2023 Cambridge IAL Mathematics (9709) Practice Paper with Answers

Using the formula for the sum to infinity of a geometric progression: \( S_\infty = \frac{a}{1-r} = 8 \implies a = 8(1-r) \). Using the formula for the sum of the first three terms: \( S_3 = a + ar + ar^2 = 7 \implies a(1+r+r^2) = 7 \). Substituting the expression for \( a \) into the second equation: \( 8(1-r)(1+r+r^2) = 7 \). Using the identity \( (1-r)(1+r+r^2) = 1-r^3 \), we get: \( 8(1-r^3) = 7 \implies 1-r^3 = \frac{7}{8} \implies r^3 = \frac{1}{8} \implies r = \frac{1}{2} \). Substituting \( r = \frac{1}{2} \) back into the expression for \( a \): \( a = 8(1 - \frac{1}{2}) = 4 \).

M1: Use the sum to infinity formula to express \( a \) in terms of \( r \) (or vice versa). M1: Substitute their expression for \( a \) into the sum of first three terms formula. A1: Solve the resulting cubic equation to obtain \( r = \frac{1}{2} \) (or \( 0.5 \)). A1: Obtain \( a = 4 \).

Method 1: The equation of the circle can be rewritten in the form \( (x-2)^2 + (y+3)^2 = 25 \), so the center is \( C(2, -3) \) and the radius is \( R = 5 \). The line with gradient \( m \) passing through \( (0, 2) \) has equation \( y - 2 = m(x - 0) \implies mx - y + 2 = 0 \). The perpendicular distance from the center \( (2, -3) \) to this line must equal the radius: \( \frac{|m(2) - (-3) + 2|}{\sqrt{m^2 + (-1)^2}} = 5 \implies \frac{|2m + 5|}{\sqrt{m^2 + 1}} = 5 \). Squaring both sides: \( (2m+5)^2 = 25(m^2+1) \implies 4m^2 + 20m + 25 = 25m^2 + 25 \implies 21m^2 - 20m = 0 \implies m(21m - 20) = 0 \). Thus, \( m = 0 \) or \( m = \frac{20}{21} \). Method 2: Substitute \( y = mx + 2 \) into the circle equation: \( x^2 + (mx+2)^2 - 4x + 6(mx+2) - 12 = 0 \implies (1+m^2)x^2 + (10m-4)x + 4 = 0 \). Setting the discriminant \( B^2 - 4AC = 0 \) for tangency: \( (10m-4)^2 - 16(1+m^2) = 0 \implies 100m^2 - 80m + 16 - 16 - 16m^2 = 0 \implies 84m^2 - 80m = 0 \implies 4m(21m-20) = 0 \), which gives \( m = 0 \) or \( m = \frac{20}{21} \).

M1: Express circle center and radius, or substitute \( y = mx + 2 \) into the circle equation. M1: Apply tangency condition (perpendicular distance equals radius, or discriminant of quadratic equals zero). A1: Formulate a correct quadratic in \( m \), e.g., \( 21m^2 - 20m = 0 \) or \( 84m^2 - 80m = 0 \). A1: Find both correct values of \( m \).

We integrate the derivative to find \( y \): \( y = \int 6(3x+4)^{-\frac{1}{2}} \text{d}x \). Using the reverse chain rule: \( y = \frac{6(3x+4)^{\frac{1}{2}}}{\frac{1}{2} \times 3} + C = 4\sqrt{3x+4} + C \). To find the constant of integration \( C \), we use the coordinates of the point \( (4, 10) \): \( 10 = 4\sqrt{3(4)+4} + C \implies 10 = 4\sqrt{16} + C \implies 10 = 16 + C \implies C = -6 \). Therefore, the equation of the curve is \( y = 4\sqrt{3x+4} - 6 \).

M1: Attempt integration of \( 6(3x+4)^{-\frac{1}{2}} \) resulting in a form \( k(3x+4)^{\frac{1}{2}} \). A1: Obtain the correct integrated expression \( 4\sqrt{3x+4} \). M1: Substitute \( x = 4, y = 10 \) into their integrated expression containing a constant \( C \) to find \( C \). A1: Obtain the correct final equation \( y = 4\sqrt{3x+4} - 6 \).

