Question

If $\left[\begin{array}{ccc}4a^2& 4a& 1\\ 4b^2& 4b& 1\\ 4c^2& 4c& 1\end{array}\right]<br/>\left[\begin{array}{c}f(-1)\\ f(1)\\ f(2)\end{array}\right]=<br/>\left[\begin{array}{c}3a^2+3a\\ 3b^2+3b\\ 3c^2+3c\end{array}\right]$
f(x) is a quadratic function and its maximum value
occurs at a point V. A is a point of intersection of y = f (x)
with X-axis and point B is such th a t chord AB subtends
a right angle at V. Find the area enclosed by f(x) and
chord AB.

Answer 1

Given, $\left[\begin{array}{ccc}4a^2& 4a& 1\\ 4b^2& 4b& 1\\ 4c^2& 4c& 1\end{array}\right]<br/>\left[\begin{array}{c}f(-1)\\ f(1)\\ f(2)\end{array}\right]=<br/>\left[\begin{array}{c}3a^2+3a\\ 3b^2+3b\\ 3c^2+3c\end{array}\right]$
$\Rightarrow 4a^{2}f(-1)+4af(1)+f(2)=3a^2+3a........(i)$
$4b^{2}f(-1)+4bf(1)+f(2)=3b^2+3b.........(ii)$
and $4c^{2}f(-1)+4cf(1)+f(2)=3c^2+3c.........(iii)$
where, /(x) is quadratic expression given by,
$f(x)=a{x}^2+bx+c$ and Eqs. (i), (ii) and (iii).
$\Rightarrow 4x^{2}f(-1)+4xf(1)+f(2)=3x^2+3x$
or $\{4f(-1)-3\}{x}^2+\{4f(1)-3\}x+f(2)=0.....(iv)$
As above equation has 3 roots a, b and c.
So, above equation is identity in x.
i.e. coefficients m ust be zero.
$\Rightarrow f(-1)=\frac{3}{4},f(1)=\frac{3}{4},f(2)=0.....(v)$
$\because f(x)=a{x}^2+bx+c$
$\therefore a=-\frac{1}{4},b=0andc=1,usingEq.(v)$
Thus, $f(x)=\frac{(4-x^2)}{4}$ shown as,
Let $A(-2,0),B=(2t,-t^2+1)$
Since, AB subtends right angle at vertex V (0,1).
$\Rightarrow 22t=-1\Rightarrow t=4$
$\therefore B(8,-15)$
So, equation of chord AB is $y=\frac{-(3x+6)}{2}.$
$\therefore Requiredarea=\left|\underset{-2}{\overset{8}{\int }}\right(\frac{(4-x^2)}{4}+\frac{(3x+6)}{2}\left)dx\right|$
$=\left|\right(x-\frac{x^3}{12}+\frac{3x^2}{4}+3x{)}_{-2}^8|$
$=\left|\right[8-\frac{128}{3}+48+24-(-2+\frac{2}{3}+3-6)\left]\right|$
$=\frac{125}{3}squnits$