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(2) iusA. -1B. `(-3)/(2)`C. `(5)/(2)`D. `(7)/(2)`

Consider a polynomial y = P(x) of the least degree passing through A(-

1.	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(2) iusA. -1B. `(-3)/(2)`C. `(5)/(2)`D. `(7)/(2)`
Answer» Correct Answer - 3 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 and p(x)=1/2` `P(x) =6(x^(3))/(3)-(x^(2))/(2)+1=(x-1)^(2)(2x+1)`