We use the identity \( \tan\theta = \frac{\sin\theta}{\cos\theta} \) to rewrite the equation: \( 3\sin\theta \left(\frac{\sin\theta}{\cos\theta}\right) = 8 \implies 3\sin^2\theta = 8\cos\theta \). Using \( \sin^2\theta = 1 - \cos^2\theta \): \( 3(1-\cos^2\theta) = 8\cos\theta \implies 3\cos^2\theta + 8\cos\theta - 3 = 0 \). Factorising the quadratic equation in terms of \( \cos\theta \): \( (3\cos\theta - 1)(\cos\theta + 3) = 0 \). Since \( \cos\theta \in [-1, 1] \), the factor \( \cos\theta + 3 = 0 \) has no real solutions. Thus, \( \cos\theta = \frac{1}{3} \). Finding the solutions in the interval \( 0 \le \theta \le 2\pi \): \( \theta = \arccos\left(\frac{1}{3}\right) \approx 1.2309 \approx 1.23 \) and \( \theta = 2\pi - 1.2309 \approx 5.0522 \approx 5.05 \).

M1: Substitute \( \tan\theta = \frac{\sin\theta}{\cos\theta} \) and \( \sin^2\theta = 1 - \cos^2\theta \) to obtain a quadratic equation in \( \cos\theta \). A1: Obtain the correct quadratic equation \( 3\cos^2\theta + 8\cos\theta - 3 = 0 \) and find \( \cos\theta = \frac{1}{3} \). A1: Find one correct solution \( \theta \approx 1.23 \). A1: Find the second correct solution \( \theta \approx 5.05 \) and no other solutions in range.

For part (a): The first three terms of the geometric progression are \(a_1 = a\), \(a_2 = a + 2d\), and \(a_3 = a + 10d\). Since these terms form a geometric progression, the common ratio is constant, giving: \(\frac{a+2d}{a} = \frac{a+10d}{a+2d}\) which simplifies to \((a+2d)^2 = a(a+10d)\). Expanding both sides: \(a^2 + 4ad + 4d^2 = a^2 + 10ad \implies 4d^2 = 6ad \implies 2d(2d-3a) = 0\). Since \(r \ne 1\), the common difference \(d\) cannot be 0. Thus, we must have \(2d = 3a\), which simplifies to \(d = 1.5a\).

For part (b): The sum of the first 10 terms of the arithmetic progression is given by \(S_{10} = \frac{10}{2}(2a + 9d) = 155\), which simplifies to \(5(2a + 9d) = 155 \implies 2a + 9d = 31\). Substituting \(d = 1.5a\) into this equation: \(2a + 9(1.5a) = 31 \implies 2a + 13.5a = 31 \implies 15.5a = 31 \implies a = 2\). Substituting back to find \(d\): \(d = 1.5(2) = 3\).

Part (a):
M1: Write expressions for the 1st, 3rd, and 11th terms in terms of \(a\) and \(d\) and set up the geometric progression equation \((a+2d)^2 = a(a+10d)\).
A1: Correctly expand and simplify to \(4d^2 = 6ad\) or equivalent.
A1: Complete proof showing \(d = 1.5a\) clearly stating \(d \ne 0\).

Part (b):
M1: Set up the AP sum formula equation: \(5(2a + 9d) = 155\).
M1: Substitute \(d = 1.5a\) into their AP sum equation to solve for \(a\) or \(d\).
A1: Obtain both correct values: \(a = 2\) and \(d = 3\).

For part (a): Equating the line and the curve: \(mx + 4 = x^2 - 2x + 8 \implies x^2 - (2 + m)x + 4 = 0\). For the line to be a tangent to the curve, the discriminant of this quadratic equation must be zero: \(b^2 - 4ac = 0 \implies (-(2+m))^2 - 4(1)(4) = 0 \implies (2+m)^2 = 16\). Taking square roots: \(2+m = \pm 4\), which gives \(m = 2\) or \(m = -6\).

