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:
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)
$\mathrm{b}=\frac{2}{9} \quad \& \mathrm{c}=\frac{5}{9}$
$9(b+c)=7$
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