2023 Cambridge IGCSE Mathematics - Additional (0606) 模擬試題連答案詳解

We can rewrite each term as a logarithm to base 10:

1. \( 4\lg(2x) = \lg((2x)^4) = \lg(16x^4) \)

2. \( \frac{1}{4}\lg(16) = \lg(16^{1/4}) = \lg(2) \)

3. \( 3 = \lg(10^3) = \lg(1000) \)

Combining these using the laws of logarithms:

\( 4\lg(2x) - \frac{1}{4}\lg(16) + 3 = \lg(16x^4) - \lg(2) + \lg(1000) \)

\( = \lg\left( \frac{16x^4}{2} \right) + \lg(1000) \)

\( = \lg(8x^4 \times 1000) = \lg(8000x^4) \)

B1 for \( \lg(16x^4) \) oe
B1 for \( \lg(2) \) and \( \lg(1000) \)
B1 for \( \lg(8000x^4) \)

First, find the vector \( 2\mathbf{a} + \mathbf{b} \):

\( 2\mathbf{a} + \mathbf{b} = 2\begin{pmatrix} 3 \\ -5 \end{pmatrix} + \begin{pmatrix} 2 \\ -5 \end{pmatrix} = \begin{pmatrix} 6 \\ -10 \end{pmatrix} + \begin{pmatrix} 2 \\ -5 \end{pmatrix} = \begin{pmatrix} 8 \\ -15 \end{pmatrix} \)

Next, find the magnitude of this vector:

\( |2\mathbf{a} + \mathbf{b}| = \sqrt{8^2 + (-15)^2} = \sqrt{64 + 225} = \sqrt{289} = 17 \)

Finally, find the unit vector by dividing the vector by its magnitude:

\( \text{Unit vector} = \frac{1}{17}\begin{pmatrix} 8 \\ -15 \end{pmatrix} \)

M1 for attempt to find \( 2\mathbf{a} + \mathbf{b} \) correctly, getting \( \begin{pmatrix} 8 \\ -15 \end{pmatrix} \)
M1 for finding the magnitude of their vector, \( \sqrt{8^2 + (-15)^2} = 17 \)
A1 for \( \frac{1}{17}\begin{pmatrix} 8 \\ -15 \end{pmatrix} \) oe

Integrate each term individually:

1. \( \int 6x^2 \, dx = \frac{6x^3}{3} = 2x^3 \)

2. \( \int -2 \, dx = -2x \)

3. \( \int \frac{5}{5x-2} \, dx = \ln(5x-2) \)

Don't forget the constant of integration \( c \):

\( \int \left( 6x^2 - 2 + \frac{5}{5x-2} \right) dx = 2x^3 - 2x + \ln(5x-2) + c \)

B1 for \( 2x^3 - 2x \)
B1 for \( \ln(5x-2) \) or \( \ln|5x-2| \)
B1 for correct constant of integration \( + c \)

The perimeter \( P \) and area \( A \) of a sector are given by:

\( P = 2r + r\theta = 24 \implies r\theta = 24 - 2r \)

\( A = \frac{1}{2} r^2 \theta = 32 \)

Substitute \( r\theta \) into the area formula:

\( A = \frac{1}{2} r (r\theta) = 32 \)

\( \frac{1}{2} r (24 - 2r) = 32 \)

\( 12r - r^2 = 32 \)

\( r^2 - 12r + 32 = 0 \)

Factor the quadratic equation:

\( (r-4)(r-8) = 0 \)

Thus, the possible values of \( r \) are \( r = 4 \) or \( r = 8 \).

M1 for writing down correct formulas for perimeter and area and substituting to eliminate \( \theta \)
M1 for obtaining the quadratic equation \( r^2 - 12r + 32 = 0 \) oe
A1 for \( r = 4 \) and \( r = 8 \)

(a) At the points of intersection, we equate the two equations:
\(x^2 + (k-2)x + 4 = 2kx - 5\)

Rearranging to make one side zero:
\(x^2 + (k - 2 - 2k)x + (4 + 5) = 0\)

\(x^2 - (k+2)x + 9 = 0\) (as required).

