Let f be a polynomial function such that f 3 x = f ′ x .   f ′ ′ x , for all x &#8712…

Let f be a polynomial function such that f3x=fx. fx, for all xR. Then :
  1. f2+f2=28
  2. f2-f2=0
  3. f2-f2+f2= 10
  4. f2-f2=4

Solution

Degree of fx will be 3

f(x)=ax3+bx2+cx+d

f3x=27ax3+9bx2+3cx+d

f'(x)=3ax2+2bx+c

fx=6ax+2b

f3x=fx fx

Comparing the coefficient, we get

27a=18a2 a= 3 2

Also b=0, c=0, d=0

fx=32x3, f( 2 )=12

fx=92x2, fx=9x

Hence, f'(2)=18, f''(2)=18

Hence, f''2f'(2)=0

Asked in: JEE Main 2017 (09 Apr Online)

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