2023 Cambridge IAL Mathematics (9709) Practice Paper with Answers

To find the points of intersection, we equate the two equations:
\(x^2 + 4x + 7 = kx - 2\)

Rearranging into a standard quadratic form:
\(x^2 + (4 - k)x + 9 = 0\)

For the line and curve not to intersect, the quadratic equation must have no real roots. Thus, the discriminant must be negative:
\(b^2 - 4ac < 0\)

Substituting the coefficients:
\((4 - k)^2 - 4(1)(9) < 0\)
\((4 - k)^2 - 36 < 0\)

Expanding and simplifying:
\(16 - 8k + k^2 - 36 < 0\)
\(k^2 - 8k - 20 < 0\)

Factoring the quadratic expression:
\((k - 10)(k + 2) < 0\)

Therefore, the set of possible values of \(k\) is:
\(-2 < k < 10\)

M1: Equate the line and curve equations and form a 3-term quadratic equation in \(x\).
M1: Attempt to find the discriminant of the quadratic and set it less than 0.
A1: Identify the correct critical values of \(-2\) and \(10\).
A1: State the correct final range \(-2 < k < 10\) (or equivalent interval notation).

First, we simplify the left-hand side of the equation by finding a common denominator:
\(\frac{\sin^2 \theta + (1 + \cos \theta)^2}{\sin \theta(1 + \cos \theta)} = 4\)

Expanding the numerator:
\(\frac{\sin^2 \theta + 1 + 2\cos \theta + \cos^2 \theta}{\sin \theta(1 + \cos \theta)} = 4\)

Using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\(\frac{1 + 1 + 2\cos \theta}{\sin \theta(1 + \cos \theta)} = 4\)
\(\frac{2 + 2\cos \theta}{\sin \theta(1 + \cos \theta)} = 4\)

Factoring out 2 in the numerator:
\(\frac{2(1 + \cos \theta)}{\sin \theta(1 + \cos \theta)} = 4\)

Cancelling the common factor \((1 + \cos \theta)\) (since \(\cos \theta \ne -1\) for the equation to be defined):
\(\frac{2}{\sin \theta} = 4\)

Now we solve \(\sin \theta = \frac{1}{2}\) in the interval \(0^\circ \le \theta \le 360^\circ\):
\(\theta = 30^\circ\) and \(\theta = 180^\circ - 30^\circ = 150^\circ\)

M1: Combine terms on LHS over a common denominator and expand the numerator.
A1: Simplify the LHS to \(\frac{2}{\sin \theta}\) using the identity \(\sin^2 \theta + \cos^2 \theta = 1\).
M1: Obtain \(\sin \theta = \frac{1}{2}\) and find at least one correct angle.
A1: Provide both correct solutions \(\theta = 30^\circ\) and \(\theta = 150^\circ\) and no others in the given range.

We are given two pieces of information:
1. The sum of the first two terms:
\(S_2 = a + ar = a(1 + r) = 15\) (Equation 1)

2. The sum to infinity:
\(S_\infty = \frac{a}{1 - r} = 16\) (Equation 2)

From Equation 2, we can express \(a\) in terms of \(r\):
\(a = 16(1 - r)\)

Substitute this expression for \(a\) into Equation 1:
\(16(1 - r)(1 + r) = 15\)
\(16(1 - r^2) = 15\)
\(1 - r^2 = \frac{15}{16}\)
\(r^2 = 1 - \frac{15}{16}\)
\(r^2 = \frac{1}{16}\)

Since we are given that \(r\) is positive:
\(r = \frac{1}{4}\)

Now, substitute \(r = \frac{1}{4}\) back into the expression for \(a\):
\(a = 16\left(1 - \frac{1}{4}\right) = 16 \times \frac{3}{4} = 12\)

M1: Set up two correct equations for \(S_2 = 15\) and \(S_\infty = 16\) in terms of \(a\) and \(r\).
M1: Eliminate \(a\) (or \(r\)) to form a single equation in terms of \(r\) (or \(a\)).
A1: Solve for \(r = \frac{1}{4}\) (rejecting \(r = -\frac{1}{4}\)).
A1: Solve for \(a = 12\).

