Question

If f( {\rm{x}})=\rm{6x}f(x)=6x and b = \rm{\dfrac{a+c}{2}}2a+c​ then ,

a] f( {\rm{a - 3)}} + f {\rm{(c - 3)}}= 2f {\rm{(b - 3)}}f(a−3)+f(c−3)=2f(b−3) 

b] \dfrac{ f{\rm{(a - 3)}} - f \rm{{(b - 3)}}}{f {\rm{(b - 3)}} - f {\rm{(c - 3)}}} = 1f(b−3)−f(c−3)f(a−3)−f(b−3)​=1 

c] \dfrac{ f{\rm{(a - 3)}} - f \rm{{(c - 3)}}}{f {\rm{(b - 3)}} - f {\rm{(a - 3)}}} = 2f(b−3)−f(a−3)f(a−3)−f(c−3)​=2 

d] f({\rm{a - 3)}} - f {\rm{(c - 3)}} = \rm 6(a - c)f(a−3)−f(c−3)=6(a−c) 

NOTE : This question has more than one Correct answer.​

Answer 1

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Answer 2

Answer:

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