Cambridge IAS-Level · thinka 原创模拟试题

2026 Cambridge IAS-Level Mathematics (9709) 模拟试题及答案详解

Thinka Jun 2026 (V5) Cambridge International AS Level-Style Mock — Mathematics (9709)

125 分185 分钟2026

An original Thinka practice paper modelled on the structure and difficulty of the Jun 2026 (V5) Cambridge International AS Level Mathematics (9709) paper. Not affiliated with or reproduced from Cambridge.

卷一 Pure Mathematics 1

Answer all 12 questions. Show all necessary working clearly. Unsupported calculator answers receive no marks.

11 题目 · 66 分

题目 1 · algebraic_and_quadratic_solutions

5 分

(a) Express $ 2x^2 - 12x + 13 $ in the form $ a(x - b)^2 + c $, where $ a $, $ b $ and $ c $ are constants. [2]

(b) The equation $ 2x^2 - 12x + 13 = k $, where $ k $ is a constant, has no real roots. Find the set of possible values of $ k $. [3]

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解题

(a) Completing the square:
$ 2x^2 - 12x + 13 = 2(x^2 - 6x) + 13 $
$ = 2\left((x - 3)^2 - 9\right) + 13 $
$ = 2(x - 3)^2 - 18 + 13 $
$ = 2(x - 3)^2 - 5 $
So $ a = 2 $, $ b = 3 $, and $ c = -5 $.

(b) Method 1 (Using completed square):
The minimum value of $ 2x^2 - 12x + 13 $ is $ -5 $.
For the equation $ 2(x - 3)^2 - 5 = k $ to have no real roots, the horizontal line $ y = k $ must lie strictly below the minimum point of the parabola.
Hence, $ k < -5 $.

Method 2 (Using the discriminant):
Rearrange $ 2x^2 - 12x + (13 - k) = 0 $.
For no real roots, the discriminant must satisfy $ b^2 - 4ac < 0 $:
$ (-12)^2 - 4(2)(13 - k) < 0 $
$ 144 - 8(13 - k) < 0 $
$ 144 - 104 + 8k < 0 $
$ 40 + 8k < 0 $
$ 8k < -40 \implies k < -5 $.

评分标准

(a)
- M1: For attempting to complete the square, obtaining $ 2(x - 3)^2 + c $ or $ a=2 $ and $ b=3 $.
- A1: Fully correct expression $ 2(x - 3)^2 - 5 $ (or stating $ a=2, b=3, c=-5 $).

(b)
- M1: For setting up the discriminant $ (-12)^2 - 4(2)(13 - k) $ or using their minimum value from part (a).
- M1: For stating condition $ \Delta < 0 $ or equating expression to $ k $ and deducing condition on vertex.
- A1: Correct final inequality $ k < -5 $.

题目 2 · algebraic_and_quadratic_solutions

5 分

The line with equation $ y = mx - 4 $ is a tangent to the curve with equation $ y = 2x^2 - 5x + 4 $.

Find the two possible values of the constant $ m $, and for each value of $ m $, find the coordinates of the point of contact.

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解题

Equating the line and the curve:
$ 2x^2 - 5x + 4 = mx - 4 $
$ 2x^2 - (5 + m)x + 8 = 0 $

Since the line is a tangent to the curve, the quadratic equation must have equal roots, so the discriminant $ b^2 - 4ac = 0 $:
$ (-(5 + m))^2 - 4(2)(8) = 0 $
$ (5 + m)^2 - 64 = 0 $
$ (5 + m)^2 = 64 $
$ 5 + m = 8 \implies m = 3 $
$ 5 + m = -8 \implies m = -13 $

When $ m = 3 $:
$ 2x^2 - 8x + 8 = 0 \implies 2(x - 2)^2 = 0 \implies x = 2 $
$ y = 3(2) - 4 = 2 $
Point of contact is $ (2, 2) $.

When $ m = -13 $:
$ 2x^2 - (-8)x + 8 = 0 \implies 2x^2 + 8x + 8 = 0 \implies 2(x + 2)^2 = 0 \implies x = -2 $
$ y = -13(-2) - 4 = 26 - 4 = 22 $
Point of contact is $ (-2, 22) $.

评分标准

- M1: Equating line and curve to form a 3-term quadratic in $ x $: $ 2x^2 - (5+m)x + 8 = 0 $.
- M1: Applying the condition $ b^2 - 4ac = 0 $ to obtain an equation in $ m $.
- A1: Finding both correct values of $ m $: $ m = 3 $ and $ m = -13 $.
- M1: Finding the corresponding $ x $-values for both values of $ m $ by solving the repeated root quadratics.
- A1: Both pairs of coordinates correctly stated: $ (2, 2) $ and $ (-2, 22) $.

题目 3 · algebraic_and_quadratic_solutions

5 分

Solve the simultaneous equations
\[ 2x - y = 1 \]
\[ x^2 + 2xy - y^2 = 7 \]

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解题

From the linear equation, rearrange to express $ y $ in terms of $ x $:
$ y = 2x - 1 $

Substitute $ y = 2x - 1 $ into the quadratic equation:
$ x^2 + 2x(2x - 1) - (2x - 1)^2 = 7 $

Expand the brackets:
$ x^2 + (4x^2 - 2x) - (4x^2 - 4x + 1) = 7 $
$ x^2 + 4x^2 - 2x - 4x^2 + 4x - 1 = 7 $
$ x^2 + 2x - 1 = 7 $
$ x^2 + 2x - 8 = 0 $

Factorise the quadratic equation:
$ (x + 4)(x - 2) = 0 $
So $ x = 2 $ or $ x = -4 $.

Find the corresponding $ y $-values:
When $ x = 2 $:
$ y = 2(2) - 1 = 3 $

When $ x = -4 $:
$ y = 2(-4) - 1 = -9 $

The solutions are $ x = 2, y = 3 $ and $ x = -4, y = -9 $.

评分标准

- M1: Rearranging the linear equation to make $ y $ (or $ x $) the subject and substituting into the quadratic equation.
- A1: Correctly expanding and simplifying to obtain the 3-term quadratic equation $ x^2 + 2x - 8 = 0 $ (or equivalent in $ y $).
- M1: Valid method to solve their 3-term quadratic equation (factorisation or quadratic formula).
- A1: Correct values of $ x $ (or $ y $): $ x = 2 $ and $ x = -4 $.
- A1: Correct corresponding values of $ y $ properly paired with $ x $: $ (2, 3) $ and $ (-4, -9) $.

题目 4 · Calculus

4 分

A point $P(x, y)$ moves along the curve $y = \frac{24}{(3x-1)^2} - 2x$ for $x > \frac{1}{3}$.

