`f:RrarrR,f(x)=x^(3)+x^(2)f'(1)+xf''(2)+f'''(3)" for all "x in R.`
The value of f(1) is

Options

A. `f(0)+f(2)=f(1)`
B. `f(0)+f(3)=0`
C. `f(1)+f(3)=f(2)`
D. all the above

Answer

D

Explanation

We have,
`f(x)=x^(3)+x^(2)f'(1)+xf''(2)+f'''(3)" "…(i)`
`f'(x)=3x^(2)+2xf'(1)+f''(2)" "…(ii)`
`implies" "f''(x)=6x+2f'(1)" "…(iii)`
`impliesf'''(x)=6" "…(iv)`
Putting x=1, 2, 3 in (ii), (iii), and (iv) respectively, we get
`f'(1)=32f'(1)+f''(2)`
`f''(2)=12+2f'(1),f'''(3)=6`
Solving these equations, we get
`f'(1)=-5,f''(2)=2" and "f'''(3)=6`
`f(x)=x^(3)-5x^(2)+2x+6.`
`implies" "f(0)=6,f(1)=4,f(2)=8-20+4+6=-2`
and, `f(3)=-6.`
Clearly, these values satisfy options (a), (b) and c.