`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(x)=0` has four real roots, then the range of values of leading coefficient of polynomial is

Options

A. [4/9, 1/2]
B. [4/9, 1]
C. [1/3, 1/2]
D. none of these

Answer

A

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).`
For `f(x)=0, " we have " (x^(4))/(4)-2x^(3)+(11x^(2))/(2)-6x= -(1)/(a)`
For `x=1, -(9)/(4) = -(1)/(a) " or " a=(4)/(9)`
For `x=2, -2= -(1)/(a) " or " a=(1)/(2)`
So, `f(x)=0` has four roots if `a in [(4)/(9),(1)/(2)]`.