(a) Find an expression for $\frac{\mathrm{d}y}{\mathrm{d}x}$. [2]

(b) The $x$-coordinate of $P$ is increasing at a constant rate of $0.15\text{ units s}^{-1}$. Find the rate of change of the $y$-coordinate at the instant when $x = 1$. [2]

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解题

(a)
Rewrite the equation of the curve as:
\[y = 24(3x - 1)^{-2} - 2x\]
Differentiating with respect to $x$ using the chain rule:
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 24 \cdot (-2)(3x - 1)^{-3} \cdot 3 - 2 = -144(3x - 1)^{-3} - 2\]
\[\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{144}{(3x - 1)^3} - 2\]

(b)
When $x = 1$:
\[\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{144}{(3(1) - 1)^3} - 2 = -\frac{144}{2^3} - 2 = -\frac{144}{8} - 2 = -18 - 2 = -20\]
Using the chain rule for connected rates of change:
\[\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}x} \times \frac{\mathrm{d}x}{\mathrm{d}t}\]
Given $\frac{\mathrm{d}x}{\mathrm{d}t} = 0.15$:
\[\frac{\mathrm{d}y}{\mathrm{d}t} = -20 \times 0.15 = -3\text{ units s}^{-1}\]

评分标准

(a)
M1: Attempt to differentiate using the chain rule, obtaining an expression of the form $k(3x-1)^{-3}$ where $k \neq 0$.
A1: Correct derivative $-\frac{144}{(3x-1)^3} - 2$ or $-144(3x-1)^{-3} - 2$.

(b)
M1: Substitute $x = 1$ into their $\frac{\mathrm{d}y}{\mathrm{d}x}$ and multiply by $0.15$.
A1: $-3$ (or state that $y$ is decreasing at a rate of $3\text{ units s}^{-1}$).

题目 5 · Calculus

5 分

The equation of a curve is $y = 6\sqrt{x} + \frac{24}{x} - 7$ for $x > 0$.

(a) Find $\frac{\mathrm{d}y}{\mathrm{d}x}$. [2]

(b) Find the coordinates of the stationary point on the curve and determine whether it is a maximum or a minimum. [3]

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解题

(a)
Write the equation in index form:
\[y = 6x^{\frac{1}{2}} + 24x^{-1} - 7\]
Differentiating with respect to $x$:
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 6 \left(\frac{1}{2}\right)x^{-\frac{1}{2}} + 24(-1)x^{-2} = 3x^{-\frac{1}{2}} - 24x^{-2} = \frac{3}{\sqrt{x}} - \frac{24}{x^2}\]

(b)
At the stationary point, $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$:
\[\frac{3}{\sqrt{x}} - \frac{24}{x^2} = 0 \implies \frac{3}{\sqrt{x}} = \frac{24}{x^2}\]
\[3x^2 = 24\sqrt{x} \implies x^{\frac{3}{2}} = 8\]
\[x = 8^{\frac{2}{3}} = (\sqrt[3]{8})^2 = 2^2 = 4\]
Find the corresponding $y$-coordinate:
\[y = 6\sqrt{4} + \frac{24}{4} - 7 = 6(2) + 6 - 7 = 12 + 6 - 7 = 11\]
So the coordinates are $(4, 11)$.

To determine the nature, find the second derivative:
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 3\left(-\frac{1}{2}\right)x^{-\frac{3}{2}} - 24(-2)x^{-3} = -\frac{3}{2}x^{-\frac{3}{2}} + 48x^{-3}\]
At $x = 4$:
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\frac{3}{2}(4^{-\frac{3}{2}}) + 48(4^{-3}) = -\frac{3}{2}\left(\frac{1}{8}\right) + \frac{48}{64} = -\frac{3}{16} + \frac{3}{4} = \frac{9}{16}\]
Since $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0$, the stationary point $(4, 11)$ is a minimum.

评分标准

(a)
M1: Differentiate at least one term correctly (either $6x^{\frac{1}{2}} \to 3x^{-\frac{1}{2}}$ or $24x^{-1} \to -24x^{-2}$).
A1: Fully correct derivative $3x^{-\frac{1}{2}} - 24x^{-2}$ or equivalent.

(b)
M1: Set their $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ and solve for $x$.
A1: Both coordinates correct: $x = 4, y = 11$ (or coordinate pair $(4, 11)$).
B1: Correctly evaluate the second derivative (or examine gradient change around $x = 4$) showing $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{9}{16} > 0$ and conclude Minimum.

题目 6 · structured

7 分

(a) Find the first 3 terms in the expansion, in ascending powers of $x$, of $(2 + ax)^5$, giving each term in terms of the constant $a$.

(b) In the expansion of $(1 - 4x)(2 + ax)^5$, the coefficient of $x$ is 32 and the coefficient of $x^2$ is $c$.

(i) Find the value of $a$.

(ii) Hence find the value of $c$.

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解题

(a) Using the binomial expansion:
\[(2 + ax)^5 = 2^5 + \binom{5}{1} 2^4 (ax) + \binom{5}{2} 2^3 (ax)^2 + \dots\]
\[= 32 + 5(16)(ax) + 10(8)(a^2 x^2) = 32 + 80ax + 80a^2 x^2\]

(b)(i) Consider the product:
\[(1 - 4x)(32 + 80ax + 80a^2 x^2 + \dots)\]
The term in $x$ is given by:
\[1(80ax) - 4x(32) = (80a - 128)x\]
Given that the coefficient of $x$ is 32:
\[80a - 128 = 32\]
\[80a = 160 \implies a = 2\]

(b)(ii) The term in $x^2$ is given by:
\[1(80a^2 x^2) - 4x(80ax) = (80a^2 - 320a)x^2\]
Substitute $a = 2$:
\[c = 80(2)^2 - 320(2) = 80(4) - 640 = 320 - 640 = -320\]

评分标准

(a)
B1: 32 as the constant term.
M1: Correct binomial term structure for the $x$ or $x^2$ term, e.g. $\binom{5}{1}(2)^4(ax)$ or $\binom{5}{2}(2)^3(ax)^2$.
A1: $32 + 80ax + 80a^2 x^2$ fully simplified.

(b)(i)
M1: Forms an equation for the coefficient of $x$: $80a - 4(32) = 32$.
A1: $a = 2$.

(b)(ii)
M1: Identifies both terms contributing to $x^2$, i.e., $80a^2 - 4(80a)$, and substitutes their $a$.
A1: $c = -320$.

题目 7 · structured

8 分

An arithmetic progression has first term $a$ and common difference $d$, where $d \neq 0$. The 1st, 5th and 7th terms of this arithmetic progression are the first three terms of a geometric progression.

(a) Show that $a = -8d$.

(b) Find the common ratio of the geometric progression.

(c) State, giving a reason, why the geometric progression has a sum to infinity.

(d) Given that the sum of the first 20 terms of the arithmetic progression is 240,

(i) find the value of $a$ and the value of $d$,

(ii) find the sum to infinity of the geometric progression.

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解题

(a) The 1st, 5th and 7th terms of the arithmetic progression are:
\[u_1 = a, \quad u_5 = a + 4d, \quad u_7 = a + 6d\]
Since these form a geometric progression:
\[\frac{a + 4d}{a} = \frac{a + 6d}{a + 4d}\]
\[(a + 4d)^2 = a(a + 6d)\]
\[a^2 + 8ad + 16d^2 = a^2 + 6ad\]
\[2ad + 16d^2 = 0 \implies 2d(a + 8d) = 0\]
Since $d \neq 0$, $a + 8d = 0 \implies a = -8d$.

(b) The common ratio is:
\[r = \frac{a + 4d}{a} = \frac{-8d + 4d}{-8d} = \frac{-4d}{-8d} = \frac{1}{2}\]

(c) The geometric progression converges and has a sum to infinity because $|r| = \left|\frac{1}{2}\right| < 1$.