To find the stationary points, we first find the derivative \(\frac{dy}{dx}\).
Rewrite the equation as:
\(y = 3x + 6(2x - 1)^{-1}\)

Differentiating with respect to \(x\) using the chain rule:
\(\frac{dy}{dx} = 3 + 6(-1)(2x - 1)^{-2} \times 2\)
\(\frac{dy}{dx} = 3 - \frac{12}{(2x - 1)^2}\)

At stationary points, \(\frac{dy}{dx} = 0\):
\(3 - \frac{12}{(2x - 1)^2} = 0\)
\(3 = \frac{12}{(2x - 1)^2}\)
\((2x - 1)^2 = 4\)

Taking the square root of both sides:
\(2x - 1 = 2\) or \(2x - 1 = -2\)

Case 1:
\(2x = 3 \implies x = 1.5\)
Substitute \(x = 1.5\) into the original equation:
\(y = 3(1.5) + \frac{6}{2(1.5) - 1} = 4.5 + \frac{6}{2} = 7.5\)
So one stationary point is \((1.5, 7.5)\).

Case 2:
\(2x = -1 \implies x = -0.5\)
Substitute \(x = -0.5\) into the original equation:
\(y = 3(-0.5) + \frac{6}{2(-0.5) - 1} = -1.5 + \frac{6}{-2} = -4.5\)
So the other stationary point is \((-0.5, -4.5)\).

M1: Apply the chain rule correctly to differentiate \(\frac{6}{2x - 1}\).
A1: Obtain the correct expression for \(\frac{dy}{dx} = 3 - \frac{12}{(2x - 1)^2}\).
M1: Set \(\frac{dy}{dx} = 0\) and solve the resulting quadratic equation for \(x\).
A1: Correctly state both coordinates: \((1.5, 7.5)\) and \((-0.5, -4.5)\) (accept fractional forms \((3/2, 15/2)\) and \((-1/2, -9/2)\)).

To find the coordinates of \(A\) and \(B\), equate the equations of the curve and the line:
\(9 - x^2 = 2x + 6\)
\(x^2 + 2x - 3 = 0\)
\((x + 3)(x - 1) = 0\)

This gives \(x = -3\) and \(x = 1\).

Substitute these values back into the linear equation to find the corresponding \(y\)-coordinates:
For \(x = -3\), \(y = 2(-3) + 6 = 0\).
For \(x = 1\), \(y = 2(1) + 6 = 8\).
So, the points of intersection are \(A(-3, 0)\) and \(B(1, 8)\).

Now, find the area of the region completely enclosed by the curve and the line by integrating the difference between the upper function (the curve) and the lower function (the line) from \(x = -3\) to \(x = 1\):
\(\text{Area} = \int_{-3}^{1} \left( (9 - x^2) - (2x + 6) \right) \, dx\)
\(\text{Area} = \int_{-3}^{1} (3 - 2x - x^2) \, dx\)

Integrate each term:
\(\left[ 3x - x^2 - \frac{1}{3}x^3 \right]_{-3}^{1}\)

Substitute the upper limit \(x = 1\):
\(3(1) - (1)^2 - \frac{1}{3}(1)^3 = 3 - 1 - \frac{1}{3} = \frac{5}{3}\)

Substitute the lower limit \(x = -3\):
\(3(-3) - (-3)^2 - \frac{1}{3}(-3)^3 = -9 - 9 + 9 = -9\)

Calculate the difference:
\(\text{Area} = \frac{5}{3} - (-9) = \frac{32}{3}\)

M1: Set up quadratic equation by equating the curve and the line.
A1: Correct coordinates of \(A\) and \(B\) (both required).
M1: Set up the definite integral with correct limits and order of subtraction.
A1: Correct integration of the expression.
M1: Substitute the limits correctly into their integrated expression.
A1: Obtain final area of \(\frac{32}{3}\) (or equivalent fraction / exact decimal).

