Edexcel IAL · Thinka 原創模擬試題

2023 Edexcel IAL Pure Mathematics (YPM01) 模擬試題連答案詳解

Thinka Jan 2023 Cambridge International A Level-Style Mock — Pure Mathematics (YPM01)

300 分360 分鐘2023
分析模擬試題
An original Thinka practice paper modelled on the structure and difficulty of the Jan 2023 Cambridge International A Level Pure Mathematics (YPM01) paper. Not affiliated with or reproduced from Cambridge.

部分 WMA11/01: Pure Mathematics P1

Answer all questions. Show sufficient working. Calculators allowed (no symbolic algebra).
11 題目 · 74.7 分
題目 1 · Short Answer
4 分
Find the exact value of

\(\int_{1}^{4} \left( \frac{6}{\sqrt{x}} - \frac{1}{2}x^2 \right) \text{d}x\)
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解題

To find the exact value of the integral:

1. Rewrite the integrand in a form ready for integration:
\(\frac{6}{\sqrt{x}} - \frac{1}{2}x^2 = 6x^{-\frac{1}{2}} - \frac{1}{2}x^2\)

2. Integrate term by term:
\(\int \left( 6x^{-\frac{1}{2}} - \frac{1}{2}x^2 \right) \text{d}x = \left[ \frac{6x^{\frac{1}{2}}}{\frac{1}{2}} - \frac{\frac{1}{2}x^3}{3} \right] = \left[ 12x^{\frac{1}{2}} - \frac{1}{6}x^3 \right]\)

3. Substitute the upper limit \(x = 4\):
\(12(4)^{\frac{1}{2}} - \frac{1}{6}(4)^3 = 12(2) - \frac{64}{6} = 24 - \frac{32}{3} = \frac{40}{3}\)

4. Substitute the lower limit \(x = 1\):
\(12(1)^{\frac{1}{2}} - \frac{1}{6}(1)^3 = 12 - \frac{1}{6} = \frac{71}{6}\)

5. Subtract the lower limit result from the upper limit result:
\(\frac{40}{3} - \frac{71}{6} = \frac{80}{6} - \frac{71}{6} = \frac{9}{6} = \frac{3}{2}\)

評分準則

M1: For an attempt to integrate at least one term (power of \(x\) increased by 1).
A1: Correct integration: \(12x^{\frac{1}{2}} - \frac{1}{6}x^3\) (ignore constant of integration).
M1: For substituting both limits 4 and 1 into their integrated expression and subtracting the correct way round.
A1: Correct exact value of \(\frac{3}{2}\) or \(1.5\) (or any equivalent fraction).
題目 2 · Short Answer
5 分
The curve \(C\) has equation

\[y = \frac{4x^3 - 5}{2x}, \quad x > 0\]

Find the equation of the tangent to \(C\) at the point where \(x = 1\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers to be found.
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解題

1. Express \(y\) in a form suitable for differentiation:
\[y = \frac{4x^3}{2x} - \frac{5}{2x} = 2x^2 - \frac{5}{2}x^{-1}\]

2. Find the coordinate of the point where \(x = 1\):
\[y = 2(1)^2 - \frac{5}{2}(1)^{-1} = 2 - 2.5 = -0.5 = -\frac{1}{2}\]
So the point of contact is \((1, -0.5)\).

3. Differentiate \(y\) with respect to \(x\) to find the gradient function:
\[\frac{\text{d}y}{\text{d}x} = 4x - \left(-\frac{5}{2}\right)x^{-2} = 4x + \frac{5}{2}x^{-2}\]

4. Evaluate the gradient at \(x = 1\):
\[m = 4(1) + \frac{5}{2}(1)^{-2} = 4 + 2.5 = 6.5 = \frac{13}{2}\]

5. Find the equation of the tangent line using \(y - y_1 = m(x - x_1)\):
\[y - \left(-\frac{1}{2}\right) = \frac{13}{2}(x - 1)\]
\[y + \frac{1}{2} = \frac{13}{2}x - \frac{13}{2}\]
Multiply the entire equation by 2 to clear fractions:
\[2y + 1 = 13x - 13\]
Rearrange to the form \(ax + by + c = 0\):
\[13x - 2y - 14 = 0\]

評分準則

M1: For attempting to expand or split the fraction to express \(y\) as separate terms in powers of \(x\).
A1: Correct simplified expression for \(y = 2x^2 - \frac{5}{2}x^{-1}\) (or equivalent index form).
M1: Differentiates their expression. Look for at least one term differentiated correctly (power multiplied by coefficient and power reduced by 1).
A1: Correct derivative \(\frac{\text{d}y}{\text{d}x} = 4x + \frac{5}{2}x^{-2}\). Finds correct gradient \(m = \frac{13}{2}\) and correct coordinate \(y = -\frac{1}{2}\).
M1: Applies the tangent formula \(y - y_1 = m(x - x_1)\) using their calculated \(y\) and \(m\) at \(x=1\).
A1: Correct equation in the form \(13x - 2y - 14 = 0\) (or any integer multiple thereof, e.g., \(26x - 4y - 28 = 0\)).
題目 3 · Short Answer
4 分
Solve, for \(0 \le \theta < 360^\circ\), the equation

\[3\cos^2\theta - 5\sin\theta - 1 = 0\]

giving your answers to one decimal place.
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解題

1. Use the trigonometric identity \(\cos^2\theta = 1 - \sin^2\theta\) to rewrite the equation in terms of \(\sin\theta\):
\[3(1 - \sin^2\theta) - 5\sin\theta - 1 = 0\]
\[3 - 3\sin^2\theta - 5\sin\theta - 1 = 0\]
\[-3\sin^2\theta - 5\sin\theta + 2 = 0\]
Multiply by \(-1\) to make the leading term positive:
\[3\sin^2\theta + 5\sin\theta - 2 = 0\]

2. Factorise the quadratic equation:
\[(3\sin\theta - 1)(\sin\theta + 2) = 0\]

3. Solve for \(\sin\theta\):
- Either \(3\sin\theta - 1 = 0 \implies \sin\theta = \frac{1}{3}\)
- Or \(\sin\theta + 2 = 0 \implies \sin\theta = -2\) (which has no real solutions since \(-1 \le \sin\theta \le 1\))

4. Find the values of \(\theta\) for \(\sin\theta = \frac{1}{3}\) in the range \(0 \le \theta < 360^\circ\):
- Primary solution: \(\theta = \arcsin\left(\frac{1}{3}\right) \approx 19.47^\circ\), which rounds to \(19.5^\circ\).
- Secondary solution: \(\theta = 180^\circ - 19.47^\circ \approx 160.53^\circ\), which rounds to \(160.5^\circ\).

評分準則

M1: Substitution of \(\cos^2\theta = 1 - \sin^2\theta\) to form an equation in \(\sin\theta\).
A1: Correct three-term quadratic equation, e.g., \(3\sin^2\theta + 5\sin\theta - 2 = 0\) (or equivalent).
M1: Solves the quadratic to find a value for \(\sin\theta\) (must be \(\frac{1}{3}\)), and finds at least one correct principal value for \(\theta\) to 1 d.p.
A1: Correct final answers of \(19.5^\circ\) and \(160.5^\circ\) only (with no extra angles in the range).
題目 4 · Short Answer
5 分
The line \(l_1\) has equation \(2x + 3y - 6 = 0\).

The line \(l_2\) is perpendicular to \(l_1\) and passes through the point \(P(4, 7)\).

(a) Find an equation for \(l_2\) in the form \(y = mx + c\).

(b) Find the coordinates of the point of intersection of \(l_1\) and \(l_2\).
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解題

**(a)**
First, express the equation of \(l_1\) in gradient-intercept form:
\[2x + 3y - 6 = 0 \implies 3y = -2x + 6 \implies y = -\frac{2}{3}x + 2\]
So the gradient of \(l_1\) is \(m_1 = -\frac{2}{3}\).

Since \(l_2\) is perpendicular to \(l_1\), the gradient of \(l_2\) is:
\[m_2 = -\frac{1}{m_1} = \frac{3}{2}\]

Using the point-slope form with point \(P(4, 7)\):
\[y - 7 = \frac{3}{2}(x - 4)\]
\[y - 7 = \frac{3}{2}x - 6\]
\[y = \frac{3}{2}x + 1\]

**(b)**
To find the point of intersection, substitute \(y = \frac{3}{2}x + 1\) into the equation of \(l_1\):
\[2x + 3\left(\frac{3}{2}x + 1\right) - 6 = 0\]
\[2x + \frac{9}{2}x + 3 - 6 = 0\]
\[\frac{13}{2}x - 3 = 0\]
\[\frac{13}{2}x = 3 \implies x = \frac{6}{13}\]

Substitute \(x = \frac{6}{13}\) back into the equation of \(l_2\):
\[y = \frac{3}{2}\left(\frac{6}{13}\right) + 1 = \frac{9}{13} + 1 = \frac{22}{13}\]

So the point of intersection is \(\left(\frac{6}{13}, \frac{22}{13}\right)\).

評分準則

**(a)**
M1: Attempts to find the gradient of \(l_1\) and uses \(m_1 m_2 = -1\) to find the gradient of \(l_2\).
A1: Correct gradient of \(l_2\) is \(\frac{3}{2}\).
A1: Correct equation of \(l_2\) in the form \(y = mx + c\), which is \(y = \frac{3}{2}x + 1\).

**(b)**
M1: Eliminates one variable (either \(x\) or \(y\)) to find the intersection point.
A1: Correct coordinates \(\left(\frac{6}{13}, \frac{22}{13}\right)\) (both must be exact fractions or equivalent).
題目 5 · Structured
8.1 分
The curve \(C\) has equation \(y = x^2 - 4kx + 2\), where \(k\) is a constant.
The line \(L\) has equation \(y = 2x - k\).

(a) Show that the \(x\)-coordinates of the points of intersection of \(C\) and \(L\) satisfy the equation
\[x^2 - 2(2k+1)x + k + 2 = 0\]

(b) Given that \(C\) and \(L\) do not intersect, find the set of possible values of \(k\).
查看答案詳解

解題

(a) Equating the equations of the curve and line:
\[x^2 - 4kx + 2 = 2x - k\]
Rearranging to make the right-hand side zero:
\[x^2 - 4kx - 2x + k + 2 = 0\]
\[x^2 - 2(2k+1)x + k + 2 = 0\] as required.

(b) For the curve and line to not intersect, this quadratic equation in \(x\) must have no real roots. Therefore, the discriminant must be less than zero:
\[b^2 - 4ac < 0\]
Here, \(a = 1\), \(b = -2(2k+1)\), and \(c = k+2\).
\[[-2(2k+1)]^2 - 4(1)(k+2) < 0\]
\[4(4k^2 + 4k + 1) - 4(k+2) < 0\]
Divide both sides by 4:
\[4k^2 + 4k + 1 - (k+2) < 0\]
\[4k^2 + 3k - 1 < 0\]
Factorising the quadratic inequality:
\[(4k-1)(k+1) < 0\]
The critical values are \(k = -1\) and \(k = 0.25\).
Since the inequality is less than zero, the solution is the region between these values:
\[-1 < k < 0.25\]

評分準則

(a)
M1: Equating the curve and line expressions to set up an equation.
A1: Rearranging terms to show the given equation with correct working.

(b)
M1: Attempting to use the discriminant condition \(b^2 - 4ac < 0\).
A1: Correct substitution of \(a\), \(b\), and \(c\) into the discriminant expression.
M1: Expanding and simplifying to a three-term quadratic inequality, e.g., \(4k^2 + 3k - 1 < 0\).
M1: Finding critical values of their quadratic inequality (factorising or using formula).
A1.1: Correct set of values: \(-1 < k < 0.25\) (or equivalent notation, e.g., interval notation).
題目 6 · Structured
8.1 分
The points \(A\) and \(B\) have coordinates \((-2, 5)\) and \((6, 1)\) respectively.

(a) Find an equation of the perpendicular bisector of \(AB\), giving your answer in the form \(ay + bx + c = 0\), where \(a\), \(b\) and \(c\) are integers to be found.

The perpendicular bisector of \(AB\) crosses the \(x\)-axis at the point \(P\).