(d)(i) The sum of the first $n$ terms of an arithmetic progression is:
\[S_n = \frac{n}{2}[2a + (n - 1)d]\]
For $n = 20$:
\[S_{20} = \frac{20}{2}[2a + 19d] = 10[2(-8d) + 19d] = 10[-16d + 19d] = 10(3d) = 30d\]
Given $S_{20} = 240$:
\[30d = 240 \implies d = 8\]
\[a = -8(8) = -64\]

(d)(ii) The first term of the geometric progression is the first term of the AP, so $u_1 = a = -64$.
The sum to infinity is:
\[S_\infty = \frac{u_1}{1 - r} = \frac{-64}{1 - \frac{1}{2}} = \frac{-64}{0.5} = -128\]

评分标准

(a)
M1: Sets up the equation $(a + 4d)^2 = a(a + 6d)$ or equivalent ratio equation and expands.
A1: Correct algebraic steps showing $2ad + 16d^2 = 0$ leading legitimately to $a = -8d$.

(b)
B1: $r = \frac{1}{2}$ (or 0.5).

(c)
B1: States that $|r| < 1$ (or $-1 < \frac{1}{2} < 1$).

(d)(i)
M1: Uses $S_{20} = \frac{20}{2}[2a + 19d] = 240$ and substitutes $a = -8d$.
A1: $d = 8$ and $a = -64$.

(d)(ii)
M1: Uses $S_\infty = \frac{a}{1 - r}$ with their $a$ and $r$.
A1: $-128$.

题目 8 · standard

6 分

(a) Prove the identity \[ \frac{1}{\cos\theta - \sin\theta} - \frac{1}{\cos\theta + \sin\theta} \equiv \frac{2\sin\theta}{1 - 2\sin^2\theta}. \] [3]

(b) Hence solve the equation \[ \frac{1}{\cos\theta - \sin\theta} - \frac{1}{\cos\theta + \sin\theta} = 3\tan\theta \] for $0^\circ < \theta < 360^\circ$. [3]

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解题

(a) Starting with the left-hand side and putting over a common denominator:
\[ \text{LHS} = \frac{(\cos\theta + \sin\theta) - (\cos\theta - \sin\theta)}{(\cos\theta - \sin\theta)(\cos\theta + \sin\theta)} \]
\[ = \frac{2\sin\theta}{\cos^2\theta - \sin^2\theta} \]
Using the identity $\cos^2\theta = 1 - \sin^2\theta$:
\[ = \frac{2\sin\theta}{(1 - \sin^2\theta) - \sin^2\theta} = \frac{2\sin\theta}{1 - 2\sin^2\theta} = \text{RHS}. \]

(b) Using the result from part (a) and substituting $\tan\theta = \frac{\sin\theta}{\cos\theta}$:
\[ \frac{2\sin\theta}{1 - 2\sin^2\theta} = \frac{3\sin\theta}{\cos\theta} \]
\[ \sin\theta \left( \frac{2}{1 - 2\sin^2\theta} - \frac{3}{\cos\theta} \right) = 0 \]
From $\sin\theta = 0$, for $0^\circ < \theta < 360^\circ$:
\[ \theta = 180^\circ \]
From $2\cos\theta = 3(1 - 2\sin^2\theta)$:
Using $\sin^2\theta = 1 - \cos^2\theta$:
\[ 2\cos\theta = 3\left(1 - 2(1 - \cos^2\theta)\right) = 3(2\cos^2\theta - 1) \]
\[ 6\cos^2\theta - 2\cos\theta - 3 = 0 \]
Using the quadratic formula:
\[ \cos\theta = \frac{2 \pm \sqrt{(-2)^2 - 4(6)(-3)}}{12} = \frac{2 \pm \sqrt{76}}{12} = \frac{1 \pm \sqrt{19}}{6} \]
For $\cos\theta = \frac{1 + \sqrt{19}}{6} \approx 0.89315$:
\[ \theta = 26.7^\circ, \quad 360^\circ - 26.73^\circ = 333.3^\circ \]
For $\cos\theta = \frac{1 - \sqrt{19}}{6} \approx -0.55982$:
\[ \theta = 124.0^\circ, \quad 360^\circ - 124.04^\circ = 236.0^\circ \]
Thus, the solutions are $\theta = 26.7^\circ, 124.0^\circ, 180^\circ, 236.0^\circ, 333.3^\circ$.

评分标准

(a)
M1: Express LHS over a single common denominator $(\cos\theta - \sin\theta)(\cos\theta + \sin\theta)$.
M1: Simplify numerator to $2\sin\theta$ and denominator to $\cos^2\theta - \sin^2\theta$.
A1: Correctly substitute $\cos^2\theta = 1 - \sin^2\theta$ to complete the proof convincingly.

(b)
M1: Use identity and equate to $\frac{3\sin\theta}{\cos\theta}$, obtaining $\sin\theta = 0$ or transforming to a quadratic in $\cos\theta$.
M1: Form the 3-term quadratic $6\cos^2\theta - 2\cos\theta - 3 = 0$ and attempt to solve for $\cos\theta$.
A1: All five angles correct: $26.7^\circ, 124.0^\circ, 180^\circ, 236.0^\circ, 333.3^\circ$ (allow rounding to 1 d.p.).

题目 9 · standard

6 分

The sector $OAB$ of a circle with centre $O$ and radius $r\text{ cm}$ has angle $AOB = \theta$ radians. The point $C$ lies on the line segment $OB$ such that triangle $OAC$ is right-angled at $C$.

(a) Given that $\theta = \frac{\pi}{6}$ and the area of triangle $OAC$ is $18\sqrt{3}\text{ cm}^2$, show that $r = 12$. [2]

(b) Find the exact perimeter of the shaded region enclosed by the arc $AB$, the line segment $BC$, and the line segment $AC$. [2]

(c) Find the exact area of the shaded region. [2]

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解题

(a) In the right-angled triangle $OAC$, with hypotenuse $OA = r$ and $\angle AOC = \frac{\pi}{6}$:
\[ AC = r\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}r \]
\[ OC = r\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}r \]
The area of triangle $OAC$ is:
\[ \text{Area} = \frac{1}{2} \times AC \times OC = \frac{1}{2} \times \left(\frac{1}{2}r\right) \times \left(\frac{\sqrt{3}}{2}r\right) = \frac{\sqrt{3}}{8}r^2 \]
Equating this to $18\sqrt{3}$:
\[ \frac{\sqrt{3}}{8}r^2 = 18\sqrt{3} \implies r^2 = 144 \implies r = 12. \]

(b) The perimeter of the shaded region consists of:
- Length of arc $AB = r\theta = 12 \times \frac{\pi}{6} = 2\pi$
- Length of $AC = 12\sin\left(\frac{\pi}{6}\right) = 6$
- Length of $BC = OB - OC = 12 - 12\cos\left(\frac{\pi}{6}\right) = 12 - 6\sqrt{3}$

Adding these lengths:
\[ \text{Perimeter} = 2\pi + 6 + (12 - 6\sqrt{3}) = 2\pi + 18 - 6\sqrt{3}\text{ cm}. \]

(c) The area of sector $OAB$ is:
\[ \text{Area of sector} = \frac{1}{2}r^2\theta = \frac{1}{2}(12)^2\left(\frac{\pi}{6}\right) = 12\pi \]
The area of the shaded region is:
\[ \text{Area of shaded region} = \text{Area of sector } OAB - \text{Area of } \triangle OAC = 12\pi - 18\sqrt{3}\text{ cm}^2. \]

评分标准

(a)
M1: Form an expression for the area of triangle $OAC$ in terms of $r$ using $\frac{1}{2}ab\sin C$ or $\frac{1}{2} \times \text{base} \times \text{height}$.
A1: Equate to $18\sqrt{3}$ and show convincingly that $r = 12$.

