In the given figue vertices of `DeltaABC` lie on `y=f(x)=ax^(2)+bx+c`. The `DeltaAB` is right angled isosceles triangle whose hypotenuse `AC=4sqrt(2)` units.

Number of integral value of `lamda` for which `(lamda)/2` lies between the roots of `f(x)=0`, is

Options

A. `y = x^(2) - 2sqrt(2)`
B. `y = x^(2) - 12`
C. `y = (x^(2))/(2)-2`
D. `y = (x^(2))/(2sqrt(2)) - 2 sqrt(2)`

Answer

4

Explanation

`because AC = 4sqrt(2)`
`therefore AB = BC = (4sqrt(2))/(2) =4` units
`OB = sqrt(4^(2) -(2sqrt(2))^(2)) = 2sqrt(2)`
`therefore A(-2sqrt(2),0),C(2sqrt(2),0),B(0,-2sqrt(2))`
Since `y = ax^(2) + bx + x` passes through A,B and C, we get
`y = (x^(2))/(2sqrt(2)) - 2sqrt(2)`