Let `f(x)` be the fourth degree polynomial such that `f^(prime)(0)-6,f(0)=2a n d(lim)_(xvec1)(f(x))/((x-1)^2)=1` The value of `f(2)` is `3` b. 1 c. `0` d.`2`

Options

A. 4
B. 5
C. 6
D. 7

Answer

C

Explanation

Since `underset(xrarr1)(lim)(f(x))/((x-1)^(2))=1,f(1)=0`
`therefore" "underset(xrarr1)(lim)(f(x))/((x-1)^(2))=underset(xrarr1)(lim)(f'(x))/(2(x-1))=1`
`rArr" "f'(1)=0`
`therefore" "underset(xrarr1)(lim)(f''(x))/(2)=1rArrf''(1)=2`
Since x = 1 is root of f(x) = 0 and `f'(x) = 0.`
`f(x)=(x-1)^(2)(ax^(2)+bx+2)" "(because f(0)=2)`
`rArr" "f'(x)=2(x-1)(ax^(2)+bx+2)+(2ax+b)(x-1)^(2)`
`because" "f'(0)=-6rArr b=-2`
Using f''(1)= 2, we get `a+b=-1 rArr a=1`
`rArr" "f(x)=(x-1)^(2)(x^(2)-2x+2)`
`rArr" "f(x)=(x-1)^(4)+(x-1)^(2)`
`rArr" "f(2)=1+1=2`
`"Also, "f'(x)=4(x-1)^(3)+2(x-1)`
`rArr" "f'(2)=4+2=6`