JEE PYQ: Quadratic Equation Question 2
Question 2 - 2021 (16 Mar Shift 2)
Let $P(x) = x^2 + bx + c$ be a quadratic polynomial with real coefficients such that $\int_0^1 P(x),dx = 1$ and $P(x)$ 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
Show Answer
Answer: (3)
Solution
$\int_0^1 (x^2 + bx + c),dx = 1$
$\frac{1}{3} + \frac{b}{2} + c = 1 \Rightarrow \frac{b}{2} + c = \frac{2}{3}$
$3b + 6c = 4$
$P(2) = 5$
$4 + 2b + c = 5$
$2b + c = 1$
From (1) & (2): $b = \frac{2}{9}$ and $c = \frac{5}{9}$
$9(b + c) = 7$
BETA
