Question

If $f\left(x\right)=a+bx+c{x}^2$, $c>0$, $b^2-4ac<0$ then area enclosed by the co-ordinate axes, the line $x=2$ & the curve $y=f\left(x\right)$ is given by $\frac{1}{3}\{f\left(0\right)+\lambda f\left(1\right)+f\left(2\right)\}$ square units. Then value of $\lambda$ equals

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

Answer: (B)

$4$

Now ${\int }_0^{2}ydx={\int }_0^2(a+bx+c{x}^2)dx$
$=2a+2b+\frac{8}{3}c=\frac{1}{3}[6a+6b+8c]$ (i)
Again $f\left(x\right)=a+bx+c{x}^2$
$\therefore$ $f\left(0\right)=a$, $f\left(1\right)=a+b+c$, $f\left(2\right)=a+2b+4c$ then
$a=f\left(0\right)$, $b=\frac{4f\left(1\right)-f\left(2\right)-3f\left(0\right)}{2}$ & $c=\frac{f\left(2\right)+f\left(0\right)-2f\left(1\right)}{2}$
$\therefore$ By equation (i) we have (by using a, b, c)
$\therefore$ ${\int }_0^{2}ydx=\frac{1}{3}[6a+6b+8c]$
$=\frac{1}{3}[6f\left(0\right)+3(4f\left(1\right)-f\left(2\right)-3f\left(0\right))+4(f\left(2\right)+f\left(0\right)-2f\left(1\right))]$
$=\frac{1}{3}[(6-9+4)f\left(0\right)+(12-8)f\left(1\right)+(-3+4)f\left(2\right)]$
$=\frac{1}{3}[f\left(0\right)+4f\left(1\right)+f\left(2\right)]=\frac{1}{3}[f\left(0\right)+\lambda f\left(1\right)+f\left(2\right)]$
$\therefore$ $\lambda =4$