(b) Find the area of the triangle \(ABP\).
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解題

(a) First, find the midpoint \(M\) of \(AB\):
\[M = \left(\frac{-2+6}{2}, \frac{5+1}{2}\right) = (2, 3)\]
Next, find the gradient of \(AB\):
\[m_{AB} = \frac{1 - 5}{6 - (-2)} = \frac{-4}{8} = -\frac{1}{2}\]
The gradient of the perpendicular bisector is the negative reciprocal:
\[m_{\perp} = -\frac{1}{m_{AB}} = 2\]
Using the point-slope form with \(M(2,3)\) and \(m=2\):
\[y - 3 = 2(x - 2)\]
\[y - 3 = 2x - 4\]
Rearranging to the form \(ay + bx + c = 0\):
\[y - 2x + 1 = 0\] (or any integer multiple, e.g., \(2x - y - 1 = 0\)).

(b) The point \(P\) is on the \(x\)-axis, so let \(y = 0\) in the equation of the perpendicular bisector:
\[0 - 2x + 1 = 0 \implies x = 0.5\]
So, \(P\) has coordinates \((0.5, 0)\).
To find the area of triangle \(ABP\), we can use the coordinates of the vertices \(A(-2, 5)\), \(B(6, 1)\), and \(P(0.5, 0)\):
\[\text{Area} = \frac{1}{2} |x_A(y_B - y_P) + x_B(y_P - y_A) + x_P(y_A - y_B)|\]
\[\text{Area} = \frac{1}{2} |-2(1 - 0) + 6(0 - 5) + 0.5(5 - 1)|\]
\[\text{Area} = \frac{1}{2} |-2 - 30 + 2| = \frac{1}{2} |-30| = 15\]

評分準則

(a)
M1: Finding the midpoint of \(AB\).
M1: Finding the gradient of \(AB\).
M1: Finding the gradient of the perpendicular bisector (negative reciprocal).
M1: Applying \(y - y_1 = m(x - x_1)\) with their midpoint and perpendicular gradient.
A1: Obtaining a correct equation in the required form, e.g., \(y - 2x + 1 = 0\).

(b)
M1: Finding the coordinates of \(P\) by setting \(y = 0\).
M1: Using an appropriate method to find the area of the triangle (e.g., Shoelace formula, or finding length of base \(AB\) and perpendicular height \(PM\)).
A1.1: Correct area of \(15\).
題目 7 · Structured
8.1 分
(a) Show that the equation
\[6\sin^2 \theta - \cos \theta - 4 = 0\]
can be written as
\[6\cos^2 \theta + \cos \theta - 2 = 0\]

(b) Hence, solve for \(0 \le x < 360^\circ\), the equation
\[6\sin^2 (2x + 10^\circ) - \cos (2x + 10^\circ) - 4 = 0\]
giving your answers to 1 decimal place.
查看答案詳解

解題

(a) Using the identity \(\sin^2 \theta = 1 - \cos^2 \theta\):
\[6(1 - \cos^2 \theta) - \cos \theta - 4 = 0\]
\[6 - 6\cos^2 \theta - \cos \theta - 4 = 0\]
\[-6\cos^2 \theta - \cos \theta + 2 = 0\]
Multiplying by \(-1\):
\[6\cos^2 \theta + \cos \theta - 2 = 0\] as required.

(b) Let \(\theta = 2x + 10^\circ\). Since \(0 \le x < 360^\circ\), we have:
\[0 \le 2x < 720^\circ \implies 10^\circ \le \theta < 730^\circ\]
From part (a), \(6\cos^2 \theta + \cos \theta - 2 = 0\).
Factorising this quadratic equation:
\[(3\cos \theta + 2)(2\cos \theta - 1) = 0\]
So, \(\cos \theta = -\frac{2}{3}\) or \(\cos \theta = \frac{1}{2}\).

Case 1: \(\cos \theta = \frac{1}{2}\)
Within the interval \(10^\circ \le \theta < 730^\circ\):
\[\theta = 60^\circ, 300^\circ, 420^\circ, 660^\circ\]
Solving for \(x\) where \(x = \frac{\theta - 10^\circ}{2}\):
- \(\theta = 60^\circ \implies x = 25^\circ\)
- \(\theta = 300^\circ \implies x = 145^\circ\)
- \(\theta = 420^\circ \implies x = 205^\circ\)
- \(\theta = 660^\circ \implies x = 325^\circ\)

Case 2: \(\cos \theta = -\frac{2}{3}\)
The principal value is \(\theta = \cos^{-1}(-2/3) \approx 131.81^\circ\).
Other values in the interval \(10^\circ \le \theta < 730^\circ\) are:
\[\theta \approx 131.81^\circ, 228.19^\circ, 491.81^\circ, 588.19^\circ\]
Solving for \(x\):
- \(\theta = 131.81^\circ \implies x = 60.9^\circ\)
- \(\theta = 228.19^\circ \implies x = 109.1^\circ\)
- \(\theta = 491.81^\circ \implies x = 240.9^\circ\)
- \(\theta = 588.19^\circ \implies x = 289.1^\circ\)

評分準則

(a)
M1: Attempting to use \(\sin^2 \theta = 1 - \cos^2 \theta\) in the equation.
A1: Correct substitution and rearrangement to show the given quadratic in \(\cos \theta\).

(b)
M1: Attempting to factorise or solve the quadratic equation to find values of \(\cos(2x+10^\circ)\).
A1: Correct values of \(\cos(2x+10^\circ) = -2/3\) and \(\cos(2x+10^\circ) = 1/2\).
M1: Finding at least two correct values of \(\theta = 2x+10^\circ\) in the appropriate range.
M1: Setting up correct equations to solve for \(x\) from their values of \(\theta\).
A1: Finding the four solutions from \(\cos(2x+10^\circ) = 1/2\): \(25^\circ, 145^\circ, 205^\circ, 325^\circ\).
A1.1: Finding the four solutions from \(\cos(2x+10^\circ) = -2/3\): \(60.9^\circ, 109.1^\circ, 240.9^\circ, 289.1^\circ\) (allow rounding errors of 0.1).
題目 8 · Structured
8.1 分
The curve \(C\) has equation \(y = 2x^2 + \frac{32}{x} - 5\), \(x \neq 0\).

(a) Find \(\frac{\mathrm{d}y}{\mathrm{d}x}\).

(b) Find the equation of the tangent to \(C\) at the point \(P(4, 35)\), giving your answer in the form \(y = mx + c\).

(c) Find the coordinates of the stationary point of \(C\).
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解題

(a) Rewrite \(y\) as: \(y = 2x^2 + 32x^{-1} - 5\).
Differentiating with respect to \(x\):
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 4x - 32x^{-2} = 4x - \frac{32}{x^2}\]

(b) To find the gradient of the tangent at \(P(4, 35)\), substitute \(x = 4\) into \(\frac{\mathrm{d}y}{\mathrm{d}x}\):
\[m = 4(4) - \frac{32}{4^2} = 16 - 2 = 14\]
Using the tangent line equation: \(y - 35 = 14(x - 4)\)
\[y - 35 = 14x - 56 \implies y = 14x - 21\]

(c) For stationary points, set \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\[4x - \frac{32}{x^2} = 0\]
\[4x^3 = 32 \implies x^3 = 8 \implies x = 2\]
Substitute \(x = 2\) back into the equation of the curve to find the \(y\)-coordinate:
\[y = 2(2)^2 + \frac{32}{2} - 5 = 8 + 16 - 5 = 19\]
So the stationary point has coordinates \((2, 19)\).

評分準則

(a)
M1: Differentiating at least one term correctly (power decreased by 1).
A1: Fully correct derivative: \(4x - \frac{32}{x^2}\) (or equivalent).

(b)
M1: Substituting \(x=4\) into their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to find the gradient of the tangent.
M1: Using \(y - 35 = m(x - 4)\) with their gradient.
A1: Obtaining \(y = 14x - 21\).

(c)
M1: Setting their \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and attempting to solve for \(x\).
A1: Correct value \(x = 2\).
A1.1: Correct coordinates \((2, 19)\).
題目 9 · Structured
8.1 分
Figure 1 shows a sketch of part of the curve \(C\) with equation \(y = 9x^{\frac{1}{2}} - 3x^{\frac{3}{2}}\), \(x \ge 0\).
The curve meets the \(x\)-axis at the origin and at the point \(A(3, 0)\).
The region \(R\) is bounded by the curve \(C\) and the \(x\)-axis.

(a) Use integration to find the exact area of the region \(R\).

(b) State, with a brief reason, whether the area of the region bounded by the curve \(y = 9(x-2)^{\frac{1}{2}} - 3(x-2)^{\frac{3}{2}}
\) and the \(x\)-axis is greater than, equal to, or less than the area of \(R\).
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解題

(a) The area \(R\) is given by the definite integral:
\[R = \int_0^3 \left( 9x^{\frac{1}{2}} - 3x^{\frac{3}{2}} \right) \mathrm{d}x\]
Integrating each term:
\[\int 9x^{\frac{1}{2}} \mathrm{d}x = 9 \frac{x^{3/2}}{3/2} = 6x^{\frac{3}{2}}\]
\[\int -3x^{\frac{3}{2}} \mathrm{d}x = -3 \frac{x^{5/2}}{5/2} = -\frac{6}{5}x^{\frac{5/2}{}}\]
So, the integral is:
\[\left[ 6x^{\frac{3}{2}} - \frac{6}{5}x^{\frac{5/2}{}} \right]_0^3\]
Substituting the upper limit \(x = 3\):
\[6(3)^{\frac{3}{2}} - \frac{6}{5}(3)^{\frac{5/2}{}} = 6(3\sqrt{3}) - \frac{6}{5}(9\sqrt{3}) = 18\sqrt{3} - \frac{54}{5}\sqrt{3} = \frac{90\sqrt{3} - 54\sqrt{3}}{5} = \frac{36\sqrt{3}}{5}\]
Substituting the lower limit \(x = 0\) gives \(0\).
Thus, the exact area is \(\frac{36\sqrt{3}}{5}\).

(b) The curve \(y = 9(x-2)^{\frac{1}{2}} - 3(x-2)^{\frac{3}{2}}\) is obtained by translating the curve \(C\) by 2 units to the right (translation vector \(\begin{pmatrix} 2 \\ 0 \end{pmatrix}\)).
Since translations are rigid transformations, they preserve the size and shape of regions.
Therefore, the area of the new region is equal to the area of \(R\).

評分準則

(a)
M1: Attempting to integrate: raising the power of \(x\) by 1 on at least one term.
A1: Fully correct integrated expression: \(6x^{\frac{3}{2}} - \frac{6}{5}x^{\frac{5/2}{}}\).
M1: Substituting the limits 3 and 0 and subtracting.
M1: Attempting to simplify powers of 3 to surd form (e.g., \(3^{3/2} = 3\sqrt{3}\)).
A1: Correct exact area, e.g., \(\frac{36\sqrt{3}}{5}\) or \(7.2\sqrt{3}\).

(b)
M1: Explaining that the new curve represents a translation (by 2 units to the right).
A1: Stating that translation does not alter the area of the region.
A1.1: Concluding that the area is equal to the original area of \(R\).
題目 10 · Structured
8.1 分
The curve \(C_1\) has equation \(y = (x-2)^2(x+3)\).

(a) Sketch \(C_1\), showing the coordinates of the points where the curve meets or cuts the coordinate axes.

The curve \(C_2\) has equation \(y = x^3 - 3x^2 - 4x + 12\).

(b) Show that \(C_1\) and \(C_2\) intersect at the point where \(x = 0\).

(c) Find the coordinates of the other point of intersection of \(C_1\) and \(C_2\).
查看答案詳解

解題

(a) For \(C_1\): \(y = (x-2)^2(x+3)\)
- The \(x\)-intercepts are at \(x = 2\) (where it touches the \(x\)-axis because of the squared factor) and \(x = -3\) (where it crosses the \(x\)-axis).
- The \(y\)-intercept is at \(x = 0\), where \(y = (-2)^2(3) = 12\).
- Since the coefficient of \(x^3\) is positive, the curve has a standard positive cubic shape.

(b) At \(x = 0\):
For \(C_1\): \(y = (0-2)^2(0+3) = 4 \times 3 = 12\).
For \(C_2\): \(y = 0^3 - 3(0)^2 - 4(0) + 12 = 12\).
Since both curves have the same \(y\)-coordinate at \(x = 0\), they intersect at this point.

