A) - 4
B) 0
C) 2
D) 4
Correct Answer: A
Solution :
\[\because \]\[f(x)={{x}^{2}}+ax+b,\] where \[g'(1)=\alpha \] and \[g''(2)=b\] \[\because \] \[g(x)=(3+a+b){{x}^{2}}+ax+2\] \[\because \] \[g'(x)=2(3+a+b)x+a\] and \[g''(x)\] \[=2(3+a+b)\] \[=g''(2)\] \[\because \] \[g'(1)=2(3+a+b)+a=a\] \[\Rightarrow \,\] \[3+a+b=0\] \[\Rightarrow \,\]\[g''(2)=0=b\] and \[3+a+b=0\] \[\Rightarrow \,\]\[a=-3\]. \[\therefore \] \[f(x)={{x}^{2}}-3x\,\text{ang}\,g(x)=-3x+2\] \[\Rightarrow \,\] \[f'(x)=2x-3\] and \[g'(x)=-3\]. \[\therefore \,\,f'(1)+g'(1)=-1-3=-4\].You need to login to perform this action.
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