Question:
Let $\mathrm{P}(\mathrm{x})=\mathrm{x}^{2}+\mathrm{bx}+\mathrm{c}$ be a quadratic polynomial
with real coefficients such that $\int_{0}^{1} \mathrm{P}(\mathrm{x}) \mathrm{d} \mathrm{x}=1$ and
$\mathrm{P}(\mathrm{x})$ leaves remainder 5 when it is divided by $(x-2)$. Then the value of $9(b+c)$ is equal to:
Correct Option: , 3
Solution:
$\int_{0}^{1}\left(x^{2}+b x+c\right) d x=1$
$\frac{1}{3}+\frac{b}{2}+c=1 \quad \Rightarrow \quad \frac{b}{2}+c=\frac{2}{3}$
$3 b+6 c=4$ ......(1)
$P(2)=5$
$4+2 b+c=5$
$2 \mathrm{~b}+\mathrm{c}=1$......(2)
From (1) & (2)
$\mathrm{b}=\frac{2}{9} \quad \& \quad \mathrm{c}=\frac{5}{9}$
$9(b+c)=7$