(c) Equating \(C_1\) and \(C_2\):
\[(x-2)^2(x+3) = x^3 - 3x^2 - 4x + 12\]
Expand \(C_1\):
\[(x^2 - 4x + 4)(x+3) = x^3 + 3x^2 - 4x^2 - 12x + 4x + 12 = x^3 - x^2 - 8x + 12\]
Set the equations equal:
\[x^3 - x^2 - 8x + 12 = x^3 - 3x^2 - 4x + 12\]
Subtract \(x^3\) and 12 from both sides:
\[-x^2 - 8x = -3x^2 - 4x\]
Rearrange:
\[2x^2 - 4x = 0\]
\[2x(x - 2) = 0\]
The solutions are \(x = 0\) (from part b) and \(x = 2\).
Substituting \(x = 2\) back into either equation:
\[y = (2-2)^2(2+3) = 0\]
Thus, the other point of intersection is \((2, 0)\).

評分準則

(a)
M1: Drawing a positive cubic shape.
A1: Touching the \(x\)-axis at \(2\) and crossing at \(-3\).
A1: Labeling the \(y\)-intercept at \((0, 12)\).

(b)
M1: Finding the value of \(y\) for both curves when \(x = 0\).
A1: Showing both equal to 12 and concluding they intersect.

(c)
M1: Expanding the expression for \(C_1\) correctly to \(x^3 - x^2 - 8x + 12\).
M1: Equating the two cubic curves and simplifying to a quadratic equation.
A1.1: Solving the quadratic to find \(x = 2\) and finding the correct coordinate \((2, 0)\).
題目 11 · Structured
8.1 分
The circle \(C\) has equation
\[x^2 + y^2 - 8x + 6y - 11 = 0\]

(a) Find
(i) the coordinates of the centre of \(C\),
(ii) the radius of \(C\).

The line \(L\) has equation \(y = 2x + k\), where \(k\) is a constant.
Given that \(L\) is a tangent to \(C\),

(b) find the two possible values of \(k\), giving your answers in the form \(p \pm q\sqrt{5}\), where \(p\) and \(q\) are integers to be found.
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解題

(a) Complete the square for \(x\) and \(y\):
\[(x - 4)^2 - 16 + (y + 3)^2 - 9 - 11 = 0\]
\[(x - 4)^2 + (y + 3)^2 - 36 = 0\]
\[(x - 4)^2 + (y + 3)^2 = 36\]
(i) The centre of \(C\) is \((4, -3)\).
(ii) The radius of \(C\) is \(\sqrt{36} = 6\).

(b) Since \(L\) is a tangent to \(C\), the perpendicular distance from the centre of the circle \((4, -3)\) to the line \(L\) (which can be written as \(2x - y + k = 0\)) must equal the radius of the circle, \(6\).
Using the formula for the perpendicular distance from a point \((x_0, y_0)\) to a line \(Ax + By + C = 0\):
\[d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}\]
\[6 = \frac{|2(4) - (-3) + k|}{\sqrt{2^2 + (-1)^2}}\]
\[6 = \frac{|8 + 3 + k|}{\sqrt{5}}\]
\[|11 + k| = 6\sqrt{5}\]
This gives two possible equations:
\[11 + k = 6\sqrt{5} \implies k = -11 + 6\sqrt{5}\]
\[11 + k = -6\sqrt{5} \implies k = -11 - 6\sqrt{5}\]
Thus, the two possible values of \(k\) are \(-11 \pm 6\sqrt{5}\).

Alternatively, substituting \(y = 2x + k\) into the circle's equation yields a quadratic in \(x\) with discriminant set to zero, which simplifies to \(k^2 + 22k - 59 = 0\), solving to give the same result.

評分準則

(a)
M1: Attempting to complete the square for both \(x\) and \(y\).
A1: Correct centre of \((4, -3)\).
A1: Correct radius of \(6\).

(b)
M1: Attempting to use the distance formula from the centre \((4, -3)\) to the line \(2x - y + k = 0\) equated to their radius, or substituting \(y = 2x + k\) into the circle's equation.
A1: Obtaining the correct equation, e.g., \(\frac{|11+k|}{\sqrt{5}} = 6\) or the quadratic \(5x^2 + (4k+4)x + k^2 + 6k - 11 = 0\).
M1: Formulating a solvability condition, e.g., setting \(|11+k| = 6\sqrt{5}\) or setting the discriminant of the quadratic to 0.
A1: Obtaining a correct simplified relation for \(k\), e.g., \(k^2 + 22k - 59 = 0\) or \(|11+k| = 6\sqrt{5}\).
A1.1: Correct values: \(k = -11 \pm 6\sqrt{5}\) (or \(p = -11\) and \(q = 6\)).

部分 WMA12/01: Pure Mathematics P2

Answer all questions. Show sufficient working. Calculators allowed (no symbolic algebra).
10 題目 · 75 分
題目 1 · Short Answer
4 分
A geometric series has second term \( 6 \) and fifth term \( \frac{16}{9} \). Find the sum to infinity of this series.
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解題

The \( n \)-th term of a geometric series is given by \( u_n = a r^{n-1} \). We are given \( u_2 = ar = 6 \) and \( u_5 = ar^4 = \frac{16}{9} \). Dividing the two equations: \( \frac{ar^4}{ar} = \frac{16/9}{6} \implies r^3 = \frac{16}{54} = \frac{8}{27} \). Taking the cube root gives \( r = \frac{2}{3} \). Substituting this back into the first equation: \( a \left(\frac{2}{3}\right) = 6 \implies a = 9 \). Since \( |r| < 1 \), the sum to infinity exists and is given by: \( S_{\infty} = \frac{a}{1 - r} = \frac{9}{1 - 2/3} = \frac{9}{1/3} = 27 \).

評分準則

M1: Attempts to write two equations in \( a \) and \( r \), and divides them to obtain an equation in \( r^3 \) or \( r \). A1: Correct value of \( r = \frac{2}{3} \) (or equivalent fraction). M1: Uses their \( r \) to find \( a \) and applies the sum to infinity formula \( S_{\infty} = \frac{a}{1-r} \). A1: Correct answer of \( 27 \).
題目 2 · Short Answer
4 分
Solve the equation \( 2 \log_3(x - 2) - \log_3(x + 4) = 1 \).
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解題

Using the laws of logarithms: \( \log_3(x - 2)^2 - \log_3(x + 4) = 1 \implies \log_3\left(\frac{(x - 2)^2}{x + 4}\right) = 1 \). Converting from logarithmic to exponential form: \( \frac{(x - 2)^2}{x + 4} = 3^1 = 3 \). Expanding and rearranging: \( (x - 2)^2 = 3(x + 4) \implies x^2 - 4x + 4 = 3x + 12 \implies x^2 - 7x - 8 = 0 \). Factoring the quadratic equation: \( (x - 8)(x + 1) = 0 \), which gives \( x = 8 \) or \( x = -1 \). Since \( \log_3(x - 2) \) is only defined for \( x > 2 \), the solution \( x = -1 \) is rejected. Thus, the only valid solution is \( x = 8 \).

評分準則

M1: Applies log laws to express the equation in the form \( \log_3(f(x)) = 1 \) or equivalent. M1: Eliminates logarithms correctly to form a quadratic equation, e.g. \( \frac{(x-2)^2}{x+4} = 3 \). A1: Solves the quadratic to find \( x = 8 \) and \( x = -1 \). A1: Correctly identifies \( x = 8 \) as the only valid solution and rejects \( x = -1 \) with a valid reason.
題目 3 · Short Answer
4 分
The circle \( C \) has equation \( x^2 + y^2 - 8x + 6y = 0 \). Find the equation of the tangent to \( C \) at the point \( P(7, 1) \), giving your answer in the form \( ax + by + c = 0 \), where \( a \), \( b \) and \( c \) are integers.
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解題

Completing the square for the circle equation: \( (x - 4)^2 - 16 + (y + 3)^2 - 9 = 0 \implies (x - 4)^2 + (y + 3)^2 = 25 \). The centre of the circle is \( Q(4, -3) \). The gradient of the radius \( QP \) is \( m_r = \frac{1 - (-3)}{7 - 4} = \frac{4}{3} \). Since the tangent is perpendicular to the radius at the point of contact, the gradient of the tangent is \( m_t = -\frac{1}{m_r} = -\frac{3}{4} \). The equation of the tangent is \( y - 1 = -\frac{3}{4}(x - 7) \). Multiplying by 4: \( 4y - 4 = -3(x - 7) \implies 4y - 4 = -3x + 21 \). Rearranging into the required form: \( 3x + 4y - 25 = 0 \).

評分準則

M1: Completes the square to find the centre of the circle \( (4, -3) \). M1: Finds the gradient of the radius \( QP \) using their centre and the point \( P(7, 1) \). M1: Uses the perpendicular gradient condition \( m_t = -\frac{1}{m_r} \) and attempts the equation of the straight line. A1: Correct equation in the form \( ax + by + c = 0 \) with integer coefficients, e.g., \( 3x + 4y - 25 = 0 \) (or any non-zero integer multiple).
題目 4 · Structured
9 分

The function \(f(x)\) is defined by \(f(x) = (2 + kx)^6\), where \(k\) is a non-zero constant.

(a) Find the first 4 terms in ascending powers of \(x\) of the binomial expansion of \((2 + kx)^6\), giving each term in its simplest form.

(b) Given that the coefficient of \(x^3\) is 5 times the coefficient of \(x^2\), find the value of \(k\).

(c) Using this value of \(k\), find the coefficient of \(x\) in the expansion.

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解題

(a) Using the binomial theorem:

\((2+kx)^6 = 2^6 + \binom{6}{1} 2^5 (kx) + \binom{6}{2} 2^4 (kx)^2 + \binom{6}{3} 2^3 (kx)^3 + \dots\)

\(= 64 + 6(32)kx + 15(16)k^2x^2 + 20(8)k^3x^3 + \dots\)

\(= 64 + 192kx + 240k^2x^2 + 160k^3x^3 + \dots\)

(b) The coefficient of \(x^3\) is \(160k^3\) and the coefficient of \(x^2\) is \(240k^2\).

We are given: \(160k^3 = 5 \times 240k^2\)

\(160k^3 = 1200k^2\)

Since \(k \neq 0\), we can divide both sides by \(160k^2\):

\(k = \frac{1200}{160} = 7.5\)

(c) The coefficient of \(x\) is \(192k\).

Substituting \(k = 7.5\):

\(192 \times 7.5 = 1440\)

評分準則

(a)

M1: Attempt binomial expansion with at least 3 terms. Powers of \(2\) decreasing, powers of \(kx\) increasing.

A1: Any two terms correct and simplified.

A1: Any three terms correct and simplified.

A1: Fully correct simplified expression: \(64 + 192kx + 240k^2x^2 + 160k^3x^3\).

(b)

M1: Sets up a correct equation using their coefficients of \(x^3\) and \(x^2\): \(160k^3 = 5 \times 240k^2\).

M1: Attempts to solve the equation for \(k\) (must involve dividing by \(k^2\) or factorising).

A1: \(k = 7.5\) or \(\frac{15}{2}\).

(c)

M1: Substitutes their value of \(k\) into their coefficient of \(x\), which must be of the form \(C k\).

A1: \(1440\) (must be a single constant value).

題目 5 · Structured
9 分

(a) Solve, for \(0 \le \theta < 360^\circ\), the equation \(4\cos^2 \theta + 7\sin \theta - 7 = 0\), giving your answers to 1 decimal place where appropriate.

(b) Hence, solve for \(0 \le x < 180^\circ\) the equation \(4\cos^2 (2x - 20^\circ) + 7\sin (2x - 20^\circ) - 7 = 0\), giving your answers to 1 decimal place where appropriate.

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解題

(a) Using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\):

\(4(1 - \sin^2 \theta) + 7\sin \theta - 7 = 0\)

\(4 - 4\sin^2 \theta + 7\sin \theta - 7 = 0\)

\(4\sin^2 \theta - 7\sin \theta + 3 = 0\)

Factorising the quadratic:

\((4\sin \theta - 3)(\sin \theta - 1) = 0\)

So \\sin \theta = \frac{3}{4}\) or \(\sin \theta = 1\).

