ESAT Mathematics 2 Mock 5

At $\displaystyle x=2$ , $\displaystyle P=(2,2)$ . The tangent gradient is $\displaystyle 3(2)^2-3=9$ , so the tangent is $\displaystyle y=9x-16$ and $\displaystyle R=(0,-16)$ . The normal has gradient $\displaystyle -1/9$ , so $\displaystyle y-2=-\dfrac19(x-2)$ . Setting $\displaystyle y=0$ gives $\displaystyle Q=(20,0)$ . Therefore the area is $\displaystyle \dfrac12(20)(16)=160$ .

Step 1:

Identify Domain Constraints

For the logarithms to be defined, their arguments must be strictly positive.
1. $\displaystyle \sin x > 0$ 2. $\displaystyle 1 - \cos x > 0 \implies \cos x < 1$ The condition $\displaystyle \sin x > 0$ means that any solution must be in the interval $\displaystyle (0, \pi)$ .
The condition $\displaystyle \cos x < 1$ excludes $\displaystyle x=0$ and $\displaystyle x=2\pi$ from the given interval $\displaystyle 0 \le x \le 2\pi$ .

Step 2:

Simplify the Logarithmic Equation

Using the logarithm laws $\displaystyle \log_a M + \log_a N = \log_a(MN)$ and $\displaystyle 1 = \log_2(2)$ , we can rewrite the equation: $\displaystyle \log_2((\sin x)(3 \sin x)) = \log_2(2) + \log_2(1 - \cos x)$ $\displaystyle \log_2(3 \sin^2 x) = \log_2(2(1 - \cos x))$ Equating the arguments of the logarithms gives: $\displaystyle 3 \sin^2 x = 2(1 - \cos x)$ Step 3:

Form a Quadratic Equation in $\displaystyle \cos x$ Using the Pythagorean identity $\displaystyle \sin^2 x = 1 - \cos^2 x$ , we substitute to get an equation solely in terms of $\displaystyle \cos x$ : $\displaystyle 3(1 - \cos^2 x) = 2 - 2 \cos x$ $\displaystyle 3 - 3 \cos^2 x = 2 - 2 \cos x$ Rearranging gives a standard quadratic form: $\displaystyle 3 \cos^2 x - 2 \cos x - 1 = 0$ Step 4:

Solve the Quadratic Equation

Let $\displaystyle c = \cos x$ . The equation is $\displaystyle 3c^2 - 2c - 1 = 0$ . We can factor this quadratic: $\displaystyle (3c + 1)(c - 1) = 0$ This gives two possible values for $\displaystyle \cos x$ : $\displaystyle \cos x = -\dfrac{1}{3} \quad \text{or} \quad \cos x = 1$ Step 5:

Verify Solutions Against Domain Constraints

We must check these potential solutions against the domain constraints identified in the first step.

Case 1: $\displaystyle \cos x = 1$ This occurs at $\displaystyle x=0$ and $\displaystyle x=2\pi$ in the given interval. However, our domain requires $\displaystyle \cos x < 1$ . Therefore, these are not valid solutions.

Case 2: $\displaystyle \cos x = -\dfrac{1}{3}$ In the interval $\displaystyle 0 \le x \le 2\pi$ , there are two angles for which $\displaystyle \cos x = -\dfrac{1}{3}$ : one in the second quadrant ( $\displaystyle x_1 \in (\pi/2, \pi)$ ) and one in the third quadrant ( $\displaystyle x_2 \in (\pi, 3\pi/2)$ ).
We must check these against the other domain constraint, $\displaystyle \sin x > 0$ .
- For the solution $\displaystyle x_1$ in the second quadrant, $\displaystyle \sin x_1 > 0$ . This is a valid solution.
- For the solution $\displaystyle x_2$ in the third quadrant, $\displaystyle \sin x_2 < 0$ . This is not a valid solution.

Thus, only one of the potential solutions is valid.

Step 6:

Conclusion

There is exactly one value of $\displaystyle x$ in the interval $\displaystyle 0 \le x \le 2\pi$ that satisfies the original equation.

Let $\displaystyle n$ be the number of sides of polygon P.

The exterior angle of a regular $\displaystyle k$ -sided polygon is $\displaystyle \dfrac{360^{\circ}}{k}$ , and its interior angle is $\displaystyle 180^{\circ} - \dfrac{360^{\circ}}{k}$ .

