2025 Cambridge IGCSE International Mathematics (0607) 模擬試題連答案詳解

The terms increase by \(4\) each time, so the sequence is arithmetic with a common difference of \(4\). Therefore, the formula is of the form \(4n + c\). Since the first term is \(3\) (when \(n = 1\)), we set up the equation: \(4(1) + c = 3 \implies c = -1\). The \(n\)-th term of the sequence is \(4n - 1\).

M1 for identifying the common difference is \(4\) (e.g. writing \(4n + c\) or \(d = 4\))
A1 for \(4n - 1\) (or equivalent)

First, find the highest common factor of \(6x^2y\) and \(15xy^2\). The highest common factor of the numerical coefficients \(6\) and \(15\) is \(3\). The highest common factor of \(x^2\) and \(x\) is \(x\). The highest common factor of \(y\) and \(y^2\) is \(y\). Thus, the overall highest common factor is \(3xy\). Factoring this out from each term gives: \(6x^2y - 15xy^2 = 3xy(2x - 5y)\).

M1 for identifying a partial common factor (e.g. \(3(2x^2y - 5xy^2)\) or \(xy(6x - 15y)\))
A1 for \(3xy(2x - 5y)\)

The mapping of the point is \((x, y) \to (-x, -y)\). This transformation is a rotation of \(180^{\circ}\) about the origin \((0, 0)\). (Alternatively, it can also be described as an enlargement with scale factor \(-1\) and centre of enlargement at the origin \((0,0)\)).

M1 for 'Rotation' or 'Enlargement' with one correct parameter (e.g., angle or centre/scale factor)
A1 for 'Rotation of 180 degrees about the origin (0,0)' or 'Enlargement, scale factor -1, centre (0,0)'. Note: Award 0 marks if more than one transformation is named.

The gradient of the given line is \(m_1 = 3\). The gradient of a line perpendicular to it is \(m_2 = -\frac{1}{m_1} = -\frac{1}{3}\). Using the slope-intercept form \(y = mx + c\) with the perpendicular gradient \(m = -\frac{1}{3}\) and substituting the coordinates of the point \((6, 2)\): \(2 = -\frac{1}{3}(6) + c \implies 2 = -2 + c \implies c = 4\). Therefore, the equation of the perpendicular line is \(y = -\frac{1}{3}x + 4\).

M1 for gradient of perpendicular line \(m = -\frac{1}{3}\) (or substituting \(-\frac{1}{3}\) and point into line formula)
A1 for \(y = -\frac{1}{3}x + 4\) (or equivalent form)

Since \(AC\) is a diameter of the circle, the angle subtended by the diameter at the circumference, angle \(ABC\), is a right angle (\(90^\circ\)). The angles in triangle \(ABC\) sum to \(180^\circ\). Therefore, angle \(BCA = 180^\circ - 90^\circ - 35^\circ = 55^\circ\).

M1 for stating or using angle \(ABC = 90^\circ\)
A1 for \(55\)

To rationalise the denominator, we multiply both the numerator and the denominator by \(\sqrt{3}\): \(\frac{12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{12\sqrt{3}}{3}\). Simplifying the fraction by dividing the numerator and denominator by \(3\) yields \(4\sqrt{3}\).

M1 for multiplying numerator and denominator by \(\sqrt{3}\) to get \(\frac{12\sqrt{3}}{3}\)
A1 for \(4\sqrt{3}\)

Let's check the integers between \(50\) and \(55\):
- \(51\) is divisible by \(3\) (since \(5 + 1 = 6\), which is a multiple of 3; \(51 = 3 \times 17\))
- \(52\) is even, so it is divisible by \(2\)
- \(53\) has no integer factors other than \(1\) and \(53\), so it is a prime number
- \(54\) is even, so it is divisible by \(2\)
Therefore, the prime number in this range is \(53\).

B2 for \(53\)
(B1 for listing \(51, 52, 53, 54\) or if \(53\) is identified alongside another incorrect value)

Since \(y\) is inversely proportional to \(x\), we can write the relation as \(y = \frac{k}{x}\) where \(k\) is a constant. Using the given values \(x = 6\) and \(y = 4\), we find: \(4 = \frac{k}{6} \implies k = 24\). The equation representing the relationship is \(y = \frac{24}{x}\). Substituting \(x = 8\) into this equation gives: \(y = \frac{24}{8} = 3\).

M1 for finding the constant of proportionality \(k = 24\) (or writing \(y \times x = 24\) or \(6 \times 4 = 8 \times y\))
A1 for \(3\)

First, express each surd in its simplest form: \(\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}\), \(\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}\), and \(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\). Now substitute these into the expression: \(5\sqrt{3} - 3\sqrt{3} + 2\sqrt{3}\). Combine the like terms: \((5 - 3 + 2)\sqrt{3} = 4\sqrt{3}\).