For \(\sin \theta = 1\): \(\theta = 90^\circ\)

For \(\sin \theta = 0.75\): \(\theta = \sin^{-1}(0.75) \approx 48.59^\circ\)

Other solution: \(180^\circ - 48.59^\circ = 131.41^\circ\)

So the solutions in the interval are \(\theta = 48.6^\circ, 90^\circ, 131.4^\circ\).

(b) Let \(\theta = 2x - 20^\circ\). Since \(0 \le x < 180^\circ\), the range for \(\theta\) is \(-20^\circ \le \theta < 340^\circ\).

The solutions for \(\theta\) from part (a) all fall in this range.

Case 1: \(2x - 20^\circ = 48.59^\circ \implies 2x = 68.59^\circ \implies x \approx 34.3^\circ\)

Case 2: \(2x - 20^\circ = 90^\circ \implies 2x = 110^\circ \implies x = 55^\circ\)

Case 3: \(2x - 20^\circ = 131.41^\circ \implies 2x = 151.41^\circ \implies x \approx 75.7^\circ\)

評分準則

(a)

M1: Uses the identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to form a quadratic equation in \(\sin \theta\).

A1: Obtains a correct quadratic: \(4\sin^2 \theta - 7\sin \theta + 3 = 0\).

M1: Attempts to solve the quadratic to find values of \(\sin \theta\).

A1: Obtains \(\theta = 90^\circ\) or both \(48.6^\circ\) and \(131.4^\circ\).

A1: All three solutions correct: \(48.6^\circ, 90^\circ, 131.4^\circ\). Deduct 1 mark for extra solutions in the range.

(b)

M1: Equates \(2x - 20^\circ\) to their values of \(\theta\).

M1: Attempts to solve for \(x\) for at least one of their values of \(\theta\).

A1: Any two of \(34.3^\circ, 55^\circ, 75.7^\circ\) correct.

A1: All three of \(34.3^\circ, 55^\circ, 75.7^\circ\) and no other values in the interval.

題目 6 · Structured
9 分

The polynomial \(f(x)\) is given by \(f(x) = 2x^3 + ax^2 + bx - 10\), where \(a\) and \(b\) are constants.

Given that \((x-2)\) is a factor of \(f(x)\),

(a) show that \(2a + b = -3\).

Given also that when \(f(x)\) is divided by \((x+1)\), the remainder is \(-18\),

(b) find the value of \(a\) and the value of \(b\).

(c) Hence, factorise \(f(x)\) completely.

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解題

(a) Since \((x-2)\) is a factor, \(f(2) = 0\).

\(2(2)^3 + a(2)^2 + b(2) - 10 = 0\)

\(16 + 4a + 2b - 10 = 0\)

\(4a + 2b = -6\)

Dividing by 2 gives: \(2a + b = -3\) (as required).

(b) Since the remainder when divided by \((x+1)\) is \(-18\), \(f(-1) = -18\).

\(2(-1)^3 + a(-1)^2 + b(-1) - 10 = -18\)

\(-2 + a - b - 10 = -18\)

\(a - b = -6\)

We solve the simultaneous equations:

1) \(2a + b = -3\)

2) \(a - b = -6\)

Adding 1) and 2) gives:

\(3a = -9 \implies a = -3\)

Substituting \(a = -3\) into 2):

\(-3 - b = -6 \implies b = 3\)

(c) Substituting \(a = -3\) and \(b = 3\) gives: \(f(x) = 2x^3 - 3x^2 + 3x - 10\).

Since \((x-2)\) is a factor, we can divide by \((x-2)\) to find the quadratic factor:

\(2x^3 - 3x^2 + 3x - 10 = (x-2)(2x^2 + cx + d)\)

Comparing constant terms: \(-2d = -10 \implies d = 5\).

Comparing \(x^2\) terms: \(-4x^2 + cx^2 = -3x^2 \implies c = 1\).

So the quadratic factor is \(2x^2 + x + 5\).

For \(2x^2 + x + 5\), the discriminant is \(1^2 - 4(2)(5) = -39 < 0\), so it does not factorise further.

Thus, \(f(x) = (x-2)(2x^2 + x + 5)\).

評分準則

(a)

M1: Attempts to evaluate \(f(2) = 0\) with substitution.

A1*: Fully correct proof with no errors showing the intermediate step \(4a + 2b = -6\) leading to \(2a + b = -3\).

(b)

M1: Attempts to evaluate \(f(-1) = -18\).

A1: Obtains a correct simplified equation in \(a\) and \(b\), e.g. \(a - b = -6\).

M1: Eliminates one variable from their two equations to solve for \(a\) or \(b\).

A1: Correct values: \(a = -3\) and \(b = 3\).

(c)

M1: Attempts algebraic division or equating coefficients to find the quadratic factor of \(f(x)\).

A1: Correct quadratic factor: \(2x^2 + x + 5\).

A1: Fully factorised form: \((x-2)(2x^2 + x + 5)\).

題目 7 · Structured
9 分

(a) Find the exact value of \(x\) for which \(\log_3(x + 5) - \log_3(x - 1) = 2\).

(b) Solve the equation \(3^{2y + 1} - 10(3^y) + 3 = 0\), giving your answers in exact form.

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解題

(a) Using the subtraction law of logarithms:

\(\log_3\left(\frac{x+5}{x-1}\right) = 2\)

Removing the logarithm:

\(\frac{x+5}{x-1} = 3^2 = 9\)

\(x + 5 = 9(x - 1)\)

\(x + 5 = 9x - 9\)

\(8x = 14 \implies x = 1.75\) (or \(\frac{7}{4}\))

(b) Using index laws, write \(3^{2y+1}\) as \(3 \times 3^{2y} = 3(3^y)^2\).

Let \(u = 3^y\).

The equation becomes:

\(3u^2 - 10u + 3 = 0\)

Factorising the quadratic:

\((3u - 1)(u - 3) = 0\)

So \(u = \frac{1}{3}\) or \(u = 3\).

If \(3^y = \frac{1}{3} \implies y = -1\).

If \(3^y = 3 \implies y = 1\).

評分準則

(a)

M1: Uses the subtraction law of logarithms correctly: \(\log_3\left(\frac{x+5}{x-1}\right)\).

M1: Converts the logarithmic equation to exponential form: \(\frac{x+5}{x-1} = 3^2\).

M1: Solves the linear equation for \(x\).

A1: \(x = 1.75\) (or equivalent fraction, e.g., \(\frac{7}{4}\)).

(b)

M1: Recognises that \(3^{2y+1} = 3 \times (3^y)^2\).

M1: Substitutes \(u = 3^y\) to obtain a quadratic equation in \(u\).

A1: Correct quadratic: \(3u^2 - 10u + 3 = 0\).

M1: Solves the quadratic to find two values for \(3^y\).

A1: Both \(y = 1\) and \(y = -1\) (must be exact values).

題目 8 · Structured
9 分

The curve \(C\) has equation \(y = 2x^2 - 8x + 10\) and the line \(L\) has equation \(y = x + 6\).

(a) Show that the line \(L\) and the curve \(C\) intersect at the points with \(x\)-coordinates \(x = \frac{1}{2}\) and \(x = 4\).

(b) Find, using calculus, the exact area of the finite region bounded by the curve \(C\) and the line \(L\).

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解題

(a) To find points of intersection, set \(y_C = y_L\):

\(2x^2 - 8x + 10 = x + 6\)

\(2x^2 - 9x + 4 = 0\)

Factorising the quadratic:

\((2x - 1)(x - 4) = 0\)

This gives solutions \(x = \frac{1}{2}\) and \(x = 4\), as required.

(b) The finite region is bounded between \(x = \frac{1}{2}\) and \(x = 4\). In this interval, the line is above the curve.

\(\text{Area} = \int_{1/2}^4 (y_L - y_C) \, dx\)

\(= \int_{1/2}^4 ((x+6) - (2x^2 - 8x + 10)) \, dx\)

\(= \int_{1/2}^4 (-2x^2 + 9x - 4) \, dx\)

Integrating each term:

\(= \left[ -\frac{2}{3}x^3 + \frac{9}{2}x^2 - 4x \right]_{1/2}^4\)

Evaluating at the upper limit \(x = 4\):

\(-\frac{2}{3}(64) + \frac{9}{2}(16) - 4(4) = -\frac{128}{3} + 72 - 16 = \frac{40}{3}\)

Evaluating at the lower limit \(x = \frac{1}{2}\):

\(-\frac{2}{3}\left(\frac{1}{8}\right) + \frac{9}{2}\left(\frac{1}{4}\right) - 4\left(\frac{1}{2}\right) = -\frac{1}{12} + \frac{9}{8} - 2 = -\frac{23}{24}\)

Finding the difference:

\(\text{Area} = \frac{40}{3} - \left(-\frac{23}{24}\right) = \frac{320}{24} + \frac{23}{24} = \frac{343}{24}\)

評分準則

(a)

M1: Equates the equations of the curve and the line.

A1: Obtains a correct simplified quadratic equation: \(2x^2 - 9x + 4 = 0\).

A1*: Factorises or uses the quadratic formula to show both roots are \(x = \frac{1}{2}\) and \(x = 4\).

(b)

M1: Formulates a correct integral with appropriate limits: \(\int_{1/2}^4 (x+6 - (2x^2 - 8x + 10)) \, dx\).

M1: Integrates a quadratic expression of the form \(Ax^2 + Bx + C\), increasing the power of at least one term.

A1: Correct integrated expression: \(-\frac{2}{3}x^3 + \frac{9}{2}x^2 - 4x\) (ignore constant of integration).

M1: Substitutes both limits \(4\) and \(\frac{1}{2}\) into their integrated expression.

A1: Correct intermediate values: \(\frac{40}{3}\) and \(-\frac{23}{24}\) (or equivalent values if integrated separately).

A1: \(\frac{343}{24}\) (or exact equivalent, e.g., \(14 \frac{7}{24}\)).

題目 9 · Structured
9 分

A circle \(C\) has equation \(x^2 + y^2 - 6x + 8y = 0\).

(a) Find:

(i) the coordinates of the centre of \(C\),

(ii) the radius of \(C\).

The line \(L\) has equation \(y = \frac{3}{4}x\).

(b) Show that the line \(L\) is a tangent to the circle \(C\).

(c) Find the coordinates of the point of contact between \(L\) and \(C\).

查看答案詳解

解題

(a) Rearranging the equation by completing the square:

\((x - 3)^2 - 9 + (y + 4)^2 - 16 = 0\)

\((x - 3)^2 + (y + 4)^2 = 25\)

(i) Centre of \(C\) is \((3, -4)\).

(ii) Radius is \(\sqrt{25} = 5\).

(b) Method 1: Algebraic Substitution

Substitute \(y = \frac{3}{4}x\) into the circle equation:

\(x^2 + \left(\frac{3}{4}x\right)^2 - 6x + 8\left(\frac{3}{4}x\right) = 0\)

\(x^2 + \frac{9}{16}x^2 - 6x + 6x = 0\)

\(\frac{25}{16}x^2 = 0\)

This quadratic equation has a single repeated root, \(x = 0\), which indicates that the line touches the circle at exactly one point, meaning \(L\) is a tangent to \(C\).

Method 2: Distance from centre

The equation of \(L\) is \(3x - 4y = 0\). The perpendicular distance from the centre \((3, -4)\) to the line is:

\(d = \frac{|3(3) - 4(-4)|}{\sqrt{3^2 + (-4)^2}} = \frac{|9 + 16|}{\sqrt{25}} = \frac{25}{5} = 5\)

Since the perpendicular distance from the centre to the line is equal to the radius of the circle, the line is a tangent.

(c) Using Method 1, the repeated root occurs at \(x = 0\). Substituting \(x = 0\) into the line equation: \(y = \frac{3}{4}(0) = 0\).

So the point of contact is \((0, 0)\).

評分準則

(a)

M1: Completes the square for both \(x\) and \(y\) terms.

A1: Centre is \((3, -4)\).

A1: Radius is \(5\).

(b)

M1: Substitutes the line equation into the circle equation, OR attempts to find the perpendicular distance from the centre to the line.