(b)
M1: Calculate arc length $AB = 12 \times \frac{\pi}{6} = 2\pi$ and find $BC = 12 - 6\sqrt{3}$ and $AC = 6$.
A1: Fully correct exact perimeter $2\pi + 18 - 6\sqrt{3}$.

(c)
M1: Calculate area of sector $OAB = \frac{1}{2}(12)^2\left(\frac{\pi}{6}\right) = 12\pi$ and subtract area of triangle $OAC$.
A1: Fully correct exact area $12\pi - 18\sqrt{3}$.

题目 10 · free-response

8 分

The function $\mathrm{f}$ is defined by $\mathrm{f}(x) = 2x^2 - 12x + 13$ for the domain $x \leqslant p$, where $p$ is a constant.

(a) Express $2x^2 - 12x + 13$ in the form $a(x + b)^2 + c$, where $a$, $b$ and $c$ are constants. [2]

(b) State the greatest value of $p$ for which $\mathrm{f}$ has an inverse. [1]

(c) For this value of $p$, find an expression for $\mathrm{f}^{-1}(x)$ and state the domain of $\mathrm{f}^{-1}$. [3]

(d) The function $\mathrm{g}$ is defined by $\mathrm{g}(x) = 2x + 1$ for $x \in \mathbb{R}$. Solve the equation $\mathrm{f}(x) = \mathrm{g}(x)$ for $x \leqslant p$, where $p$ is the value found in part (b). [2]

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解题

(a)
\[ 2x^2 - 12x + 13 = 2\left(x^2 - 6x\right) + 13 \]
\[ = 2\left((x - 3)^2 - 9\right) + 13 = 2(x - 3)^2 - 18 + 13 = 2(x - 3)^2 - 5 \]
So $a = 2$, $b = -3$, $c = -5$.

(b)
The line of symmetry / vertex of the parabola occurs at $x = 3$.
For $\mathrm{f}$ to be a one-one function (and hence possess an inverse) on $x \leqslant p$, the domain cannot include values on both sides of the vertex.
Therefore, the greatest value of $p$ is $3$.

(c)
Using $y = 2(x - 3)^2 - 5$:
\[ y + 5 = 2(x - 3)^2 \implies (x - 3)^2 = \frac{y + 5}{2} \]
\[ x - 3 = \pm \sqrt{\frac{y + 5}{2}} \]
Since the domain of $\mathrm{f}$ is $x \leqslant 3$, $x - 3 \leqslant 0$, so we take the negative square root:
\[ x - 3 = -\sqrt{\frac{y + 5}{2}} \implies x = 3 - \sqrt{\frac{y + 5}{2}} \]
Hence, $\mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x + 5}{2}}$.

The range of $\mathrm{f}$ is $\mathrm{f}(x) \geqslant -5$, so the domain of $\mathrm{f}^{-1}$ is $x \geqslant -5$.

(d)
Set $\mathrm{f}(x) = \mathrm{g}(x)$:
\[ 2x^2 - 12x + 13 = 2x + 1 \]
\[ 2x^2 - 14x + 12 = 0 \]
\[ x^2 - 7x + 6 = 0 \]
\[ (x - 1)(x - 6) = 0 \]
Thus $x = 1$ or $x = 6$.
Since the domain requires $x \leqslant 3$, we reject $x = 6$.
Therefore, $x = 1$.

评分标准

(a)
- M1: Attempt to complete the square to obtain $2(x \pm 3)^2 + c$ or equivalent.
- A1: Fully correct expression $2(x - 3)^2 - 5$.

(b)
- B1: Stating $p = 3$ (or $3$).

(c)
- M1: Correct order of operations to make $(x - 3)^2$ the subject and take the square root.
- A1: Fully correct inverse function $\mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x + 5}{2}}$ (must be in terms of $x$ and have the negative sign before the surd).
- B1: Stating domain $x \geqslant -5$ (accept $[-5, \infty)$ or $\mathrm{Domain} \geqslant -5$; do not accept $y \geqslant -5$).

(d)
- M1: Equating $\mathrm{f}(x)$ and $\mathrm{g}(x)$ to form a 3-term quadratic and attempting to solve by factorisation or formula.
- A1: $x = 1$ only (must reject $x = 6$).

题目 11 · Integration & Area Determination

7 分

The equation of a curve is $ y = \frac{8}{\sqrt{3x + 4}} $ and the equation of a line is $ x + 2y = 8 $. The curve and the line intersect at the points $ (0, 4) $ and $ (4, 2) $.

(a) Find $ \int \frac{8}{\sqrt{3x + 4}} \, \mathrm{d}x $. [2]

(b) Find the exact area of the region enclosed between the curve and the line. [5]

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解题

(a)
Rewrite the integrand in index form:
\[ \int 8(3x + 4)^{-\frac{1}{2}} \, \mathrm{d}x \]
Integrate using the reverse chain rule:
\[ = \frac{8(3x + 4)^{\frac{1}{2}}}{\frac{1}{2} \times 3} + c = \frac{16}{3}(3x + 4)^{\frac{1}{2}} + c \]

(b)
The region is bounded by $ x = 0 $ and $ x = 4 $.

First, find the area under the line $ y = 4 - \frac{1}{2}x $:
\[ \text{Area}_{\text{line}} = \int_0^4 \left(4 - \frac{1}{2}x\right) \mathrm{d}x = \left[ 4x - \frac{1}{4}x^2 \right]_0^4 = \left(4(4) - \frac{1}{4}(4)^2\right) - 0 = 16 - 4 = 12 \]
(Alternatively, using the area of a trapezium: $ \frac{1}{2}(4 + 2) \times 4 = 12 $).

Next, find the area under the curve between $ x = 0 $ and $ x = 4 $:
\[ \text{Area}_{\text{curve}} = \left[ \frac{16}{3}(3x + 4)^{\frac{1}{2}} \right]_0^4 \]
\[ = \frac{16}{3}\sqrt{3(4) + 4} - \frac{16}{3}\sqrt{3(0) + 4} = \frac{16}{3}\sqrt{16} - \frac{16}{3}\sqrt{4} \]
\[ = \frac{16}{3}(4) - \frac{16}{3}(2) = \frac{64}{3} - \frac{32}{3} = \frac{32}{3} \]

The area of the enclosed region is:
\[ \text{Area} = \text{Area}_{\text{line}} - \text{Area}_{\text{curve}} = 12 - \frac{32}{3} = \frac{4}{3} \]

评分标准

(a)
- M1: For obtaining an integral of the form $ k(3x+4)^{\frac{1}{2}} $, where $ k $ is a constant (power increased by 1 and divided by the coefficient of $ x $).
- A1: For $ \frac{16}{3}(3x+4)^{\frac{1}{2}} $ or $ \frac{16}{3}\sqrt{3x+4} $ (condone omission of $ + c $).

