CSEC Additional Mathematics Paper 1 May/June 2019 - Free CXC Past Paper with Answers

The expression x - 2 is a factor of

$2x^3 + 2x^2 - 4x - 8$

$3x^4 - 10x^3 - 5x^2 + 4$

When $x^3 - 7x^2 + 2x - 1$ is divided by $x + 2$ , the quotient is

Given that $f(x) = ax^2 + bx + c$ , $f(x)$ can be expressed in the form

$a\left(x + \frac{b}{a}\right)^2 + \frac{ac - b^2}{a^2}$

Which of the following graphs BEST represents $f(x) = x(1-x)$ ?

The values of $x$ for which $(x + 15)^2 = 64x$ are

The range of values for which $x^2 - 7x + 10 < 0$ is

x < 2 \text{ or } x > 5

$x < -5$ and $x > -5$

The set of values of $x$ for which $\frac{2x+1}{x-1} \ge 0$ is

$x \ge -\frac{1}{2}$

$x \ge -\frac{1}{2}$ and $x \ge 1$

$x \le -\frac{1}{2}$ or $x > 1$

If $f(x) = -\frac{2}{9}x^3$ , $-3 \le x \le 3$ , $x \in \mathbb{R}$ , then

The function $fg(x)$ is given by

$\frac{x+1}{x}, x \ne 0, x \in \mathbb{R}$

$\frac{x-2}{x+1}, x \ne -1, x \in \mathbb{R}$

$\frac{x+1}{x+2}, x \ne -2, x \in \mathbb{R}$

$\frac{x-1}{2x+1}, x \ne -\frac{1}{2}, x \in \mathbb{R}$

If function $m: x \mapsto 5+2x$ , then $m(4-2a)$ is

If $f^{-1}: x \mapsto x^2-1, x \ge 0$ , then

$f:x \mapsto 1-x^2, x \in \mathbb{R}$

$\frac{2^{-1}}{8^{\frac{1}{3}}}}$ simplifies to

$\frac{1}{\sqrt{2}}$

Given that $\log_p X = 6$ and $\log_p Y = 4$ , the value of $\log_p \left(\frac{X}{Y}\right)$ is

$\frac{\log_p 6}{\log_p 4}$

Given that 3 × 27^{2x} = 9^x, the value of x is

The value of $x$ for which $4^{x+1} = 2$ is

The value of $x$ such that $\log_2 (5x+1) - \log_2 (3x-5) = 2$ is

The series $-2 + \frac{4}{3} - \frac{8}{9} + \dots$ converges to the limit

The sum of the first $n$ terms of a series is given by $\sum_{r=1}^{n} (5-3r)$ . The sum of the first 10 terms is

The sum of $\sum_{k=1}^{3} \frac{1}{k}$ is

The first four terms of a convergent geometric progression (GP) are 500, 200, 80 and 32 respectively. The sum to infinity of this GP is

A line L passes through the point (6, 5) and is perpendicular to a line whose equation is $3x+4y-7=0$ . The equation of L is

The lines $7x-4y+25=0$ and $3x-y-5=0$ intersect at the point P. The coordinates of P are

A circle C has centre (3, -2) and radius 4. The equation of C is

Two vectors are equal if they

have the same magnitude and different directions

have the same magnitude and same direction

are parallel and are in different directions

have different magnitudes and are in the same direction

The triangle OAB has vertices given by $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ , $\begin{pmatrix} -3 \\ a \end{pmatrix}$ and $\begin{pmatrix} 6 \\ 2 \end{pmatrix}$ respectively. Given that the angle AOB is $\frac{\pi}{2}$ , then $a=$

Given that $OA = \begin{pmatrix} -17 \\ 25 \end{pmatrix}$ and $OB = \begin{pmatrix} 4 \\ 5 \end{pmatrix}$ , then the vector $AB=$

$\begin{pmatrix} -13 \\ 30 \end{pmatrix}$

$\begin{pmatrix} -13 \\ -20 \end{pmatrix}$

Which of the following formulae are used to calculate the area of a sector, A, with radius, r, arc length, l, and angle, q, in radians? I. $A = \frac{r^2 q}{2}$ II. $A = \frac{l^2}{2q}$ III. $A = \frac{lr}{2}$

Item 28 refers to the following triangle (not drawn to scale).

The size of the missing angle, X, measured in radians, is

Which of the following graphs represents $y = \sin x$ ?

