`y= f(x)` is a polynomial function passing through point (0, 1) and which increases in the intervals `(1, 2) and (3, oo)` and decreases in the intervals `(oo,1) and (2, 3).`
If `f(1) = -8,` then the value of `f(2)` is

Options

A. -3
B. -6
C. -20
D. -7

Answer

D

Explanation

From the given data, we can conclude that `(dy)/(dx)=0 " at " x=1,2,3.`
Hence, `f'(x)=a(x-1)(x-2)(x-3), a gt 0`
`implies f(x)=int a(x^(3)-6x^(2)+11x-6)dx`
`=a((x^(4))/(4)-2x^(3)+(11x^(2))/(2)-6x)+C`
Also, `f(0)=1 implies c=1`
` :. f(x)=a((x^(4))/(4)-2x^(3)+(11x^(2))/(2)-6x)+1 " (1)" `
So, graph is symmetrical about line `x=2` and range is `[f(1), oo) or [f(3),oo).`
`f(1)=8`
`implies a(-(9)/(4))+1=-8`
`impliesa=4`
`:. f(2)= -7`