ESAT Mathematics 2 Mock 1

The sum of the first $\displaystyle n$ terms of a geometric progression is $\displaystyle S_n = \dfrac{a(1-r^n)}{1-r}$ . We notice that the term $\displaystyle (1-r^6)$ in $\displaystyle S_6$ can be factored as a difference of squares, $\displaystyle (1-r^3)(1+r^3)$ . This allows us to express $\displaystyle S_6$ in terms of $\displaystyle S_3$ :Substituting the given values $\displaystyle S_3 = 8$ and $\displaystyle S_6 = 72$ :Dividing by $\displaystyle 8$ gives $\displaystyle 9 = 1+r^3$ , which simplifies to $\displaystyle r^3 = 8$ . Therefore, the common ratio is $\displaystyle r=2$ .

For a non-degenerate triangle with side lengths $\displaystyle a = x + 3$ , $\displaystyle b = 2x - 1$ , and $\displaystyle c = 11 - x$ to exist, all side lengths must be strictly positive and the triangle inequalities must be satisfied.

1. Positivity of side lengths:
- $\displaystyle x + 3 > 0 \implies x > -3$ - $\displaystyle 2x - 1 > 0 \implies x > \dfrac{1}{2}$ - $\displaystyle 11 - x > 0 \implies x < 11$ Thus, $\displaystyle \dfrac{1}{2} < x < 11$ .

2. Triangle inequalities ( $\displaystyle a + b > c$ , $\displaystyle a + c > b$ , $\displaystyle b + c > a$ ):
- $\displaystyle (x + 3) + (2x - 1) > 11 - x \implies 3x + 2 > 11 - x \implies 4x > 9 \implies x > \dfrac{9}{4}$ - $\displaystyle (x + 3) + (11 - x) > 2x - 1 \implies 14 > 2x - 1 \implies 2x < 15 \implies x < \dfrac{15}{2}$ - $\displaystyle (2x - 1) + (11 - x) > x + 3 \implies x + 10 > x + 3 \implies 10 > 3$ , which holds for all real $\displaystyle x$ .

Taking the intersection of all conditions: $\displaystyle x > \dfrac{9}{4}$ (since $\displaystyle \dfrac{9}{4} = 2.25 > \dfrac{1}{2}$ ) $\displaystyle x < \dfrac{15}{2}$ (since $\displaystyle \dfrac{15}{2} = 7.5 < 11$ )

Therefore, the complete set of values of $\displaystyle x$ is $\displaystyle \dfrac{9}{4} < x < \dfrac{15}{2}$ .

1. Find the radius of the circular arc:
In the regular hexagon, the interior angle at each vertex is $\displaystyle 120^\circ$ .
Consider $\displaystyle \triangle ABC$ , where $\displaystyle AB = BC = a$ and $\displaystyle \angle ABC = 120^\circ$ .
By the cosine rule: $\displaystyle AC^2 = a^2 + a^2 - 2a^2\cos 120^\circ = 2a^2 - 2a^2\left(-\dfrac{1}{2}\right) = 3a^2 \implies AC = a\sqrt{3}$ Similarly, $\displaystyle AE = a\sqrt{3}$ . Thus, the circular arc has radius $\displaystyle R = a\sqrt{3}$ .