- Polygon P has $\displaystyle n$ sides, so its exterior angle is $\displaystyle E_\text{P} = \dfrac{360^{\circ}}{n}$ .
- Polygon Q has $\displaystyle 2n$ sides, so its exterior angle is $\displaystyle E_\text{Q} = \dfrac{360^{\circ}}{2n} = \dfrac{1}{2}E_\text{P}$ .
- Polygon R has $\displaystyle 3 \times 2n = 6n$ sides, so its exterior angle is $\displaystyle E_\text{R} = \dfrac{360^{\circ}}{6n} = \dfrac{1}{6}E_\text{P}$ .

The difference between the interior angles of R and P equals the difference between their exterior angles: $\displaystyle I_\text{R} - I_\text{P} = E_\text{P} - E_\text{R} = E_\text{P} - \dfrac{1}{6}E_\text{P} = \dfrac{5}{6}E_\text{P}$ We are given that $\displaystyle I_\text{R} - I_\text{P} = 25^{\circ}$ : $\displaystyle \dfrac{5}{6}E_\text{P} = 25^{\circ} \implies E_\text{P} = 25^{\circ} \times \dfrac{6}{5} = 30^{\circ}$ Then the exterior angle of Q is: $\displaystyle E_\text{Q} = \dfrac{1}{2} E_\text{P} = 15^{\circ}$ The difference between the interior angle of Q and the interior angle of P is: $\displaystyle I_\text{Q} - I_\text{P} = E_\text{P} - E_\text{Q} = 30^{\circ} - 15^{\circ} = 15^{\circ}$

The function $\displaystyle f(x) = 2x - x^2 = x(2-x)$ has roots at $\displaystyle x=0$ and $\displaystyle x=2$ . Since $\displaystyle k > 2$ , the curve lies above the axis for $\displaystyle 0 < x < 2$ and below the axis for $\displaystyle 2 < x < k$ .

We first calculate the area of the positive region:Let $\displaystyle N$ be the area of the region below the axis (where $\displaystyle N > 0$ ). We can express the definite integral $\displaystyle I$ and the total geometric area $\displaystyle A$ as:Adding these equations eliminates $\displaystyle N$ , giving $\displaystyle A + I = \dfrac{8}{3}$ . We are given $\displaystyle A = 3I$ , so:We now equate this value to the integral formula in terms of $\displaystyle k$ :Setting $\displaystyle k^2 - \dfrac{k^3}{3} = \dfrac{2}{3}$ leads to the cubic equation $\displaystyle k^3 - 3k^2 + 2 = 0$ . By inspection, $\displaystyle k=1$ is a root. We factorise the cubic:The roots are $\displaystyle k=1$ and $\displaystyle k = \dfrac{2 \pm \sqrt{4 - 4(1)(-2)}}{2} = 1 \pm \sqrt{3}$ . Since we require $\displaystyle k > 2$ , the only valid solution is $\displaystyle k = 1 + \sqrt{3}$ .

Let the two points be $\displaystyle P(1, 1)$ and $\displaystyle Q(3, 5)$ . The square of the distance, $\displaystyle d^2$ , between them is:A regular hexagon with side length $\displaystyle s$ is composed of 6 equilateral triangles of side length $\displaystyle s$ . The area of one such triangle is $\displaystyle \dfrac{1}{2} s^2 \sin(60^\circ) = \dfrac{\sqrt{3}}{4}s^2$ . Therefore, the area of the hexagon is $\displaystyle A = 6 \times \dfrac{\sqrt{3}}{4}s^2 = \dfrac{3\sqrt{3}}{2}s^2$ .

There are three distinct geometric relationships for two vertices on a regular hexagon, leading to three possible hexagon sizes.

Case 1: The points are adjacent vertices.
The distance between them is the side length, $\displaystyle s_1$ . So, $\displaystyle s_1^2 = d^2 = 20$ .
The area of this hexagon is $\displaystyle A_1 = \dfrac{3\sqrt{3}}{2} s_1^2 = \dfrac{3\sqrt{3}}{2}(20) = 30\sqrt{3}$ .

Case 2: The points are opposite vertices.
The distance between them is the long diagonal, which has length $\displaystyle 2s_2$ . So, $\displaystyle (2s_2)^2 = d^2$ , which means $\displaystyle 4s_2^2 = 20$ , and $\displaystyle s_2^2 = 5$ .
The area of this hexagon is $\displaystyle A_2 = \dfrac{3\sqrt{3}}{2} s_2^2 = \dfrac{3\sqrt{3}}{2}(5) = \dfrac{15\sqrt{3}}{2}$ .