For part (b): Using the positive value, \(m = 2\). The quadratic equation from part (a) becomes: \(x^2 - (2+2)x + 4 = 0 \implies x^2 - 4x + 4 = 0 \implies (x-2)^2 = 0 \implies x = 2\). Substitute \(x = 2\) back into the line equation: \(y = 2(2) + 4 = 8\). Therefore, the coordinates of the point of contact are \((2, 8)\).

Part (a):
M1: Equate the line and curve and collect terms into a three-term quadratic.
M1: Use the discriminant condition \(b^2 - 4ac = 0\).
A1: Find both correct values \(m = 2\) and \(m = -6\).

Part (b):
M1: Substitute \(m = 2\) back into their quadratic equation.
A1: Solve for \(x\) to obtain \(x = 2\).
A1: Find the corresponding \(y\)-coordinate and state the point \((2, 8)\).

For part (a): Using the trigonometric identity \(\sin^2 x = 1 - \cos^2 x\), substitute this into the given equation: \(4(1 - \cos^2 x) - 5\cos x - 5 = 0 \implies 4 - 4\cos^2 x - 5\cos x - 5 = 0 \implies -4\cos^2 x - 5\cos x - 1 = 0\). Multiplying the entire equation by \(-1\) yields the required form: \(4\cos^2 x + 5\cos x + 1 = 0\).

For part (b): Let \(u = \cos x\). The quadratic equation becomes \(4u^2 + 5u + 1 = 0\). Factoring this gives: \((4u+1)(u+1) = 0 \implies u = -0.25\) or \(u = -1\). Therefore, \(cos x = -0.25\) or \(\cos x = -1\). If \(\cos x = -1\): \(x = 180^\circ\). If \(\cos x = -0.25\): The basic angle is \(\alpha = \cos^{-1}(0.25) \approx 75.52^\circ\). Since cosine is negative, the solutions lie in the second and third quadrants: \(x = 180^\circ - 75.52^\circ = 104.5^\circ\) (to 1 d.p.) and \(x = 180^\circ + 75.52^\circ = 255.5^\circ\) (to 1 d.p.).

Part (a):
M1: Apply the identity \(\sin^2 x = 1 - \cos^2 x\) correctly.
A1: Show all algebraic steps clearly to arrive at the given target equation.

Part (b):
M1: Factorise or use the quadratic formula to solve for \(\cos x\).
A1: Identify the roots \(\cos x = -0.25\) and \(\cos x = -1\).
A1: Obtain \(180^\circ\) and at least one other correct angle.
A1: Obtain both \(104.5^\circ\) and \(255.5^\circ\) (and no extra solutions inside the range).

For part (a): We differentiate the equation of the curve with respect to \(x\) using the chain rule: \(\frac{\text{d}y}{\text{d}x} = 3(2x-5)^2 \cdot 2 - 24 = 6(2x-5)^2 - 24\). To find the stationary points, we set \(\frac{\text{d}y}{\text{d}x} = 0 \implies 6(2x-5)^2 = 24 \implies (2x-5)^2 = 4\). Taking square roots: \(2x-5 = 2 \implies 2x = 7 \implies x = 3.5\) or \(2x-5 = -2 \implies 2x = 3 \implies x = 1.5\).

For part (b): To find the \(y\)-coordinates, substitute the \(x\)-values back into the original curve equation: For \(x = 1.5\): \(y = (2(1.5)-5)^3 - 24(1.5) = (-3)^3 - 36 = -8 - 36 = -44\). For \(x = 3.5\): \(y = (2(3.5)-5)^3 - 24(3.5) = (2)^3 - 84 = 8 - 84 = -76\). To determine the nature, find the second derivative: \(\frac{\text{d}^2y}{\text{d}x^2} = 12(2x-5) \cdot 2 = 24(2x-5)\). At \(x = 1.5\): \(\frac{\text{d}^2y}{\text{d}x^2} = 24(3-5) = -48 < 0\), which is a maximum. At \(x = 3.5\): \(\frac{\text{d}^2y}{\text{d}x^2} = 24(7-5) = 48 > 0\), which is a minimum.

