bx+11\] or \[f'\left( 2+\frac{1}{\sqrt{3}} \right)=3a{{\left( 2+\frac{1}{\sqrt{3}} \right)}^{2}}+2b\left( 2+\frac{1}{\sqrt{3}} \right)+11=0\]or \[3a\left( 4+\frac{1}{3}+\frac{1}{\sqrt{3}} \right)+2b\left( 2+\frac{1}{\sqrt{3}} \right)+11=0\] ...(2) From equations (1) and (2), We get \[a=1,\text{ }b=-\,6.\]
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