(b)
- M1: Attempt to find the area under the line from $ x = 0 $ to $ x = 4 $ by integration or using the formula for the area of a trapezium.
- A1: Correct area under the line $ = 12 $.
- M1: Correct application of limits 0 and 4 to their integrated curve expression from part (a).
- A1: Correct area under the curve $ = \frac{32}{3} $.
- A1: Correct exact enclosed area $ \frac{4}{3} $ (or $ 1\frac{1}{3} $).

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卷二 Pure Mathematics 2

Answer all 8 questions. Show all necessary working clearly. Non-exact numerical answers correct to 3 s.f. or 1 d.p. for angles.

47 题目 · 322.5 分

题目 1 · Exponential & Logarithmic Integration

5.5 分

(a) Find $\int \left( \frac{6}{2x + 1} + 4\mathrm{e}^{2x - 1} \right) \mathrm{d}x$.

(b) Hence find the exact value of $\int_{\frac{1}{2}}^{1} \left( \frac{6}{2x + 1} + 4\mathrm{e}^{2x - 1} \right) \mathrm{d}x$, giving your answer in the form $a \ln\left(\frac{3}{2}\right) + b\mathrm{e} + c$, where $a, b$ and $c$ are integers.

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解题

(a) Integrating each term separately:
\[ \int \frac{6}{2x + 1} \, \mathrm{d}x = 6 \times \frac{1}{2} \ln(2x + 1) = 3\ln(2x + 1) \]
\[ \int 4\mathrm{e}^{2x - 1} \, \mathrm{d}x = 4 \times \frac{1}{2} \mathrm{e}^{2x - 1} = 2\mathrm{e}^{2x - 1} \]
Thus, the indefinite integral is:
\[ 3\ln(2x + 1) + 2\mathrm{e}^{2x - 1} + C \]

(b) Applying the limits $x = 1$ and $x = \frac{1}{2}$:
\[ \left[ 3\ln(2x + 1) + 2\mathrm{e}^{2x - 1} \right]_{\frac{1}{2}}^{1} \]
At $x = 1$:
\[ 3\ln(2(1) + 1) + 2\mathrm{e}^{2(1) - 1} = 3\ln 3 + 2\mathrm{e}^1 = 3\ln 3 + 2\mathrm{e} \]
At $x = \frac{1}{2}$:
\[ 3\ln\left(2\left(\frac{1}{2}\right) + 1\right) + 2\mathrm{e}^{2\left(\frac{1}{2}\right) - 1} = 3\ln 2 + 2\mathrm{e}^0 = 3\ln 2 + 2 \]
Subtracting the lower limit from the upper limit:
\[ (3\ln 3 + 2\mathrm{e}) - (3\ln 2 + 2) = 3(\ln 3 - \ln 2) + 2\mathrm{e} - 2 \]
Using the logarithm law $\ln 3 - \ln 2 = \ln\left(\frac{3}{2}\right)$:
\[ = 3\ln\left(\frac{3}{2}\right) + 2\mathrm{e} - 2 \]
Here, $a = 3$, $b = 2$, and $c = -2$.

评分标准

(a)
- M1: For integrating to obtain at least one term of the correct form, i.e. $k\ln(2x + 1)$ or $p\mathrm{e}^{2x - 1}$.
- A1: Fully correct expression $3\ln(2x + 1) + 2\mathrm{e}^{2x - 1}$ (condone omission of $+ C$).

(b)
- M1: Correct substitution of both limits $x = 1$ and $x = \frac{1}{2}$ into an integrated expression of the form $A\ln(2x + 1) + B\mathrm{e}^{2x - 1}$ and subtraction attempted.
- M1: Correct use of the logarithm quotient rule $\ln 3 - \ln 2 = \ln\left(\frac{3}{2}\right)$ and simplifying $\mathrm{e}^0 = 1$.
- A1.5: Obtain $3\ln\left(\frac{3}{2}\right) + 2\mathrm{e} - 2$ in exact form with integers $a = 3, b = 2, c = -2$.

题目 2 · Exponential & Logarithmic Integration

5.5 分

(a) Use the trapezium rule with 2 intervals of equal width to find an approximation to the value of $\int_0^4 \left( 6\mathrm{e}^{-\frac{1}{2}x} + \frac{2}{3x + 4} \right) \mathrm{d}x$, giving your answer correct to 3 decimal places.

(b) Find the exact value of $\int_0^4 \left( 6\mathrm{e}^{-\frac{1}{2}x} + \frac{2}{3x + 4} \right) \mathrm{d}x$, giving your answer in the form $p + q\mathrm{e}^{-2} + r\ln 2$, where $p, q$ and $r$ are rational numbers.

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解题

(a) The width of each interval is $h = \frac{4 - 0}{2} = 2$.
The $x$-values are $x_0 = 0$, $x_1 = 2$, and $x_2 = 4$.

Calculate the corresponding $y$-values:
\[ y_0 = 6\mathrm{e}^{0} + \frac{2}{3(0) + 4} = 6 + 0.5 = 6.5 \]
\[ y_1 = 6\mathrm{e}^{-1} + \frac{2}{3(2) + 4} = \frac{6}{\mathrm{e}} + 0.2 \approx 2.20728 + 0.2 = 2.40728 \]
\[ y_2 = 6\mathrm{e}^{-2} + \frac{2}{3(4) + 4} = \frac{6}{\mathrm{e}^2} + \frac{2}{16} \approx 0.81201 + 0.125 = 0.93701 \]

Using the trapezium rule:
\[ \text{Area} \approx \frac{1}{2} \times 2 \times \left[ y_0 + y_2 + 2y_1 \right] \]
\[ = 1 \times \left[ 6.5 + 0.93701 + 2(2.40728) \right] \]
\[ = 6.5 + 0.93701 + 4.81456 = 12.25157 \approx 12.252 \]

(b) Integrating the expression analytically:
\[ \int \left( 6\mathrm{e}^{-\frac{1}{2}x} + \frac{2}{3x + 4} \right) \mathrm{d}x = \frac{6}{-\frac{1}{2}}\mathrm{e}^{-\frac{1}{2}x} + \frac{2}{3}\ln(3x + 4) = -12\mathrm{e}^{-\frac{1}{2}x} + \frac{2}{3}\ln(3x + 4) \]
Evaluating between $x = 0$ and $x = 4$:
At $x = 4$:
\[ -12\mathrm{e}^{-2} + \frac{2}{3}\ln(16) = -12\mathrm{e}^{-2} + \frac{2}{3}\ln(2^4) = -12\mathrm{e}^{-2} + \frac{8}{3}\ln 2 \]
At $x = 0$:
\[ -12\mathrm{e}^{0} + \frac{2}{3}\ln(4) = -12 + \frac{2}{3}\ln(2^2) = -12 + \frac{4}{3}\ln 2 \]
Subtracting the lower limit from the upper limit:
\[ \left(-12\mathrm{e}^{-2} + \frac{8}{3}\ln 2\right) - \left(-12 + \frac{4}{3}\ln 2\right) = 12 - 12\mathrm{e}^{-2} + \left(\frac{8}{3} - \frac{4}{3}\right)\ln 2 = 12 - 12\mathrm{e}^{-2} + \frac{4}{3}\ln 2 \]
Hence, $p = 12$, $q = -12$, and $r = \frac{4}{3}$.