Part (a):
M1: Attempt differentiation using chain rule to get \(k(2x-5)^2 - 24\).
A1: Correct first derivative \(6(2x-5)^2 - 24\).
A1: Set derivative to zero and solve to obtain \(x = 1.5\) and \(x = 3.5\).

Part (b):
M1: Substitute both \(x\)-coordinates into the original equation to find \(y\)-coordinates.
A1: Correct coordinates: \((1.5, -44)\) and \((3.5, -76)\).
A1: Evaluate the second derivative (or use gradient test) to correctly identify \((1.5, -44)\) as a maximum and \((3.5, -76)\) as a minimum.

For part (a): To find the equation of the curve, integrate \(\frac{\text{d}y}{\text{d}x}\): \(y = \int 6(3x+4)^{-1/2} \text{d}x\). Using the reverse chain rule: \(y = 6 \cdot \frac{(3x+4)^{1/2}}{(1/2) \cdot 3} + C = 4(3x+4)^{1/2} + C\). Substitute the point \((4, 15)\) to find \(C\): \(15 = 4\sqrt{3(4)+4} + C \implies 15 = 4(4) + C \implies 15 = 16 + C \implies C = -1\). The equation of the curve is \(y = 4\sqrt{3x+4} - 1\).

For part (b): At \(x = 7\), find the \(y\)-coordinate: \(y = 4\sqrt{3(7)+4} - 1 = 4(5) - 1 = 19\). Find the gradient of the tangent at \(x = 7\): \(m = \frac{\text{d}y}{\text{d}x}\Big|_{x=7} = \frac{6}{\sqrt{3(7)+4}} = \frac{6}{5} = 1.2\). The equation of the tangent is: \(y - 19 = 1.2(x - 7) \implies y = 1.2x - 8.4 + 19 \implies y = 1.2x + 10.6\) (or \(5y = 6x + 53\)).

Part (a):
M1: Attempt to integrate \(6(3x+4)^{-1/2}\) to get \(k(3x+4)^{1/2}\).
A1: Correctly integrated term \(4(3x+4)^{1/2}\).
M1: Substitute the coordinates \((4, 15)\) to evaluate the constant of integration \(C\).
A1: Obtain the correct equation \(y = 4\sqrt{3x+4} - 1\).

Part (b):
M1: Find both the correct \(y\)-coordinate (19) and gradient (1.2) at \(x=7\), and attempt the straight line equation.
A1: Obtain the correct equation of the tangent, e.g., \(y = 1.2x + 10.6\) or \(5y = 6x + 53\).

(a) Write down expressions for the terms of the arithmetic progression (AP) and geometric progression (GP):

For the AP:
\( u_2 = a + d = 4 + d \)
\( u_5 = a + 4d = 4 + 4d \)

For the GP:
\( v_2 = ar = 4r \)
\( v_3 = ar^2 = 4r^2 \)

Set up the equations given in the question:
1) \( 4 + d = 4r \implies d = 4r - 4 \)
2) \( 4 + 4d = 4r^2 \implies 1 + d = r^2 \)

Substitute \( d = 4r - 4 \) into the second equation:
\( 1 + (4r - 4) = r^2 \)
\( r^2 - 4r + 3 = 0 \)
\( (r - 1)(r - 3) = 0 \)

This gives \( r = 1 \) or \( r = 3 \).

If \( r = 1 \), then \( d = 4(1) - 4 = 0 \). But we are given that \( d \ne 0 \), so we reject this solution.

Therefore, \( r = 3 \).
Using this value to find \( d \):
\( d = 4(3) - 4 = 8 \).

(b)(i) Use the sum formula for an AP, \( S_n = \frac{n}{2}[2a + (n-1)d] \):
\( S_{20} = \frac{20}{2}[2(4) + (20-1)8] \)
\( S_{20} = 10[8 + 19(8)] \)
\( S_{20} = 10[8 + 152] = 10(160) = 1600 \).

