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If [(4a^2,4a,1),(4b^2,4b,1),(4c^2,4c,1)...

If `[(4a^2,4a,1),(4b^2,4b,1),(4c^2,4c,1)][(f(-1)),(f(1)),(f(2))][(3a^2+3a),(3b^2+3b),(3c^2+3c)] ` ,`f(x)` is a quadratic function and its maximum valueoccurs at a point V.A is a point of intersection of `y = f (x)` with X-axis and point B is such that chord AB subtendsa right angle at V. Find the area enclosed by f(x) andchord AB.

Text Solution

Verified by Experts

The correct Answer is:
`(125)/(3)` sq units
Given, `[(4a^(2),4a,1),(4b^(2),4b,1),(4c^(2),4c,1)][(f(-1)),(f(1)),(f(2))]=[(3a^(2)+3a),(3b^(2)+3b),(3c^(2)+3c)]`
`rArr 4a^(2) f(-1)+4af(1)+f(2)=3a^(2)+3a, " " ` ...(i)
`4b^(2) f(-1)+4bf(1)+f(2)=3b^(2)+3b, " " ` ...(ii)
`and 4c^(2) f(-1)+4cf(1)+f(2) =3c^(2) + 3c " "` ...(iii)
where, f(x) is quadratic expression given by,
`f(x)=ax^(2) + bx+c `and Eqs. (i), (ii), (iii).
`rArr 4x^(2)f(-1)+4x f(1)+f(2)=3x^(2)+3x`
`or {4f(-1)-3} x^(2)+{4(1)-3}x+f(2) =0 " " ` ...(iv)
As above equation has 3 roots a, b and c.
So, above equation is identity in x.
i.e. coefficients must be zero.
`rArr f(-1) =3//4, f(1)=3//4, f(2)=0 " " ` (v)
` because f(x)=ax^(2) +bx+c`
`therefore a= -1//4, b= 0 and c=1`, using Eq. (v)
Thus, `f(x) =(4-x^(2))/(4)` shown as,
Let `A(-2,0), B(2t,-t^(2)+1)`
Since, AB subtends right angle at vertex V (0, 1).
`rArr (1)/(2) *(-t^(2))/(2t)= -1`
`rArr t=4`
`therefore B((8, -15)`
So, equation of chord AB is `y=(-(3x+6))/(2)`
`therefore" Required area "=|int_(-2)^(8)((4-x^(2))/(4)+(3x+6)/(2)) dx|`
`=|(x-(x^(2))/(12)+(3x^(2))/(4)+3x)_(-2)^(8)|`
`=|[8-(128)/(3)+48+24-(-2+(2)/(3)+3-6)]|`
`=(125)/(3)` sq units
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