Đề thi Giữa kì 2 Toán 12 năm 2025-2026 trường THPT Số 1 Lê Hồng Phong (Đắk Lắk)

$2x^2 + 3x + C$ .

$x^2 + 3x + C$ .

$x^2 - 3x + C$ .

$2x^2 + 3x$ .

$S=\displaystyle\int_0^1 |e^x|\mathrm{d}x$

$S=\displaystyle\int_1^2 e^x \mathrm{d}x$

$S=\displaystyle\int_0^2 |e^{2x}|\mathrm{d}x$

$S=\displaystyle\int_1^2 \pi e^x\mathrm{d}x$

$f(x) = x^2 + e^x$ .

$f(x) = \dfrac{x^4}{12} + e^x$ .

$f(x) = \dfrac{x^4}{3} + e^x$ .

$f(x) = 3x^2 + e^x$ .

$\displaystyle \int \frac{1}{\cos^{2}x} dx = \tan x + C$

$\displaystyle \int \frac{1}{\sin^{2} x} dx = - \cot x + C$

$\displaystyle \int \cos x dx = \sin x + C$

$\displaystyle \int \sin x dx = \cos x + C$

$2 \ln 3$

$3 \ln 2 - 1$

$1 - \ln 2$

$3 \ln 2$

$\int_{a}^{b} f(x) dx = F(b) + F(a)$

$\int_{a}^{b} f(x) dx = F^{2}(b) - F^{2}(a)$

$\int_{a}^{b} f(x) dx = F(a) - F(b)$

$\int_{a}^{b} f(x) dx = F(b) - F(a)$

$\vec{n} = (2;4;3)$

$\vec{n} = (-3;4;2)$

$\vec{n} = (2;4;-3)$

$\vec{n} = (2;-4;-3)$

$-2021$

$-4035$

$-4053$

$-2$

$x = 0$

$z = 0$

$y + z = 0$

$y = 0$