2023 Cambridge IGCSE Mathematics - Additional (0606) Practice Paper with Answers

Since the vectors are parallel, one is a scalar multiple of the other, which means the ratio of their components must be equal: \(\frac{2}{k-1} = \frac{k}{6}\). Cross-multiplying gives: \(12 = k(k-1)\) which simplifies to the quadratic equation: \(k^2 - k - 12 = 0\). Factoring the quadratic, we get: \((k-4)(k+3) = 0\). This gives the possible values of \(k\) as \(k = 4\) or \(k = -3\).

M1: For setting up a correct relation representing the parallel vectors, e.g., \(\frac{2}{k-1} = \frac{k}{6}\).
M1: For expanding and forming the quadratic equation \(k^2 - k - 12 = 0\).
A1: For both correct values \(k = 4\) and \(k = -3\).

By the Factor Theorem, since \(x-2\) is a factor, we have \(p(2) = 0\): \(2(2)^3 + a(2)^2 + b(2) - 6 = 0 \implies 16 + 4a + 2b - 6 = 0 \implies 2a + b = -5\) (Equation 1). By the Remainder Theorem, since dividing by \(x+1\) gives a remainder of \(-9\), we have \(p(-1) = -9\): \(2(-1)^3 + a(-1)^2 + b(-1) - 6 = -9 \implies -2 + a - b - 6 = -9 \implies a - b = -1\) (Equation 2). Adding Equation 1 and Equation 2 gives \(3a = -6 \implies a = -2\). Substituting \(a = -2\) into Equation 2 gives \(-2 - b = -1 \implies b = -1\).

M1: For applying the Factor Theorem with \(p(2) = 0\) to obtain a linear equation in \(a\) and \(b\) (e.g., \(2a + b = -5\)).
M1: For applying the Remainder Theorem with \(p(-1) = -9\) to obtain a second linear equation in \(a\) and \(b\) (e.g., \(a - b = -1\)).
A1: For solving the simultaneous equations to get both \(a = -2\) and \(b = -1\).

The perimeter of the sector is given by \(P = 2r + r\theta = 30 \implies r\theta = 30 - 2r\). The area of the sector is given by \(A = \frac{1}{2}r^2\theta = 50\). Substituting the expression for \(r\theta\) into the area formula gives: \ \frac{1}{2}r(r\theta) = 50 \implies \frac{1}{2}r(30 - 2r) = 50 \implies 15r - r^2 = 50\). Rearranging this yields the quadratic equation: \(r^2 - 15r + 50 = 0\). Factoring the quadratic gives: \((r-5)(r-10) = 0\). Thus, the possible values of \(r\) are \(r = 5\) or \(r = 10\).

M1: For using sector formulas to write the system of equations: \(2r + r\theta = 30\) and \(\frac{1}{2}r^2\theta = 50\).
M1: For substituting one equation into the other to eliminate \(\theta\) and forming a quadratic equation in \(r\), e.g., \(r^2 - 15r + 50 = 0\).
A1: For obtaining both correct values \(r = 5\) and \(r = 10\).

First, find the \(y\)-coordinate of the point of contact: when \(x = -2\), \(y = \frac{2(-2) - 1}{-2 + 3} = -5\). Next, differentiate the curve using the quotient rule: \(\frac{dy}{dx} = \frac{(x+3)(2) - (2x-1)(1)}{(x+3)^2} = \frac{2x+6-2x+1}{(x+3)^2} = \frac{7}{(x+3)^2}\). At \(x = -2\), the gradient of the tangent is \(m_{\text{tangent}} = \frac{7}{(-2+3)^2} = 7\). The gradient of the normal is the negative reciprocal of the tangent gradient: \(m_{\text{normal}} = -\frac{1}{7}\). Using the point-gradient formula with point \((-2, -5)\), the equation of the normal is: \(y - (-5) = -\frac{1}{7}(x - (-2)) \implies y + 5 = -\frac{1}{7}x - \frac{2}{7} \implies y = -\frac{1}{7}x - \frac{37}{7}\), which can also be written as \(x + 7y + 37 = 0\).

