I dont understand this 2 parts Please show me how do you sol

I dont understand this 2 parts.. Please show me how do you solve this and inlcude your answer to these both parts.. Thank you!

Solution

(a) C={ f(x) P : f \'(0)=0}

example: let f(x) = 2x² ...[ because f \' (x) = 4x => f \' (0) = 0]

OR f(x) = 2x³-3x²

D={ f(x) P | f(0) = 0} : example f(x) = 2x OR f(x)= 2x³-3x

E= { f(x)=a₀ | a₀ R} : example f(x) = 2 , f(x) = 3

F= { f(x) P | f(0)=f(1)=f(2)=0} : example f(x) = (x-2)(x-1)x ; f(x) = (x-2)(x-1)x²

G= {f(x) P | f \'\'(0) = f \'(0)=0}: example f(x)= 2x^{3 ...} [ because f \'(x)= 6x² and f \'\'(x)= 12x => f \'\'(0)=f \'(0)=0]

also f(x) = 3x⁴-2x³ G

(b) iii) C E : False

C={ f(x) P : f \'(0)=0} & E= { f(x)=a₀ | a₀ R}

since f(x) = 2x² C but it does not belong in E.

iv) E C : True

let f(x) E => f(x) = a₀ for some a₀ R

=> f \'(x)=0 for all x . therefore in particular, f \' (0) =0

=>f(x) C.

v) C   F : False

F= { f(x) P | f(0)=f(1)=f(2)=0} and C={ f(x) P : f \'(0)=0}

let f(x)= 2x² . then f(x) C

but f(x) F [ because f(1)= 2]

vi) F C : False

f(x) = x(x-1)(x-2) F

=> f(x) = x( x²-3x+2) = x³-3x²+2x ( on simplifying)

=> f \'(x) = 3x²-6x+2

but f \'(0) 0 . therefore f(x) C.

vi) D F : False

D={ f(x) P | f(0) = 0} and F= { f(x) P | f(0)=f(1)=f(2)=0}

since f(x) = 2x²-x D

but f(x) F [ because f(1) = 1]

xii) F D: true

since if f(x) F => f(0)=f(1)=f(2)=0

=> f(x) D [ we need only f(0)= 0 for it to belong in D & clearly f satisfies it ]

xvii) E G : true

E= { f(x)=a₀ | a₀ R} and G= {f(x) P | f \'\'(0) = f \'(0)=0}

let f(x) E => f(x) = a₀ for some a₀ R

=> f \'(x)=0 for all x . therefore in particular, f \' (0) =0

Also, f \'\'(x) = 0 for all x, in particular true for x=0

=> f(x) G.

xviii) G E : False

f(x) =  2x³ G

but  f(x) E [ since 2x³ is not constant]