2. **Find the angle of sector $\displaystyle ACE$ :**
In isosceles triangle $\displaystyle ABC$ , $\displaystyle \angle BAC = \dfrac{180^\circ - 120^\circ}{2} = 30^\circ$ .
Similarly, in $\displaystyle \triangle AFE$ , $\displaystyle \angle FAE = 30^\circ$ .
Since the interior angle $\displaystyle \angle FAB = 120^\circ$ : $\displaystyle \angle CAE = \angle FAB - \angle BAC - \angle FAE = 120^\circ - 30^\circ - 30^\circ = 60^\circ$ 3. **Calculate the area of sector $\displaystyle ACE$ :** $\displaystyle \text{Area}(\text{sector } ACE) = \dfrac{60^\circ}{360^\circ} \pi R^2 = \dfrac{1}{6} \pi (3a^2) = \dfrac{\pi}{2}a^2$ 4. **Calculate the area of quadrilateral $\displaystyle ACDE$ :**
The total area of the regular hexagon is: $\displaystyle \text{Area}(ABCDEF) = 6 \times \left(\dfrac{\sqrt{3}}{4}a^2\right) = \dfrac{3\sqrt{3}}{2}a^2$ The areas of $\displaystyle \triangle ABC$ and $\displaystyle \triangle AFE$ are: $\displaystyle \text{Area}(\triangle ABC) = \text{Area}(\triangle AFE) = \dfrac{1}{2}a^2\sin 120^\circ = \dfrac{\sqrt{3}}{4}a^2$ Thus: $\displaystyle \text{Area}(ACDE) = \text{Area}(ABCDEF) - 2 \times \left(\dfrac{\sqrt{3}}{4}a^2\right) = \dfrac{3\sqrt{3}}{2}a^2 - \dfrac{\sqrt{3}}{2}a^2 = \sqrt{3}a^2$ 5. Calculate the shaded area: $\displaystyle \text{Area}(\text{shaded}) = \text{Area}(ACDE) - \text{Area}(\text{sector } ACE) = \sqrt{3}a^2 - \dfrac{\pi}{2}a^2 = \left(\dfrac{2\sqrt{3} - \pi}{2}\right)a^2$

The trapezium-rule estimate is $\displaystyle \dfrac12[2+11+2(3+6)]=\dfrac{31}{2}$ . Since $\displaystyle f''(x)>0$ , the curve is convex, so the straight-line chords lie above the curve and the trapezium rule gives an overestimate.

The problem asks for the area of the region between a parabolic segment and an inscribed triangle.

1. Find the vertices of the triangle.
The parabola is given by $\displaystyle y = kx - x^2$ . The points where it meets the $\displaystyle x$ -axis are found by setting $\displaystyle y=0$ : $\displaystyle x(k-x) = 0$ , which gives $\displaystyle x=0$ and $\displaystyle x=k$ . So, two vertices of the triangle are $\displaystyle (0,0)$ and $\displaystyle (k,0)$ .
The base of the triangle is the distance between these points, which is $\displaystyle k$ .

The vertex of the parabola occurs at the midpoint of the roots, by symmetry. The $\displaystyle x$ -coordinate is $\displaystyle x = \dfrac{0+k}{2} = \dfrac{k}{2}$ .
The corresponding $\displaystyle y$ -coordinate is $\displaystyle y = \dfrac{k}{2}(k - \dfrac{k}{2}) = \dfrac{k}{2} \cdot \dfrac{k}{2} = \dfrac{k^2}{4}$ .
The vertex of the parabola, which is the third vertex of the triangle, is at $\displaystyle (\dfrac{k}{2}, \dfrac{k^2}{4})$ . The height of the triangle is this $\displaystyle y$ -coordinate, $\displaystyle \dfrac{k^2}{4}$ .

2. Calculate the area of the triangle.3. Calculate the area under the parabolic arc.
This is the area of the region bounded by the curve $\displaystyle y=x(k-x)$ and the $\displaystyle x$ -axis, from $\displaystyle x=0$ to $\displaystyle x=k$ . This is found by a definite integral:4. Find the required area.
The total area of the regions between the arc and the triangle sides is the difference between the area under the parabola and the area of the triangle.Therefore the correct answer is $\displaystyle \dfrac{k^3}{24}$ .

The region is bounded below by the V-shaped graph of $\displaystyle y = |2x + 4|$ and bounded above by the horizontal line $\displaystyle y = 8$ . This forms a triangle.

First, we find the vertices of this triangle.

The lowest vertex of the region is the vertex of the graph $\displaystyle y = |2x + 4|$ . This occurs when the expression inside the modulus is zero: $\displaystyle 2x + 4 = 0 \implies x = -2$ .
At this point, $\displaystyle y = |0| = 0$ . So, the lower vertex is at $\displaystyle (-2, 0)$ .