M1 for writing at least two of the surds in the form \(k\sqrt{3}\) (e.g. \(5\sqrt{3}\), \(3\sqrt{3}\), or \(2\sqrt{3}\)). A1 for the correct final answer \(4\sqrt{3}\).

Find the first differences of the sequence: \(8 - 3 = 5\), \(15 - 8 = 7\), \(24 - 15 = 9\). Find the second differences: \(7 - 5 = 2\) and \(9 - 7 = 2\). Since the second differences are constant and equal to 2, the sequence is quadratic with leading term \(\frac{2}{2}n^2 = n^2\). Subtracting \(n^2\) (which is 1, 4, 9, 16, ...) from each term of the sequence yields: \(3 - 1 = 2\), \(8 - 4 = 4\), \(15 - 9 = 6\), \(24 - 16 = 8\). The sequence of differences is 2, 4, 6, 8, ... which has the \(n\)th term \(2n\). Therefore, the overall \(n\)th term is \(n^2 + 2n\).

M1 for finding second differences are 2, or writing an expression of the form \(n^2 + bn + c\). A1 for \(n^2 + 2n\) or \(n(n+2)\).

(a) Group the terms: \( 6ax - 15ay - 4bx + 10by = 3a(2x - 5y) - 2b(2x - 5y) = (3a - 2b)(2x - 5y) \). (b) Factorise the numerator and the denominator: Numerator: \( x^2 - 9 = (x - 3)(x + 3) \). Denominator: \( 2x^2 + 5x - 3 = (2x - 1)(x + 3) \). Divide both by the common factor \( (x + 3) \) to get \( \frac{x - 3}{2x - 1} \).

(a) M1 for partial factorisation such as \( 3a(2x - 5y) \) or \( -2b(2x - 5y) \), A1 for final correct factorised form \( (3a - 2b)(2x - 5y) \). (b) M1 for factorising the numerator to \( (x - 3)(x + 3) \), M1 for factorising the denominator to \( (2x - 1)(x + 3) \), A1 for final fraction \( \frac{x - 3}{2x - 1} \).

(a) Since \( OA \) and \( OB \) are radii of the circle, triangle \( OAB \) is isosceles with \( OA = OB \). Thus, angle \( OBA = 28^\circ \). The sum of angles in triangle \( OAB \) is \( 180^\circ \), so angle \( AOB = 180^\circ - 2(28^\circ) = 124^\circ \). (b) The angle subtended by an arc at the centre is twice the angle subtended at the circumference. Therefore, angle \( ACB = \frac{1}{2} \times \text{angle } AOB = \frac{124^\circ}{2} = 62^\circ \). (c) Since \( ABCD \) is a cyclic quadrilateral, opposite angles sum to \( 180^\circ \). Therefore, angle \( ABC = 180^\circ - \text{angle } ADC = 180^\circ - 115^\circ = 65^\circ \).

(a) M1 for \( 180 - 2 \times 28 \, \), A1 for \( 124^\circ \). (b) B1 for \( 62^\circ \) (or \( \frac{1}{2} \times \text{their } AOB \)). (c) M1 for \( 180 - 115 \, \), A1 for \( 65^\circ \).

(a) First, find the gradient of \( L_1 \): \( m_1 = \frac{2 - 5}{4 - (-2)} = \frac{-3}{6} = -\frac{1}{2} \). Use the point-slope formula with point \( B(4, 2) \): \( y - 2 = -\frac{1}{2}(x - 4) \Rightarrow y - 2 = -\frac{1}{2}x + 2 \Rightarrow y = -\frac{1}{2}x + 4 \). (b) The gradient of the perpendicular line \( L_2 \) is \( m_2 = -\frac{1}{m_1} = 2 \). Since \( L_2 \) passes through \( P(3, -1) \): \( y - (-1) = 2(x - 3) \Rightarrow y + 1 = 2x - 6 \Rightarrow 2x - y - 7 = 0 \).

(a) M1 for gradient of \( L_1 \) is \( -\frac{1}{2} \), A1 for \( y = -\frac{1}{2}x + 4 \). (b) M1 for perpendicular gradient \( m_2 = 2 \) (or \( -1 / \text{their } m_1 \)), M1 for substituting \( (3, -1) \) into their line formula, A1 for \( 2x - y - 7 = 0 \) or any equivalent integer equation in the form \( ax + by + d = 0 \).