For $0^\circ < x < 360^\circ$ , the equation $3 \sin^2 x = 8 \cos x$ has only one real solution given by

$\sin x = -\frac{1}{3}$

$\cos x = \frac{1}{3}$

sin ( $\alpha + 45^\circ$ ) is equal to

$\frac{1}{\sqrt{2}}(\sin \alpha + \cos \alpha)$

$\frac{1}{\sqrt{2}}(\cos \alpha - \sin \alpha)$

$\frac{1}{2}(\sin \alpha - \cos \alpha)$

$\frac{1}{2}(\cos \alpha - \sin \alpha)$

tan ( $2x + \frac{\pi}{4}$ ) is equal to

$\frac{\tan 2x - 1}{1 + \tan 2x}$

$\frac{\tan 2x + 1}{1 - \tan 2x}$

$\frac{2 \tan 2x}{1 - \tan^2 2x}$

If sin ( $x + 20^\circ$ ) = cos x, then the value of x is

For $0 \le \theta \le \frac{\pi}{2}$ , $\tan \theta = \frac{4}{3}$ . The exact value of tan $2\theta$ is

$\frac{d}{dx} \sqrt{7x^2 + 4} =$

$\frac{14x}{\sqrt{7x^2 + 4}}$

At the point (7, 4) on the curve $y = f(x)$ , $\frac{dy}{dx} = 0$ and $\frac{d^2 y}{dx^2} = -5$ . The point (7, 4) is

a point of inflexion

a minimum turning point

a maximum turning point

The gradient at $x = \frac{\pi}{6}$ on the curve $y = \cos x$ is

$-\frac{\sqrt{3}}{2}$

$\frac{\sqrt{3}}{2}$

Given that $y = \cos 2x$ , then $\frac{dy}{dx} =$

$-\frac{1}{2} \sin 2x$

The gradient function $\frac{dy}{dx}$ of the curve $y = \sin (2x^2 + 1)$ is

$4x \cos (2x^2 + 1)$

The region bounded by the curve $y = x^2$ , the x-axis and the lines $x = 0$ and $x = 1$ is rotated $360^\circ$ about the x-axis. The volume of the solid generated can be found from

$\pi \int_0^1 x^2 dx$

$\int (2x-5)^3 dx = $

$\frac{(2x-5)^4}{4} + c$

$\frac{(2x-5)^2}{8} + c$

$\frac{(2x-5)^4}{8} + c$

$\frac{2(2x-5)}{4} + c$

$\int (\sin x - \cos x)dx$

$2 \sin x - \cos x + C$

$-\cos x - 2 \sin x + C$

$\cos x - 2 \sin x + C$

$-\cos x - \sin x + C$

If $\int_{2}^{a} (6+3x)dx = 72$ where $a > 2$ , then $a = $

If $X = \int_{a}^{b} f(x)dx$ and $a < c < b$ , then $X = $

$\int_{0}^{a} f(x)dx + \int_{0}^{b} f(x)dx$

The region R is enclosed by the x-axis, the curve $y = x^2 + 2x - 1$ , the lines $x = 2$ and $x = 3$ . The area of R in units² is

$a\left(x + \frac{b}{2a}\right)^2 + \frac{ac - b^2}{a^2}$

$a\left(x + \frac{b}{2a}\right)^2 + \frac{4ac - b^2}{4a}$

$a\left(x + \frac{b}{2a}\right)^2 + \frac{4ac - b^2}{4a^2}$

$f:x \mapsto \sqrt{x-1}, x \ge 1$

$f:x \mapsto \frac{2}{x-1}, x \ne \pm 1$

$\begin{pmatrix} -21 \\ 20 \end{pmatrix}$

$\begin{pmatrix} 21 \\ -20 \end{pmatrix}$

$\sin x = \frac{1}{3}$

$\cos x = -\frac{1}{3}$

$\frac{7x}{\sqrt{7x^2 + 4}}$

$\frac{7x}{2\sqrt{7x^2 + 4}}$

$\frac{7}{\sqrt{7x^2 + 4}}$

$-\frac{1}{4x} \cos (2x^2 + 1)$

$-4x \cos (2x^2 + 1)$

$\pi \int_0^1 x^4 dx$

$\int_{0}^{c} f(x)dx + \int_{c}^{b} f(x)dx$

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

$\int_{0}^{a} f(x)dx + \int_{0}^{c} f(x)dx - \int_{0}^{b} f(x)dx$

CSEC Additional Mathematics Paper 1 May/June 2019 - Free CXC Past Paper with Answers

CSEC Additional Mathematics Paper 1 May/June 2019 Questions