ESAT Mathematics 2 Mock 4

Let $\displaystyle f(x) = x^3 - 3x^2 + 5$ . The number of distinct real solutions to $\displaystyle f(x)=0$ is the number of times the graph of $\displaystyle y=f(x)$ intersects the x-axis.

To understand the shape of the curve, we find its stationary points by differentiating:The stationary points are at $\displaystyle x=0$ and $\displaystyle x=2$ .

Next, we evaluate the function at these points to find the y-coordinates of the turning points: $\displaystyle f(0) = 0^3 - 3(0)^2 + 5 = 5$ $\displaystyle f(2) = 2^3 - 3(2)^2 + 5 = 8 - 12 + 5 = 1$ The stationary points are a local maximum at $\displaystyle (0, 5)$ and a local minimum at $\displaystyle (2, 1)$ .

Since both the local maximum and local minimum values are positive, the turning points are both above the x-axis. The lowest point the curve reaches for $\displaystyle x>0$ is $\displaystyle y=1$ .

However, $\displaystyle f(x)$ is a cubic with a positive leading coefficient, so as $\displaystyle x \to -\infty$ , $\displaystyle f(x) \to -\infty$ . Since the function comes from negative infinity and its local minimum is positive, it must cross the x-axis exactly once (for some $\displaystyle x < 0$ ).

Therefore, there is only one distinct real solution.

The circle $\displaystyle x^2 + y^2 = 50$ is centred at the origin $\displaystyle (0, 0)$ .

The line $\displaystyle 7x + y = 1$ can be written as $\displaystyle y = -7x + 1$ . Its gradient is $\displaystyle m_{line} = -7$ .

The tangent to the circle at point $\displaystyle P$ is parallel to this line, so the gradient of the tangent is also $\displaystyle m_{tangent} = -7$ .

The radius from the origin to the point of tangency $\displaystyle P(x_0, y_0)$ is perpendicular to the tangent at $\displaystyle P$ .

The gradient of the radius is $\displaystyle m_{radius} = \dfrac{y_0 - 0}{x_0 - 0} = \dfrac{y_0}{x_0}$ .

For perpendicular lines, the product of their gradients is $\displaystyle -1$ . So, $\displaystyle m_{radius} \times m_{tangent} = -1$ .This gives the relationship $\displaystyle x_0 = 7y_0$ .

The point $\displaystyle P(x_0, y_0)$ lies on the circle, so its coordinates must satisfy the circle's equation:Substitute $\displaystyle x_0 = 7y_0$ into the equation:This gives $\displaystyle y_0 = 1$ or $\displaystyle y_0 = -1$ .

The question states that point $\displaystyle P$ has positive $\displaystyle x$ and $\displaystyle y$ coordinates. We must choose $\displaystyle y_0 = 1$ .

Using $\displaystyle x_0 = 7y_0$ , we find $\displaystyle x_0 = 7(1) = 7$ .

So the coordinates of $\displaystyle P$ are $\displaystyle (7, 1)$ .

Therefore the correct answer is D.

Let the cubic polynomial be $\displaystyle p(x)$ .

The stationary points of the polynomial occur where its derivative, $\displaystyle p'(x)$ , is equal to zero. Since the stationary points are at $\displaystyle x=1$ and $\displaystyle x=3$ , the derivative $\displaystyle p'(x)$ must have roots at these two values.

As $\displaystyle p(x)$ is a cubic polynomial, $\displaystyle p'(x)$ is a quadratic polynomial. We can express $\displaystyle p'(x)$ in terms of its roots:for some non-zero constant $\displaystyle k$ . Expanding this gives:To find the polynomial $\displaystyle p(x)$ , we integrate $\displaystyle p'(x)$ with respect to $\displaystyle x$ :where $\displaystyle C$ is the constant of integration.

The polynomial can be written as $\displaystyle p(x) = A(x^3 - 6x^2 + 9x) + C$ , where $\displaystyle A = k/3$ . The constants $\displaystyle A$ (representing a vertical scaling) and $\displaystyle C$ (representing a vertical shift) are unknown and cannot be determined from the given information.

We are asked to find the value of $\displaystyle p(2)$ :Alternatively, using the form $\displaystyle p(x) = A(x^3 - 6x^2 + 9x) + C$ :Since the values of the constants $\displaystyle A$ (or $\displaystyle k$ ) and $\displaystyle C$ are unknown, the value of $\displaystyle p(2)$ cannot be determined.