(b)(ii) Use the sum formula for a GP, \( S_n = \frac{a(r^n - 1)}{r - 1} \):
\( S_n = \frac{4(3^n - 1)}{3 - 1} = 2(3^n - 1) \)

We require:
\( 2(3^n - 1) > 10^6 \)
\( 3^n - 1 > 500000 \)
\( 3^n > 500001 \)

Taking logarithms on both sides:
\( n \ln(3) > \ln(500001) \)
\( n > \frac{\ln(500001)}{\ln(3)} \approx 11.94 \)

Since \( n \) must be an integer, the least value of \( n \) is 12.

(a)
- M1: Attempt to write down equations for \( u_2 = v_2 \) and \( u_5 = v_3 \) in terms of \( d \) and \( r \).
- A1: Obtain correct equations: e.g., \( 4 + d = 4r \) and \( 4 + 4d = 4r^2 \).
- M1: Attempt to solve the simultaneous equations by eliminating one variable to form a quadratic in \( r \) or \( d \).
- A1: Obtain \( r = 3 \) (clearly rejecting \( r = 1 \) due to \( d \ne 0 \)).
- A1: Obtain \( d = 8 \).

(b)(i)
- M1: Correct use of the AP sum formula with \( a = 4 \), \( d = 8 \), and \( n = 20 \).
- A1: Obtain 1600.

(b)(ii)
- M1: Correct use of the GP sum formula and setting up the inequality \( S_n > 10^6 \).
- M1: Attempt to solve the inequality using logarithms.
- A1: Obtain \( n = 12 \) (must be an integer).

(a) First find the derivative of \( y \) with respect to \( x \):
\( \frac{dy}{dx} = 6x^2 - 18x + 12 \)

At stationary points, \( \frac{dy}{dx} = 0 \):
\( 6x^2 - 18x + 12 = 0 \)
\( 6(x^2 - 3x + 2) = 0 \)
\( 6(x - 1)(x - 2) = 0 \)

This gives \( x = 1 \) and \( x = 2 \).

Find the corresponding \( y \)-coordinates:
- For \( x = 1 \):
\( y = 2(1)^3 - 9(1)^2 + 12(1) - 3 = 2 - 9 + 12 - 3 = 2 \)
So, one stationary point is \( (1, 2) \).

- For \( x = 2 \):
\( y = 2(2)^3 - 9(2)^2 + 12(2) - 3 = 16 - 36 + 24 - 3 = 1 \)
So, the second stationary point is \( (2, 1) \).

To determine the nature, find the second derivative:
\( \frac{d^2y}{dx^2} = 12x - 18 \)

- At \( x = 1 \):
\( \frac{d^2y}{dx^2} = 12(1) - 18 = -6 < 0 \)
Since the second derivative is negative, \( (1, 2) \) is a local maximum.

- At \( x = 2 \):
\( \frac{d^2y}{dx^2} = 12(2) - 18 = 6 > 0 \)
Since the second derivative is positive, \( (2, 1) \) is a local minimum.

(b) First find the coordinate and gradient at \( x = 3 \):
- For \( x = 3 \):
\( y = 2(3)^3 - 9(3)^2 + 12(3) - 3 = 54 - 81 + 36 - 3 = 6 \)
So the point on the curve is \( (3, 6) \).

- Gradient of the tangent at \( x = 3 \):
\( m_t = 6(3)^2 - 18(3) + 12 = 54 - 54 + 12 = 12 \)

- Gradient of the normal at \( x = 3 \):
\( m_n = -\frac{1}{m_t} = -\frac{1}{12} \)

Using the point-slope form, the equation of the normal is:
\( y - 6 = -\frac{1}{12}(x - 3) \)

Multiply the entire equation by 12 to eliminate fractions:
\( 12y - 72 = -(x - 3) \)
\( 12y - 72 = -x + 3 \)
\( 12y + x = 75 \)

(a)
- M1: Differentiate to find \( \frac{dy}{dx} \) and set equal to 0.
- A1: Find both critical values \( x = 1 \) and \( x = 2 \).
- A1: Obtain correct \( y \)-coordinates: \( (1, 2) \) and \( (2, 1) \).
- M1: Use a valid method (such as second derivative) to determine nature.
- A1: Correctly identify \( (1, 2) \) as a local maximum and \( (2, 1) \) as a local minimum.