M1: For differentiating \(y = \frac{2x - 1}{x + 3}\) using the quotient rule to find \(\frac{dy}{dx} = \frac{7}{(x+3)^2}\).
M1: For finding the \(y\)-coordinate \((-5)\), the tangent gradient \(7\), and the corresponding normal gradient \(-\frac{1}{7}\).
A1: For a correct equation of the normal in any equivalent form, such as \(y = -\frac{1}{7}x - \frac{37}{7}\) or \(x + 7y + 37 = 0\).

First, find the y-coordinate at \( x = 0 \): \( y = (3(0) - 2)\ln(2(0) + 1) = -2\ln(1) = 0 \). The point is \( (0, 0) \). Now differentiate \( y = (3x - 2)\ln(2x + 1) \) using the product rule: \( \frac{dy}{dx} = 3\ln(2x + 1) + (3x - 2)\frac{2}{2x + 1} \). Substitute \( x = 0 \) to find the gradient of the tangent: \( m_t = 3\ln(1) + (3(0) - 2)\frac{2}{2(0) + 1} = -4 \). The gradient of the normal, \( m_n \), is perpendicular to the tangent, so \( m_n = -\frac{1}{-4} = \frac{1}{4} \). The equation of the normal is \( y - 0 = \frac{1}{4}(x - 0) \), which simplifies to \( x - 4y = 0 \).

M1: For attempt to differentiate using the product rule. A1: For correct derivative \( \frac{dy}{dx} = 3\ln(2x + 1) + \frac{2(3x - 2)}{2x + 1} \). M1: For substituting \( x = 0 \) to find the gradient of the tangent. M1: For finding the gradient of the normal. M1: For using the point \( (0, 0) \) and their normal gradient to form an equation of a straight line. A1: For correct final equation in the required form: \( x - 4y = 0 \) or any integer multiple (e.g., \( 4y - x = 0 \)).

For a geometric progression, the common ratio \( r \) is constant: \( \frac{k - 4}{k + 2} = \frac{k - 8}{k - 4} \). Cross-multiplying gives \( (k - 4)^2 = (k + 2)(k - 8) \). Expanding both sides: \( k^2 - 8k + 16 = k^2 - 6k - 16 \). Simplifying, we get: \( -8k + 16 = -6k - 16 \implies 2k = 32 \implies k = 16 \). Using \( k = 16 \), the first term is \( a = 16 + 2 = 18 \) and the second term is \( 16 - 4 = 12 \). The common ratio is \( r = \frac{12}{18} = \frac{2}{3} \). Since \( |r| < 1 \), the sum to infinity exists and is given by \( S_{\infty} = \frac{a}{1 - r} = \frac{18}{1 - 2/3} = \frac{18}{1/3} = 54 \).

M1: For setting up the ratio equation \( \frac{k - 4}{k + 2} = \frac{k - 8}{k - 4} \). M1: For correct algebraic expansion to a linear equation: \( k^2 - 8k + 16 = k^2 - 6k - 16 \). A1: For finding \( k = 16 \). M1: For calculating the first term \( a = 18 \) and the common ratio \( r = \frac{2}{3} \). M1: For using the sum to infinity formula \( S_{\infty} = \frac{a}{1 - r} \). A1: For \( 54 \).

To find the center and radius, we complete the square for the circle's equation: \( (x - 3)^2 - 9 + (y - 2)^2 - 4 - 12 = 0 \implies (x - 3)^2 + (y - 2)^2 = 25 \). Thus, the center is \( (3, 2) \) and the radius is \( \sqrt{25} = 5 \). The gradient of the radius from the center \( (3, 2) \) to the point \( (6, 6) \) is \( m_r = \frac{6 - 2}{6 - 3} = \frac{4}{3} \). The tangent is perpendicular to the radius, so its gradient is \( m_t = -\frac{3}{4} \). Using the point-slope form with \( (6, 6) \): \( y - 6 = -\frac{3}{4}(x - 6) \implies 4y - 24 = -3x + 18 \implies 3x + 4y = 42 \).

M1: For attempting to complete the square. A1: For correct center \( (3, 2) \) and radius \( 5 \). M1: For finding the gradient of the radius \( \frac{4}{3} \). M1: For finding the gradient of the tangent \( -\frac{3}{4} \) (negative reciprocal). M1: For substituting \( (6, 6) \) and their tangent gradient into a line equation. A1: For correct tangent equation, e.g., \( 3x + 4y = 42 \) or equivalent.