Case 3: The points are separated by one vertex.
The distance between them is the short diagonal, which has length $\displaystyle s_3\sqrt{3}$ . So, $\displaystyle (s_3\sqrt{3})^2 = d^2$ , which means $\displaystyle 3s_3^2 = 20$ , and $\displaystyle s_3^2 = \dfrac{20}{3}$ .
The area of this hexagon is $\displaystyle A_3 = \dfrac{3\sqrt{3}}{2} s_3^2 = \dfrac{3\sqrt{3}}{2} \left(\dfrac{20}{3}\right) = 10\sqrt{3}$ .

The sum of the areas of all possible hexagons is $\displaystyle A_1 + A_2 + A_3$ .Thus, the correct answer is $\displaystyle \dfrac{95\sqrt{3}}{2}$ .

We can solve for $\displaystyle \angle OCA$ using circle theorems in two steps:

1. **Find $\displaystyle \angle BAC$ and $\displaystyle \angle ABC$ :**
- By the alternate segment theorem, the angle subtended by chord $\displaystyle AB$ at the circumference in the alternate segment is $\displaystyle \angle ACB = \angle YAB = 65^\circ$ .
- Since $\displaystyle AB = BC$ , triangle $\displaystyle ABC$ is isosceles, so $\displaystyle \angle BAC = \angle BCA = 65^\circ$ .
- Therefore, $\displaystyle \angle ABC = 180^\circ - 2(65^\circ) = 50^\circ$ .

2. **Find $\displaystyle \angle OCA$ :
- Method 1 (via angle at centre):** The angle subtended by chord $\displaystyle AC$ at the centre is $\displaystyle \angle AOC = 2 \times \angle ABC = 2 \times 50^\circ = 100^\circ$ . Since $\displaystyle OA = OC$ (both are radii), triangle $\displaystyle OAC$ is isosceles, so $\displaystyle \angle OCA = \dfrac{180^\circ - 100^\circ}{2} = 40^\circ$ .
- Method 2 (via radius perpendicular to tangent): The radius $\displaystyle OA$ is perpendicular to tangent $\displaystyle XY$ , so $\displaystyle \angle OAY = 90^\circ$ . Thus, $\displaystyle \angle OAB = 90^\circ - 65^\circ = 25^\circ$ . Then $\displaystyle \angle OAC = \angle BAC - \angle OAB = 65^\circ - 25^\circ = 40^\circ$ . Since $\displaystyle OA = OC$ , $\displaystyle \angle OCA = \angle OAC = 40^\circ$ .

Hence, the correct option is C.

Let the arithmetic sequence be denoted by $\displaystyle u_n$ .

We are given the first two terms: $\displaystyle u_1 = 1$ $\displaystyle u_2 = \dfrac{1}{6}$ The common difference, $\displaystyle d$ , is the difference between consecutive terms:We need the third term, $\displaystyle u_3$ , of the arithmetic sequence:We also need the sum of the first three terms, $\displaystyle S_3$ :Now, let the geometric sequence have first term $\displaystyle a$ and common ratio $\displaystyle r$ .

According to the problem statement:
- The first term $\displaystyle a$ is equal to $\displaystyle u_3$ . So, $\displaystyle a = -\dfrac{2}{3}$ .
- The common ratio $\displaystyle r$ is equal to $\displaystyle S_3$ . So, $\displaystyle r = \dfrac{1}{2}$ .

Since $\displaystyle |r| = |\dfrac{1}{2}| < 1$ , the sum to infinity of the geometric sequence exists.
The formula for the sum to infinity is $\displaystyle S_\infty = \dfrac{a}{1-r}$ .

Substituting the values of $\displaystyle a$ and $\displaystyle r$ :Therefore, the correct answer is $\displaystyle -\dfrac{4}{3}$ .

$\displaystyle k=[x^3+x]_0^2=8+2=10$ . The sum to infinity is $\displaystyle 10/(1-1/3)=10/(2/3)=15$ . The correct option is C.

We can express the function entirely in terms of $\displaystyle \cos x$ using the identity $\displaystyle \sin^2 x = 1 - \cos^2 x$ . Let the expression be $\displaystyle y$ .Let $\displaystyle u = \cos x$ . Since $\displaystyle x$ is a real number, the possible values for $\displaystyle u$ are in the interval $\displaystyle [-1, 1]$ . We need to find the maximum value of the quadratic function $\displaystyle f(u) = -u^2 + 3u + 1$ on this restricted domain.

The parabola $\displaystyle f(u)$ opens downwards, and its vertex is at $\displaystyle u = -\dfrac{3}{2(-1)} = 1.5$ .