(b) For the line to not intersect the curve, this quadratic equation must have no real roots. Therefore, the discriminant must be less than zero:
\(b^2 - 4ac < 0\)

\([-(k+2)]^2 - 4(1)(9) < 0\)

\((k+2)^2 - 36 < 0\)

\((k+2)^2 < 36\)

\(-6 < k+2 < 6\)

\(-8 < k < 4\)

M1: Equating the line and curve equations
A1: Correctly simplifying to the given quadratic equation
M1: Setting the discriminant of their quadratic equation to be less than zero
M1: Correct expansion/factorisation of the quadratic inequality
A1: Obtaining the critical values -8 and 4
A1: Correct range: -8 < k < 4

(a) Let \(u = 3^x\). The equation becomes:
\(2u^2 - 7u - 4 = 0\)

\((2u + 1)(u - 4) = 0\)

So \(u = -0.5\) or \(u = 4\).

Since \(3^x > 0\) for all real \(x\), \(3^x = -0.5\) has no real solution.

Thus, \(3^x = 4\), which gives \(x = \log_3 4\).

(b) Comparing the equation in (b) with the one in (a), we substitute \(x = y+1\).

Therefore,
\(y + 1 = \log_3 4\)

\(y = \log_3 4 - 1\)

M1: Using an appropriate substitution (e.g., u = 3^x) to form a quadratic equation
A1: Finding the roots of the quadratic equation (u = -0.5 and u = 4)
B1: Rejecting the negative root with a valid reason
A1: Finding the final exact answer x = \log_3 4
M1: Identifying that x = y + 1
A1: Correct final exact answer y = \log_3 4 - 1 (or \log_3(4/3))

(a) Since \((x - 2)\) is a factor of \(P(x)\), by the factor theorem, \(P(2) = 0\):
\(2(2)^3 + a(2)^2 + b(2) - 6 = 0\)

\(16 + 4a + 2b - 6 = 0 \Rightarrow 4a + 2b = -10 \Rightarrow 2a + b = -5\) --- (Equation 1)

By the remainder theorem, when \(P(x)\) is divided by \((x + 1)\), the remainder is \(P(-1) = -12\):
\(2(-1)^3 + a(-1)^2 + b(-1) - 6 = -12\)

\(-2 + a - b - 6 = -12 \Rightarrow a - b = -4\) --- (Equation 2)

From Equation 2, \(b = a + 4\). Substitute into Equation 1:
\(2a + (a + 4) = -5 \Rightarrow 3a = -9 \Rightarrow a = -3\)

Then \(b = -3 + 4 = 1\).

(b) With \(a = -3\) and \(b = 1\), \(P(x) = 2x^3 - 3x^2 + x - 6\).

Dividing \(P(x)\) by \((x - 2)\) using algebraic long division or synthetic division:
\(2x^3 - 3x^2 + x - 6 = (x - 2)(2x^2 + kx + 3)\)

Comparing coefficients of \(x^2\):
\(-4 + k = -3 \Rightarrow k = 1\)

So, \(P(x) = (x - 2)(2x^2 + x + 3)\).

M1: Applying the factor theorem P(2) = 0 to form a linear equation in a and b
M1: Applying the remainder theorem P(-1) = -12 to form a second linear equation in a and b
M1: Correct attempt to solve the simultaneous equations
A1: Obtaining a = -3 and b = 1
M1: Attempt to find the quadratic factor by division or comparing coefficients
A1: Fully correct expression: (x - 2)(2x^2 + x + 3)

(a) Using the quotient rule, where \(u = x^2\) and \(v = 3x - 1\):
\(\frac{du}{dx} = 2x\) and \(\frac{dv}{dx} = 3\)

\(\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\)

\(\frac{dy}{dx} = \frac{(3x - 1)(2x) - (x^2)(3)}{(3x - 1)^2}\)

\(\frac{dy}{dx} = \frac{6x^2 - 2x - 3x^2}{(3x - 1)^2} = \frac{3x^2 - 2x}{(3x - 1)^2}\)

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

\(x(3x - 2) = 0\)

So \(x = 0\) or \(x = \frac{2}{3}\).

When \(x = 0\):
\(y = \frac{0^2}{3(0)-1} = 0\)

Stationary point: \((0, 0)\).

When \(x = \frac{2}{3}\):
\(y = \frac{(2/3)^2}{3(2/3)-1} = \frac{4/9}{2-1} = \frac{4}{9}\)

Stationary point: \((\frac{2}{3}, \frac{4}{9})\).

