Question

If $\left[\begin{array}{ccc}4a^2& 4a& 1\\ 4b^2& 4b& 1\\ 4c^2& 4c& 1\end{array}\right]\left[\begin{array}{c}f(-1)\\ f\left(1\right)\\ f\left(2\right)\end{array}\right]=\left[\begin{array}{c}3a^2+3a\\ 3b^2+3b\\ 3c^2+3c\end{array}\right],f\left(x\right)$ 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 that chord AB subtends a right angle at V. Find the area enclosed by f (x) and chord AB.

• A. $\frac{125}{3}$ sq. units
  
• B. $\frac{175}{3}$ sq. units
• C. $\frac{135}{3}$ sq. units
• D. None of these
  

Answer 1

The correct option is A $\frac{125}{3}$ sq. units

Let we have
${4a}^{2}f(-1)+4af\left(1\right)+f\left(2\right)={3a}^2+3a$ ...(1)
${4b}^{2}f(-1)+4bf\left(1\right)+f\left(2\right)={3b}^2+3b$ ...(2)
${4c}^{2}f(-1)+4cf\left(1\right)+f\left(2\right)={3c}^2+3c$ ...(3)
consider a quadratic equation
${4x}^{2}f(-1)+4xf\left(1\right)+f\left(2\right)={3x}^2+3x$
$\Rightarrow [4f(-1)-3]x^2+[4f(-1)-3]x+f\left(2\right)=0$ ...(4)
As equation (4) has three roots $\Rightarrow x=a,b,c$
It is an identity
$\Rightarrow f(-1)=\frac{3}{4},f\left(1\right)=\frac{3}{4}$ and $f\left(2\right)=0$
$\Rightarrow f\left(x\right)=\frac{\left({4-x}^2\right)}{4}$ ...(5)
Let point $A$ be $(-2,0)$ and $B$ be $(2t,-t^2+1)$
Now as $AB$ subtends a right angle at the vertex $V(0,1)$
$\frac{1}{2}\times \frac{{-t}^2}{2t}=-1\Rightarrow =4\Rightarrow B=(8,15)$
$\therefore$ Area $={\int }_{-2}^8(\frac{{4-x}^2}{4}+\frac{3x+6}{2})=\frac{125}{3}$ sq.units.