A1: Obtains the correct simplified quadratic equation \(\frac{25}{16}x^2 = 0\) OR calculates distance as \(\frac{25}{5}\).

M1: Identifies that there is a repeated root \(x = 0\) OR compares the distance \(5\) with the radius \(5\).

A1: Clear conclusion stating that since there is a repeated root OR distance equals radius, \(L\) is a tangent.

(c)

M1: Solves for the coordinates of the intersection point.

A1: Point of contact is \((0, 0)\).

題目 10 · Structured
9 分

An open-topped rectangular box is to be made from thin sheet metal. The box has a rectangular base of length \(2x\) cm and width \(x\) cm, and a height of \(h\) cm. Given that the volume of the box is \(36\text{ cm}^3\):

(a) Show that the surface area of the box, \(A\text{ cm}^2\), is given by \(A = 2x^2 + \frac{108}{x}\).

(b) Use calculus to find the value of \(x\) for which \(A\) is a minimum, justifying that your value of \(x\) gives a minimum.

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解題

(a) The volume of the box is given by:

\(V = \text{length} \times \text{width} \times \text{height} = (2x)(x)(h) = 2x^2h\)

We are given \(V = 36\), so:

\(2x^2h = 36 \implies h = \frac{18}{x^2}\)

The box has no top. The surface area \(A\) consists of the base and four vertical sides:

\(A = \text{Base area} + 2(\text{front/back area}) + 2(\text{side area})\)

\(A = (2x)(x) + 2(2xh) + 2(xh)\)

\(A = 2x^2 + 6xh\)

Substitute \(h = \frac{18}{x^2}\) into the equation for \(A\):

\(A = 2x^2 + 6x\left(\frac{18}{x^2}\right)\)

\(A = 2x^2 + \frac{108}{x}\) (as required).

(b) To find stationary points, differentiate \(A\) with respect to \(x\):

\(A = 2x^2 + 108x^{-1}\)

\(\frac{dA}{dx} = 4x - 108x^{-2} = 4x - \frac{108}{x^2}\)

Set \(\frac{dA}{dx} = 0\):

\(4x - \frac{108}{x^2} = 0\)

\(4x^3 = 108\)

\(x^3 = 27 \implies x = 3\)

To justify that this gives a minimum, find the second derivative:

\(\frac{d^2A}{dx^2} = 4 + 216x^{-3} = 4 + \frac{216}{x^3}\)

When \(x = 3\):

\(\frac{d^2A}{dx^2} = 4 + \frac{216}{27} = 4 + 8 = 12\)

Since \(\frac{d^2A}{dx^2} = 12 > 0\), the value of \(x = 3\) gives a minimum value for the surface area.

評分準則

(a)

M1: Expresses the volume in terms of \(x\) and \(h\): \(V = 2x^2h\).

M1: Uses \(V = 36\) to express \(h\) in terms of \(x\): \(h = \frac{18}{x^2}\).

M1: Formulates a correct expression for the total surface area of an open-topped box: \(A = 2x^2 + 6xh\).

A1*: Substitutes \(h\) and simplifies to obtain the given formula \(A = 2x^2 + \frac{108}{x}\) with no errors.

(b)

M1: Differentiates \(A\) with respect to \(x\) (at least one term correct).

A1: Correct derivative: \(4x - \frac{108}{x^2}\).

M1: Sets their \(\frac{dA}{dx} = 0\) and solves for \(x\).

A1: \(x = 3\).

M1: Finds the second derivative and evaluates it at \(x = 3\) (or uses another valid method) to show that it is a minimum.

部分 WMA13/01: Pure Mathematics P3

Answer all questions. Show sufficient working. Calculators allowed (no symbolic algebra).
10 題目 · 75 分
題目 1 · Short Answer
4 分
The function \( f(x) = \ln(3x + 1) - x^2 + 2 \) has a single positive root \(\alpha\). (a) Show that \(\alpha\) lies in the interval \([1.5, 2.0]\). (b) Use the iteration formula \( x_{n+1} = \sqrt{\ln(3x_n + 1) + 2} \) with \( x_0 = 1.7 \) to find the value of \( x_3 \) to 3 decimal places.
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解題

(a) Evaluating the function at the boundaries: \( f(1.5) = \ln(3(1.5) + 1) - (1.5)^2 + 2 = \ln(5.5) - 2.25 + 2 \approx 1.7047 - 0.25 = 1.455 \) (to 3 d.p.). \( f(2.0) = \ln(3(2) + 1) - (2)^2 + 2 = \ln(7) - 4 + 2 \approx 1.9459 - 2 = -0.054 \) (to 3 d.p.). Since there is a change of sign and \( f(x) \) is continuous in the interval \([1.5, 2.0]\), a root \(\alpha\) lies in this interval. (b) Using the iteration formula: \( x_1 = \sqrt{\ln(3(1.7) + 1) + 2} = \sqrt{\ln(6.1) + 2} \approx 1.95148 \), \( x_2 = \sqrt{\ln(3(1.95148) + 1) + 2} = \sqrt{\ln(6.85444) + 2} \approx 1.98114 \), \( x_3 = \sqrt{\ln(3(1.98114) + 1) + 2} = \sqrt{\ln(6.94342) + 2} \approx 1.98439 \). Thus, to 3 decimal places, \( x_3 = 1.984 \).

評分準則

(a) M1: Attempts to evaluate both \( f(1.5) \) and \( f(2.0) \) (at least one correct value to 1 d.p.). A1: Finds correct values of \( f(1.5) \approx 1.45 \) and \( f(2.0) \approx -0.05 \), and concludes with a reason (change of sign and continuity). (b) M1: Attempts to find \( x_1 \) using the iteration formula. A1: Correctly calculates \( x_3 = 1.984 \) to 3 decimal places.
題目 2 · Short Answer
4 分
The function \( \mathrm{g} \) is defined by \( \mathrm{g}(x) = \frac{3e^x - 2}{e^x + 4} \), \( x \in \mathbb{R} \). (a) Find an expression for \( \mathrm{g}^{-1}(x) \). (b) State the domain of \( \mathrm{g}^{-1} \).
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解題

(a) Let \( y = \frac{3e^x - 2}{e^x + 4} \). Rearranging to make \( e^x \) the subject: \( y(e^x + 4) = 3e^x - 2 \implies y e^x + 4y = 3e^x - 2 \implies 4y + 2 = e^x(3 - y) \implies e^x = \frac{4y+2}{3-y} \). Taking the natural logarithm of both sides: \( x = \ln\left(\frac{4y+2}{3-y}\right) \). Replacing \( y \) with \( x \), we obtain: \( \mathrm{g}^{-1}(x) = \ln\left(\frac{4x+2}{3-x}\right) \). (b) The domain of \( \mathrm{g}^{-1} \) is the range of \( \mathrm{g} \). Since \( e^x > 0 \), \( \mathrm{g}(x) = 3 - \frac{14}{e^x + 4} \). As \( x \to -\infty \), \( e^x \to 0 \), so \( \mathrm{g}(x) \to -\frac{2}{4} = -\frac{1}{2} \). As \( x \to \infty \), \( \mathrm{g}(x) \to 3 \). Thus, the range of \( \mathrm{g} \) is \( -\frac{1}{2} < \mathrm{g}(x) < 3 \). Therefore, the domain of \( \mathrm{g}^{-1} \) is \( -\frac{1}{2} < x < 3 \).

評分準則

(a) M1: Starts process to make \( e^x \) the subject by multiplying out and collecting terms. M1: Correctly isolates \( e^x = \frac{4y+2}{3-y} \) or equivalent. A1: Correct expression \( \mathrm{g}^{-1}(x) = \ln\left(\frac{4x+2}{3-x}\right) \) or equivalent, with correct notation. (b) B1: Correctly states \( -\frac{1}{2} < x < 3 \) (or equivalent interval notation).
題目 3 · Short Answer
4 分
The curve \( C \) has equation \( y = \frac{\ln(2x - 1)}{x} \), \( x > \frac{1}{2} \). Find the equation of the tangent to \( C \) at the point where \( x = 1 \).
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解題

To find the equation of the tangent, we first find the coordinates of the point of contact and the gradient at that point. At \( x = 1 \), \( y = \frac{\ln(2(1) - 1)}{1} = \frac{\ln(1)}{1} = 0 \). So the point is \( (1, 0) \). To find the gradient, we use the quotient rule: let \( u = \ln(2x-1) \implies \frac{\mathrm{d}u}{\mathrm{d}x} = \frac{2}{2x-1} \) and \( v = x \implies \frac{\mathrm{d}v}{\mathrm{d}x} = 1 \). Then \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v \frac{\mathrm{d}u}{\mathrm{d}x} - u \frac{\mathrm{d}v}{\mathrm{d}x}}{v^2} = \frac{x \left(\frac{2}{2x-1}\right) - \ln(2x-1)}{x^2} \). Substituting \( x = 1 \): \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1 \left(\frac{2}{1}\right) - \ln(1)}{1^2} = 2 \). The equation of the tangent is given by \( y - y_1 = m(x - x_1) \), which becomes \( y - 0 = 2(x - 1) \implies y = 2x - 2 \).

評分準則

M1: Applies the quotient rule (or product rule) to differentiate the function, with correct structure. A1: Correct derivative \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x \left(\frac{2}{2x-1}\right) - \ln(2x-1)}{x^2} \) or equivalent. M1: Finds the value of \( y \) at \( x = 1 \) (which is 0) and attempts to evaluate their \( \frac{\mathrm{d}y}{\mathrm{d}x} \) at \( x = 1 \). A1: Correct equation of the tangent, e.g., \( y = 2x - 2 \) or equivalent.
題目 4 · Structured
9 分
The function \( f \) is defined by
\[ f(x) = 3|x - 2| - 4, \quad x \in \mathbb{R} \]

(a) Sketch the graph of \( y = f(x) \), showing the coordinates of the vertex and the \( y \)-intercept. (3)

(b) Solve the equation \( f(x) = 2x + 1 \). (4)

(c) State the range of \( f(x) \). (2)
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解題

(a) The graph of \( y = 3|x - 2| - 4 \) is a V-shape.
The vertex is located at \( (2, -4) \).
The \( y \)-intercept is found by setting \( x = 0 \):
\( y = 3|0 - 2| - 4 = 3(2) - 4 = 2 \). So the \( y \)-intercept is at \( (0, 2) \).

(b) To solve \( 3|x - 2| - 4 = 2x + 1 \):
Case 1: \( x \ge 2 \)
\( 3(x - 2) - 4 = 2x + 1 \Rightarrow 3x - 10 = 2x + 1 \Rightarrow x = 11 \)
Since \( 11 \ge 2 \), this is a valid solution.

Case 2: \( x < 2 \)
\( -3(x - 2) - 4 = 2x + 1 \Rightarrow -3x + 2 = 2x + 1 \Rightarrow 5x = 1 \Rightarrow x = 0.2 \)
Since \( 0.2 < 2 \), this is a valid solution.
So the solutions are \( x = 11 \) and \( x = 0.2 \).

(c) The minimum value of \( f(x) \) is \( -4 \), and it goes to positive infinity.
So the range is \( f(x) \ge -4 \).

評分準則

(a)
M1: For a V-shaped graph in the correct orientation.
A1: Vertex at \( (2, -4) \) correctly labelled or coordinates stated.
A1: \( y \)-intercept at \( (0, 2) \) correctly labelled or coordinates stated.

(b)
M1: Formulates the equation for \( x \ge 2 \): \( 3(x-2) - 4 = 2x + 1 \).
A1: Obtains \( x = 11 \).
M1: Formulates the equation for \( x < 2 \): \( -3(x-2) - 4 = 2x + 1 \).
A1: Obtains \( x = 0.2 \) (or \( \frac{1}{5} \)).