Use the identity \( \sin^2 \theta = 1 - \cos^2 \theta \) to rewrite the equation in terms of \( \cos \theta \): \( 3(1 - \cos^2 \theta) + 5\cos \theta - 5 = 0 \implies 3 - 3\cos^2 \theta + 5\cos \theta - 5 = 0 \implies 3\cos^2 \theta - 5\cos \theta + 2 = 0 \). Factorizing the quadratic equation: \( (3\cos\theta - 2)(\cos\theta - 1) = 0 \). This yields two cases: 1) \( \cos\theta = 1 \implies \theta = 0^{\circ}, 360^{\circ} \). 2) \( \cos\theta = \frac{2}{3} \implies \theta = \cos^{-1}(\frac{2}{3}) \approx 48.2^{\circ} \). The other solution in the range \( 0^{\circ} \le \theta \le 360^{\circ} \) is \( 360^{\circ} - 48.2^{\circ} = 311.8^{\circ} \). Thus, the complete set of solutions is \( \theta = 0^{\circ}, 48.2^{\circ}, 311.8^{\circ}, 360^{\circ} \).

M1: For using the identity \( \sin^2 \theta = 1 - \cos^2 \theta \). A1: For forming the correct quadratic \( 3\cos^2 \theta - 5\cos \theta + 2 = 0 \). M1: For solving the quadratic to get \( \cos\theta = \frac{2}{3} \) and \( \cos\theta = 1 \). A1: For \( \theta = 0^{\circ}, 360^{\circ} \). A1: For \( 48.2^{\circ} \) (accept 48.19). A1: For \( 311.8^{\circ} \) (accept 311.81).

From the first equation, rewrite the logarithm in exponential form: \( 2x - y = 3^2 \implies 2x - y = 9 \). From the second equation, express both sides with base 2: \( (2^2)^x \times 2^y = 2^{11} \implies 2^{2x + y} = 2^{11} \implies 2x + y = 11 \). Now we have a system of linear equations: 1) \( 2x - y = 9 \), 2) \( 2x + y = 11 \). Adding the two equations: \( 4x = 20 \implies x = 5 \). Substituting \( x = 5 \) into the second equation: \( 2(5) + y = 11 \implies y = 1 \). Since \( 2(5) - 1 = 9 > 0 \), the logarithmic term is valid. Thus, the solution is \( x = 5, y = 1 \).

M1: For converting \( \log_3(2x - y) = 2 \) to \( 2x - y = 9 \). M1: For expressing terms in \( 4^x \times 2^y = 2048 \) as powers of 2. A1: For obtaining \( 2x + y = 11 \). M1: For attempting to solve the simultaneous equations. A1: For \( x = 5 \). A1: For \( y = 1 \).

(i) Using the product rule on \(y = (2x - 5)(2x+1)^{1/2}\):
Let \(u = 2x - 5 \Rightarrow \frac{\text{d}u}{\text{d}x} = 2\)
Let \(v = (2x+1)^{1/2} \Rightarrow \frac{\text{d}v}{\text{d}x} = \frac{1}{2}(2x+1)^{-1/2} \cdot 2 = (2x+1)^{-1/2}\)

\(\frac{\text{d}y}{\text{d}x} = u \frac{\text{d}v}{\text{d}x} + v \frac{\text{d}u}{\text{d}x}\)
\(= (2x - 5)(2x+1)^{-1/2} + 2(2x+1)^{1/2}\)
\(= \frac{2x - 5}{\sqrt{2x+1}} + 2\sqrt{2x+1}\)
\(= \frac{(2x - 5) + 2(2x+1)}{\sqrt{2x+1}}\)
\(= \frac{6x - 3}{\sqrt{2x+1}}\) as required.

(ii) At the stationary point, \(\frac{\text{d}y}{\text{d}x} = 0\):
\(6x - 3 = 0 \Rightarrow x = 0.5\)
Substitute \(x = 0.5\) into the equation of the curve:
\(y = (2(0.5) - 5)\sqrt{2(0.5)+1} = -4\sqrt{2}\)
So the exact coordinates are \((0.5, -4\sqrt{2})\).