评分标准

(a)
- M1: State or use correct interval width $h = 2$ and evaluate the function at $x = 0, 2, 4$.
- M1: Apply the trapezium rule structure correctly: $\frac{1}{2} \times 2 \times [y_0 + y_2 + 2y_1]$.
- A0.5: Obtain $12.252$ (must be rounded to 3 decimal places).

(b)
- M1: Obtain integral of the form $k\mathrm{e}^{-\frac{1}{2}x} + c\ln(3x + 4)$ where $k, c \neq 0$.
- A1: Correct indefinite integral $-12\mathrm{e}^{-\frac{1}{2}x} + \frac{2}{3}\ln(3x + 4)$.
- A1: Substitute limits correctly and use power rule for logarithms to obtain $12 - 12\mathrm{e}^{-2} + \frac{4}{3}\ln 2$ (or exact rational equivalents).

题目 3 · Trigonometric Equations & Identities

4 分

Solve the equation
\[ 2\cot^2\theta - 5\operatorname{cosec}\theta + 4 = 0 \]
for $ 0^\circ < \theta < 360^\circ $.

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解题

Use the identity $ 1 + \cot^2\theta = \operatorname{cosec}^2\theta $, so $ \cot^2\theta = \operatorname{cosec}^2\theta - 1 $.

Substitute into the given equation:
\[ 2(\operatorname{cosec}^2\theta - 1) - 5\operatorname{cosec}\theta + 4 = 0 \]
\[ 2\operatorname{cosec}^2\theta - 2 - 5\operatorname{cosec}\theta + 4 = 0 \]
\[ 2\operatorname{cosec}^2\theta - 5\operatorname{cosec}\theta + 2 = 0 \]

Factorise the quadratic equation:
\[ (2\operatorname{cosec}\theta - 1)(\operatorname{cosec}\theta - 2) = 0 \]

This gives:
\[ \operatorname{cosec}\theta = \frac{1}{2} \quad \text{or} \quad \operatorname{cosec}\theta = 2 \]

Since $ \operatorname{cosec}\theta = \frac{1}{\sin\theta} $:
- If $ \operatorname{cosec}\theta = \frac{1}{2} $, then $ \sin\theta = 2 $, which has no real solutions as $ -1 \leqslant \sin\theta \leqslant 1 $.
- If $ \operatorname{cosec}\theta = 2 $, then $ \sin\theta = \frac{1}{2} $.

For $ 0^\circ < \theta < 360^\circ $:
\[ \theta = 30^\circ, \quad 180^\circ - 30^\circ = 150^\circ \]

评分标准

M1: Use the identity $ \cot^2\theta = \operatorname{cosec}^2\theta - 1 $ to form a three-term quadratic in $ \operatorname{cosec}\theta $ (or in $ \sin\theta $).

A1: Obtain correct quadratic equation $ 2\operatorname{cosec}^2\theta - 5\operatorname{cosec}\theta + 2 = 0 $ (or $ 2\sin^2\theta - 5\sin\theta + 2 = 0 $).

M1: Solve the 3-term quadratic in $ \operatorname{cosec}\theta $ or $ \sin\theta $ to reach $ \sin\theta = \frac{1}{2} $ and find at least one value for $ \theta $.

A1: Obtain $ \theta = 30^\circ $ and $ \theta = 150^\circ $ and no extra solutions in the given range.

题目 4 · structured

5 分

The variables $x$ and $y$ satisfy the equation $y = A x^n$, where $A$ and $n$ are constants. The graph of $\ln y$ against $\ln x$ is a straight line passing through the points $(1.25, 4.10)$ and $(3.65, 0.74)$.

(a) Find the values of $n$ and $A$, giving the value of $A$ correct to 3 significant figures. [4]

(b) Hence find the value of $y$ when $x = 5.2$. [1]

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解题

(a)
Taking the natural logarithm of both sides of $y = A x^n$:
\[ \ln y = \ln(A x^n) = \ln A + n \ln x \]
This is in the linear form $Y = mX + c$, where $Y = \ln y$, $X = \ln x$, gradient $m = n$, and intercept $c = \ln A$.

Calculate the gradient $n$:
\[ n = \frac{0.74 - 4.10}{3.65 - 1.25} = \frac{-3.36}{2.40} = -1.4 \]

Substitute $n = -1.4$ and the point $(1.25, 4.10)$ into the linear equation:
\[ 4.10 = \ln A + (-1.4)(1.25) \]
\[ 4.10 = \ln A - 1.75 \implies \ln A = 4.10 + 1.75 = 5.85 \]
\[ A = \mathrm{e}^{5.85} \approx 347.23... = 347 \text{ (to 3 s.f.)} \]

(b)
When $x = 5.2$:
\[ \ln y = 5.85 - 1.4 \ln(5.2) = 5.85 - 1.4(1.64866...) \approx 3.54188 \]
\[ y = \mathrm{e}^{3.54188} \approx 34.5 \]

评分标准

(a)
- M1: State or imply the linear form $\ln y = \ln A + n \ln x$
- M1: Attempt to find gradient using $\frac{y_2 - y_1}{x_2 - x_1}$
- A1: Obtain $n = -1.4$ (or $-\frac{7}{5}$)
- A1: Obtain $A = 347$ (accept 347.2 from unrounded $\ln A$)

(b)
- B1: Obtain $y = 34.5$ (allow 34.5 to 34.6)

题目 5 · Differentiation (Quotient & Parametric)

7.5 分

The equation of a curve is $ y = \frac{\mathrm{e}^{2x}}{3x - 1} $ for $ x > \frac{1}{3} $.

(a) Find $ \frac{\mathrm{d}y}{\mathrm{d}x} $ and hence determine the exact coordinates of the stationary point on the curve.

(b) Find the equation of the tangent to the curve at the point where $ x = 1 $, giving your answer in the form $ y = mx + c $, where $ m $ and $ c $ are expressed in terms of $ \mathrm{e} $.

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解题

(a) Using the quotient rule with $ u = \mathrm{e}^{2x} $ and $ v = 3x - 1 $:
\[ \frac{\mathrm{d}u}{\mathrm{d}x} = 2\mathrm{e}^{2x}, \quad \frac{\mathrm{d}v}{\mathrm{d}x} = 3 \]
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2\mathrm{e}^{2x}(3x - 1) - 3\mathrm{e}^{2x}}{(3x - 1)^2} = \frac{\mathrm{e}^{2x}(6x - 2 - 3)}{(3x - 1)^2} = \frac{\mathrm{e}^{2x}(6x - 5)}{(3x - 1)^2} \]

At a stationary point, $ \frac{\mathrm{d}y}{\mathrm{d}x} = 0 $:
Since $ \mathrm{e}^{2x} \neq 0 $, we have $ 6x - 5 = 0 \implies x = \frac{5}{6} $.

Substitute $ x = \frac{5}{6} $ into the curve equation:
\[ y = \frac{\mathrm{e}^{2(5/6)}}{3(5/6) - 1} = \frac{\mathrm{e}^{5/3}}{\frac{5}{2} - 1} = \frac{\mathrm{e}^{5/3}}{\frac{3}{2}} = \frac{2}{3}\mathrm{e}^{5/3} \]
So the stationary point is $ \left(\frac{5}{6}, \frac{2}{3}\mathrm{e}^{5/3}\right) $.

