ESAT Mathematics 2 Mock 2

We observe that the interval $\displaystyle [-2, 2]$ is symmetric about the origin. This suggests checking the parity of the terms in the integrand.

The term $\displaystyle x^3 \cos x$ is the product of an odd function ( $\displaystyle x^3$ ) and an even function ( $\displaystyle \cos x$ ), making it an odd function. The integral of an odd function over a symmetric interval is zero.

The remaining terms, $\displaystyle 3x^2 + 5$ , form an even function. We evaluate the integral by doubling the result over $\displaystyle [0, 2]$ :Calculating the antiderivative:This simplifies to $\displaystyle 2(8 + 10) = 2(18) = 36$ .

The length of the rectangle is given by $\displaystyle L = \dfrac{A}{W} = A \div W$ .

Substituting the given expressions: $\displaystyle L = \dfrac{3x^2 - 11x - 4}{9x^2 - 1} \div \dfrac{x^2 - 2x - 8}{3x^2 + 14x - 5}$ Factorising each quadratic:
1. $\displaystyle 3x^2 - 11x - 4 = (3x + 1)(x - 4)$ 2. $\displaystyle 9x^2 - 1 = (3x - 1)(3x + 1)$ 3. $\displaystyle x^2 - 2x - 8 = (x - 4)(x + 2)$ 4. $\displaystyle 3x^2 + 14x - 5 = (3x - 1)(x + 5)$ Rewriting the division as multiplication by the reciprocal: $\displaystyle L = \dfrac{(3x + 1)(x - 4)}{(3x - 1)(3x + 1)} \times \dfrac{(3x - 1)(x + 5)}{(x - 4)(x + 2)}$ Cancelling common factors:
- Cancel $\displaystyle (3x + 1)$ in the first fraction: $\displaystyle L = \dfrac{x - 4}{3x - 1} \times \dfrac{(3x - 1)(x + 5)}{(x - 4)(x + 2)}$ - Cancel $\displaystyle (3x - 1)$ and $\displaystyle (x - 4)$ : $\displaystyle L = \dfrac{x + 5}{x + 2}$ Hence, the correct option is E.

Let $\displaystyle A_+$ be the area of the region above the $\displaystyle x$ -axis and $\displaystyle A_-$ be the area of the region below it.

The definite integral gives the signed area, so we can write:The total area is the sum of the (positive) areas of these regions:We now have a pair of simultaneous equations. Subtracting the first equation from the second eliminates $\displaystyle A_+$ :This simplifies to:The area of the region lying below the $\displaystyle x$ -axis is 5.

To find the overall mean, compute the sum of all scores and divide by the total number of items.

1. Calculate the sum of scores for each group:
- Group A: $\displaystyle \text{sum}_A = n \times 12 = 12n$ - Group B: $\displaystyle \text{sum}_B = 2n \times 18 = 36n$ - Group C: $\displaystyle \text{sum}_C = k \times 30 = 30k$ 2. The total sum of all scores is: $\displaystyle \text{Total sum} = 12n + 36n + 30k = 48n + 30k$ 3. The total number of items is: $\displaystyle \text{Total items} = n + 2n + k = 3n + k$ 4. Set up the equation for the overall mean: $\displaystyle \dfrac{48n + 30k}{3n + k} = 20$ 5. Solve for the ratio $\displaystyle \dfrac{k}{n}$ : $\displaystyle 48n + 30k = 20(3n + k)$ $\displaystyle 48n + 30k = 60n + 20k$ $\displaystyle 30k - 20k = 60n - 48n$ $\displaystyle 10k = 12n$ $\displaystyle \dfrac{k}{n} = \dfrac{12}{10} = \dfrac{6}{5}$ Thus, the ratio $\displaystyle k : n$ is $\displaystyle 6 : 5$ .

First, find the x-coordinates of the stationary points by setting $\displaystyle f'(x) = 0$ . $\displaystyle 3x^2 - 12 = 0 \implies 3x^2 = 12 \implies x^2 = 4 \implies x = \pm 2$ .
So, the x-coordinates of the stationary points are $\displaystyle x_1 = -2$ and $\displaystyle x_2 = 2$ .

Next, integrate $\displaystyle f'(x)$ to find $\displaystyle f(x)$ : $\displaystyle f(x) = \int (3x^2 - 12) dx = x^3 - 12x + C$ .
The y-intercept of the curve is $\displaystyle f(0) = C$ .

