Let, f : R → R be a function such that f x = x 3 + x 2 f ' 1 + x f '' 2 + f ''' 3 ,   ∀ x…

Let, f:RR be a function such that fx=x3+x2f'1+xf''2+f'''3, xR. Then f2 equals
  1. 30
  2. 8
  3. -4
  4. -2

Solution

Let, fx=x3+ax2+bx+c

f'x=3x2+2ax+b

f''x=6x+2a

f'''x=6

According to the question, 

a=f'1=3+2a+ba+b=-3 ...i

b=f''2=12+2a2a-b=-12 ...ii

c=f'''3c=6

Solving equations i & ii, we get, a= -5 & b=2

fx=x3-5x2+2x+6

 f2=8-20+4+6=-2.

Asked in: JEE Main 2019 (10 Jan Shift 1)

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