For example, if we assume $\displaystyle A=1, C=0$ , then $\displaystyle p(2)=2$ . If we assume $\displaystyle A=1, C=-4$ , then $\displaystyle p(2)=-2$ . Both underlying polynomials satisfy the condition of having stationary points at $\displaystyle x=1$ and $\displaystyle x=3$ .

Therefore, the value of $\displaystyle p(2)$ cannot be determined from the information given.

For at least two red balls to be drawn, the first two balls drawn must both be red.

The probability of drawing a red ball on the first draw is $\displaystyle \dfrac{4}{12} = \dfrac{1}{3}$ .
Given that the first ball is red, the probability of drawing a red ball on the second draw is $\displaystyle \dfrac{3}{11}$ .
Thus, the probability that the first two balls are both red is: $\displaystyle P(\text{first two are red}) = \dfrac{4}{12} \times \dfrac{3}{11} = \dfrac{1}{11}$ After two red balls have been removed, the box contains 2 red balls, 5 blue balls, and 3 green balls. Drawing any further red balls merely continues the process without stopping it. The process terminates as soon as the first non-red ball is drawn.

Among the 8 remaining non-red balls (5 blue and 3 green), each is equally likely to be the first non-red ball drawn. Therefore, the probability that the stopping ball is blue is: $\displaystyle P(\text{stops on blue} \mid \text{first two are red}) = \dfrac{5}{5 + 3} = \dfrac{5}{8}$ Multiplying these probabilities gives the total probability: $\displaystyle P = \dfrac{1}{11} \times \dfrac{5}{8} = \dfrac{5}{88}$

Step Analyse the number of solutions for $\displaystyle x$ based on the value of $\displaystyle s$ .:

For a given value of $\displaystyle s$ in the range $\displaystyle [-1, 1]$ , the number of solutions to $\displaystyle \sin(x) = s$ in $\displaystyle 0 \le x < 2\pi$ is:
- One solution if $\displaystyle s=1$ (at $\displaystyle x=\dfrac{\pi}{2}$ ) or $\displaystyle s=-1$ (at $\displaystyle x=\dfrac{3\pi}{2}$ ).
- Two solutions if $\displaystyle -1 < s < 1$ .
- No solutions if $\displaystyle |s| > 1$ .

Step Determine the condition on the roots of the quadratic for three solutions.:

To obtain exactly three distinct solutions for $\displaystyle x$ , the quadratic in $\displaystyle s$ must have two distinct real roots, $\displaystyle s_1$ and $\displaystyle s_2$ , such that one root gives one solution for $\displaystyle x$ and the other gives two solutions. This means one root must be $\displaystyle 1$ or $\displaystyle -1$ , and the other must be in the interval $\displaystyle (-1, 1)$ .

Step Case 1: One root is $\displaystyle s=1$ .:

Substitute $\displaystyle s=1$ into the quadratic $\displaystyle 2s^2 + 3s + (k-2) = 0$ : $\displaystyle 2(1)^2 + 3(1) + (k-2) = 0$ $\displaystyle 2 + 3 + k - 2 = 0$ $\displaystyle k+3 = 0 \implies k = -3$ If $\displaystyle k=-3$ , the quadratic is $\displaystyle 2s^2 + 3s - 5 = 0$ . Factoring gives $\displaystyle (2s+5)(s-1)=0$ , so the roots are $\displaystyle s=1$ and $\displaystyle s=-\dfrac{5}{2}$ . The root $\displaystyle s=-\dfrac{5}{2}$ is outside the valid range $\displaystyle [-1, 1]$ , so it yields no solutions for $\displaystyle x$ . The root $\displaystyle s=1$ yields one solution. Thus, for $\displaystyle k=-3$ , there is only one solution in total. This case is incorrect.

