A) AP
B) GP
C) HP
D) arithmetico-geometric progression
Correct Answer: A
Solution :
Let \[f(x)=A{{x}^{2}}+B{{x}^{2}}+C\] \[\therefore \] \[f(1)=A+B+C\] and \[f(-1)=A-B+C\] \[\because \] \[f(1)=f(-1)\] \[\Rightarrow \] \[A+B+C=A-B+C\] \[\Rightarrow \] \[2B=0\] \[\Rightarrow \] \[B=0\] \[\therefore f(x)=A{{x}^{2}}+C\] Now, \[f'(x)=2Ax\] \[\therefore f'(a)=2\,Aa,\,f'(b)=2Ab,f'(c)=2Ac\] Also. a, b. c are in AP. \[\therefore 2\,Aa,\,2\,Ab,\,2\,Ac\] are in AP. Hence, \[f'(a),\,f'(b)f'(c)\] are in AP.You need to login to perform this action.
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