JEE MAIN - Mathematics (2021 - 16th March Evening Shift - No. 9)
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 :
9
11
7
15
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
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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