Now, find the y-coordinates of the stationary points:
For $\displaystyle x_1 = -2$ : $\displaystyle y_1 = f(-2) = (-2)^3 - 12(-2) + C = -8 + 24 + C = 16 + C$ .
For $\displaystyle x_2 = 2$ : $\displaystyle y_2 = f(2) = (2)^3 - 12(2) + C = 8 - 24 + C = -16 + C$ .
The stationary points are $\displaystyle P_1(-2, 16+C)$ and $\displaystyle P_2(2, -16+C)$ .

Calculate the distance between these two points using the distance formula $\displaystyle D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ : $\displaystyle D = \sqrt{(2 - (-2))^2 + ((-16 + C) - (16 + C))^2}$ $\displaystyle D = \sqrt{(4)^2 + (-16 + C - 16 - C)^2}$ $\displaystyle D = \sqrt{16 + (-32)^2}$ $\displaystyle D = \sqrt{16 + 1024}$ $\displaystyle D = \sqrt{1040}$ .

The problem states that the distance between the two stationary points is $\displaystyle 4\sqrt{65}$ .
Let's verify this: $\displaystyle 4\sqrt{65} = \sqrt{16 \times 65} = \sqrt{1040}$ .
The given distance matches the calculated distance.

Crucially, the constant of integration $\displaystyle C$ (which represents the y-intercept) cancels out during the distance calculation. This means that the distance between the stationary points is independent of the value of $\displaystyle C$ . Therefore, the information about the distance ( $\displaystyle 4\sqrt{65}$ ) does not provide any means to determine the value of $\displaystyle C$ .
Thus, the y-intercept of the curve cannot be determined from the given information.

1. Find the gradient of line $\displaystyle L_2$ using its intercepts $\displaystyle (0, 4)$ and $\displaystyle (6, 0)$ : $\displaystyle m_2 = \dfrac{0 - 4}{6 - 0} = -\dfrac{4}{6} = -\dfrac{2}{3}$ 2. Since $\displaystyle L_1$ is perpendicular to $\displaystyle L_2$ , the product of their gradients is $\displaystyle -1$ : $\displaystyle m_1 = -\dfrac{1}{m_2} = -\dfrac{1}{-\dfrac{2}{3}} = \dfrac{3}{2}$ 3. Express the gradient of $\displaystyle L_1$ using the two given points $\displaystyle (2 - k, 3k - 1)$ and $\displaystyle (k + 4, k + 5)$ : $\displaystyle m_1 = \dfrac{(k + 5) - (3k - 1)}{(k + 4) - (2 - k)} = \dfrac{6 - 2k}{2k + 2} = \dfrac{3 - k}{k + 1}$ 4. Equate the two expressions for $\displaystyle m_1$ and solve for $\displaystyle k$ : $\displaystyle \dfrac{3 - k}{k + 1} = \dfrac{3}{2}$ $\displaystyle 2(3 - k) = 3(k + 1)$ $\displaystyle 6 - 2k = 3k + 3$ $\displaystyle 5k = 3 \implies k = \dfrac{3}{5}$

We use the double-angle identity $\displaystyle \sin(2x) = 2\sin x \cos x$ . The equation becomes:Rearranging and factorising gives:This yields two cases:
1. $\displaystyle \cos x = 0$ , which gives $\displaystyle x = \dfrac{\pi}{2}$ in the interval $\displaystyle 0 \le x \le \pi$ .
2. $\displaystyle \sin x = \dfrac{1}{2}$ , which gives $\displaystyle x = \dfrac{\pi}{6}$ and $\displaystyle x = \dfrac{5\pi}{6}$ .

The sum of these solutions is:

1. Find the side length of the original rhombus:
The diagonals of a rhombus bisect each other at right angles. The half-diagonals have lengths $\displaystyle \dfrac{2\sqrt{5}}{2} = \sqrt{5}\text{ cm}$ and $\displaystyle \dfrac{4}{2} = 2\text{ cm}$ .

Using Pythagoras' theorem for one of the four right-angled triangles: $\displaystyle \text{side length } s = \sqrt{(\sqrt{5})^2 + 2^2} = \sqrt{5 + 4} = \sqrt{9} = 3\text{ cm}$ 2. Find the area and perimeter of the original rhombus: $\displaystyle \text{Area}_1 = \dfrac{1}{2} d_1 d_2 = \dfrac{1}{2} (2\sqrt{5})(4) = 4\sqrt{5}\text{ cm}^2$ $\displaystyle \text{Perimeter}_1 = 4s = 4 \times 3 = 12\text{ cm}$ 3. **Determine the linear scale factor $\displaystyle k$ :**
The area scales by $\displaystyle k^2$ : $\displaystyle k^2 = \dfrac{\text{Area}_2}{\text{Area}_1} = \dfrac{80\sqrt{5}}{4\sqrt{5}} = 20$ $\displaystyle k = \sqrt{20} = 2\sqrt{5}$ 4. Calculate the perimeter of the enlarged rhombus: $\displaystyle \text{Perimeter}_2 = k \times \text{Perimeter}_1 = 2\sqrt{5} \times 12 = 24\sqrt{5}\text{ cm}$

