Let P(x) = x2 + 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 :
Solution
$(x - 2)Q(x) + 5 = {x^2} + bx + c$<br><br>Put x = 2<br><br>5 = 2b + c + 4 .... (1)<br><br>$\int_0^1 {({x^2} + bx + c)dx} = 1$<br><br>$\Rightarrow {1 \over 3} + {b \over 2} + c = 1$<br><br>${b \over 2} + c = {2 \over 3}$ .... (2)<br><br>Solve (1) & (2)<br><br>$b = {2 \over 9}$<br><br>$c = {5 \over 9}$<br><br>9(b + c) = 7
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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