(iii) At \(x = 4\):
\(y = (2(4) - 5)\sqrt{2(4)+1} = 3(3) = 9\)
Gradient of the tangent, \(m_T = \frac{6(4)-3}{\sqrt{2(4)+1}} = \frac{21}{3} = 7\)
Gradient of the normal, \(m_N = -\frac{1}{7}\)
Equation of the normal:
\(y - 9 = -\frac{1}{7}(x - 4)\)
\(7(y - 9) = -x + 4\)
\(x + 7y = 67\)

(i)
M1: For applying product rule or quotient rule correctly.
A1: For correct derivatives of individual components: \(2\) and \((2x+1)^{-1/2}\).
M1: For putting over a common denominator and simplifying.
A1: For fully correct proof showing all steps.

(ii)
M1: For setting their derivative to 0 and solving for \(x\).
A1: For \(x = 0.5\) or equivalent.
A0.5: For exact \(y = -4\sqrt{2}\).

(iii)
M1: For finding \(y = 9\) and substituting \(x = 4\) into \(\frac{\text{d}y}{\text{d}x}\) to find tangent gradient 7.
M1: For using negative reciprocal to find normal gradient and applying straight line formula.
A1: For correct final equation in the required form: \(x + 7y = 67\).

(i) Terms of AP: \(u_1 = a\), \(u_3 = a + 2d\), \(u_7 = a + 6d\).
Terms of GP: \(v_1 = a\), \(v_2 = ar\), \(v_3 = ar^2\).
Given \(u_1 = v_1 \Rightarrow a = a\).
\(u_3 = v_2 \Rightarrow a + 2d = ar\) [Equation 1]
\(u_7 = v_3 \Rightarrow a + 6d = ar^2\) [Equation 2]
From Equation 1, \(r = \frac{a+2d}{a} = 1 + \frac{2d}{a}\).
Substitute into Equation 2:
\(a + 6d = a\left(1 + \frac{2d}{a}\right)^2\)
\(a + 6d = a\left(1 + \frac{4d}{a} + \frac{4d^2}{a^2}\right)\)
\(a + 6d = a + 4d + \frac{4d^2}{a}\)
\(2d = \frac{4d^2}{a}\)
Since \(d \neq 0\), we can divide by \(2d\):
\(1 = \frac{2d}{a} \Rightarrow a = 2d\) as required.

(ii) The sum of the first 12 terms of the AP is:
\(S_{12} = \frac{12}{2}(2a + 11d) = 360\)
\(6(2a + 11d) = 360\)
\(2a + 11d = 60\)
Substitute \(a = 2d\):
\(2(2d) + 11d = 60 \Rightarrow 15d = 60 \Rightarrow d = 4\)
Then \(a = 2(4) = 8\).

(iii) For the GP:
First term is \(a = 8\).
Common ratio \(r = \frac{a+2d}{a} = \frac{8+2(4)}{8} = 2\).
Sum of the first 6 terms of the GP:
\(S_6 = \frac{a(r^6 - 1)}{r - 1} = \frac{8(2^6 - 1)}{2 - 1} = 8(64 - 1) = 8(63) = 504\)

(i)
M1: For writing down expressions for the terms of the AP and GP.
M1: For equating ratios to eliminate \(r\), e.g., \(\frac{a+6d}{a+2d} = \frac{a+2d}{a}\).
M1: For expanding and simplifying the algebraic expression.
A1: For arriving correctly at \(a = 2d\).

(ii)
M1: For using the sum formula of AP to set up an equation: \(6(2a+11d) = 360\).
M1: For substituting \(a=2d\) (or equivalent) and solving for one variable.
A0.5: For both \(a=8\) and \(d=4\).

(iii)
M1: For finding the common ratio \(r = 2\).
M1: For using the sum formula of a GP with \(n = 6\).
A1: For correct final answer of 504.

