Prove the following
Question:

Let $\mathrm{P}(\mathrm{x})=\mathrm{x}^{2}+\mathrm{bx}+\mathrm{c}$ be a quadratic polynomial with real coefficients such that $\int_{0}^{1} \mathrm{P}(\mathrm{x}) \mathrm{dx}=1$ 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:

1. (1) 9

2. (2) 15

3. (3) 7

4. (4) 11

Correct Option: , 3

Solution:

$\int_{0}^{1}\left(\mathrm{x}^{2}+\mathrm{bx}+\mathrm{c}\right) \mathrm{dx}=1$

$\frac{1}{3}+\frac{b}{2}+c=1 \quad \Rightarrow \quad \frac{b}{2}+c=\frac{2}{3}$

$3 b+6 c=4$

$\mathrm{P}(2)=5$

$4+2 b+c=5$

$2 b+c=1$

From (1)\&(2)

$\mathrm{b}=\frac{2}{9} \quad \& \mathrm{c}=\frac{5}{9}$

$9(b+c)=7$