M1: Applying the quotient rule formula correctly with their derivatives
A1: Correct numerator expansion: (3x-1)(2x) - 3x^2
A1: Correct simplified derivative: (3x^2 - 2x) / (3x-1)^2
M1: Setting their numerator equal to zero and solving the quadratic equation
A1: Correct x-coordinates (x = 0 and x = 2/3)
A1: Correct y-coordinates matching the x-coordinates: (0, 0) and (2/3, 4/9)

(a) The displacement vector \(\vec{AB}\) is given by:
\(\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} 8 \\ 1 \end{pmatrix} - \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}\).

The magnitude of \(\vec{AB}\) is:
\(|\vec{AB}| = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \approx 7.21\).

(b) The displacement vector \(\vec{AC}\) is:
\(\vec{AC} = \vec{OC} - \vec{OA} = \begin{pmatrix} k \\ 5 \end{pmatrix} - \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} k-2 \\ 8 \end{pmatrix}\).

Since \(A\), \(B\), and \(C\) are collinear, \(\vec{AC}\) is a scalar multiple of \(\vec{AB}\):
\(\vec{AC} = \lambda \vec{AB}\)

\(\begin{pmatrix} k-2 \\ 8 \end{pmatrix} = \lambda \begin{pmatrix} 6 \\ 4 \end{pmatrix}\).

From the \(y\)-components:
\(8 = 4\lambda \Rightarrow \lambda = 2\).

From the \(x\)-components:
\(k-2 = 6\lambda \Rightarrow k-2 = 12 \Rightarrow k = 14\).

B1: Finding displacement vector AB = (6, 4)^T (or equivalent column vector)
B1: Calculating magnitude |AB| = sqrt(52) or 2*sqrt(13) (accept awrt 7.21)
M1: Finding displacement vector AC = (k-2, 8)^T
M1: Setting up the collinearity condition AC = lambda * AB
A1: Finding lambda = 2
A1: Correctly solving for k = 14

(a) Given \( y = (3x-1)(2x+3)^3 \).
Using the product rule:
\( u = 3x - 1 \implies u' = 3 \)
\( v = (2x+3)^3 \implies v' = 3(2x+3)^2 \cdot 2 = 6(2x+3)^2 \)

\( \frac{\mathrm{d}y}{\mathrm{d}x} = u'v + uv' = 3(2x+3)^3 + 6(3x-1)(2x+3)^2 \)
Factor out \( 3(2x+3)^2 \):
\( \frac{\mathrm{d}y}{\mathrm{d}x} = 3(2x+3)^2 [ (2x+3) + 2(3x-1) ] \)
\( \frac{\mathrm{d}y}{\mathrm{d}x} = 3(2x+3)^2 [ 2x + 3 + 6x - 2 ] \)
\( \frac{\mathrm{d}y}{\mathrm{d}x} = 3(2x+3)^2 (8x + 1) \)
\( \frac{\mathrm{d}y}{\mathrm{d}x} = (2x+3)^2 (24x + 3) \)
So \( A = 24 \) and \( B = 3 \).

(b) For stationary points, set \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \):
\( (2x+3)^2(24x+3) = 0 \)
This gives \( x = -1.5 \) or \( x = -0.125 \).

When \( x = -1.5 \):
\( y = (3(-1.5) - 1)(2(-1.5) + 3)^3 = 0 \)
Coordinates of the first point: \( (-1.5, 0) \).

When \( x = -0.125 \):
\( y = (3(-0.125) - 1)(2(-0.125) + 3)^3 = (-1.375)(2.75)^3 = -\frac{14641}{512} \approx -28.6 \)
Coordinates of the second point: \( (-0.125, -28.6) \).

(c) The curve crosses the \( y \)-axis at \( x = 0 \).
When \( x = 0 \):
\( y = (3(0) - 1)(2(0) + 3)^3 = -27 \)
Point: \( (0, -27) \).

Gradient of the tangent at \( x = 0 \):
\( m_T = (2(0)+3)^2(24(0)+3) = 9 \times 3 = 27 \)

Gradient of the normal:
\( m_N = -\frac{1}{27} \)

Equation of the normal:
\( y - (-27) = -\frac{1}{27}(x - 0) \)
\( y + 27 = -\frac{1}{27}x \implies x + 27y + 729 = 0 \)

(a)
- M1: Attempt at product rule differentiation resulting in two terms.
- A1: Correct unsimplified derivative: \( 3(2x+3)^3 + 6(3x-1)(2x+3)^2 \).
- M1: Factorising out \( (2x+3)^2 \) or similar.
- A1: Correct final form with \( A=24, B=3 \).

