Let P(x) = x squre + bx + c be a quadratic polynomial

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:

  1. 9

  2. 15

  3. 7

  4. 11


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)

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

$9(b+c)=7$

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