(b) When $ x = 1 $:
\[ y = \frac{\mathrm{e}^{2(1)}}{3(1) - 1} = \frac{\mathrm{e}^2}{2} = \frac{1}{2}\mathrm{e}^2 \]
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{e}^{2}(6(1) - 5)}{(3(1) - 1)^2} = \frac{\mathrm{e}^2(1)}{2^2} = \frac{1}{4}\mathrm{e}^2 \]

The equation of the tangent is:
\[ y - \frac{1}{2}\mathrm{e}^2 = \frac{1}{4}\mathrm{e}^2(x - 1) \]
\[ y = \frac{1}{4}\mathrm{e}^2 x - \frac{1}{4}\mathrm{e}^2 + \frac{1}{2}\mathrm{e}^2 \]
\[ y = \frac{1}{4}\mathrm{e}^2 x + \frac{1}{4}\mathrm{e}^2 \]

评分标准

(a) [4.5 marks]
- M1: Apply the quotient rule correctly with $ u = \mathrm{e}^{2x} $ and $ v = 3x - 1 $ (or product rule with $ (3x - 1)^{-1} $).
- A1: Obtain fully correct unsimplified or simplified derivative $ \frac{2\mathrm{e}^{2x}(3x - 1) - 3\mathrm{e}^{2x}}{(3x - 1)^2} $.
- M1: Set derivative (or numerator) equal to 0 and attempt to solve for $ x $.
- A1: Obtain $ x = \frac{5}{6} $.
- A0.5: Obtain exact $ y = \frac{2}{3}\mathrm{e}^{5/3} $.

(b) [3 marks]
- M1: Find the value of $ y $ and the gradient $ \frac{\mathrm{d}y}{\mathrm{d}x} $ at $ x = 1 $.
- A1: Obtain $ y = \frac{1}{2}\mathrm{e}^2 $ and $ m = \frac{1}{4}\mathrm{e}^2 $.
- A1: Form the correct tangent equation in the form $ y = mx + c $, obtaining $ y = \frac{1}{4}\mathrm{e}^2 x + \frac{1}{4}\mathrm{e}^2 $.

题目 6 · structured

7 分

The equation of a curve is
\[ y = \frac{2\mathrm{e}^{2x}}{3\mathrm{e}^x - 1}, \quad \text{for } x > -\ln 3. \]

(a) Find an expression for $ \frac{\mathrm{d}y}{\mathrm{d}x} $, expressing your answer in factorised form. [3]

(b) Hence find the exact coordinates of the stationary point on the curve. [4]

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解题

(a)
Using the quotient rule with $ u = 2\mathrm{e}^{2x} $ and $ v = 3\mathrm{e}^x - 1 $:
\[ \frac{\mathrm{d}u}{\mathrm{d}x} = 4\mathrm{e}^{2x}, \quad \frac{\mathrm{d}v}{\mathrm{d}x} = 3\mathrm{e}^x \]

\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v\frac{\mathrm{d}u}{\mathrm{d}x} - u\frac{\mathrm{d}v}{\mathrm{d}x}}{v^2} = \frac{(3\mathrm{e}^x - 1)(4\mathrm{e}^{2x}) - (2\mathrm{e}^{2x})(3\mathrm{e}^x)}{(3\mathrm{e}^x - 1)^2} \]
\[ = \frac{12\mathrm{e}^{3x} - 4\mathrm{e}^{2x} - 6\mathrm{e}^{3x}}{(3\mathrm{e}^x - 1)^2} = \frac{6\mathrm{e}^{3x} - 4\mathrm{e}^{2x}}{(3\mathrm{e}^x - 1)^2} \]
Factoring the numerator gives:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2\mathrm{e}^{2x}(3\mathrm{e}^x - 2)}{(3\mathrm{e}^x - 1)^2} \]

(b)
At a stationary point, $ \frac{\mathrm{d}y}{\mathrm{d}x} = 0 $:
\[ \frac{2\mathrm{e}^{2x}(3\mathrm{e}^x - 2)}{(3\mathrm{e}^x - 1)^2} = 0 \]
Since $ 2\mathrm{e}^{2x} \neq 0 $ for all real $ x $:
\[ 3\mathrm{e}^x - 2 = 0 \implies \mathrm{e}^x = \frac{2}{3} \implies x = \ln\left(\frac{2}{3}\right) \]

Substitute $ \mathrm{e}^x = \frac{2}{3} $ into the curve equation to find $ y $:
\[ y = \frac{2\left(\frac{2}{3}\right)^2}{3\left(\frac{2}{3}\right) - 1} = \frac{2 \left(\frac{4}{9}\right)}{2 - 1} = \frac{8}{9} \]
Thus, the exact coordinates are $ \left(\ln\left(\frac{2}{3}\right), \frac{8}{9}\right) $ or $ \left(-\ln\left(\frac{3}{2}\right), \frac{8}{9}\right) $.

评分标准

(a)
- M1: Valid application of the quotient rule (or product rule with $ (3\mathrm{e}^x - 1)^{-1} $), showing correct differentiation of $ 2\mathrm{e}^{2x} $ and $ 3\mathrm{e}^x - 1 $.
- A1: Correct unsimplified expression, e.g. $ \frac{(3\mathrm{e}^x - 1)(4\mathrm{e}^{2x}) - 2\mathrm{e}^{2x}(3\mathrm{e}^x)}{(3\mathrm{e}^x - 1)^2} $.
- A1: Correct fully factorised final expression $ \frac{2\mathrm{e}^{2x}(3\mathrm{e}^x - 2)}{(3\mathrm{e}^x - 1)^2} $.

(b)
- M1: Sets their numerator equal to 0 and attempts to solve for $ \mathrm{e}^x $.
- A1: Obtains $ x = \ln\left(\frac{2}{3}\right) $ or $ x = -\ln\left(\frac{3}{2}\right) $.
- M1: Substitutes their $ \mathrm{e}^x $ value into the original expression for $ y $.
- A1: Obtains exact $ y = \frac{8}{9} $ and states coordinates clearly.

题目 7 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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解题

(a) Squaring both sides of the inequality:
\[(3x - 2)^2 < (x + 4)^2\]
\[9x^2 - 12x + 4 < x^2 + 8x + 16\]
\[8x^2 - 20x - 12 < 0\]
Dividing by 4:
\[2x^2 - 5x - 3 < 0\]
\[(2x + 1)(x - 3) < 0\]
The critical values are $x = -\frac{1}{2}$ and $x = 3$.
Therefore, the solution set is $-\frac{1}{2} < x < 3$.

(b) Using the power law of logarithms on $2\ln y$:
\[\ln(2y + 3) = \ln(y^2) + \ln 2\]
Combining the logarithms on the right-hand side using the addition rule:
\[\ln(2y + 3) = \ln(2y^2)\]
Equating arguments:
\[2y + 3 = 2y^2\]
\[2y^2 - 2y - 3 = 0\]
Applying the quadratic formula:
\[y = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(2)(-3)}}{2(2)} = \frac{2 \pm \sqrt{4 + 24}}{4} = \frac{2 \pm \sqrt{28}}{4} = \frac{1 \pm \sqrt{7}}{2}\]
Since $\ln y$ is only defined for $y > 0$, the negative root $y = \frac{1 - \sqrt{7}}{2} \approx -0.823$ must be rejected.

Thus, $y = \frac{1 + \sqrt{7}}{2} \approx 1.82$.