(b)
- M1: Setting their \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \) and solving for at least one value of \( x \).
- A1: Correct coordinates \( (-1.5, 0) \).
- A1: Correct coordinates \( (-0.125, -28.6) \) or \( \left(-\frac{1}{8}, -\frac{14641}{512}\right) \).

(c)
- B1: Coordinates of the point on the \( y \)-axis \( (0, -27) \).
- M1: Substituting \( x = 0 \) into their derivative and finding the negative reciprocal of the gradient.
- A1: Correct equation of the normal in any equivalent linear form.

(a) Using the properties of logarithms:
\( \log_2((x + 3)(x - 3)) = 4 \)
\( (x + 3)(x - 3) = 2^4 \)
\( x^2 - 9 = 16 \)
\( x^2 = 25 \implies x = 5 \text{ or } x = -5 \)

Since we must have \( x + 3 > 0 \) and \( x - 3 > 0 \), we reject \( x = -5 \).
Thus, \( x = 5 \).

(b) From the first equation:
\( 3^x \cdot (3^2)^y = 3^5 \)
\( 3^{x + 2y} = 3^5 \implies x + 2y = 5 \quad \text{--- (1)} \)

From the second equation:
\( 2x - y = 5^1 \implies y = 2x - 5 \quad \text{--- (2)} \)

Substitute (2) into (1):
\( x + 2(2x - 5) = 5 \)
\( x + 4x - 10 = 5 \)
\( 5x = 15 \implies x = 3 \)

Using (2) to find \( y \):
\( y = 2(3) - 5 = 1 \)

Since \( 2x - y = 5 > 0 \), the solution is valid.
Thus, \( x = 3 \) and \( y = 1 \).

(a)
- M1: Combining logarithms to get \( \log_2(x^2 - 9) = 4 \).
- M1: Removing logarithms to get \( x^2 - 9 = 16 \).
- A1: Finding \( x = 5 \) and \( x = -5 \).
- A1: Rejecting \( x = -5 \) to leave only \( x = 5 \).

(b)
- M1: Expressing the first equation as a linear equation: \( x + 2y = 5 \).
- M1: Expressing the second equation as a linear equation: \( 2x - y = 5 \).
- M1: Attempting to solve the simultaneous linear equations.
- A1: Correct value of \( x = 3 \).
- A1: Correct value of \( y = 1 \).

(a) For the AP:
\( u_3 = a + 2d = 11 \quad \text{--- (1)} \)
\( S_{10} = \frac{10}{2}(2a + 9d) = 5(2a + 9d) = 235 \implies 2a + 9d = 47 \quad \text{--- (2)} \)

From (1), multiply by 2:
\( 2a + 4d = 22 \quad \text{--- (3)} \)

Subtract (3) from (2):
\( 5d = 25 \implies d = 5 \)

Substitute \( d = 5 \) back into (1):
\( a + 2(5) = 11 \implies a = 1 \)

(b) For the GP:
\( A = a = 1 \)
\( S_\infty = \frac{A}{1-R} = 32 \)
\( \frac{1}{1-R} = 32 \implies 1-R = \frac{1}{32} \implies R = 1 - \frac{1}{32} = \frac{31}{32} = 0.96875 \)

(c) We want \( S_n > 31.9 \):
\( \frac{A(1-R^n)}{1-R} > 31.9 \)
\( 32(1 - 0.96875^n) > 31.9 \)
\( 1 - 0.96875^n > \frac{31.9}{32} = 0.996875 \)
\( 0.96875^n < 1 - 0.996875 = 0.003125 \)

Taking natural logarithms:
\( n \ln(0.96875) < \ln(0.003125) \)
Since \( \ln(0.96875) < 0 \), the inequality sign flips:
\( n > \frac{\ln(0.003125)}{\ln(0.96875)} \)
\( n > \frac{-5.76832}{-0.031749} \approx 181.68 \)

Thus, the least integer value of \( n \) is 182.