(i) Left-Hand Side (LHS):
\(\frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta}\)
\(= \frac{\sin^2 \theta + (1 + \cos \theta)^2}{\sin \theta(1 + \cos \theta)}\)
\(= \frac{\sin^2 \theta + 1 + 2\cos \theta + \cos^2 \theta}{\sin \theta(1 + \cos \theta)}\)
Since \(\sin^2 \theta + \cos^2 \theta = 1\):
\(= \frac{1 + 1 + 2\cos \theta}{\sin \theta(1 + \cos \theta)}\)
\(= \frac{2 + 2\cos \theta}{\sin \theta(1 + \cos \theta)}\)
\(= \frac{2(1 + \cos \theta)}{\sin \theta(1 + \cos \theta)}\)
\(= \frac{2}{\sin \theta} = 2\text{cosec }\theta\) as required.

(ii) Using the proven identity with \(\theta = 2x\), the equation becomes:
\(2\text{cosec } 2x = 5 - \text{cosec } 2x\)
\(3\text{cosec } 2x = 5\)
\(\text{cosec } 2x = \frac{5}{3}\)
\(\sin 2x = \frac{3}{5} = 0.6\)

Since \(0^\circ \le x \le 180^\circ\), the range for \(2x\) is \(0^\circ \le 2x \le 360^\circ\).
Find the principal value:
\(2x = \sin^{-1}(0.6) \approx 36.87^\circ\)
Other angle in the range:
\(2x = 180^\circ - 36.87^\circ = 143.13^\circ\)

Divide by 2 to solve for \(x\):
\(x = 18.4^\circ\) and \(x = 71.6^\circ\) (to 1 d.p.)

(i)
M1: For combining fractions with a common denominator.
M1: For expanding the numerator to \(\sin^2\theta + 1 + 2\cos\theta + \cos^2\theta\).
M1: For using \(\sin^2\theta + \cos^2\theta = 1\) and factoring out 2.
A0.5: For completing proof to get \(2\text{cosec }\theta\).

(ii)
M1: For identifying and substituting the identity to get \(2\text{cosec } 2x = 5 - \text{cosec } 2x\).
M1: For simplifying to \(\sin 2x = 0.6\).
M1: For finding the two correct values of \(2x\) (36.87 and 143.13).
M1: For dividing by 2 to find values of \(x\).
A1: For \(x = 18.4^\circ\) (accept 18.43 or 18.4).
A1: For \(x = 71.6^\circ\) (accept 71.57 or 71.6).

(i) Rearrange the circle's equation by completing the square:
\(x^2 - 6x + y^2 + 4y = 12\)
\((x - 3)^2 - 9 + (y + 2)^2 - 4 = 12\)
\((x - 3)^2 + (y + 2)^2 = 25\)
So, the centre of \(C\) is \((3, -2)\) and the radius is \(\sqrt{25} = 5\).

(ii) Method: Using perpendicular distance
The line is \(mx - y + 3 = 0\).
The perpendicular distance from the centre \((3, -2)\) to this line must equal the radius, 5:
\(\frac{|m(3) - (-2) + 3|}{\sqrt{m^2 + (-1)^2}} = 5\)
\(\frac{|3m + 5|}{\sqrt{m^2 + 1}} = 5\)
\(|3m + 5| = 5\sqrt{m^2 + 1}\)
Square both sides:
\((3m + 5)^2 = 25(m^2 + 1)\)
\(9m^2 + 30m + 25 = 25m^2 + 25\)
\(16m^2 - 30m = 0\)
\(2m(8m - 15) = 0\)
Therefore, \(m = 0\) or \(m = \frac{15}{8}\).

(i)
M1: For attempting to complete the square for both \(x\) and \(y\).
A1: For correct centre \((3, -2)\).
A1: For correct radius \(5\).

(ii)
M1: For setting up the perpendicular distance equation or substituting \(y = mx + 3\) into the circle equation.
M1: For simplifying the expression to a single quadratic equation in terms of \(x\) or clearing the square root.
M1: For applying either squaring both sides or setting discriminant \(b^2 - 4ac = 0\).
M1: For simplifying to a quadratic equation in terms of \(m\), e.g., \(16m^2 - 30m = 0\) or \(64m^2 - 120m = 0\).
A1.5: For solving and obtaining both correct values: \(m = 0\) and \(m = \frac{15}{8}\) (or 1.875).