Using the sine rule, $\displaystyle \sin B=\dfrac{4\sqrt3\sin30^\circ}{4}=\dfrac{\sqrt3}{2}$ , so $\displaystyle B=60^\circ$ or $\displaystyle 120^\circ$ . Hence $\displaystyle C=90^\circ$ or $\displaystyle 30^\circ$ . The corresponding areas are $\displaystyle 8\sqrt3$ and $\displaystyle 4\sqrt3$ , giving a total of $\displaystyle 12\sqrt3$ .

We can express each given relationship in index notation:

1. $\displaystyle u \propto v^{-1/2}$ 2. $\displaystyle v \propto w^3$ 3. $\displaystyle w \propto x^{-2}$ Substitute the expression for $\displaystyle w$ into the relation for $\displaystyle v$ : $\displaystyle v \propto (x^{-2})^3 = x^{-6}$ Now substitute the expression for $\displaystyle v$ into the relation for $\displaystyle u$ : $\displaystyle u \propto (x^{-6})^{-1/2} = x^{(-6) \times (-1/2)} = x^3$ Thus, $\displaystyle u$ is directly proportional to $\displaystyle x^3$ .

First, we express the common ratio $\displaystyle r$ in terms of $\displaystyle p$ .

Increasing 1 by $\displaystyle p\%$ gives the multiplier $\displaystyle (1 + \dfrac{p}{100})$ .
Decreasing the new value by $\displaystyle p\%$ gives the multiplier $\displaystyle (1 - \dfrac{p}{100})$ .

So, the common ratio $\displaystyle r$ is the product of these multipliers applied to the initial value of 1:This is a difference of two squares:A geometric series has a sum to infinity if and only if its common ratio $\displaystyle r$ satisfies the condition $\displaystyle |r| < 1$ .

Substituting our expression for $\displaystyle r$ gives:This absolute value inequality is equivalent to the compound inequality:We solve this as two separate inequalities.

First inequality:Since $\displaystyle p > 0$ is given, this inequality is always true.

Second inequality:Since $\displaystyle p > 0$ , we can take the positive square root of both sides:Combining this with the given condition $\displaystyle p > 0$ , the complete range of values for $\displaystyle p$ is $\displaystyle 0 < p < 100\sqrt{2}$ .

Therefore, the correct answer is B.

The expansion of $\displaystyle (3x + 2)^6$ has $\displaystyle 6 + 1 = 7$ terms. Using the binomial theorem, the coefficient of $\displaystyle x^k$ is:For a coefficient to be divisible by $\displaystyle 6$ , it must be divisible by both $\displaystyle 2$ and $\displaystyle 3$ . We analyze the prime factors provided by the powers:

- The term $\displaystyle 3^k$ is a multiple of $\displaystyle 3$ for all $\displaystyle k \geq 1$ . For $\displaystyle k=0$ , the coefficient simplifies to $\displaystyle 1 \cdot 3^0 \cdot 2^6 = 64$ , which is not divisible by $\displaystyle 3$ .
- The term $\displaystyle 2^{6-k}$ is a multiple of $\displaystyle 2$ for all $\displaystyle k \leq 5$ . For $\displaystyle k=6$ , the coefficient simplifies to $\displaystyle 1 \cdot 3^6 \cdot 2^0 = 729$ , which is not divisible by $\displaystyle 2$ .

For the intermediate indices $\displaystyle 1 \leq k \leq 5$ , the coefficient contains at least one factor of $\displaystyle 3$ (from $\displaystyle 3^k$ ) and one factor of $\displaystyle 2$ (from $\displaystyle 2^{6-k}$ ), ensuring divisibility by $\displaystyle 6$ .

This leaves the set of indices $\displaystyle k \in \{1, 2, 3, 4, 5\}$ , so there are $\displaystyle 5$ such coefficients.

The derivative of the function $\displaystyle f(x) = ax^2 + bx + c$ is $\displaystyle f'(x) = 2ax + b$ .