(a)
- M1: Setting up the equation for the 3rd term: \( a + 2d = 11 \).
- M1: Setting up the equation for the sum of 10 terms: \( 5(2a + 9d) = 235 \).
- A1: Correct value of \( d = 5 \).
- A1: Correct value of \( a = 1 \).

(b)
- M1: Setting up the sum to infinity equation: \( \frac{1}{1-R} = 32 \).
- A1: Correct ratio \( R = \frac{31}{32} \) or \( 0.96875 \).

(c)
- M1: Using the sum of a GP formula with their \( A \) and \( R \) set up as an inequality/equation: \( 32(1 - 0.96875^n) > 31.9 \).
- M1: Rearranging to obtain \( 0.96875^n < 0.003125 \) or equivalent.
- M1: Correct use of logarithms to solve for \( n \).
- A1: Correct least value \( n = 182 \).

(a) \( \overrightarrow{OQ} = \overrightarrow{OA} + \overrightarrow{AQ} = \mathbf{a} + \lambda \overrightarrow{AB} \)
Since \( \overrightarrow{AB} = \mathbf{b} - \mathbf{a} \):
\( \overrightarrow{OQ} = \mathbf{a} + \lambda(\mathbf{b} - \mathbf{a}) = (1-\lambda)\mathbf{a} + \lambda\mathbf{b} \)

(b) \( \overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} \)
\( \overrightarrow{PQ} = (1-\lambda)\mathbf{a} + \lambda\mathbf{b} - \frac{1}{4}\mathbf{a} = \left(\frac{3}{4}-\lambda\right)\mathbf{a} + \lambda\mathbf{b} \)

(c) For \( P, Q, R \) to be collinear, \( \overrightarrow{PQ} = k \overrightarrow{PR} \) for some scalar \( k \).
First, find \( \overrightarrow{PR} \):
\( \overrightarrow{PR} = \overrightarrow{OR} - \overrightarrow{OP} = 2\mathbf{b} - \frac{1}{4}\mathbf{a} \)

So:
\( \left(\frac{3}{4}-\lambda\right)\mathbf{a} + \lambda\mathbf{b} = k \left(-\frac{1}{4}\mathbf{a} + 2\mathbf{b}\right) \)

Equating coefficients of \( \mathbf{a} \) and \( \mathbf{b} \):
From \( \mathbf{b} \): \( \lambda = 2k \implies k = \frac{\lambda}{2} \)
From \( \mathbf{a} \): \( \frac{3}{4}-\lambda = -\frac{1}{4}k \)

Substitute \( k = \frac{\lambda}{2} \):
\( \frac{3}{4}-\lambda = -\frac{\lambda}{8} \)
Multiply through by 8:
\( 6 - 8\lambda = -\lambda \)
\( 7\lambda = 6 \implies \lambda = \frac{6}{7} \)

Now, to find the ratio \( PQ:QR \):
\( k = \frac{\lambda}{2} = \frac{3}{7} \)
So \( \overrightarrow{PQ} = \frac{3}{7} \overrightarrow{PR} \).
This means that point \( Q \) lies \( \frac{3}{7} \) of the way along the line segment \( PR \).
Thus, \( PQ = \frac{3}{7}PR \) and \( QR = \frac{4}{7}PR \).
Therefore, the ratio \( PQ:QR = 3:4 \).

(a)
- M1: Writing \( \overrightarrow{OQ} = \overrightarrow{OA} + \lambda\overrightarrow{AB} \) or equivalent vector sum.
- A1: Obtaining \( (1-\lambda)\mathbf{a} + \lambda\mathbf{b} \).

(b)
- M1: Writing \( \overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} \).
- A1: Obtaining \( \left(\frac{3}{4}-\lambda\right)\mathbf{a} + \lambda\mathbf{b} \).

(c)
- B1: Correct expression for \( \overrightarrow{PR} = 2\mathbf{b} - \frac{1}{4}\mathbf{a} \).
- M1: Setting up the equation \( \overrightarrow{PQ} = k \overrightarrow{PR} \) and equating coefficients of \( \mathbf{a} \) and \( \mathbf{b} \).
- A1: Correct value of \( \lambda = \frac{6}{7} \).
- M1: Determining the scalar factor \( k = \frac{3}{7} \) to establish the position of \( Q \) on \( PR \).
- A1: Concluding that \( PQ:QR = 3:4 \).