A) - 1, 6
B) - 2, 1
C) 1, -6
D) \[-1,\,\,\frac{1}{2}\]
Correct Answer: C
Solution :
By Rolles theorem, \[f(1)=f(3)\] \[\Rightarrow \] \[a+b+11-6=27a+9b+33-6\] \[\Rightarrow \] \[26a+8b+22=0\] \[13a+4b+11=0\] ... (i) Now, \[f(x)=3a\,\,{{x}^{2}}+2b\,\,{{x}^{2}}+11\] \[\Rightarrow \] \[f\left( 2+\frac{1}{\sqrt{3}} \right)\] \[=3a{{\left( 2+\frac{1}{\sqrt{3}} \right)}^{2}}+2b\left( 2+\frac{1}{\sqrt{3}} \right)+11\] \[\,\,\Rightarrow \,\,0=3a\,\left( 4+\frac{1}{3}+\frac{4}{\sqrt{3}} \right)+4b+\frac{2b}{\sqrt{3}}+11\] \[\Rightarrow \] \[13a+4b+\frac{12a}{\sqrt{3}}+\frac{2b}{\sqrt{3}}+11=0\] \[\Rightarrow \] \[-11+\frac{12a}{\sqrt{3}}+\frac{2b}{\sqrt{3}}+11=0\] [from Eq.(i)] \[\Rightarrow \] \[6a+b=0\] ... (ii) On solving Eqs. (i) and (ii), we get \[a=1,\,\,b=-6\]
You need to login to perform this action.
You will be redirected in
3 sec
