Let $$f\left( x \right)$$  be a polynomial function of the second degree. If $$f\left( 1 \right) = f\left( { - 1} \right)$$    and $${a_1},\,{a_2},\,{a_3}$$   are in AP then $$f'\left( {{a_1}} \right),\,f'\left( {{a_2}} \right),\,f'\left( {{a_3}} \right)$$     are in :

Solution :
$$\eqalign{ & {\text{Let }}f\left( x \right) = \lambda {x^2} + \mu x + v \cr & {\text{Then }}f'\left( x \right) = 2\lambda x + \mu \cr & {\text{Also,}}\,\,f\left( 1 \right) = f\left( { - 1} \right) \cr & \Rightarrow \lambda + \mu + v = \lambda - \mu + v \cr & \Rightarrow \mu = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\therefore f'\left( x \right) = 2\lambda x \cr & \therefore f'\left( {{a_1}} \right) = 2\lambda {a_1}\,\,\,\,f'\left( {{a_2}} \right) = 2\lambda {a_2}\,\,\,\,f'\left( {{a_3}} \right) = 2\lambda {a_3} \cr & {\text{As }}{a_1},\,{a_2},\,{a_3}{\text{ are in AP, }}f'\left( {{a_1}} \right),\,f'\left( {{a_2}} \right),\,f'\left( {{a_3}} \right){\text{ are in AP}}{\text{.}} \cr} $$