The other two vertices are the points of intersection between $\displaystyle y = |2x + 4|$ and $\displaystyle y = 8$ . We solve the equation $\displaystyle |2x + 4| = 8$ .

This gives two cases:
1. $\displaystyle 2x + 4 = 8 \implies 2x = 4 \implies x = 2$ .
The intersection point is $\displaystyle (2, 8)$ .

2. $\displaystyle -(2x + 4) = 8 \implies 2x + 4 = -8 \implies 2x = -12 \implies x = -6$ .
The intersection point is $\displaystyle (-6, 8)$ .

The three vertices of the triangular region are $\displaystyle (-2, 0)$ , $\displaystyle (2, 8)$ , and $\displaystyle (-6, 8)$ .

The base of the triangle is the horizontal distance between $\displaystyle (2, 8)$ and $\displaystyle (-6, 8)$ , which is $\displaystyle 2 - (-6) = 8$ .

The height of the triangle is the vertical distance from the vertex $\displaystyle (-2, 0)$ to the line $\displaystyle y=8$ , which is $\displaystyle 8 - 0 = 8$ .

The area of the triangle is given by the formula $\displaystyle \dfrac{1}{2} \times \text{base} \times \text{height}$ .

Area $\displaystyle = \dfrac{1}{2} \times 8 \times 8 = 32$ .

Let the given equation be $\displaystyle x\sqrt{1-x^2} = \dfrac{1}{3}$ First, we must determine the domain of the variable $\displaystyle x$ . The term under the square root must be non-negative, so $\displaystyle 1-x^2 \ge 0$ , which implies $\displaystyle x^2 \le 1$ , or $\displaystyle -1 \le x \le 1$ .

Furthermore, the right-hand side of the equation, $\displaystyle \dfrac{1}{3}$ , is positive. The term $\displaystyle \sqrt{1-x^2}$ is always non-negative. For the product $\displaystyle x\sqrt{1-x^2}$ to be positive, $\displaystyle x$ must also be positive. Therefore, any real roots of the equation must lie in the interval $\displaystyle 0 < x \le 1$ .

To solve the equation, we square both sides: $\displaystyle (x\sqrt{1-x^2})^2 = \left(\dfrac{1}{3}\right)^2$ $\displaystyle x^2(1-x^2) = \dfrac{1}{9}$ This equation is a quartic in $\displaystyle x$ , but it can be simplified by making the substitution $\displaystyle y = x^2$ . The equation becomes: $\displaystyle y(1-y) = \dfrac{1}{9}$ $\displaystyle y - y^2 = \dfrac{1}{9}$ Rearranging this into a standard quadratic equation form ( $\displaystyle ay^2+by+c=0$ ): $\displaystyle y^2 - y + \dfrac{1}{9} = 0$ The roots of this quadratic equation, let's call them $\displaystyle y_1$ and $\displaystyle y_2$ , represent the possible values for $\displaystyle x^2$ . The question asks for the sum of the squares of the real roots of the original equation. Let the real roots of the original equation be $\displaystyle x_1, x_2, \dots$ . We are looking for the value of $\displaystyle x_1^2 + x_2^2 + \dots$ .

Let's check if the roots $\displaystyle y_1, y_2$ are valid. The discriminant of the quadratic in $\displaystyle y$ is $\displaystyle \Delta = b^2 - 4ac = (-1)^2 - 4(1)(\dfrac{1}{9}) = 1 - \dfrac{4}{9} = \dfrac{5}{9}$ . Since $\displaystyle \Delta > 0$ , there are two distinct real roots for $\displaystyle y$ .

