Question

If $f\left(x\right)$ is qadratic expression in $x$ and $6{\int }_0^{1}f\left(x\right)dx=(f\left(0\right)+4f\left(\frac{1}{2}\right))=kf\left(1\right)$ then $k$ equals

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is C $1$

Let $f\left(x\right)=a{x}^2+bx+c$
Then
${\int }_0^1\left(f\right(x\left)\right)dx$
$=[\frac{a{x}^3}{3}+\frac{b{x}^2}{2}+cx{]}_0^1$
$=\frac{a}{3}+\frac{b}{2}+c$
Multiplying by 6, we get
$=2a+3b+6c$
Now
$2a+3b+6c=f\left(0\right)+4\left(f\frac{1}{2}\right)$
$=c+4(\frac{a}{4}+\frac{b}{2}+c)$
$=c+a+2b+4c$
Hence
$2a+3b+6c=a+2b+5c$
$a+b+c=0$
$f\left(1\right)=0$
Hence 1 is a root of the following quadratic.