∫01(x2+bx+c)dx=1
31+2b+c=1⇒2b+c=32
3b+6c=4...(1)
P(2)=5
4+2b+c=5
2b+c=1...(2)
From (1)&(2)
b=\frac{2}{9}&c=\frac{5}{9}
9(b+c)=7
Let P(x)=x2+bx+c be a quadratic polynomial with real coefficients such that ∫01P(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:
Held on 16 Mar 2021 · Verified 6 Jul 2026.
9
15
7
11
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