(a) For the function \( y = a\cos(bx) \), the amplitude is \( |a| = 3 \) and the period is \( \frac{360^\circ}{b} = \frac{360^\circ}{2} = 180^\circ \). (b) Set \( 3\cos(2x) = 0 \Rightarrow \cos(2x) = 0 \). For \( 0^\circ \le x \le 180^\circ \), we have \( 0^\circ \le 2x \le 360^\circ \). The angles where cosine is zero are \( 2x = 90^\circ \) and \( 2x = 270^\circ \). Dividing by 2, we get \( x = 45^\circ \) and \( x = 135^\circ \).

(a) B1 for amplitude = 3, B1 for period = \( 180^\circ \). (b) M1 for setting \( 2x = 90^\circ \) or \( 2x = 270^\circ \), A1 for \( x = 45^\circ \), A1 for \( x = 135^\circ \).

(a) Simplify each surd: \( \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} \); \( 2\sqrt{27} = 2\sqrt{9 \times 3} = 2(3\sqrt{3}) = 6\sqrt{3} \); \( \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \). Substituting these back: \( 5\sqrt{3} - 6\sqrt{3} + 2\sqrt{3} = (5 - 6 + 2)\sqrt{3} = \sqrt{3} \). (b) Rationalise the denominator by multiplying the numerator and denominator by the conjugate \( 3 + \sqrt{5} \): \( \frac{(1 + \sqrt{5})(3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})} = \frac{3 + \sqrt{5} + 3\sqrt{5} + 5}{3^2 - (\sqrt{5})^2} = \frac{8 + 4\sqrt{5}}{9 - 5} = \frac{8 + 4\sqrt{5}}{4} = 2 + \sqrt{5} \).

(a) M1 for simplifying at least two surds correctly (e.g., finding \( 5\sqrt{3} \), \( 6\sqrt{3} \), or \( 2\sqrt{3} \)), A1 for \( \sqrt{3} \). (b) M1 for multiplying numerator and denominator by \( 3 + \sqrt{5} \), M1 for expanding the numerator to \( 8 + 4\sqrt{5} \) or the denominator to \( 4 \), A1 for \( 2 + \sqrt{5} \).

(a) A stretch of scale factor 3 parallel to the \( y \)-axis transforms \( y = f(x) \) to \( y = 3f(x) \). A translation by the vector \( \begin{pmatrix} 2 \\ -5 \end{pmatrix} \) shifts the graph 2 units to the right and 5 units down, giving the equation \( y = 3f(x - 2) - 5 \). (b) Substituting \( f(x) = x^2 \) into the equation from part (a): \( y = 3(x - 2)^2 - 5 \). Expanding the squared term: \( (x - 2)^2 = x^2 - 4x + 4 \). Multiplying by 3: \( 3(x^2 - 4x + 4) = 3x^2 - 12x + 12 \). Subtracting 5: \( 3x^2 - 12x + 12 - 5 = 3x^2 - 12x + 7 \). This matches the required form.

(a) M1 for either \( 3f(x) \) or \( f(x - 2) \) seen, A1 for \( y = 3f(x - 2) - 5 \). (b) M1 for substituting \( f(x) = x^2 \) to get \( y = 3(x - 2)^2 - 5 \), M1 for expanding \( (x - 2)^2 \) correctly, A1 for fully simplifying to obtain \( 3x^2 - 12x + 7 \) with all steps shown.

(a) Let us find the differences between successive terms: \( 11 - 3 = 8 \), \( 25 - 11 = 14 \), \( 45 - 25 = 20 \). The first differences are \( 8, 14, 20 \). The difference between these first differences is \( 14 - 8 = 6 \) and \( 20 - 14 = 6 \). The next first difference is \( 20 + 6 = 26 \). Therefore, the next term is \( 45 + 26 = 71 \). (b) Since the second difference is constant and equals 6, the formula is quadratic of the form \( an^2 + bn + c \) where \( a = \frac{6}{2} = 3 \). Subtracting \( 3n^2 \) from each term: For \( n = 1 \): \( 3 - 3(1)^2 = 0 \); For \( n = 2 \): \( 11 - 3(2)^2 = -1 \); For \( n = 3 \): \( 25 - 3(3)^2 = -2 \); For \( n = 4 \): \( 45 - 3(4)^2 = -3 \). The sequence of differences is \( 0, -1, -2, -3, \dots \), which is arithmetic with first term \( 0 \) and common difference \( -1 \). This linear sequence has the formula \( -n + 1 \). Combining these, the \( n \)-th term of the original sequence is \( 3n^2 - n + 1 \).