Using Vieta's formulas on the quadratic $\displaystyle y^2 - y + \dfrac{1}{9} = 0$ , the sum of the roots is: $\displaystyle y_1 + y_2 = -\dfrac{b}{a} = -\dfrac{-1}{1} = 1$ And the product of the roots is: $\displaystyle y_1 y_2 = \dfrac{c}{a} = \dfrac{1/9}{1} = \dfrac{1}{9}$ Since the sum ( $\displaystyle 1$ ) and product ( $\displaystyle 1/9$ ) of the roots are both positive, both roots $\displaystyle y_1$ and $\displaystyle y_2$ must be positive. This means that for each $\displaystyle y$ value, we can find a real value for $\displaystyle x = \pm\sqrt{y}$ .

From our initial analysis, the roots of the original equation must be in the interval $\displaystyle 0 < x \le 1$ . This means we must take the positive square roots, $\displaystyle x_1 = \sqrt{y_1}$ and $\displaystyle x_2 = \sqrt{y_2}$ . We should also confirm that $\displaystyle y_1, y_2 \le 1$ . The roots are $\displaystyle y = \dfrac{1 \pm \sqrt{5/9}}{2} = \dfrac{3 \pm \sqrt{5}}{6}$ . Since $\displaystyle \sqrt{5} \approx 2.236$ , we have $\displaystyle y_1 = \dfrac{3 - \sqrt{5}}{6} \approx \dfrac{3 - 2.236}{6} = \dfrac{0.764}{6} \approx 0.127$ and $\displaystyle y_2 = \dfrac{3 + \sqrt{5}}{6} \approx \dfrac{3 + 2.236}{6} = \dfrac{5.236}{6} \approx 0.873$ . Both values are positive and less than 1. Thus, there are two distinct real roots for the original equation, $\displaystyle x_1 = \sqrt{y_1}$ and $\displaystyle x_2 = \sqrt{y_2}$ , both satisfying $\displaystyle 0 < x \le 1$ .

The squares of these real roots are $\displaystyle x_1^2 = y_1$ and $\displaystyle x_2^2 = y_2$ .
Therefore, the sum of the squares of the real roots of the original equation is $\displaystyle x_1^2 + x_2^2 = y_1 + y_2$ .
From Vieta's formulas applied to the quadratic in $\displaystyle y$ , we found that $\displaystyle y_1 + y_2 = 1$ .
Thus, the sum of the squares of the real roots is indeed 1.

Let the profit in the first year be $\displaystyle a$ . A decrease of $\displaystyle p\%$ per year means the profit is multiplied by a common ratio $\displaystyle r = 1 - \dfrac{p}{100}$ each year.

The annual profits form a geometric sequence: $\displaystyle a, ar, ar^2, \dots$ The sum of the first two terms is $\displaystyle S_2 = a + ar = a(1+r)$ .

The sum to infinity is $\displaystyle S_\infty = \dfrac{a}{1-r}$ . The condition $\displaystyle p>0$ ensures that $\displaystyle |r|<1$ , so the sum to infinity exists.

The problem states that $\displaystyle S_2 = 0.36 \times S_\infty$ .

Substituting the expressions for $\displaystyle S_2$ and $\displaystyle S_\infty$ :Since the initial profit $\displaystyle a$ must be non-zero, we can divide both sides by $\displaystyle a$ :Multiplying both sides by $\displaystyle (1-r)$ gives:Solving for $\displaystyle r^2$ :Since profit is decreasing, $\displaystyle r$ must be positive, so we take the positive square root:Finally, we find $\displaystyle p$ using the relationship $\displaystyle r = 1 - \dfrac{p}{100}$ :

Let $\displaystyle k$ be the number of gold tokens drawn before the game ends.

- If a silver token is drawn after $\displaystyle k$ gold tokens, the final score is $\displaystyle k - 2$ .
- For the final score to be strictly greater than 0, we must have $\displaystyle k - 2 > 0$ , which means $\displaystyle k \ge 3$ .
- If all 7 gold tokens are drawn without a silver token, $\displaystyle k = 7$ and the score is $\displaystyle 7 > 0$ .

Thus, the player finishes with a score strictly greater than 0 if and only if at least 3 gold tokens are drawn before the first silver token.

