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 :
JEE · Math · previous-year question
- A.9
- B.11
- C.7correct
- D.15
Answer
C. 7
Explanation
$$(x - 2)Q(x) + 5 = {x^2} + bx + c$$ Put x = 2 5 = 2b + c + 4 .... (1) $$\int_0^1 {({x^2} + bx + c)dx} = 1$$ $$ \Rightarrow {1 \over 3} + {b \over 2} + c = 1$$ $${b \over 2} + c = {2 \over 3}$$ .... (2) Solve (1) & (2) $$b = {2 \over 9}$$ $$c = {5 \over 9}$$ 9(b + c) = 7
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