JEE Main 2021 (Online) 16th March Evening ShiftQuestion 1
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 :
Quadratic Equation · Miscellaneous
Solution
Concept. Two linear equations in $b$ and $c$. Formulas used. (1) $\int_0^1(x^2+bx+c)\,dx=\frac13+\frac b2+c$; remainder on division by $x-2$ is $P(2)$Step 1 - Equations. By (1): $$\frac13+\frac b2+c=1,\qquad4+2b+c=5\ \Longrightarrow\ c=1-2b.$$ Step 2 - Solve. $\frac13+\frac b2+1-2b=1$ gives $b=\frac29$, so $c=\frac59$. Step 3 - Value. $9(b+c)=9\cdot\frac79=7$. Answer: (A)