评分标准

(a)
- M1: Attempt to solve by squaring both sides to obtain a 3-term quadratic, or by setting up and solving the two linear equations $3x - 2 = x + 4$ and $3x - 2 = -(x + 4)$.
- A1: Obtain both correct critical values $x = -\frac{1}{2}$ (or $-0.5$) and $x = 3$.
- A1: State the correct inequality $-\frac{1}{2} < x < 3$ (or $-0.5 < x < 3$).

(b)
- M1: Use the power law $2\ln y = \ln(y^2)$.
- M1: Use product rule $\ln(y^2) + \ln 2 = \ln(2y^2)$ and remove logarithms to obtain an algebraic equation in $y$.
- A1: Obtain the correct 3-term quadratic equation $2y^2 - 2y - 3 = 0$ (or equivalent).
- A1: Solve the quadratic and discard the invalid negative root, stating $y = 1.82$ (or exact $\frac{1 + \sqrt{7}}{2}$).

题目 8 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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解题

评分标准

题目 9 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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解题

评分标准

题目 10 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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解题

评分标准

题目 11 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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解题

评分标准

题目 12 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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解题

评分标准

题目 13 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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解题

评分标准

题目 14 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 15 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 16 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 17 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 18 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 19 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 20 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 21 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 22 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 23 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 24 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 25 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 26 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 27 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 28 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 29 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 30 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 31 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 32 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 33 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 34 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 35 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 36 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 37 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 38 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 39 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 40 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 41 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 42 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 43 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 44 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 45 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 46 · structured

7 分

(a) Solve the inequality $|3x - 2| < |x + 4|$.

(b) Solve the equation $\ln(2y + 3) = 2\ln y + \ln 2$, giving your answer correct to 3 significant figures.

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题目 47 · Polynomials & Iterative Methods

8 分

The polynomial $ \mathrm{p}(x) $ is defined by
\[ \mathrm{p}(x) = 3x^3 + ax^2 + bx - 4, \]
where $ a $ and $ b $ are constants. It is given that $ (x + 2) $ is a factor of $ \mathrm{p}(x) $, and that when $ \mathrm{p}(x) $ is divided by $ (x - 1) $, the remainder is $ -9 $.

(a) Find the values of $ a $ and $ b $. [3]

(b)
(i) Using the values of $ a $ and $ b $ found in part (a), show that the equation $ \mathrm{p}(x) = 0 $ has a root between $ 1.5 $ and $ 2.0 $. [1]

(ii) Show that the equation $ \mathrm{p}(x) = 0 $ can be rearranged in the form
\[ x = \sqrt{\frac{10x + 4}{3x + 2}}. \] [1]

(iii) Use an iterative formula based on the rearrangement in part (b)(ii),
\[ x_{n+1} = \sqrt{\frac{10x_n + 4}{3x_n + 2}}, \]
with initial value $ x_1 = 1.7 $, to determine the root correct to 3 decimal places. Give the value of each iteration to 5 decimal places. [3]

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(a)
Since $ (x + 2) $ is a factor of $ \mathrm{p}(x) $, by the Factor Theorem:
\[ \mathrm{p}(-2) = 0 \]
\[ 3(-2)^3 + a(-2)^2 + b(-2) - 4 = 0 \]
\[ 3(-8) + 4a - 2b - 4 = 0 \]
\[ 4a - 2b - 28 = 0 \implies 2a - b = 14 \quad \text{--- (1)} \]

Since the remainder when divided by $ (x - 1) $ is $ -9 $, by the Remainder Theorem:
\[ \mathrm{p}(1) = -9 \]
\[ 3(1)^3 + a(1)^2 + b(1) - 4 = -9 \]
\[ 3 + a + b - 4 = -9 \]
\[ a + b - 1 = -9 \implies a + b = -8 \quad \text{--- (2)} \]

Adding equations (1) and (2):
\[ (2a - b) + (a + b) = 14 + (-8) \]
\[ 3a = 6 \implies a = 2 \]

Substituting $ a = 2 $ into (2):
\[ 2 + b = -8 \implies b = -10 \]

---

(b)(i)
With $ a = 2 $ and $ b = -10 $, the polynomial is:
\[ \mathrm{p}(x) = 3x^3 + 2x^2 - 10x - 4 \]
Evaluating at the endpoints:
\[ \mathrm{p}(1.5) = 3(1.5)^3 + 2(1.5)^2 - 10(1.5) - 4 = 10.125 + 4.5 - 15 - 4 = -4.375 < 0 \]
\[ \mathrm{p}(2.0) = 3(2)^3 + 2(2)^2 - 10(2) - 4 = 24 + 8 - 20 - 4 = 8 > 0 \]
Since $ \mathrm{p}(x) $ is continuous and there is a change of sign between $ x = 1.5 $ and $ x = 2.0 $, the equation $ \mathrm{p}(x) = 0 $ has a root in the interval $ (1.5, 2.0) $.

---

(b)(ii)
From $ \mathrm{p}(x) = 0 $:
\[ 3x^3 + 2x^2 - 10x - 4 = 0 \]
\[ x^2(3x + 2) - (10x + 4) = 0 \]
\[ x^2(3x + 2) = 10x + 4 \]
\[ x^2 = \frac{10x + 4}{3x + 2} \]
Taking the positive square root (since the root is positive):
\[ x = \sqrt{\frac{10x + 4}{3x + 2}} \]

---

(b)(iii)
Using $ x_1 = 1.7 $:
\[ x_2 = \sqrt{\frac{10(1.7) + 4}{3(1.7) + 2}} = \sqrt{\frac{21}{7.1}} \approx 1.71981 \]
\[ x_3 = \sqrt{\frac{10(1.71981) + 4}{3(1.71981) + 2}} \approx 1.72072 \]
\[ x_4 = \sqrt{\frac{10(1.72072) + 4}{3(1.72072) + 2}} \approx 1.72076 \]
\[ x_5 = \sqrt{\frac{10(1.72076) + 4}{3(1.72076) + 2}} \approx 1.72076 \]

Hence, the root correct to 3 decimal places is $ 1.721 $.

评分标准

(a)
- M1: Use $ \mathrm{p}(-2) = 0 $ and $ \mathrm{p}(1) = -9 $ to form two linear equations in $ a $ and $ b $.
- A1: Obtain correct pair of equations, e.g. $ 2a - b = 14 $ and $ a + b = -8 $ (or unsimplified equivalents).
- A1: Solve simultaneously to obtain $ a = 2 $ and $ b = -10 $.

(b)(i)
- B1: Calculate $ \mathrm{p}(1.5) = -4.375 $ (or $ -\frac{35}{8} $) and $ \mathrm{p}(2) = 8 $, and state conclusion based on sign change.

(b)(ii)
- B1: Show complete and correct algebraic steps from $ 3x^3 + 2x^2 - 10x - 4 = 0 $ to the given rearrangement.

(b)(iii)
- M1: Apply the iteration formula correctly with $ x_1 = 1.7 $ to find $ x_2 $.
- A1: Obtain $ x_2 \approx 1.71981 $, $ x_3 \approx 1.72072 $, and $ x_4 \approx 1.72076 $ (to at least 5 decimal places).
- A1: State final answer $ 1.721 $ (correct to 3 d.p.) with sufficient iterations shown to justify accuracy.

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