Using similar triangles in the cross-section of the cone and cylinder, the height of the cone above the cylinder is \(12 - h\). The ratio of this height to the cylinder's radius \(r\) is equal to the ratio of the total cone height to the cone's base radius:
\(\frac{12 - h}{r} = \frac{12}{6}\)
\(12 - h = 2r \implies h = 12 - 2r\)

The volume of the cylinder, \(V\), is given by:
\(V = \pi r^2 h = \pi r^2 (12 - 2r) = 12\pi r^2 - 2\pi r^3\) (as shown).

To find the maximum volume, differentiate \(V\) with respect to \(r\):
\(\frac{dV}{dr} = 24\pi r - 6\pi r^2\)

For a maximum or minimum, set \(\frac{dV}{dr} = 0\):
\(6\pi r(4 - r) = 0\)

Since \(r > 0\) for a physical cylinder, the only valid solution is \(r = 4\).

To verify this is a maximum, check the second derivative:
\(\frac{d^2V}{dr^2} = 24\pi - 12\pi r\)

At \(r = 4\):
\(\frac{d^2V}{dr^2} = 24\pi - 48\pi = -24\pi < 0\), which confirms a maximum.

Now substitute \(r = 4\) back into the volume equation:
\(V = 12\pi(4^2) - 2\pi(4^3) = 192\pi - 128\pi = 64\pi\text{ cm}^3\).

M1: Use similar triangles to establish a relationship between \(h\) and \(r\).
A1: Correct expression \(h = 12 - 2r\) substituted into the cylinder volume formula to obtain the given result.
M1: Differentiate the volume formula to find \(\frac{dV}{dr}\).
A1: Equate to zero and solve to obtain \(r = 4\).
M1: Use second derivative (or alternative method) to show that \(r = 4\) yields a maximum volume.
A1: Calculate the correct maximum volume of \(64\pi\).

First, find the coordinates of the point on the curve where \(\theta = \frac{\pi}{4}\):
\(x = 2\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{2}\right) = \frac{\pi}{2} - 1\)
\(y = 1 - \cos\left(\frac{\pi}{2}\right) = 1 - 0 = 1\)

Next, find the derivatives \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\):
\(\frac{dx}{d\theta} = 2 - 2\cos(2\theta)\)
\(\frac{dy}{d\theta} = 2\sin(2\theta)\)

Use the chain rule to find \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2\sin(2\theta)}{2(1 - \cos(2\theta))} = \frac{\sin(2\theta)}{1 - \cos(2\theta)}
\)

Evaluate the gradient of the tangent at \(\theta = \frac{\pi}{4}\):
\(\frac{dy}{dx} = \frac{\sin(\pi/2)}{1 - \cos(\pi/2)} = \frac{1}{1 - 0} = 1\)

The gradient of the normal, \(m_{\text{normal}}\), is the negative reciprocal of the tangent gradient:
\(m_{\text{normal}} = -1\)

Using the equation of a line with the point \(\left(\frac{\pi}{2} - 1, 1\right)\) and gradient \(-1\):
\(y - 1 = -1\left(x - \left(\frac{\pi}{2} - 1\right)\right)\)
\(y - 1 = -x + \frac{\pi}{2} - 1\)
\(y = -x + \frac{\pi}{2}\)

M1: Find the numerical/exact values of \(x\) and \(y\) at \(\theta = \frac{\pi}{4}\).
A1: Obtain the correct coordinates \(\left(\frac{\pi}{2} - 1, 1\right)\).
M1: Correctly differentiate \(x\) and \(y\) with respect to \(\theta\) and apply the chain rule to find \(\frac{dy}{dx}\).
A1: Find the gradient of the tangent is \(1\).
M1: Find the negative reciprocal gradient for the normal and use the straight line equation.
A1: Obtain the correct final equation \(y = -x + \frac{\pi}{2}\).

The volume of revolution \(V\) is given by:
\(V = \pi \int_{0}^{1} y^2 \, dx = \pi \int_{0}^{1} x^2 e^{-2x} \, dx\)

We evaluate \(\int x^2 e^{-2x} \, dx\) using integration by parts, where \(\int u \, dv = uv - \int v \, du\).
Let \(u = x^2\) and \(dv = e^{-2x} \, dx\).
Then \(du = 2x \, dx\) and \(v = -\frac{1}{2}e^{-2x}\).