We are given two conditions on the gradient.
The gradient at $\displaystyle x=0$ is $\displaystyle 12$ , which means $\displaystyle f'(0) = 12$ . Substituting into the derivative gives $\displaystyle f'(0) = 2a(0) + b = b$ , so we find $\displaystyle b=12$ .

A stationary point at $\displaystyle x=2$ means that $\displaystyle f'(2) = 0$ . Using our value for $\displaystyle b$ , we have: $\displaystyle f'(2) = 2a(2) + 12 = 4a + 12$ .
Setting this to zero gives $\displaystyle 4a + 12 = 0$ , which solves to $\displaystyle a = -3$ .

The function is therefore $\displaystyle f(x) = -3x^2 + 12x + c$ .

Finally, we use the condition $\displaystyle f(1) = 15$ to find $\displaystyle c$ . $\displaystyle f(1) = -3(1)^2 + 12(1) + c = -3 + 12 + c = 9 + c$ .
Since $\displaystyle f(1) = 15$ , we have $\displaystyle 9 + c = 15$ , which gives $\displaystyle c=6$ .

The equation of the circle is given as $\displaystyle x^2 + y^2 - 2x + 2y - 23 = 0$ .

First, we find the centre of the circle by completing the square:So, the centre of the circle is $\displaystyle C(1, -1)$ and the radius is $\displaystyle 5$ .

The point of tangency is given as $\displaystyle P(4, 3)$ . We can verify this point lies on the circle: $\displaystyle (4-1)^2 + (3+1)^2 = 3^2 + 4^2 = 9 + 16 = 25$ .

The gradient of the radius connecting the centre $\displaystyle C(1, -1)$ to the point $\displaystyle P(4, 3)$ is:The tangent line at point $\displaystyle P$ is perpendicular to the radius $\displaystyle CP$ . Therefore, the gradient of the tangent is the negative reciprocal of the gradient of the radius:Now we find the equation of the tangent line using the point-gradient form, with point $\displaystyle P(4, 3)$ and gradient $\displaystyle m_{\text{tangent}} = -\dfrac{3}{4}$ :Multiply by 4 to clear the denominator:Rearrange into the general form:To find the area of the triangle bounded by this line and the coordinate axes, we find the $\displaystyle x$ and $\displaystyle y$ intercepts.

For the $\displaystyle x$ -intercept, set $\displaystyle y=0$ : $\displaystyle 3x = 24 \implies x = 8$ .
For the $\displaystyle y$ -intercept, set $\displaystyle x=0$ : $\displaystyle 4y = 24 \implies y = 6$ .

The triangle formed by the line $\displaystyle 3x+4y=24$ and the coordinate axes is a right-angled triangle with vertices at $\displaystyle (0,0)$ , $\displaystyle (8,0)$ , and $\displaystyle (0,6)$ . Its base is 8 units and its height is 6 units.

The area is:Thus, the area of the triangle is 24 square units.

The function is defined piecewise, so we must find the maximum value on each interval and then compare them.

For the interval $\displaystyle 0 \le x \le 3$ , the function is $\displaystyle f(x) = 11 + 2x - x^2$ .
To find any turning points, we differentiate: $\displaystyle f'(x) = 2 - 2x$ .
Setting $\displaystyle f'(x) = 0$ gives a stationary point at $\displaystyle x=1$ . This point lies within the interval $\displaystyle [0, 3]$ .
The value at this point is $\displaystyle f(1) = 11 + 2(1) - 1^2 = 12$ .
We also check the endpoints of this interval: $\displaystyle f(0) = 11$ and $\displaystyle f(3) = 11 + 6 - 9 = 8$ .
The maximum value on this first interval is therefore $\displaystyle 12$ .

For the interval $\displaystyle 3 < x \le 6$ , the function is $\displaystyle f(x) = 8x - x^2 - 6$ .
Differentiating gives $\displaystyle f'(x) = 8 - 2x$ .
Setting $\displaystyle f'(x) = 0$ gives a stationary point at $\displaystyle x=4$ . This point lies within the interval $\displaystyle (3, 6]$ .
The value at this point is $\displaystyle f(4) = 8(4) - 4^2 - 6 = 32 - 16 - 6 = 10$ .
The value at the endpoint is $\displaystyle f(6) = 8(6) - 6^2 - 6 = 48 - 36 - 6 = 6$ .
The maximum value on this second interval is $\displaystyle 10$ .

Comparing the maximum values from both intervals, $\displaystyle 12$ and $\displaystyle 10$ , the overall maximum value of $\displaystyle f(x)$ is the larger of these, which is $\displaystyle 12$ .