This condition is satisfied if and only if the first 3 tokens drawn from the bag are all gold tokens (since if the first 3 are gold, the game will necessarily end with $\displaystyle k \ge 3$ gold tokens regardless of subsequent draws).

The probability that the first 3 tokens drawn without replacement are all gold is: $\displaystyle \text{P}(\text{1st is gold}) \times \text{P}(\text{2nd is gold} \mid \text{1st is gold}) \times \text{P}(\text{3rd is gold} \mid \text{1st and 2nd are gold})$ $\displaystyle = \dfrac{7}{12} \times \dfrac{6}{11} \times \dfrac{5}{10} = \dfrac{7}{12} \times \dfrac{6}{11} \times \dfrac{1}{2} = \dfrac{7}{44}$

Since the particle returns to its starting position, the sum of the displacement vectors must be the zero vector: $\displaystyle \mathbf{d}_1 + \mathbf{d}_2 + \mathbf{d}_3 = \mathbf{0}$ .

Let's use a coordinate system where the positive $\displaystyle x$ -axis represents East and the positive $\displaystyle y$ -axis represents North. We can write the first two displacements in component form.
A bearing of $\displaystyle 090^\circ$ is purely East, so $\displaystyle \mathbf{d}_1 = (\sqrt{3}, 0)$ .
A bearing of $\displaystyle 180^\circ$ is purely South, so $\displaystyle \mathbf{d}_2 = (0, -1)$ .

From the condition $\displaystyle \mathbf{d}_1 + \mathbf{d}_2 + \mathbf{d}_3 = \mathbf{0}$ , we must have $\displaystyle \mathbf{d}_3 = -(\mathbf{d}_1 + \mathbf{d}_2)$ .
The resultant of the first two displacements is $\displaystyle \mathbf{d}_1 + \mathbf{d}_2 = (\sqrt{3}, -1)$ .
Therefore, the third displacement is $\displaystyle \mathbf{d}_3 = -(\sqrt{3}, -1) = (-\sqrt{3}, 1)$ .

This vector has a component of $\displaystyle -\sqrt{3}$ in the East direction (i.e., $\displaystyle \sqrt{3}$ West) and a component of $\displaystyle 1$ in the North direction. It lies in the North-West quadrant.

To find the bearing, we need the angle measured clockwise from North. Let $\displaystyle \alpha$ be the angle between the North direction (positive $\displaystyle y$ -axis) and the vector $\displaystyle \mathbf{d}_3$ . From a right-angled triangle with sides $\displaystyle 1$ (North) and $\displaystyle \sqrt{3}$ (West), we have:This gives $\displaystyle \alpha = 60^\circ$ . Since the vector is in the North-West quadrant, the bearing is $\displaystyle 360^\circ - \alpha$ .

Bearing $\displaystyle = 360^\circ - 60^\circ = 300^\circ$

Let the expression inside the main brackets be $\displaystyle B(x)$ .The second factor is a finite geometric series with first term $\displaystyle a=1$ , common ratio $\displaystyle r=2x^2$ , and 4 terms.
We can see this by writing the terms as $\displaystyle 1, (2x^2)^1, (2x^2)^2, (2x^2)^3$ .

This product is in the form $\displaystyle (1-r)(1+r+r^2+r^3)$ , which simplifies to $\displaystyle 1-r^4$ .
Substituting $\displaystyle r = 2x^2$ into this identity gives:Therefore, the full expression is $\displaystyle (1 - 16x^8)^5$ .

We need to find the coefficient of $\displaystyle x^{24}$ in the binomial expansion of this expression.
The general term in the expansion of $\displaystyle (a+b)^n$ is $\displaystyle \binom{n}{k} a^{n-k} b^k$ .
Here, $\displaystyle a=1$ , $\displaystyle b=-16x^8$ , and $\displaystyle n=5$ .
The general term is:We want the term where the power of $\displaystyle x$ is 24, so we set $\displaystyle 8k=24$ , which gives $\displaystyle k=3$ .