Therefore:
\(\int x^2 e^{-2x} \, dx = -\frac{1}{2}x^2 e^{-2x} - \int \left(-\frac{1}{2}e^{-2x}\right) (2x) \, dx = -\frac{1}{2}x^2 e^{-2x} + \int x e^{-2x} \, dx\)

To evaluate \(\int x e^{-2x} \, dx\), we use integration by parts again.
Let \(u = x\) and \(dv = e^{-2x} \, dx\).
Then \(du = dx\) and \(v = -\frac{1}{2}e^{-2x}\).

So:
\(\int x e^{-2x} \, dx = -\frac{1}{2}x e^{-2x} - \int \left(-\frac{1}{2} e^{-2x}\right) \, dx = -\frac{1}{2}x e^{-2x} - \frac{1}{4}e^{-2x}\)

Substituting this back into the first equation yields:
\(\int x^2 e^{-2x} \, dx = \left[ -\frac{1}{2}x^2 e^{-2x} - \frac{1}{2}x e^{-2x} - \frac{1}{4}e^{-2x} \right]\)

Now, substitute the limits \(0\) and \(1\):
At \(x = 1\):
\(\left( -\frac{1}{2}(1)^2 - \frac{1}{2}(1) - \frac{1}{4} \right) e^{-2} = -\frac{5}{4}e^{-2}\)

At \(x = 0\):
\(\left( 0 - 0 - \frac{1}{4} \right) e^0 = -\frac{1}{4}\)

Subtracting lower limit from upper limit:
\(\left(-\frac{5}{4}e^{-2}\right) - \left(-\frac{1}{4}\right) = \frac{1}{4} - \frac{5}{4}e^{-2}\)

Finally, multiply by \(\pi\):
\(V = \pi \left( \frac{1}{4} - \frac{5}{4}e^{-2} \right) = \frac{\pi}{4}(1 - 5e^{-2})\)

Comparing to the required form, \(a = 4\), \(b = 1\), and \(c = 5\).

M1: State or imply volume formula \(V = \pi \int y^2 \, dx\) with limits \(0\) and \(1\).
M1: Apply integration by parts once to find the structure \(Ax^2 e^{-2x} + B\int x e^{-2x} \, dx\).
A1: Obtain correct first-stage integral \(-\frac{1}{2}x^2 e^{-2x} + \int x e^{-2x} \, dx\).
M1: Apply integration by parts a second time to \(\int x e^{-2x} \, dx\).
A1: Obtain correct complete integrated expression \(e^{-2x}\left(-\frac{1}{2}x^2 - \frac{1}{2}x - \frac{1}{4}\right)\).
A1: Substitute limits correctly and obtain the exact volume \(\frac{\pi}{4}(1 - 5e^{-2})\).

(i) By completing the square on the circle's equation, we get: \((x - 4)^2 - 16 + (y - 6)^2 - 36 + 27 = 0\), which simplifies to \((x - 4)^2 + (y - 6)^2 = 25\). Therefore, the center of the circle is \((4, 6)\) and the radius is \(\sqrt{25} = 5\). (ii) The perpendicular distance from the center \((4, 6)\) to the line \(2x - y + c = 0\) must equal the radius \(5\). Using the distance formula: \(\frac{|2(4) - 6 + c|}{\sqrt{2^2 + (-1)^2}} = 5\), which gives \(\frac{|c + 2|}{\sqrt{5}} = 5\). This simplifies to \(|c + 2| = 5\sqrt{5}\), so \(c = -2 \pm 5\sqrt{5}\). Alternatively, substituting \(y = 2x + c\) into the circle equation and setting the discriminant of the resulting quadratic \(5x^2 + (4c - 32)x + c^2 - 12c + 27 = 0\) to zero yields the same values.

(i) M1: Attempt to complete the square for both x and y. A1: Correct center (4, 6). A1: Correct radius 5. (ii) M1: Set up perpendicular distance equation or substitute and use discriminant = 0. M1: Form a correct equation in terms of c. A1: Simplify to a correct quadratic or absolute value equation. M1: Solve for c. A1: Obtain correct exact values c = -2 \pm 5\sqrt{5}.