Step Case 2: One root is $\displaystyle s=-1$ .:

Substitute $\displaystyle s=-1$ into the quadratic $\displaystyle 2s^2 + 3s + (k-2) = 0$ : $\displaystyle 2(-1)^2 + 3(-1) + (k-2) = 0$ $\displaystyle 2 - 3 + k - 2 = 0$ $\displaystyle k-3 = 0 \implies k = 3$ If $\displaystyle k=3$ , the quadratic is $\displaystyle 2s^2 + 3s + 1 = 0$ . Factoring gives $\displaystyle (2s+1)(s+1)=0$ , so the roots are $\displaystyle s=-1$ and $\displaystyle s=-\dfrac{1}{2}$ .
The root $\displaystyle s=-1$ gives one solution for $\displaystyle x$ . The root $\displaystyle s=-\dfrac{1}{2}$ is in the interval $\displaystyle (-1, 1)$ and gives two solutions for $\displaystyle x$ . The total number of solutions is $\displaystyle 1+2=3$ . This satisfies the condition.

The value of $\displaystyle k$ for which there are exactly three distinct solutions is $\displaystyle 3$ .

The relationship is given by $\displaystyle y = Ax^n$ . To analyse the graph of $\displaystyle \log_2 y$ against $\displaystyle \log_2 x$ , we take logarithms to base 2 of this equation:This equation is in the form of a straight line, $\displaystyle Y = mX + c$ , where $\displaystyle Y = \log_2 y$ , $\displaystyle X = \log_2 x$ , the gradient is $\displaystyle m=n$ , and the vertical intercept is $\displaystyle c = \log_2 A$ .

From the problem description, the gradient is 2 and the intercept is 4.
Therefore, $\displaystyle n=2$ .
The intercept gives $\displaystyle \log_2 A = 4$ , which implies $\displaystyle A = 2^4 = 16$ .

Substituting these values for $\displaystyle A$ and $\displaystyle n$ back into the original equation gives $\displaystyle y = 16x^2$ .

To find the coordinates of the minimum point, we complete the square for the quadratic equation: $\displaystyle y = (x - a)^2 - a^2 - 4a + 5$ Since the coefficient of $\displaystyle (x - a)^2$ is positive ( $\displaystyle +1$ ), the minimum point occurs when $\displaystyle x - a = 0$ , giving the vertex coordinates: $\displaystyle (x_v, y_v) = (a, -a^2 - 4a + 5)$ For the vertex to lie strictly in the second quadrant, both coordinates must satisfy:
1. $\displaystyle x_v < 0 \implies a < 0$ 2. $\displaystyle y_v > 0 \implies -a^2 - 4a + 5 > 0$ Solving the quadratic inequality for $\displaystyle y_v$ : $\displaystyle -a^2 - 4a + 5 > 0 \iff a^2 + 4a - 5 < 0$ $\displaystyle (a + 5)(a - 1) < 0$ $\displaystyle -5 < a < 1$ Combining both conditions ( $\displaystyle a < 0$ and $\displaystyle -5 < a < 1$ ): $\displaystyle -5 < a < 0$ Thus, the correct option is C.

The sequence is an arithmetic progression. Let the first term be $\displaystyle a$ and the common difference be $\displaystyle d$ . The sum of the first $\displaystyle n$ terms is given by the formula $\displaystyle S_n = \dfrac{n}{2}(2a + (n-1)d)$ .

We are given that the sum of the first 10 terms is 200:Dividing by 5 gives our first equation: $\displaystyle 2a + 9d = 40$ .

The sum of the next 10 terms is 600, which means the sum of the first 20 terms is $\displaystyle S_{20} = 200 + 600 = 800$ . We can write a second equation for $\displaystyle S_{20}$ :Dividing by 10 gives our second equation: $\displaystyle 2a + 19d = 80$ .

We now have a pair of simultaneous equations:
1. $\displaystyle 2a + 9d = 40$ 2. $\displaystyle 2a + 19d = 80$ Subtracting the first equation from the second gives:This gives a common difference of $\displaystyle d = 4$ . Substituting this value back into the first equation:Therefore, the first term is $\displaystyle a=2$ .

Let the first term be $\displaystyle a$ and the common ratio be $\displaystyle r$ . The sum to infinity exists, so $\displaystyle |r| < 1$ .

The sum of the first two terms is $\displaystyle S_2 = a + ar = a(1+r)$ .
The sum to infinity is $\displaystyle S_\infty = \dfrac{a}{1-r}$ .

We are given the condition $\displaystyle S_2 = \dfrac{5}{9} S_\infty$ . Substituting the formulae:Since all terms are non-zero, $\displaystyle a eq 0$ , so we can divide by $\displaystyle a$ :This gives two possible values for the common ratio: $\displaystyle r = \dfrac{2}{3}$ or $\displaystyle r = -\dfrac{2}{3}$ .