(c)
M1: Realises that the range is dependent on the y-coordinate of the vertex.
A1: Correct range: \( f(x) \ge -4 \) (accept \( y \ge -4 \), do not accept \( x \ge -4 \)).
題目 5 · Structured
9 分
(a) Express \( 5\cos\theta - 12\sin\theta \) in the form \( R\cos(\theta + \alpha) \), where \( R > 0 \) and \( 0 < \alpha < \frac{\pi}{2} \). Give the value of \( \alpha \) in radians to 4 decimal places. (3)

(b) Hence, solve for \( 0 \le \theta < 2\pi \), the equation
\[ 5\cos\theta - 12\sin\theta = 6.5 \]
giving your answers to 3 significant figures. (4)

(c) Find the minimum value of
\[ \frac{30}{17 + (5\cos\theta - 12\sin\theta)^2} \] (2)
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解題

(a) Let \( 5\cos\theta - 12\sin\theta = R\cos(\theta + \alpha) = R\cos\theta\cos\alpha - R\sin\theta\sin\alpha \).
Comparing coefficients:
\( R\cos\alpha = 5 \)
\( R\sin\alpha = 12 \)
\( R^2 = 5^2 + 12^2 = 169 \Rightarrow R = 13 \).
\( \tan\alpha = \frac{12}{5} \Rightarrow \alpha = \arctan(2.4) \approx 1.1760 \) radians.
So, \( 13\cos(\theta + 1.1760) \).

(b) \( 13\cos(\theta + 1.1760) = 6.5 \Rightarrow \cos(\theta + 1.1760) = 0.5 \).
Let \( X = \theta + 1.1760 \). Since \( 0 \le \theta < 2\pi \), we have \( 1.1760 \le X < 2\pi + 1.1760 \approx 7.4592 \).
The solutions to \( \cos X = 0.5 \) in this range are:
\( X = \frac{5\pi}{3} \approx 5.2360 \)
\( X = \frac{7\pi}{3} \approx 7.3304 \)
So:
\( \theta = 5.2360 - 1.1760 = 4.0600 \approx 4.06 \)
\( \theta = 7.3304 - 1.1760 = 6.1544 \approx 6.15 \)

(c) The expression is \( \frac{30}{17 + [13\cos(\theta + \alpha)]^2} = \frac{30}{17 + 169\cos^2(\theta + \alpha)} \).
To minimize this expression, we maximize the denominator.
The maximum value of \( \cos^2(\theta + \alpha) \) is 1.
Maximum denominator \( = 17 + 169(1) = 186 \).
So the minimum value is \( \frac{30}{186} = \frac{5}{31} \).

評分準則

(a)
B1: \( R = 13 \).
M1: \( \tan\alpha = \pm \frac{12}{5} \) or \( \pm \frac{5}{12} \).
A1: \( \alpha \approx 1.1760 \) (must be 4 d.p.).

(b)
M1: Sets \( \cos(\theta + \alpha) = \frac{6.5}{13} = 0.5 \).
M1: Finds at least one correct value for \( \theta + \alpha \) (e.g., \( \frac{5\pi}{3} \) or \( 5.24 \)).
A1: One correct solution, \( \theta \approx 4.06 \).
A1: Second correct solution, \( \theta \approx 6.15 \). (Accept answers that round to 4.06 and 6.15).

(c)
M1: Recognises that the minimum value of the expression occurs when \( \cos^2(\theta + \alpha) = 1 \).
A1: \( \frac{5}{31} \) (or exact equivalent).
題目 6 · Structured
9 分
The mass, \( M \) grams, of a radioactive substance \( t \) years after it was first observed is modeled by the equation
\[ M = A \mathrm{e}^{-kt} \]
where \( A \) and \( k \) are positive constants.

Given that the initial mass of the substance was 80 grams,
(a) write down the value of \( A \). (1)

Given also that the mass of the substance was 50 grams after 6 years,
(b) find the value of \( k \), giving your answer to 4 decimal places. (3)

(c) Find the rate of decrease of the mass of the substance when \( t = 10 \), giving your answer in grams per year to 3 significant figures. (3)

(d) Find the value of \( t \) when the mass is 10 grams, giving your answer to 1 decimal place. (2)
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解題

(a) At \( t = 0 \), \( M = 80 \Rightarrow A \mathrm{e}^0 = 80 \Rightarrow A = 80 \).

(b) When \( t = 6 \), \( M = 50 \).
\( 50 = 80 \mathrm{e}^{-6k} \Rightarrow \mathrm{e}^{-6k} = 0.625 \)
\( -6k = \ln(0.625) \Rightarrow k = -\frac{\ln(0.625)}{6} \approx 0.07833 \approx 0.0783 \).

(c) The rate of decrease is given by \( -\frac{\mathrm{d}M}{\mathrm{d}t} \).
\( \frac{\mathrm{d}M}{\mathrm{d}t} = -k A \mathrm{e}^{-kt} = -0.07833 \times 80 \mathrm{e}^{-0.07833 t} \).
When \( t = 10 \):
Rate of decrease \( = 0.07833 \times 80 \mathrm{e}^{-0.07833 \times 10} \approx 6.2664 \mathrm{e}^{-0.7833} \approx 2.86 \) grams per year.

(d) When \( M = 10 \):
\( 10 = 80 \mathrm{e}^{-0.07833 t} \Rightarrow \mathrm{e}^{-0.07833 t} = 0.125 \)
\( -0.07833 t = \ln(0.125) \Rightarrow t = \frac{\ln(8)}{0.07833} \approx 26.5 \) years.

評分準則

(a)
B1: \( A = 80 \).

(b)
M1: Substitutes \( M = 50 \), \( t = 6 \), and their \( A \) into the model.
M1: Correct use of logarithms to solve for \( k \).
A1: \( k = 0.0783 \) (must be 4 d.p.).

(c)
M1: Differentiates \( M \) with respect to \( t \) to find \( \frac{\mathrm{d}M}{\mathrm{d}t} = -k A \mathrm{e}^{-kt} \).
M1: Substitutes \( t = 10 \) into their derivative.
A1: \( 2.86 \) (accept \( 2.86 \) or \( 2.87 \) depending on rounding of \( k \)).

(d)
M1: Sets \( M = 10 \) and solves for \( t \) using their \( A \) and \( k \).
A1: \( t = 26.5 \) (accept 26.5 or 26.6).
題目 7 · Structured
9 分
A curve \( C \) has equation
\[ y = \frac{x^2 - 3}{2x + 5}, \quad x \neq -2.5 \]

(a) Show that \( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2x^2 + 10x + 6}{(2x + 5)^2} \). (4)

(b) Find the exact coordinates of the stationary points of \( C \). (5)
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解題

(a) Let \( u = x^2 - 3 \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} = 2x \).
Let \( v = 2x + 5 \Rightarrow \frac{\mathrm{d}v}{\mathrm{d}x} = 2 \).
Using the quotient rule:
\( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v \frac{\mathrm{d}u}{\mathrm{d}x} - u \frac{\mathrm{d}v}{\mathrm{d}x}}{v^2} \)
\( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(2x + 5)(2x) - (x^2 - 3)(2)}{(2x + 5)^2} \)
\( \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4x^2 + 10x - 2x^2 + 6}{(2x + 5)^2} = \frac{2x^2 + 10x + 6}{(2x + 5)^2} \). (Shown)

(b) At stationary points, \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \):
\( 2x^2 + 10x + 6 = 0 \Rightarrow x^2 + 5x + 3 = 0 \).
Using the quadratic formula:
\( x = \frac{-5 \pm \sqrt{25 - 12}}{2} = \frac{-5 \pm \sqrt{13}}{2} \).
To find the y-coordinates, we can use \( x^2 = -5x - 3 \):
\( y = \frac{x^2 - 3}{2x + 5} = \frac{-5x - 6}{2x + 5} \).
For \( x = \frac{-5 + \sqrt{13}}{2} \):
\( 2x + 5 = \sqrt{13} \).
\( -5x - 6 = \frac{25 - 5\sqrt{13}}{2} - 6 = \frac{13 - 5\sqrt{13}}{2} \).
\( y = \frac{13 - 5\sqrt{13}}{2\sqrt{13}} = \frac{\sqrt{13} - 5}{2} \).
For \( x = \frac{-5 - \sqrt{13}}{2} \):
\( 2x + 5 = -\sqrt{13} \).
\( -5x - 6 = \frac{25 + 5\sqrt{13}}{2} - 6 = \frac{13 + 5\sqrt{13}}{2} \).
\( y = \frac{13 + 5\sqrt{13}}{-2\sqrt{13}} = \frac{-\sqrt{13} - 5}{2} \).
So the exact coordinates are:
\( \left(\frac{-5 + \sqrt{13}}{2}, \frac{\sqrt{13} - 5}{2}\right) \) and \( \left(\frac{-5 - \sqrt{13}}{2}, \frac{-\sqrt{13} - 5}{2}\right) \).

評分準則

(a)
M1: Applies quotient rule with \( u = x^2 - 3 \) and \( v = 2x + 5 \).
A1: Correct derivative of \( u \) is \( 2x \), and of \( v \) is \( 2 \).
M1: Correct structure of quotient rule: \( \frac{v u' - u v'}{v^2} \).
A1*: Simplifies numerator to \( 2x^2 + 10x + 6 \) with no errors seen.

(b)
M1: Sets \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \) to get \( 2x^2 + 10x + 6 = 0 \).
A1: Solves to find \( x = \frac{-5 \pm \sqrt{13}}{2} \) (or exact equivalents).
M1: Substitutes at least one exact \( x \)-value back into the original curve equation \( y = \frac{x^2 - 3}{2x + 5} \).
A1: One correct exact \( y \)-coordinate: \( y = \frac{\sqrt{13} - 5}{2} \) (or equivalent).
A1: Second correct exact \( y \)-coordinate: \( y = \frac{-\sqrt{13} - 5}{2} \) (or equivalent).
題目 8 · Structured
9 分
A curve \( C \) has equation
\[ y = \mathrm{e}^{2x} \sin(3x) \]

(a) Find \( \frac{\mathrm{d}y}{\mathrm{d}x} \). (3)

(b) Show that the \( x \)-coordinates of the stationary points of \( C \) satisfy the equation \( \tan(3x) = -1.5 \). (3)

(c) Find the equation of the tangent to \( C \) at the point where \( x = \pi \). Give your answer in the form \( y = mx + c \), where \( m \) and \( c \) are exact constants. (3)
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解題

(a) Using the product rule:
\( u = \mathrm{e}^{2x} \Rightarrow u' = 2\mathrm{e}^{2x} \)
\( v = \sin(3x) \Rightarrow v' = 3\cos(3x) \)
\( \frac{\mathrm{d}y}{\mathrm{d}x} = u'v + uv' = 2\mathrm{e}^{2x}\sin(3x) + 3\mathrm{e}^{2x}\cos(3x) = \mathrm{e}^{2x}(2\sin(3x) + 3\cos(3x)) \).

(b) At stationary points, \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \).
Since \( \mathrm{e}^{2x} \neq 0 \) for all \( x \):
\( 2\sin(3x) + 3\cos(3x) = 0 \).
Divide both sides by \( \cos(3x) \):
\( 2\tan(3x) + 3 = 0 \Rightarrow \tan(3x) = -1.5 \). (Shown)

(c) When \( x = \pi \):
\( y = \mathrm{e}^{2\pi}\sin(3\pi) = 0 \).
The gradient of the tangent is:
\( m = \left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{x=\pi} = \mathrm{e}^{2\pi}(2\sin(3\pi) + 3\cos(3\pi)) = \mathrm{e}^{2\pi}(0 - 3) = -3\mathrm{e}^{2\pi} \).
The equation of the tangent is:
\( y - 0 = -3\mathrm{e}^{2\pi}(x - \pi) \Rightarrow y = -3\mathrm{e}^{2\pi}x + 3\pi\mathrm{e}^{2\pi} \).

評分準則

(a)
M1: Correctly applies product rule: \( \frac{\mathrm{d}y}{\mathrm{d}x} = A\mathrm{e}^{2x}\sin(3x) + B\mathrm{e}^{2x}\cos(3x) \).
A1: One term correct: \( 2\mathrm{e}^{2x}\sin(3x) \) or \( 3\mathrm{e}^{2x}\cos(3x) \).
A1: Fully correct derivative: \( \mathrm{e}^{2x}(2\sin(3x) + 3\cos(3x)) \).

(b)
M1: Sets their \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \).
M1: Recognises \( \mathrm{e}^{2x} \neq 0 \) and uses \( \tan(3x) = \frac{\sin(3x)}{\cos(3x)} \).
A1*: Fully correct proof of \( \tan(3x) = -1.5 \) with no errors.