(b)
- B1: Calculate the correct coordinates of the point at \( x = 3 \), which is \( (3, 6) \).
- M1: Substitute \( x = 3 \) into their \( \frac{dy}{dx} \) to find the tangent gradient.
- M1: Use the negative reciprocal relationship to find the normal gradient and attempt to form the equation of the line.
- A1: Obtain \( m_n = -\frac{1}{12} \).
- A1: Express final answer in the required integer form: \( 12y + x = 75 \) (or equivalent with integer coefficients, e.g. \( x + 12y = 75 \)).

(a) To find the equation of the curve, integrate the gradient function:
\( y = \int \left( 8 - 6(2x-3)^{1/2} \right) dx \)

Integrate term by term, using the reverse chain rule for the linear composite term:
\( y = 8x - \frac{6(2x-3)^{3/2}}{\frac{3}{2} \times 2} + C \)
\( y = 8x - 2(2x-3)^{3/2} + C \)

Substitute the point \( (2, 10) \) to determine the constant of integration \( C \):
\( 10 = 8(2) - 2(2(2)-3)^{3/2} + C \)
\( 10 = 16 - 2(1)^{3/2} + C \)
\( 10 = 14 + C \implies C = -4 \)

Thus, the equation of the curve is:
\( y = 8x - 2(2x-3)^{3/2} - 4 \)

(b) Now, find the exact value of the definite integral:
\( \int_{2}^{6} \left( 8x - 2(2x-3)^{3/2} - 4 \right) dx \)

Integrate each term:
\( \int \left( 8x - 2(2x-3)^{3/2} - 4 \right) dx = \left[ 4x^2 - \frac{2(2x-3)^{5/2}}{\frac{5}{2} \times 2} - 4x \right] \)
\( = \left[ 4x^2 - \frac{2}{5}(2x-3)^{5/2} - 4x \right]_{2}^{6} \)

Substitute the upper limit \( x = 6 \):
\( \left( 4(6)^2 - \frac{2}{5}(2(6)-3)^{5/2} - 4(6) \right) = 144 - \frac{2}{5}(9)^{5/2} - 24 \)
Since \( 9^{5/2} = (\sqrt{9})^5 = 3^5 = 243 \):
\( = 120 - \frac{2}{5}(243) = 120 - 97.2 = 22.8 \)

Substitute the lower limit \( x = 2 \):
\( \left( 4(2)^2 - \frac{2}{5}(2(2)-3)^{5/2} - 4(2) \right) = 16 - \frac{2}{5}(1)^{5/2} - 8 \)
\( = 8 - 0.4 = 7.6 \)

Subtract the lower limit evaluation from the upper limit evaluation:
\( 22.8 - 7.6 = 15.2 \) (or \( \frac{76}{5} \)).

(a)
- M1: Attempt to integrate \( \frac{dy}{dx} \) with at least one term integrated correctly.
- A1: Obtain correct integration of the composite term: \( -2(2x-3)^{3/2} \).
- M1: Substitute \( x = 2 \) and \( y = 10 \) into an expression with a constant of integration \( C \).
- A1: Obtain correct final equation: \( y = 8x - 2(2x-3)^{3/2} - 4 \).

(b)
- M1: Attempt to integrate their \( y \) equation term by term.
- A2: Obtain correct integrated expression: \( 4x^2 - \frac{2}{5}(2x-3)^{5/2} - 4x \) (A1 for two terms correct, A1 for the remaining term correct).
- M1: Show correct substitution of limits \( x = 6 \) and \( x = 2 \) into their integrated expression.
- A1: Obtain final exact value of \( 15.2 \) (or \( \frac{76}{5} \) or \( 15\frac{1}{5} \)).