The problem states that 'the terms in the series do not all have the same sign'.
If $\displaystyle r > 0$ , all terms would have the same sign as the first term $\displaystyle a$ .
If $\displaystyle r < 0$ , the terms will alternate in sign.
Therefore, we must have $\displaystyle r < 0$ , so we choose $\displaystyle r = -\dfrac{2}{3}$ .

We need to find the third term, $\displaystyle T_3$ , as a fraction of the sum to infinity, $\displaystyle S_\infty$ . $\displaystyle T_3 = ar^2$ The required fraction is $\displaystyle \dfrac{T_3}{S_\infty}$ :Now, we substitute $\displaystyle r = -\dfrac{2}{3}$ into this expression:So, the fraction is:Thus, the correct option is C.

Taking $\displaystyle \log_{10}$ of the equation $\displaystyle y = kx^n$ yields the linear form:This represents a straight line $\displaystyle Y = nX + c$ where $\displaystyle Y = \log_{10} y$ , $\displaystyle X = \log_{10} x$ , and the vertical intercept $\displaystyle c = \log_{10} k$ .

The graph passes through the $\displaystyle Y$ -intercept $\displaystyle (0, 2)$ , so $\displaystyle c=2$ . This implies:The gradient of the line passing through $\displaystyle (0, 2)$ and $\displaystyle (4, 0)$ is:Substituting $\displaystyle k=100$ and $\displaystyle n=-0.5$ into the original equation gives $\displaystyle y = 100 x^{-1/2}$ , which simplifies to $\displaystyle y = \dfrac{100}{\sqrt{x}}$ .

To find the range on the closed interval $\displaystyle [-2, 2]$ , we evaluate $\displaystyle f(x)$ at the endpoints and at any stationary points lying within the domain.

Differentiating $\displaystyle f(x)$ gives:The stationary points are $\displaystyle x=3$ and $\displaystyle x=-1$ . Since the domain is restricted to $\displaystyle -2 \le x \le 2$ , we discard $\displaystyle x=3$ as it lies outside the interval.

We evaluate $\displaystyle f(x)$ at the valid stationary point $\displaystyle x=-1$ and the endpoints $\displaystyle x=-2$ and $\displaystyle x=2$ :Comparing these values, the minimum is $\displaystyle 8$ and the maximum is $\displaystyle 35$ . Thus, the range is $\displaystyle 8 \le f(x) \le 35$ .

We group the terms as $\displaystyle ((1+2x) + 2x^2)^5$ . The general term of the binomial expansion is:We require the coefficient of $\displaystyle x^4$ . Since the factor $\displaystyle (2x^2)^k$ contributes $\displaystyle x^{2k}$ , we must find the coefficient of $\displaystyle x^{4-2k}$ in the expansion of $\displaystyle (1+2x)^{5-k}$ . We sum the contributions for each valid integer $\displaystyle k$ .

For $\displaystyle k=0$ , we need the coefficient of $\displaystyle x^4$ in $\displaystyle (1+2x)^5$ :For $\displaystyle k=1$ , we need the coefficient of $\displaystyle x^2$ in $\displaystyle (1+2x)^4$ , multiplied by the coefficient from the outer term $\displaystyle \binom{5}{1}(2x^2)^1$ :For $\displaystyle k=2$ , we need the constant term in $\displaystyle (1+2x)^3$ , multiplied by the coefficient from $\displaystyle \binom{5}{2}(2x^2)^2$ :Any term with $\displaystyle k \ge 3$ produces a power of $\displaystyle x$ greater than 4. Summing the contributions gives $\displaystyle 80 + 240 + 40 = 360$ .

The given equation is $\displaystyle \tan\left(\dfrac{4\pi}{x}\right) = \sqrt{3}$ .

The principal value for which $\displaystyle \tan(\theta) = \sqrt{3}$ is $\displaystyle \theta = \dfrac{\pi}{3}$ .

The general solution for $\displaystyle \tan(A) = c$ is $\displaystyle A = \arctan(c) + n\pi$ , where $\displaystyle n$ is an integer.