(c)
M1: Finds the value of \( y \) and \( \frac{\mathrm{d}y}{\mathrm{d}x} \) at \( x = \pi \).
M1: Uses \( y - y_1 = m(x - x_1) \) with their coordinates and gradient.
A1: \( y = -3\mathrm{e}^{2\pi}x + 3\pi\mathrm{e}^{2\pi} \) (or exact equivalent).
題目 9 · Structured
9 分
Find:

(a) \( \int (4x + 1) \mathrm{e}^{2x^2 + x} \mathrm{d}x \) (3)

(b) the exact value of \( \int_{0}^{\pi/6} (\sin(3x) + \cos(2x)) \mathrm{d}x \) (3)

(c) Show that \( \int_{2}^{4} \frac{8}{2x - 3} \mathrm{d}x = \ln(k) \), where \( k \) is a constant to be found. (3)
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解題

(a) Let \( u = 2x^2 + x \Rightarrow \frac{\mathrm{d}u}{\mathrm{d}x} = 4x + 1 \).
So the integral becomes:
\( \int \mathrm{e}^u \mathrm{d}u = \mathrm{e}^u + C = \mathrm{e}^{2x^2 + x} + C \).

(b) \( \int_{0}^{\pi/6} (\sin(3x) + \cos(2x)) \mathrm{d}x = \left[ -\frac{1}{3}\cos(3x) + \frac{1}{2}\sin(2x) \right]_{0}^{\pi/6} \)
Evaluating at upper bound \( \pi/6 \):
\( -\frac{1}{3}\cos(\pi/2) + \frac{1}{2}\sin(\pi/3) = 0 + \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{4} \).
Evaluating at lower bound \( 0 \):
\( -\frac{1}{3}\cos(0) + \frac{1}{2}\sin(0) = -\frac{1}{3} \).
Subtracting:
\( \frac{\sqrt{3}}{4} - \left(-\frac{1}{3}\right) = \frac{3\sqrt{3} + 4}{12} \).

(c) \( \int_{2}^{4} \frac{8}{2x - 3} \mathrm{d}x = \left[ 4\ln|2x - 3| \right]_{2}^{4} \)
Evaluating at 4: \( 4\ln|5| = 4\ln(5) \).
Evaluating at 2: \( 4\ln|1| = 0 \).
So the integral value is \( 4\ln(5) = \ln(5^4) = \ln(625) \).
Thus \( k = 625 \).

評分準則

(a)
M1: Recognises reverse chain rule or substitutes \( u = 2x^2 + x \).
A1: \( \mathrm{e}^{2x^2 + x} \).
A1: Correct constant of integration \( + C \).

(b)
M1: Integrates to find \( -A\cos(3x) + B\sin(2x) \).
A1: Correct integration: \( -\frac{1}{3}\cos(3x) + \frac{1}{2}\sin(2x) \).
A1: Evaluates and obtains \( \frac{3\sqrt{3} + 4}{12} \) (or exact equivalent).

(c)
M1: Integrates to find \( A\ln|2x - 3| \).
A1: Correct integration: \( 4\ln|2x - 3| \).
A1: Correctly applies laws of logs to show \( k = 625 \).
題目 10 · Structured
9 分
The curve \( C \) has equation \( y = x^3 - 5x + 1 \).

(a) Show that \( C \) has a root \( \alpha \) in the interval \( [2, 2.5] \). (2)

(b) Show that the equation \( x^3 - 5x + 1 = 0 \) can be rearranged into the iterative formula
\[ x_{n+1} = \sqrt{5 - \frac{1}{x_n}} \] (2)

(c) Using the iterative formula with \( x_0 = 2 \), find the values of \( x_1 \), \( x_2 \), and \( x_3 \), giving each answer to 4 decimal places. (3)

(d) By choosing a suitable interval, prove that \( \alpha = 2.128 \) to 3 decimal places. (2)
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解題

(a) Let \( f(x) = x^3 - 5x + 1 \).
\( f(2) = 2^3 - 5(2) + 1 = 8 - 10 + 1 = -1 \).
\( f(2.5) = (2.5)^3 - 5(2.5) + 1 = 15.625 - 12.5 + 1 = 4.125 \).
Since \( f(x) \) is continuous and there is a change of sign between \( f(2) < 0 \) and \( f(2.5) > 0 \), a root \( \alpha \) lies in the interval \( [2, 2.5] \).

(b) \( x^3 - 5x + 1 = 0 \Rightarrow x^3 = 5x - 1 \).
Since \( x \neq 0 \) in the interval, dividing both sides by \( x \):
\( x^2 = 5 - \frac{1}{x} \).
Taking the positive square root (since \( x > 0 \)):
\( x = \sqrt{5 - \frac{1}{x}} \).
This yields \( x_{n+1} = \sqrt{5 - \frac{1}{x_n}} \).

(c) With \( x_0 = 2 \):
\( x_1 = \sqrt{5 - \frac{1}{2}} = \sqrt{4.5} \approx 2.1213 \) (4 d.p.)
\( x_2 = \sqrt{5 - \frac{1}{2.12132}} \approx 2.1280 \) (4 d.p.)
\( x_3 = \sqrt{5 - \frac{1}{2.12805}} \approx 2.1284 \) (4 d.p.)

(d) To show \( \alpha = 2.128 \) to 3 d.p., evaluate \( f(x) \) at the bounds \( 2.1275 \) and \( 2.1285 \):
\( f(2.1275) = (2.1275)^3 - 5(2.1275) + 1 \approx -0.0083 \)
\( f(2.1285) = (2.1285)^3 - 5(2.1285) + 1 \approx 0.0003 \)
Since there is a sign change and \( f(x) \) is continuous, the root lies in \( (2.1275, 2.1285) \), so \( \alpha = 2.128 \) to 3 d.p.

評分準則

(a)
M1: Evaluates \( f(2) \) and \( f(2.5) \) with at least one correct calculation.
A1: Both \( f(2) = -1 \) and \( f(2.5) = 4.125 \) correct, with a statement mentioning sign change and continuity.

(b)
M1: Realises division by \( x \) is required or makes \( x^2 \) the subject first.
A1*: Fully correct rearrangement showing all intermediate steps without errors.

(c)
M1: Correct substitution of \( x_0 = 2 \) into the formula.
A1: \( x_1 = 2.1213 \) and \( x_2 = 2.1280 \).
A1: \( x_3 = 2.1284 \).

(d)
M1: Selects the correct interval bounds: \( [2.1275, 2.1285] \).
A1: Obtains \( f(2.1275) \approx -0.0083 \) and \( f(2.1285) \approx 0.0003 \), and draws the correct conclusion based on the sign change.

部分 WMA14/01: Pure Mathematics P4

Answer all questions. Show sufficient working. Calculators allowed (no symbolic algebra).
8 題目 · 75.10000000000001 分
題目 1 · Short Answer
4.7 分
Find the coefficient of \(x^2\) in the binomial expansion of \(\frac{1+3x}{\sqrt{4-x}}\), in ascending powers of \(x\).
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解題

First, rewrite the term as \((1 + 3x)(4 - x)^{-\frac{1}{2}}\). Next, expand \((4 - x)^{-\frac{1}{2}} = 4^{-\frac{1}{2}}\left(1 - \frac{x}{4}\right)^{-\frac{1}{2}} = \frac{1}{2}\left(1 - \frac{x}{4}\right)^{-\frac{1}{2}}\). Using the binomial expansion formula: \(\left(1 - \frac{x}{4}\right)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)\left(-\frac{x}{4}\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}\left(-\frac{x}{4}\right)^2 + \dots = 1 + \frac{x}{8} + \frac{3}{128}x^2 + \dots\). Multiplying by \(\frac{1}{2}\) gives \((4 - x)^{-\frac{1}{2}} = \frac{1}{2} + \frac{x}{16} + \frac{3}{256}x^2 + \dots\). Now expand \((1+3x)\left(\frac{1}{2} + \frac{x}{16} + \frac{3}{256}x^2 + \dots\)\). The term in \(x^2\) is given by \(1 \cdot \frac{3}{256}x^2 + 3x \cdot \frac{x}{16} = \left(\frac{3}{256} + \frac{3}{16}\right)x^2 = \left(\frac{3}{256} + \frac{48}{256}\right)x^2 = \frac{51}{256}x^2\). Thus, the coefficient of \(x^2\) is \(\frac{51}{256}\).

評分準則

M1: Attempts to write as \(k\left(1 - \frac{x}{4}\right)^{-\frac{1}{2}}\) with a correct value of \(k = \frac{1}{2}\). A1: Correct simplified term in \(x\) or \(x^2\) of the expansion of \(\left(1 - \frac{x}{4}\right)^{-\frac{1}{2}}\), e.g., \(\frac{x}{8}\) or \(\frac{3}{128}x^2\). M1: Multiplies their expansion by \((1+3x)\) and identifies the relevant terms to find the coefficient of \(x^2\). A1.7: Correct final coefficient of \(\frac{51}{256}\) (or exact equivalent).
題目 2 · Short Answer
4.7 分
A curve \(C\) has equation \(2x^2 - 3xy + y^2 + 4x = 11\). Find the value of \(\frac{dy}{dx}\) at the point \((2, 1)\) on \(C\).
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解題

Differentiate the equation implicitly with respect to \(x\). Differentiating term by term: \(\frac{d}{dx}(2x^2) = 4x\). Using the product rule for \(-3xy\): \(\frac{d}{dx}(-3xy) = -3y - 3x\frac{dy}{dx}\). Differentiating the remaining terms: \(\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}\) and \(\frac{d}{dx}(4x) = 4\), and \(\frac{d}{dx}(11) = 0\). Putting it all together: \(4x - 3y - 3x\frac{dy}{dx} + 2y\frac{dy}{dx} + 4 = 0\). Substitute \(x = 2\) and \(y = 1\) into the differentiated equation: \(4(2) - 3(1) - 3(2)\frac{dy}{dx} + 2(1)\frac{dy}{dx} + 4 = 0\). Simplify to get: \(8 - 3 - 6\frac{dy}{dx} + 2\frac{dy}{dx} + 4 = 0\), which gives \(9 - 4\frac{dy}{dx} = 0\). Rearranging gives \(\frac{dy}{dx} = \frac{9}{4}\).

評分準則

M1: Attempts implicit differentiation with at least one term correct involving \(\frac{dy}{dx}\), such as \(2y\frac{dy}{dx}\) or via product rule on \(-3xy\). A1: Completely correct differentiation: \(4x - 3y - 3x\frac{dy}{dx} + 2y\frac{dy}{dx} + 4 = 0\). M1: Substitutes \(x = 2\) and \(y = 1\) into their differentiated expression to solve for \(\frac{dy}{dx}\). A1.7: Correct value of \(\frac{9}{4}\) or \(2.25\).
題目 3 · Short Answer
4.7 分
Find the exact value of \(\int_{0}^{\ln 2} \frac{e^{2x}}{e^x + 1} \, dx\).
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解題

We can write the integrand as \(\frac{e^{2x}}{e^x + 1} = \frac{e^x(e^x+1) - e^x}{e^x + 1} = e^x - \frac{e^x}{e^x + 1}\). Alternatively, using the substitution \(u = e^x + 1\) with \(du = e^x \, dx\). The integral of \(e^x\) is \(e^x\) and the integral of \(\frac{e^x}{e^x + 1}\) is \(\ln(e^x + 1)\). Thus, the indefinite integral is \(e^x - \ln(e^x + 1) + C\). Evaluating this between the limits of \(0\) and \(\ln 2\): at the upper limit \(x = \ln 2\), we have \(e^{\ln 2} - \ln(e^{\ln 2} + 1) = 2 - \ln(2+1) = 2 - \ln 3\). At the lower limit \(x = 0\), we have \(e^0 - \ln(e^0 + 1) = 1 - \ln(1+1) = 1 - \ln 2\). Subtracting the lower limit value from the upper limit value: \((2 - \ln 3) - (1 - \ln 2) = 1 - \ln 3 + \ln 2 = 1 - \ln\left(\frac{3}{2}\right)\).