We are given the function $\displaystyle f(x) = \log_3\left(\dfrac{ax+b}{x}\right)$ .

First, we use the given conditions $\displaystyle f(1)=2$ and $\displaystyle f(4)=1$ to find the constants $\displaystyle a$ and $\displaystyle b$ .

Condition 1: $\displaystyle f(1)=2$ Substitute $\displaystyle x=1$ into the function definition: $\displaystyle f(1) = \log_3\left(\dfrac{a(1)+b}{1}\right) = \log_3(a+b)$ Since $\displaystyle f(1)=2$ , we have: $\displaystyle \log_3(a+b) = 2$ By the definition of logarithm, $\displaystyle a+b = 3^2$ $\displaystyle a+b = 9$ (Equation 1)

Condition 2: $\displaystyle f(4)=1$ Substitute $\displaystyle x=4$ into the function definition: $\displaystyle f(4) = \log_3\left(\dfrac{a(4)+b}{4}\right) = \log_3\left(\dfrac{4a+b}{4}\right)$ Since $\displaystyle f(4)=1$ , we have: $\displaystyle \log_3\left(\dfrac{4a+b}{4}\right) = 1$ By the definition of logarithm, $\displaystyle \dfrac{4a+b}{4} = 3^1$ $\displaystyle \dfrac{4a+b}{4} = 3$ $\displaystyle 4a+b = 12$ (Equation 2)

Now we have a system of two linear equations:
1) $\displaystyle a+b = 9$ 2) $\displaystyle 4a+b = 12$ Subtract Equation 1 from Equation 2: $\displaystyle (4a+b) - (a+b) = 12 - 9$ $\displaystyle 3a = 3$ $\displaystyle a = 1$ Substitute $\displaystyle a=1$ into Equation 1: $\displaystyle 1+b = 9$ $\displaystyle b = 8$ So, the function is $\displaystyle f(x) = \log_3\left(\dfrac{1x+8}{x}\right) = \log_3\left(\dfrac{x+8}{x}\right)$ .

Next, we need to find the value of $\displaystyle x$ such that $\displaystyle f(x)=-1$ .
Set the function equal to $\displaystyle -1$ : $\displaystyle \log_3\left(\dfrac{x+8}{x}\right) = -1$ By the definition of logarithm, $\displaystyle \dfrac{x+8}{x} = 3^{-1}$ $\displaystyle \dfrac{x+8}{x} = \dfrac{1}{3}$ Now, solve for $\displaystyle x$ : $\displaystyle 3(x+8) = 1(x)$ $\displaystyle 3x + 24 = x$ $\displaystyle 24 = x - 3x$ $\displaystyle 24 = -2x$ $\displaystyle x = \dfrac{24}{-2}$ $\displaystyle x = -12$ Thus, the value of $\displaystyle x$ such that $\displaystyle f(x)=-1$ is $\displaystyle -12$ .

The equation rearranges to $\displaystyle \cos\left(2x + \dfrac{\pi}{3}\right) = \dfrac{1}{2}$ .

Let $\displaystyle u = 2x + \dfrac{\pi}{3}$ . We must transform the given interval for $\displaystyle x$ , which is $\displaystyle 0 \le x \le 2\pi$ , into an interval for $\displaystyle u$ .So we need to find the number of solutions to $\displaystyle \cos(u) = \dfrac{1}{2}$ in the interval $\displaystyle \left[\dfrac{\pi}{3}, \dfrac{13\pi}{3}\right]$ .

The general solution is $\displaystyle u = 2k\pi \pm \dfrac{\pi}{3}$ for any integer $\displaystyle k$ . We list the solutions that fall within our interval: $\displaystyle u = \dfrac{\pi}{3}$ (for $\displaystyle k=0$ ) $\displaystyle u = 2\pi - \dfrac{\pi}{3} = \dfrac{5\pi}{3}$ (for $\displaystyle k=1$ ) $\displaystyle u = 2\pi + \dfrac{\pi}{3} = \dfrac{7\pi}{3}$ (for $\displaystyle k=1$ ) $\displaystyle u = 4\pi - \dfrac{\pi}{3} = \dfrac{11\pi}{3}$ (for $\displaystyle k=2$ ) $\displaystyle u = 4\pi + \dfrac{\pi}{3} = \dfrac{13\pi}{3}$ (for $\displaystyle k=2$ )

There are 5 distinct solutions for $\displaystyle u$ . Since each corresponds to a unique value of $\displaystyle x$ , there are 5 solutions in total.