(i) The center of \(C_1\) is the midpoint of \(AB\), which is \(\left(\frac{2+6}{2}, \frac{1+9}{2}\right) = (4, 5)\). The radius \(R_1\) of \(C_1\) is the distance from the center to \(A(2,1)\), so \(R_1 = \sqrt{(4-2)^2 + (5-1)^2} = \sqrt{20} = 2\sqrt{5}\). Thus, the equation of \(C_1\) is \((x-4)^2 + (y-5)^2 = 20\). (ii) The distance \(d\) between the centers of \(C_1\) at \((4,5)\) and \(C_2\) at \((10,3)\) is \(d = \sqrt{(10-4)^2 + (3-5)^2} = \sqrt{40} = 2\sqrt{10}\). For the circles to intersect at two distinct points, the radius \(r\) of \(C_2\) must satisfy \(|r - R_1| < d < r + R_1\). This yields \(r + 2\sqrt{5} > 2\sqrt{10} \implies r > 2\sqrt{10} - 2\sqrt{5}\), and \(r - 2\sqrt{5} < 2\sqrt{10} \implies r < 2\sqrt{10} + 2\sqrt{5}\) (the other boundary \(r > 2\sqrt{5} - 2\sqrt{10}\) is negative and thus automatically satisfied since \(r > 0\)). Combining these, we obtain \(2\sqrt{10} - 2\sqrt{5} < r < 2\sqrt{10} + 2\sqrt{5}\).

(i) M1: Find midpoint of AB to obtain the center. M1: Calculate radius of C_1. A1: Correct equation (x-4)^2 + (y-5)^2 = 20. (ii) M1: Calculate distance between centers as 2\sqrt{10}. M1: State or apply the geometric condition for intersection of two circles. A1: Find correct upper bound r < 2\sqrt{10} + 2\sqrt{5}. A1: Find correct lower bound r > 2\sqrt{10} - 2\sqrt{5}. A1: Combine to express the final correct range.

(i) The equation of the circle \(C\) is \((x - 2)^2 + (y + 4)^2 = 15\). Substituting the line equation \(y = 3x - 5\) into the circle's equation gives \((x - 2)^2 + (3x - 5 + 4)^2 = 15\), which simplifies to \((x - 2)^2 + (3x - 1)^2 = 15\). Expanding and simplifying yields \(x^2 - 4x + 4 + 9x^2 - 6x + 1 - 15 = 0\), which reduces to \(10x^2 - 10x - 10 = 0\) or \(x^2 - x - 1 = 0\). The discriminant of this quadratic is \(\Delta = (-1)^2 - 4(1)(-1) = 5\). Since \(\Delta > 0\), there are two distinct real intersection points, \(P\) and \(Q\). (ii) The perpendicular distance \(d\) from the center \((2, -4)\) to the line \(3x - y - 5 = 0\) is \(d = \frac{|3(2) - (-4) - 5|}{\sqrt{3^2 + (-1)^2}} = \frac{5}{\sqrt{10}}\). In the right-angled triangle formed by the center of the circle, the midpoint of the chord, and a point on the circle, the half-chord length is given by \(\sqrt{R^2 - d^2} = \sqrt{15 - 2.5} = \sqrt{12.5} = \frac{5}{\sqrt{2}}\). Thus, the full length of the chord \(PQ\) is \(2 \times \frac{5}{\sqrt{2}} = 5\sqrt{2}\). Alternatively, solving \(x^2 - x - 1 = 0\) gives \(x = \frac{1 \pm \sqrt{5}}{2}\), and using the distance formula between the two coordinates yields \(\sqrt{10} |x_2 - x_1| = \sqrt{10}\sqrt{5} = 5\sqrt{2}\).

(i) M1: Substitute the line equation into the circle equation. A1: Form a correct simplified quadratic equation. A1: Calculate discriminant and conclude that \(\Delta > 0\) implies two distinct intersection points. (ii) M1: Calculate the perpendicular distance from the center to the line (or find the coordinates of P and Q). A1: Correct distance 5/\sqrt{10} (or correct coordinates). M1: Apply Pythagoras' theorem to find half-chord length (or apply distance formula between P and Q). A1: Show clear intermediate step. A1: Correct exact chord length 5\sqrt{2}.