Since the vertex is outside the domain $\displaystyle [-1, 1]$ , the maximum value on this interval must occur at an endpoint. As the vertex is to the right of the interval, the function is strictly increasing on $\displaystyle [-1, 1]$ . The maximum value therefore occurs at $\displaystyle u=1$ .

Substituting $\displaystyle u=1$ into the expression gives:The maximum value is 3.

The particle's motion is described by $\displaystyle x = 5 \cos(3t)$ . This is simple harmonic motion with amplitude $\displaystyle A=5$ and angular frequency $\displaystyle \omega=3$ .

The period of one full oscillation is $\displaystyle T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{3}$ seconds.

The time interval is from $\displaystyle t=0$ to $\displaystyle t=\dfrac{5\pi}{6}$ . The duration of this interval is $\displaystyle \dfrac{5\pi}{6}$ . To find how many oscillations this corresponds to, we can divide the duration by the period:So the particle completes $\displaystyle 1\dfrac{1}{4}$ oscillations.

In one full oscillation, the particle travels a distance of $\displaystyle 4A = 4 \times 5 = 20$ m.

The particle starts at its maximum positive displacement, $\displaystyle x(0) = 5\cos(0) = 5$ . After one full oscillation, it returns to this position. The remaining quarter of an oscillation takes the particle from this extreme position ( $\displaystyle x=5$ ) to the equilibrium position ( $\displaystyle x=0$ ). The distance covered in this segment is the amplitude, $\displaystyle A=5$ m.

The total distance travelled is the sum of these parts: $\displaystyle 20 + 5 = 25$ m.

We use the identities $\displaystyle \sin(2x) \equiv 2\sin(x)\cos(x)$ and $\displaystyle \tan(x) \equiv \dfrac{\sin(x)}{\cos(x)}$ . The equation becomes:Multiplying by $\displaystyle \cos(x)$ (valid since $\displaystyle \cos(x) eq 0$ ) and rearranging yields:This gives two cases:

1. $\displaystyle \sin(x) = 0$ . In the interval $\displaystyle 0 \le x \le 2\pi$ , the solutions are $\displaystyle 0, \pi, 2\pi$ (3 solutions).

2. $\displaystyle 2\cos^2(x) - 1 = 0 \implies \cos(x) = \pm \dfrac{1}{\sqrt{2}}$ .
* $\displaystyle \cos(x) = \dfrac{1}{\sqrt{2}}$ gives $\displaystyle x = \dfrac{\pi}{4}, \dfrac{7\pi}{4}$ .
* $\displaystyle \cos(x) = -\dfrac{1}{\sqrt{2}}$ gives $\displaystyle x = \dfrac{3\pi}{4}, \dfrac{5\pi}{4}$ .
This contributes 4 solutions.

Total solutions: $\displaystyle 3 + 4 = 7$ .

Since the sum of probabilities for the distribution of $\displaystyle Y$ must equal 1: $\displaystyle \text{P}(Y = 1) + \text{P}(Y = 2) + \text{P}(Y = 3) = 1 \implies 2p + q = 1$ Because $\displaystyle X$ and $\displaystyle Y$ both take values from $\displaystyle \{1, 2, 3\}$ , the only pair of values that gives $\displaystyle X + Y = 6$ is $\displaystyle X = 3$ and $\displaystyle Y = 3$ .
Since $\displaystyle X$ and $\displaystyle Y$ are independent: $\displaystyle \text{P}(X + Y = 6) = \text{P}(X = 3) \times \text{P}(Y = 3) = \dfrac{1}{6} \times q = \dfrac{q}{6}$ We are given $\displaystyle \text{P}(X + Y = 6) = \dfrac{1}{12}$ , so: $\displaystyle \dfrac{q}{6} = \dfrac{1}{12} \implies q = \dfrac{1}{2}$ Substituting $\displaystyle q = \dfrac{1}{2}$ into $\displaystyle 2p + q = 1$ gives: $\displaystyle 2p = 1 - \dfrac{1}{2} = \dfrac{1}{2} \implies p = \dfrac{1}{4}$ The event $\displaystyle X + Y = 3$ occurs when $\displaystyle (X, Y) = (1, 2)$ or $\displaystyle (X, Y) = (2, 1)$ :
- $\displaystyle \text{P}(X = 1, Y = 2) = \text{P}(X = 1) \times \text{P}(Y = 2) = \dfrac{1}{2} \times \dfrac{1}{4} = \dfrac{1}{8}$ - $\displaystyle \text{P}(X = 2, Y = 1) = \text{P}(X = 2) \times \text{P}(Y = 1) = \dfrac{1}{3} \times \dfrac{1}{4} = \dfrac{1}{12}$ Adding these probabilities: $\displaystyle \text{P}(X + Y = 3) = \dfrac{1}{8} + \dfrac{1}{12} = \dfrac{3}{24} + \dfrac{2}{24} = \dfrac{5}{24}$