So, for our equation, the general solution for the argument is:We can divide the entire equation by $\displaystyle \pi$ :Combining the terms on the right-hand side gives:Now, we solve for $\displaystyle x$ by taking the reciprocal of both sides and multiplying by 4:We are given the constraint that $\displaystyle x > 0$ . Since the numerator 12 is positive, the denominator must also be positive:Since $\displaystyle n$ must be an integer, the possible values for $\displaystyle n$ are $\displaystyle 0, 1, 2, 3, \dots$ .

The solutions for $\displaystyle x$ are generated by these values of $\displaystyle n$ . As $\displaystyle n$ increases, the denominator $\displaystyle 1+3n$ increases, so the value of $\displaystyle x$ decreases.

Therefore, the solutions in decreasing order $\displaystyle x_1 > x_2 > x_3 > \dots$ correspond to increasing values of $\displaystyle n$ starting from $\displaystyle n=0$ .

- The largest solution, $\displaystyle x_1$ , corresponds to $\displaystyle n=0$ : $\displaystyle x_1 = \dfrac{12}{1 + 3(0)} = 12$ .
- The second largest solution, $\displaystyle x_2$ , corresponds to $\displaystyle n=1$ : $\displaystyle x_2 = \dfrac{12}{1 + 3(1)} = \dfrac{12}{4} = 3$ .
- The third largest solution, $\displaystyle x_3$ , corresponds to $\displaystyle n=2$ : $\displaystyle x_3 = \dfrac{12}{1 + 3(2)} = \dfrac{12}{7}$ .

Therefore, the correct answer is $\displaystyle \dfrac{12}{7}$ .

Let $\displaystyle r$ be the radius and $\displaystyle s$ be the arc length of the sector. The perimeter is fixed at $\displaystyle 60$ , so we have:The area of a sector is given by $\displaystyle A = \dfrac{1}{2}rs$ . Substituting our expression for $\displaystyle s$ allows us to write the area solely in terms of $\displaystyle r$ :We wish to maximise this area. Since $\displaystyle A(r)$ is a downward-opening quadratic, the maximum occurs at the stationary point. Differentiating with respect to $\displaystyle r$ :Setting $\displaystyle \dfrac{dA}{dr} = 0$ yields $\displaystyle 2r = 30$ , which gives $\displaystyle r = 15$ .

Let the point $\displaystyle P$ on the $\displaystyle y$ -axis have coordinates $\displaystyle (0, c)$ .
The distance from $\displaystyle P(0,c)$ to a point $\displaystyle (x,0)$ on the $\displaystyle x$ -axis is given by $\displaystyle \sqrt{(x-0)^2 + (0-c)^2} = \sqrt{x^2 + c^2}$ .

We are given two distances:
1. The distance from $\displaystyle P$ to $\displaystyle (\alpha, 0)$ is $\displaystyle 5$ . This gives the equation:Squaring both sides gives:2. The distance from $\displaystyle P$ to $\displaystyle (\beta, 0)$ is $\displaystyle 3$ . This gives the equation:Squaring both sides gives:To find a relationship between $\displaystyle \alpha$ and $\displaystyle \beta$ , we can eliminate $\displaystyle c^2$ by subtracting equation (2) from equation (1):We are given the relation $\displaystyle \alpha = 3\beta$ . Substituting this into the equation above:The question asks for the value of the product $\displaystyle \alpha\beta$ . We can express this product in terms of $\displaystyle \beta$ using the given relation:Substituting the value we found for $\displaystyle \beta^2$ :Therefore the correct answer is 6.

The volume $\displaystyle V$ of a right circular cylinder is given by: $\displaystyle V = \pi r^2 h$ Rearranging for height $\displaystyle h$ : $\displaystyle h = \dfrac{V}{\pi r^2}$ The new radius after a $\displaystyle 20\%$ decrease is $\displaystyle r' = (1 - 0.20)r = 0.80r$ .
The new volume after a $\displaystyle 28\%$ increase is $\displaystyle V' = (1 + 0.28)V = 1.28V$ .

The new height $\displaystyle h'$ is: $\displaystyle h' = \dfrac{V'}{\pi (r')^2} = \dfrac{1.28V}{\pi (0.80r)^2} = \dfrac{1.28V}{0.64\pi r^2} = \dfrac{1.28}{0.64}\left(\dfrac{V}{\pi r^2}\right) = 2h$ The percentage increase in $\displaystyle h$ is: $\displaystyle \dfrac{h' - h}{h} \times 100\% = \dfrac{2h - h}{h} \times 100\% = 100\%$