A) AGP
B) AP
C) GP
D) HP
Correct Answer: B
Solution :
Let \[f(1)=f(-1)\]. Then, \[\Rightarrow \] \[A+B+C=A-B-C\] \[\Rightarrow \] \[B=0\] \[\therefore \] \[f(x)=A{{x}^{2}}+C\] \[\Rightarrow \] \[f'(x)=2Ax\] \[\Rightarrow \]\[f'(a)=2Aa,f'(b)=2Ab\,and\,\,f'(c)=2Ac\] Since, a, b, c are in AP. \[\therefore \] 2Aa, 2Ab, 2Ac 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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