A) A.P
B) G.P.
C) H.P.
D) None of these
Correct Answer: A
Solution :
Let\[f(x)=a{{x}^{2}}+bx+c\] Then \[f'(x)=2ax+b\] also, \[f(1)=f(-1)\] \[a+b+c=a-b+c\] Þ b = 0 \[\therefore \] \[{f}'(x)=2ax\]; \[\therefore \] \[{f}'({{a}_{1}})=2a{{a}_{1}}\] \[{f}'(a{{ & }_{2}})=2a{{a}_{2}}\], \[{f}'({{a}_{3}})=2a{{a}_{3}}\] As \[{{a}_{1}},{{a}_{2}},{{a}_{3}}\] are in A.P. \[{f}'({{a}_{1}}),\,{f}'({{a}_{2}}),\,{f}'({{a}_{3}})\] are in A.P.You need to login to perform this action.
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