There are two effective methods to determine the equivalent single transformation:

Method 1: Geometric reasoning
The composition of reflections in two intersecting lines $\displaystyle L_1$ and $\displaystyle L_2$ meeting at point $\displaystyle P$ with an angle $\displaystyle \theta$ between them is equivalent to a rotation about $\displaystyle P$ through an angle $\displaystyle 2\theta$ .
- The lines $\displaystyle y = x$ and $\displaystyle y = 2 - x$ intersect at the point $\displaystyle (1, 1)$ because $\displaystyle 1 = 2 - 1$ .
- The gradients of the two lines are $\displaystyle m_1 = 1$ and $\displaystyle m_2 = -1$ . Since $\displaystyle m_1 m_2 = -1$ , the two lines are perpendicular (the angle between them is $\displaystyle \theta = 90^{\circ}$ ).
- Therefore, the composite transformation is a rotation through $\displaystyle 2 \times 90^{\circ} = 180^{\circ}$ about the point of intersection $\displaystyle (1, 1)$ .

Method 2: Coordinate mapping
Let $\displaystyle (x, y)$ be an arbitrary point on shape $\displaystyle S$ :
1. Reflecting $\displaystyle (x, y)$ in the line $\displaystyle y = x$ gives $\displaystyle (y, x)$ .
2. Reflecting a point $\displaystyle (u, v)$ in the line $\displaystyle x + y = 2$ gives $\displaystyle (2 - v, 2 - u)$ . Applying this to $\displaystyle (u, v) = (y, x)$ yields: $\displaystyle (x', y') = (2 - x, 2 - y)$ A rotation of $\displaystyle 180^{\circ}$ about a centre $\displaystyle (a, b)$ maps $\displaystyle (x, y) \to (2a - x, 2b - y)$ . Matching coefficients gives $\displaystyle 2a = 2 \implies a = 1$ and $\displaystyle 2b = 2 \implies b = 1$ , confirming a rotation of $\displaystyle 180^{\circ}$ about $\displaystyle (1, 1)$ .

First, we calculate the area $\displaystyle A$ under the curve $\displaystyle y = 3x^2 + 1$ from $\displaystyle x=0$ to $\displaystyle x=2$ .The area of the region under the line $\displaystyle y=mx$ is given as $\displaystyle 10\%$ of $\displaystyle A$ . Let's call this area $\displaystyle A_L$ .Now, we calculate the area $\displaystyle A_L$ by integrating $\displaystyle y=mx$ from $\displaystyle x=0$ to $\displaystyle x=2$ . This area can also be seen as a triangle with base 2 and height $\displaystyle 2m$ .We are given that $\displaystyle A_L = 1$ . Therefore, we can set up the equation:The correct option is C.

The general term in the expansion of $\displaystyle \left(2x^2 - \dfrac{1}{x}\right)^6$ is given by:To find the term independent of $\displaystyle x$ , we need the total power of $\displaystyle x$ to be zero. The power of $\displaystyle x$ in the general term is $\displaystyle 2(6-r) - r = 12 - 3r$ .

Setting this power to zero gives:We now calculate the coefficient for the term with $\displaystyle r=4$ :The term independent of $\displaystyle x$ is 60.

To find the mean of the combined dataset, calculate the total sum of all scores divided by the total number of students: $\displaystyle \text{Total students} = 15 + 25 + 10 = 50$ $\displaystyle \text{Total sum of scores} = (15 \times 18) + (25 \times 24) + (10 \times 30) = 270 + 600 + 300 = 1170$ $\displaystyle \text{Combined mean} = \dfrac{1170}{50} = 23.4$ For the range, let $\displaystyle S = X \cup Y \cup Z$ be the combined dataset. The range is given by $\displaystyle \text{range}(S) = \max(S) - \min(S)$ . Since class Y is a subset of $\displaystyle S$ , the maximum of $\displaystyle S$ must be at least the maximum of Y, and the minimum of $\displaystyle S$ must be at most the minimum of Y: $\displaystyle \max(S) \ge \max(Y) \quad \text{and} \quad \min(S) \le \min(Y)$ Subtracting these gives: $\displaystyle \text{range}(S) = \max(S) - \min(S) \ge \max(Y) - \min(Y) = 14$ Therefore, $\displaystyle \text{mean} = 23.4$ and $\displaystyle \text{range} \ge 14$ .