Consider a polynomial y = P(x) of the least degree passing through A(-1,1) and whose graph has two points of inflection B(1,2) and C with abscissa 0 at which the curve is inclined to the positive axis of abscissa at an angle of `sec^(-1)sqrt(2)`
The value of P(0) is

Options

A. 1
B. 0
C. `3/4`
D. `1/2`

Answer

4

Explanation

Since two points of inflection occur at x=1 and x=0
p(1)=P(0)=0
`therefore P(x)=a(x^(2)-x)`
or `P(x)=a(x^(3)/(3)-x^(2)/(3))`
also given `(dy)/(dx)_(x=0)=sec^(-1)sqrt(2)=tan^(-1)1`
Hence P(0)=1 so b =1 thus
`P(x)=a(x^(3)/(3)-x^(2)/(2))+1`
`therefore P(x) =a(x^(4)/(12)-x^(3)/(6))=x+c`
As P(1)=2 ,we have
`a((1)/(12)-(1)/(6))+1+c=1`
or `(a)/(12)+c=0`
solving (1) and (2) we have `a =6 and c=1/2`
`P(x) =6(x^(4)/(12)-x^(3)/(6))+x+1/2`
`P(2)=5/2 land p(x)=1/2`
`P(x) =6(x^(3)/(3))-(x^(2)/(2))+1=(x-1)^(2)(2x+1)`