Sample Math Placement Exam (MPE) Practice Flashcards

Vocabulary-style flashcards covering the sample questions for Parts I, II, and III of the Math Placement Exam (MPE).

Slope of the line $2x = 3y - 5$

$$2/3$$

Point of intersection for $x + 2y = 5$ and $2x - 3y = -4$

$$(1, 2)$$

Minimum value of the function $x^2 - 2x + 3$

$$2$$

Value of $\ln(e^3/e^4)$

$$-1$$

Solutions to the absolute value equation $|x - 2| = 5$

$-3$ and $7$

$$\lim_{x\to 2} \frac{4 - x^2}{x^2 - 3x + 2}$$

$$-4$$

The equation of the tangent line to the graph of $y = x - \frac{1}{x}$ at $x = 1$

$$y = (x - 1)$$

Implicit derivative $\frac{dy}{dx}$ at the point $x = -1$, $y = -2$ for $y^2 + \sin(x^2 - 1) = 4$

$$-1/2$$

Maximum of $x^3 - 6x^2 + 9x + 1$ on the interval $[0, 2]$

$$5$$

The derivative calculation: $\frac{d}{dx} e^x \ln(x)$

$$e^x \ln(x) + e^x/x$$

Fundamental Theorem of Calculus evaluation: $\frac{d}{dx} \int_0^x \sin(t^2)dt$

$$\sin(x^2)$$

Evaluation of $\int_0^2 [\frac{d}{dt} \cos(t^2)]dt$

$$(\cos(4)) - 1$$

Area of the triangle with vertices $(0, 0)$, $(1, 3)$, and $(2, 2)$

$$2$$

Antiderivative of $\int (3x^2 - 2x) dx$

$$x^3 - x^2 + C$$

Area under the graph of $y = 1/x$ from $x = 1$ to $x = 2$

$$\ln(2)$$

Result of integration by parts for $\int x \cos(x) dx$

$$x \sin(x) + \cos(x) + C$$

Volume generated by revolving the region under $3x - x^2$ from $x = 0$ to $x = 3$ about the x-axis

$$\pi \int_0^3 (3x - x^2)^2 dx$$

Work done by force $F(x) = e^x$ moving an object from $x = 1$ to $x = 2$

$$e^2 - e$$

$$\int \frac{x dx}{1 + 4x^2}$$

$$\frac{1}{8} \ln(1 + 4x^2) + C$$

Convergence condition for $\int_0^1 \frac{dx}{x^p}$

$$\{p : p < 1\}$$

$$\lim_{x\to 0} \frac{\sin^2(x)}{1 - \cos(x)}$$

$$2$$

Sum of the infinite series $\sum_{n=0}^\infty \frac{x^n}{3^{2n}}$

$$\frac{1}{1 - x/9}$$

The coefficient of $x^4$ in the Maclaurin series for $f(x) = e^{-x/2}$

$$1/384$$

One of the square roots of the complex number $i$

$$(1 + i)/\sqrt{2}$$

The value of $y(2)$ for the differential equation $\frac{dy}{dx} = 2y/x$ given $y(1) = 3$

$$12$$