Substituting $\displaystyle k=3$ into the general term gives the required coefficient:First, calculate the binomial coefficient:Next, calculate $\displaystyle (-16)^3$ :We know $\displaystyle 2^{10} = 1024$ , so $\displaystyle 2^{12} = 2^{10} \times 2^2 = 1024 \times 4 = 4096$ .
So, $\displaystyle (-16)^3 = -4096$ .

Finally, the coefficient is:

The definite integral measures the net signed area bounded by the curve and the $\displaystyle x$ -axis. If $\displaystyle A_{\text{above}}$ and $\displaystyle A_{\text{below}}$ denote the geometric areas above and below the axis respectively, we have:We are given that the integral is $\displaystyle -12$ and the area above the axis is $\displaystyle 5$ . Substituting these values yields:Rearranging to solve for $\displaystyle A_{\text{below}}$ :

We begin by simplifying the rational expression inside the logarithm:The general term of the sum is therefore $\displaystyle \log_4 \left( \dfrac{n}{n+1} \right)$ , which can be written as $\displaystyle \log_4 n - \log_4 (n+1)$ . Writing out the sum reveals a telescoping pattern:All intermediate terms cancel, leaving only the first and last components:Since $\displaystyle \log_4 1 = 0$ and $\displaystyle 64 = 4^3$ , this evaluates to $\displaystyle 0 - 3 = -3$ .

Let the straight line $\displaystyle L$ have equation $\displaystyle y - 3 = m(x - 4)$ , where $\displaystyle m < 0$ .

First, we find the intercepts of $\displaystyle L$ with the positive coordinate axes.
Let the $\displaystyle x$ -intercept be $\displaystyle a$ and the $\displaystyle y$ -intercept be $\displaystyle c$ . $\displaystyle x$ -intercept (set $\displaystyle y=0$ ): $\displaystyle 0 - 3 = m(a - 4) \implies -3 = ma - 4m \implies ma = 4m - 3 \implies a = 4 - \dfrac{3}{m}$ . $\displaystyle y$ -intercept (set $\displaystyle x=0$ ): $\displaystyle c - 3 = m(0 - 4) \implies c - 3 = -4m \implies c = 3 - 4m$ .

Since $\displaystyle m<0$ , both $\displaystyle a$ and $\displaystyle c$ are positive, so the triangle is in the first quadrant as required.
The area of the large triangle, $\displaystyle A$ , is given by:The line $\displaystyle y=2x$ divides this triangle into two smaller triangles. Let the intersection of $\displaystyle L$ and $\displaystyle y=2x$ be the point $\displaystyle P(x_P, y_P)$ .

The two smaller triangles are:
1. $\displaystyle T_x$ , with vertices at the origin, the $\displaystyle x$ -intercept $\displaystyle (a,0)$ , and $\displaystyle P(x_P, y_P)$ .
2. $\displaystyle T_y$ , with vertices at the origin, the $\displaystyle y$ -intercept $\displaystyle (0,c)$ , and $\displaystyle P(x_P, y_P)$ .

The area of $\displaystyle T_x$ is $\displaystyle \dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} a y_P$ .
The area of $\displaystyle T_y$ is $\displaystyle \dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} c x_P$ .

We are given that the ratio of these areas is 3:Since $\displaystyle P$ lies on the line $\displaystyle y=2x$ , we have $\displaystyle y_P = 2x_P$ . Substituting this into the ratio equation:So, $\displaystyle 2a = 3c$ . Now we substitute our expressions for $\displaystyle a$ and $\displaystyle c$ in terms of $\displaystyle m$ :Multiplying the entire equation by $\displaystyle m$ (we know $\displaystyle m eq 0$ ):We can factorise this quadratic equation:This gives two possible values for $\displaystyle m$ : $\displaystyle m = \dfrac{3}{4}$ or $\displaystyle m = -\dfrac{2}{3}$ .
The problem states that the line has a negative gradient, so we must choose $\displaystyle m = -\dfrac{2}{3}$ .

Finally, we calculate the total area $\displaystyle A$ using this value of $\displaystyle m$ .