Area Ques 10
- If $\left[\begin{array}{ccc}4 a^2 & 4 a & 1 \\ 4 b^2 & 4 b & 1 \\ 4 c^2 & 4 c & 1\end{array}\right]\left[\begin{array}{c}f(-1) \\ f(1) \\ f(2)\end{array}\right]=\left[\begin{array}{c}3 a^2+3 a \\ 3 b^2+3 b \\ 3 c^2+3 c\end{array}\right]$,
$f(x)$ 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 that chord $A B$ subtends a right angle at $V$. Find the area enclosed by $f(x)$ and chord $A B$.
(2005, 5M)
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Answer:
Correct Answer: 10.($\frac{125} {3}$ sq units)
Solution: Given, $\left[\begin{array}{lll}4 a^2 & 4 a & 1 \\ 4 b^2 & 4 b & 1 \\ 4 c^2 & 4 c & 1\end{array}\right]\left[\begin{array}{l}f(-1) \\ f(1) \\ f(2)\end{array}\right]=\left[\begin{array}{l}3 a^2+3 a \\ 3 b^2+3 b \\ 3 c^2+3 c\end{array}\right]$
$\Rightarrow \quad 4 a^2 f(-1)+4 a f(1)+f(2)=3 a^2+3 a$,
$ \quad 4 b^2 f(-1)+4 b f(1)+f(2)=3 b^2+3 b$
and $\quad 4 c^2 f(-1)+4 c f(1)+f(2)=3 c^2+3 c$
where, $f(x)$ is quadratic expression given by,
$ \begin{aligned} & f(x)=a x^2+b x+c \text { and Eqs. (i), (ii) and (iii). } \\ & \Rightarrow \quad 4 x^2 f(-1)+4 x f(1)+f(2)=3 x^2+3 x \\ & \text { or }\{4 f(-1)-3\} x^2+\{4 f(1)-3\} x+f(2)=0 \end{aligned} $
As above equation has $3$ roots $a, b$ and $c$.
So, above equation is identity in $x$.
i.e. coefficients must be zero.
$\Rightarrow \quad f(-1)=3 / 4, f(1)=3 / 4, f(2)=0 $
$\because \quad f(x)=a x^2+b x+c $
$\therefore \quad a=-1 / 4, b=0 \text { and } c=1, \text { using Eq. (v) }$
Thus, $\quad f(x)=\frac{4-x^2}{4}$ shown as,
Let $A(-2,0), B=\left(2 t,-t^2+1\right)$
Since, $A B$ subtends right angle at vertex $V(0,1)$.
$\Rightarrow \quad \frac{1}{2} \cdot \frac{-t^2}{2 t}=-1$
$\Rightarrow \quad t=4 $
$\therefore \quad B(8,-15)$
So, equation of chord $A B$ is $y=\frac{-(3 x+6)}{2}$.
$ \begin{aligned} \therefore \text { Required area } & =\left|\int_{-2}^8\left(\frac{4-x^2}{4}+\frac{3 x+6}{2}\right) d x\right| \\ & =\left|\left(x-\frac{x^3}{12}+\frac{3 x^2}{4}+3 x\right)_{-2}^8\right| \\ & =\left|\left[8-\frac{128}{3}+48+24-\left(-2+\frac{2}{3}+3-6\right)\right]\right| \\ & =\frac{125}{3} \text { sq units } \end{aligned} $
BETA