(a) B1 for \( 71 \). (b) M1 for recognizing a quadratic sequence with \( a = 3 \), M1 for finding the remaining linear sequence \( 0, -1, -2, -3 \) or setting up simultaneous equations, M1 for solving for \( b = -1 \) and \( c = 1 \), A1 for \( 3n^2 - n + 1 \).

(a) Since \( y \) is inversely proportional to the square root of \( x \), we can write \( y = \frac{k}{\sqrt{x}} \) where \( k \) is a constant. Substitute \( x = 16 \) and \( y = 3 \) to find \( k \): \( 3 = \frac{k}{\sqrt{16}} \Rightarrow 3 = \frac{k}{4} \Rightarrow k = 12 \). Thus, the equation is \( y = \frac{12}{\sqrt{x}} \). (b) Substitute \( x = \frac{9}{4} \) into the equation: \( y = \frac{12}{\sqrt{9/4}} = \frac{12}{3/2} = 12 \times \frac{2}{3} = 8 \).

(a) M1 for \( y = \frac{k}{\sqrt{x}} \), M1 for substituting \( x = 16, y = 3 \) to solve for \( k \), A1 for \( y = \frac{12}{\sqrt{x}} \). (b) M1 for substituting \( x = \frac{9}{4} \) into their equation from (a), A1 for \( 8 \).

(a) The angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc at the circumference. Therefore, the reflex angle \(AOC = 2 \times \angle ABC\). This gives the equation: \(5y + 10 = 2(2y + 15)\). Solving this: \(5y + 10 = 4y + 30 \Rightarrow y = 20\). (b) Since \(ABCD\) is a cyclic quadrilateral, its opposite angles sum to \(180^\circ\). Thus, \(\angle ADC = 180^\circ - \angle ABC\). First, calculate \(\angle ABC = 2(20) + 15 = 55^\circ\). Then, \(\angle ADC = 180^\circ - 55^\circ = 125^\circ\).

Part (a): [3 marks] M1 for setting up the equation \(5y + 10 = 2(2y + 15)\). M1 for expanding to \(5y + 10 = 4y + 30\). A1 for \(y = 20\). Part (b): [2 marks] M1 for calculating \(\angle ABC = 55^\circ\) and using the cyclic quadrilateral property \(\angle ADC = 180^\circ - \angle ABC\). A1 for \(125^\circ\).

1. Find the midpoint of \(AB\): \(M = \left(\frac{2 + 6}{2}, \frac{-3 + 5}{2}\right) = (4, 1)\). 2. Find the gradient of \(AB\): \(m = \frac{5 - (-3)}{6 - 2} = \frac{8}{4} = 2\). 3. Find the gradient of the perpendicular bisector: \(m_{\perp} = -\frac{1}{m} = -\frac{1}{2}\). 4. Use the point-slope formula with point \((4, 1)\) and gradient \(-\frac{1}{2}\): \(y - 1 = -\frac{1}{2}(x - 4)\). Multiply both sides by 2: \(2(y - 1) = -(x - 4) \Rightarrow 2y - 2 = -x + 4\). Rearrange into the form \(ax + by = c\): \(x + 2y = 6\).

M1 for finding the midpoint \((4, 1)\). M1 for finding the gradient of \(AB\) as 2. M1 for finding the perpendicular gradient as \(-\frac{1}{2}\). M1 for substituting their midpoint and perpendicular gradient into the straight line equation. A1 for the correct final equation \(x + 2y = 6\) (or any integer multiple if simplified to have no common factors and \(a > 0\)).

First, factor each quadratic expression completely: 1. \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\) 2. \(4x^2 - 1 = (2x - 1)(2x + 1)\) (difference of two squares) 3. \(x^2 - 9 = (x - 3)(x + 3)\) (difference of two squares) 4. \(2x^2 + 5x - 3 = (2x - 1)(x + 3)\) Next, rewrite the division as multiplication by the reciprocal of the second fraction: \(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)} \times \frac{(2x - 1)(x + 3)}{(x - 3)(x + 3)}\). Now, cancel common terms from the numerator and denominator: the term \((2x+1)\) cancels, \((2x-1)\) cancels, \((x-3)\) cancels, and \((x+3)\) cancels. The expression simplifies completely to \(1\).

M1 for factoring \(2x^2 - 5x - 3\) as \((2x + 1)(x - 3)\). M1 for factoring \(4x^2 - 1\) as \((2x - 1)(2x + 1)\). M1 for factoring \(2x^2 + 5x - 3\) as \((2x - 1)(x + 3)\). M1 for factoring \(x^2 - 9\) as \((x - 3)(x + 3)\) and inverting the division to multiplication. A1 for the correct fully simplified answer of 1.