Lí do lựa chọn phương án

Vị trí thả 1

Ta có \({\rm{ }}f(0) + f(1) = \frac{{ - 7}}{5} + \frac{{ - 3}}{5} =  - 2 \Rightarrow a =  - 2.{\rm{ }}\)

Vị trí thả 2

Ta có \(\int_0^1 f (x)dx =  - 1 \Rightarrow b =  - 1.\)

Vị trí thả 3

Ta có \(\int_0^1 x f'(x)dx = \left. {x.f(x)} \right|_1^0 - \int_0^1 f (x)dx = f(1) - \int_0^1 f (x)dx = \frac{{ - 3}}{5} + 1 = \frac{2}{5}\)

Khi đó \(\frac{c}{d} = \frac{2}{5} \Rightarrow c + d = 2 + 5 = 7\)

-3

Nhiễu: Cho rằng

 \(\int_0^1 x f'(x)dx = \left. {x.f(x)} \right|_1^0 - \int_0^1 f (x)dx = f(1) - \int_0^1 f (x)dx = \frac{{ - 3}}{5} - 1 = \frac{{ - 8}}{5}\)

Khi đó \(\frac{c}{d} = \frac{{ - 8}}{5} \Rightarrow c + d =  - 8 + 5 =  - 3\)

1

Nhiễu: Cho rằng \(\int_0^1 f (x)dx =  - \int_0^1 f (1 - x)dx\) nên

\(2f(x) - 3f(1 - x) = 4x - 1\)

\( =  > \int\limits_0^1 {\left[ {2f\left( x \right) - 3f\left( {1 - x} \right)} \right]dx = \int\limits_0^1 {\left( {4x - 1} \right)dx} } \)

\( \Leftrightarrow 5\int_0^1 f (x)dx = 1 \Leftrightarrow \int_0^1 f (x)dx = \frac{1}{5}\)

Khi đó

\(\int_0^1 x f'(x)dx = \left. {x.f(x)} \right|_0^1 - \int_0^1 f (x)dx\) \( = f(1) - \int_0^1 f (x)dx = \frac{{ - 3}}{5} - \frac{1}{5} = \frac{{ - 4}}{5}.\)

Khi đó \(\frac{c}{d} = \frac{{ - 4}}{5} \Rightarrow c + d =  - 4 + 5 = 1\).