Question

If $f\left(x\right)$ be a quadratic polynomial and ${\int }_{-1}^{1}f\left(x\right)dx=pf(-\frac{1}{2})+qf\left(0\right)+rf\left(\frac{1}{2}\right),$ then $|p+q+r|$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

Let quadratic polynomial $=a{x}^2+bx+c=f\left(x\right)$

${\int }_{-1}^{1}f\left(x\right)dx={\int }_{-1}^1(a{x}^2+bx+c)dx=[a\frac{x^3}{3}+b\frac{x^2}{2}+cx{]}_{-1}^1=2(\frac{a}{3}+c\left)f\right(-\frac{1}{2})=\frac{a}{4}-\frac{b}{2}+cf(0)=cf(\frac{1}{2})=\frac{a}{4}+\frac{b}{2}+c$

As given

$\frac{2}{3}a+2c=P(\frac{a}{4}-\frac{b}{2}+c)+qc+r(\frac{a}{4}+\frac{b}{2}+c)\frac{2}{3}a+2c=a(\frac{P}{4}+\frac{1}{4})+(\frac{-P}{2}+\frac{r}{2})b+(p+q+r)c$

Compare co-efficients

$p+q+r=2$