CSEC Additional Mathematics Paper 1 May/June 2017 - Free CXC Past Paper with Answers

The function $f(x) = 2x^3 - x^2 + hx - 6$ can be expressed as . What is the value of ?

The expression $\frac{1}{x+3} + \frac{3}{x^2-9}$ expressed as a single fraction is

If $x^2-6x+13 = a(x+h)^2+k$ , then

$a=1 \quad h=3 \quad k=4$

$a=1 \quad h=-3 \quad k=4$

The roots of the equation $3x^2-6x-5=0$ are

distinct and not real

real and not distinct

A quadratic equation is such that the sum of its roots is $\frac{2}{3}$ and the product of its roots is $\frac{3}{4}$ . The quadratic equation is

The set of values of $x$ for which $\frac{5x-2}{2-3x} \ge 0$ is given by

$\left\{x : x \le \frac{2}{5} \cup x > \frac{2}{3}\right\}$

$\left\{x : x \le \frac{2}{5} \cup x < \frac{2}{3}\right\}$

$\left\{x : \frac{2}{3} < x \le \frac{2}{5}\right\}$

$\left\{x : \frac{2}{5} \le x < \frac{2}{3}\right\}$

The set of values of $x$ for which $3x+2 > x-2$ is

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

The functions $f$ and $g$ are defined as follows: $f: x \to \frac{x+1}{x-1}, x \neq 1, x \in \mathbb{R}$ $g: x \to 2x+1, x \neq \frac{1}{2}, x \in \mathbb{R}$ The function $fg(x)$ is given by

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

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

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

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

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

$\sqrt[n]{3 \times 27^m}$ is equal to

$3^{\frac{3m+1}{n}}$

The value of $2^z$ where $z = 5 + \log_2 3$ is

Given that $\log_2 x^3 = 6$ , then the value of $x$ is

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

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

The common ratio of the geometric sequence 8, 12, 18, ... is

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 expression $\sum_{r=1}^{5} (-1)^r (3)^{r+1}$ =

The sum of the odd integers between 10 and 50 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

$x^2 + y^2 + 6x - 4y + 3 = 0$

$x^2 + y^2 - 6x + 4y - 3 = 0$

$x^2 + y^2 + 3x - 2y - 3 = 0$

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 vector a is given as 5i + 12j. A unit vector parallel to a is

$\frac{1}{13} (5i + 12j)$

$\frac{3}{13} (5i + 12j)$

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

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

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

The exact value of $\frac{\sin 150^\circ}{\cos 150^\circ}$ is given as

$-\frac{1}{\sqrt{3}}$

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

$\frac{4\pi}{5}$ radians converted to degrees is

The smallest positive angle in radians, within the range $0 \le \theta \le 2\pi$ that satisfies the equation $(2 \cos \theta - 1)(\cos \theta - 2) = 0$ is

sin (α + 45°) is equal to

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

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

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

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

If $\sin \theta = \frac{5}{13}$ and $\theta$ is obtuse, then $\tan \theta = $

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

The trigonometrical expression $\frac{\sin x}{1-\cos x} + \frac{\sin x}{1+\cos x}$ is identical to

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

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

At the point (7, 4) on the curve $y = f(x)$ , $\frac{dy}{dx} = 0$ and $\frac{d^2y}{dx^2} = -5$ . 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 $y = 2x^2 + 3 \cos x$ , then $\frac{dy}{dx}=$

The curve C with equation $y = f(x)$ has a stationary point at $(-2, 5)$ . If $f''(x) = x^4 - 15$ , then the point $(-2, 5)$ is

a maximum turning point

a minimum turning point

If $\int_1^4 f(x)dx = 6$ , then $\int_1^4 4f(x)dx + 5 =$

$\int (\cos x - 2 \sin x) dx =$

$- \cos x + \sin x + c$

$\cos x - \sin x + c$

$\cos x + 2 \sin x + c$

$2 \cos x + \sin x + c$

The region bounded by the curve $y = 2x^2$ , 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 calculated by

If $y = 3x^2 + \cos x$ then $\int y dx$

$\int_0^1 (1 + x + x^2) dx =$

The region R is enclosed by the x-axis, the curve $y = -x^2 + 2$ and the lines $x = 0$ and $x = 1$ . The area of R is

$\frac{x^2-6}{(x^2-9)(x+3)}$

$a=1 \quad h=-4 \quad k=3$

$a=-1 \quad h=4 \quad k=3$

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

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

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

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

$f: x \to \frac{2}{x^2-1}, x \neq \pm 1$

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

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

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

$\frac{1}{\sqrt{3}}$

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

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

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

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

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

CSEC Additional Mathematics Paper 1 May/June 2017 - Free CXC Past Paper with Answers

CSEC Additional Mathematics Paper 1 May/June 2017 Questions