評分準則

M1: Realises the need to split the fraction or use a suitable substitution like \(u = e^x + 1\) or \(u = e^x\). A1: Obtains a correct integrated expression, such as \(e^x - \ln(e^x + 1)\) or \(u - \ln u\). M1: Applies the limits of \(0\) and \(\ln 2\) (or corresponding limits \(u = 2\) and \(u = 3\) if using substitution) to their integrated function. A1.7: Correct exact answer in the form \(1 - \ln\left(\frac{3}{2}\right)\) or \(1 + \ln\left(\frac{2}{3}\right)\) or equivalent.
題目 4 · Structured
12.2 分
Answer all questions. Show sufficient working. Calculators allowed (no symbolic algebra).

(a) Find the binomial expansion of \((4 + 3x)^{-\frac{1}{2}}\), up to and including the term in \(x^3\), simplifying each coefficient. (5 marks)

(b) Hence, or otherwise, find the expansion of \(\frac{2-x}{\sqrt{4+3x}}\) in ascending powers of \(x\) up to and including the term in \(x^2\). (5 marks)

(c) Find the set of values of \(x\) for which this expansion is valid. (2.2 marks)
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解題

(a) Rewrite the expression:
\[ (4+3x)^{-\frac{1}{2}} = 4^{-\frac{1}{2}}\left(1 + \frac{3}{4}x\right)^{-\frac{1}{2}} = \frac{1}{2}\left(1 + \frac{3}{4}x\right)^{-\frac{1}{2}} \]
Apply the binomial theorem:
\[ \left(1 + \frac{3}{4}x\right)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)\left(\frac{3}{4}x\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}\left(\frac{3}{4}x\right)^2 + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!}\left(\frac{3}{4}x\right)^3 + \dots \]
\[ = 1 - \frac{3}{8}x + \frac{27}{128}x^2 - \frac{135}{1024}x^3 + \dots \]
Multiply by the constant term \(\frac{1}{2}\):
\[ \frac{1}{2}\left(1 - \frac{3}{8}x + \frac{27}{128}x^2 - \frac{135}{1024}x^3\right) = \frac{1}{2} - \frac{3}{16}x + \frac{27}{256}x^2 - \frac{135}{2048}x^3 \]

(b) Multiply \((2-x)\) by the expansion found in (a):
\[ \frac{2-x}{\sqrt{4+3x}} = (2-x)\left(\frac{1}{2} - \frac{3}{16}x + \frac{27}{256}x^2 + \dots\right) \]
Expand up to the term in \(x^2\):
\[ = 2\left(\frac{1}{2} - \frac{3}{16}x + \frac{27}{256}x^2\right) - x\left(\frac{1}{2} - \frac{3}{16}x\right) \]
\[ = 1 - \frac{3}{8}x + \frac{27}{128}x^2 - \frac{1}{2}x + \frac{3}{16}x^2 \]
Combine like terms:
\[ = 1 - \left(\frac{3}{8} + \frac{1}{2}\right)x + \left(\frac{27}{128} + \frac{24}{128}\right)x^2 \]
\[ = 1 - \frac{7}{8}x + \frac{51}{128}x^2 \]

(c) The expansion is valid when:
\[ \left|\frac{3}{4}x\right| < 1 \implies |x| < \frac{4}{3} \]

評分準則

(a)
B1: For factorising out \(4^{-1/2}\) or \(\frac{1}{2}\).
M1: For a correct attempt at the binomial expansion with fractional power \(-\frac{1}{2}\) showing at least two terms containing \(\frac{3}{4}x\).
A1: For correct simplified first and second terms: \(\frac{1}{2} - \frac{3}{16}x\).
A1: For the third term: \(\frac{27}{256}x^2\).
A1: For the fourth term: \(-\frac{135}{2048}x^3\).

(b)
M1: For an attempt to multiply their expansion by \((2-x)\).
M1: For collecting terms up to \(x^2\) (must see at least two multiplication steps correctly handled).
A1: For \(1 - \frac{7}{8}x\).
A1: For \(\frac{51}{128}x^2\).

(c)
M1: Recognises that the validity requires \(| \frac{3}{4}x | < 1\) or equivalent.
A1.2: Correct range of validity, written as \(|x| < \frac{4}{3}\) or \(-\frac{4}{3} < x < \frac{4}{3}\).
題目 5 · Structured
12.2 分
Answer all questions. Show sufficient working. Calculators allowed (no symbolic algebra).

A curve \(C\) has parametric equations
\[x = 3t^2 + 2, \quad y = 2t^3 - 3t\]

(a) Find \(\frac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(t\). (3 marks)

(b) Find the equation of the normal to \(C\) at the point where \(t = 2\). (5.2 marks)

(c) Find the coordinates of the points on \(C\) where the tangent is parallel to the x-axis. (4 marks)
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解題

(a) First, find the derivatives of \(x\) and \(y\) with respect to \(t\):
\[ \frac{\mathrm{d}x}{\mathrm{d}t} = 6t \]
\[ \frac{\mathrm{d}y}{\mathrm{d}t} = 6t^2 - 3 \]
Using the chain rule:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t} = \frac{6t^2 - 3}{6t} = \frac{2t^2 - 1}{2t} \]

(b) At \(t = 2\):
\[ x = 3(2)^2 + 2 = 14 \]
\[ y = 2(2)^3 - 3(2) = 10 \]
The gradient of the tangent at \(t = 2\) is:
\[ m_t = \frac{2(2)^2 - 1}{2(2)} = \frac{7}{4} \]
Therefore, the gradient of the normal is:
\[ m_n = -\frac{1}{m_t} = -\frac{4}{7} \]
The equation of the normal is:
\[ y - 10 = -\frac{4}{7}(x - 14) \]
\[ 7y - 70 = -4x + 56 \implies 4x + 7y - 126 = 0 \]

(c) The tangent is parallel to the x-axis when \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\[ \frac{2t^2 - 1}{2t} = 0 \implies 2t^2 - 1 = 0 \implies t^2 = \frac{1}{2} \implies t = \pm\frac{1}{\sqrt{2}} \]
When \(t = \frac{1}{\sqrt{2}}\):
\[ x = 3\left(\frac{1}{2}\right) + 2 = 3.5 \]
\[ y = 2\left(\frac{1}{2\sqrt{2}}\right) - \frac{3}{\sqrt{2}} = \frac{1}{\sqrt{2}} - \frac{3}{\sqrt{2}} = -\sqrt{2} \]
When \(t = -\frac{1}{\sqrt{2}}\):
\[ x = 3\left(\frac{1}{2}\right) + 2 = 3.5 \]
\[ y = 2\left(-\frac{1}{2\sqrt{2}}\right) + \frac{3}{\sqrt{2}} = -\frac{1}{\sqrt{2}} + \frac{3}{\sqrt{2}} = \sqrt{2} \]
So the coordinates are \((3.5, -\sqrt{2})\) and \((3.5, \sqrt{2})\).

評分準則

(a)
M1: Differentiates \(x\) and \(y\) with respect to \(t\) correctly to find \(\frac{\mathrm{d}x}{\mathrm{d}t}\) and \(\frac{\mathrm{d}y}{\mathrm{d}t}\).
M1: Applies the chain rule correctly to obtain \(\frac{\mathrm{d}y}{\mathrm{d}x}\).
A1: Correct expression \(\frac{2t^2 - 1}{2t}\) or equivalent.

(b)
M1: Substitutes \(t = 2\) to find coordinates \((14, 10)\).
M1: Substitutes \(t = 2\) into their expression for \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to find the tangent gradient.
M1: Uses the negative reciprocal of their tangent gradient to obtain the normal gradient.
M1: Applies the straight line equation using their point and normal gradient.
A1.2: Correct equation of the normal, e.g., \(4x + 7y - 126 = 0\) or \(y = -\frac{4}{7}x + 18\).

(c)
M1: Equates the numerator of their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to zero and solves for \(t\).
A1: Obtains \(t = \pm \frac{1}{\sqrt{2}}\) or equivalent.
M1: Substitutes at least one \(t\) value back to find both \(x\) and \(y\) coordinates.
A1: Correct coordinates: \((3.5, -\sqrt{2})\) and \((3.5, \sqrt{2})\) (or exact equivalents).
題目 6 · Structured
12.2 分
Answer all questions. Show sufficient working. Calculators allowed (no symbolic algebra).

A container is being filled with water. The volume of water in the container, \(V\text{ cm}^3\), at time \(t\text{ seconds}\) satisfies the differential equation
\[\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{k(200 - V)}{t+4}\]
where \(k\) is a positive constant.
Given that when \(t = 0\), \(V = 50\), and when \(t = 4\), \(V = 125\):

(a) Show that \(V = 200 - 150\left(\frac{4}{t+4}\right)^k\). (7.2 marks)

(b) Find the value of \(k\). (3 marks)

(c) Find the rate at which the volume of water is increasing when \(t = 12\). (2 marks)
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解題

(a) Separating the variables in the differential equation:
\[ \int \frac{1}{200 - V}\,\mathrm{d}V = \int \frac{k}{t+4}\,\mathrm{d}t \]
Integrating both sides yields:
\[ -\ln|200 - V| = k\ln|t+4| + C \]
\[ \ln|200 - V| = -k\ln(t+4) - C \]
Exponentiate both sides:
\[ 200 - V = A(t+4)^{-k} \quad \text{where } A = \mathrm{e}^{-C} \]
Using the boundary condition \(t = 0\), \(V = 50\):
\[ 200 - 50 = A(4)^{-k} \implies 150 = A \cdot 4^{-k} \implies A = 150 \cdot 4^k \]
Substitute \(A\) back into the equation:
\[ 200 - V = 150 \cdot 4^k \cdot (t+4)^{-k} = 150 \left(\frac{4}{t+4}\right)^k \]
\[ V = 200 - 150 \left(\frac{4}{t+4}\right)^k \]

(b) Use the condition \(t = 4\), \(V = 125\):
\[ 125 = 200 - 150 \left(\frac{4}{4+4}\right)^k \]
\[ 125 = 200 - 150 \left(\frac{1}{2}\right)^k \]
\[ 150 \left(\frac{1}{2}\right)^k = 75 \implies \left(\frac{1}{2}\right)^k = 0.5 \implies k = 1 \]

(c) With \(k = 1\), the volume equation is:
\[ V = 200 - 150\left(\frac{4}{t+4}\right) \]
When \(t = 12\):
\[ V = 200 - 150\left(\frac{4}{16}\right) = 200 - 37.5 = 162.5 \]
Substitute \(t = 12\), \(V = 162.5\), and \(k = 1\) into the differential equation:
\[ \frac{\mathrm{d}V}{\mathrm{d}t} = \frac{1(200 - 162.5)}{12+4} = \frac{37.5}{16} = 2.34375 \text{ cm}^3/\text{s} \text{ (or } \frac{75}{32}\text{)} \]

評分準則

(a)
M1: Separates variables correctly.
M1: Integrates both sides to obtain log terms, allowing for a constant of integration.
A1: Correct integration: \(-\ln|200 - V| = k\ln(t+4) + C\) or equivalent.
M1: Applies exponential rules correctly to express \(200 - V\) as a function of \(t\).
M1: Substitutes boundary condition \(t = 0\), \(V = 50\) to evaluate their constant of integration.
A1: Obtains the correct expression for \(A = 150 \cdot 4^k\).
A1.2: Completes the algebraic steps to show the given answer cleanly.

(b)
M1: Substitutes \(t = 4\), \(V = 125\) into the given equation.
M1: Simplifies to find \((0.5)^k = 0.5\) or uses logarithms to solve for \(k\).
A1: Correct value \(k = 1\).

(c)
M1: Substitutes \(t = 12\) into the equation to find \(V\) (or substitutes \(k = 1\) and \(t = 12\) into an expression for \(\frac{\mathrm{d}V}{\mathrm{d}t}\)).
A1: Correct rate \(2.34\) (or exact fraction \(\frac{75}{32}\)).
題目 7 · Structured
12.2 分
Answer all questions. Show sufficient working. Calculators allowed (no symbolic algebra).

The line \(l_1\) has vector equation
\[\mathbf{r} = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}\]
and the line \(l_2\) has vector equation
\[\mathbf{r} = \begin{pmatrix} 4 \\ 2 \\ 4 \end{pmatrix} + \mu \begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix}\]

(a) Show that the lines \(l_1\) and \(l_2\) intersect, and find the coordinates of their point of intersection, \(P\). (5 marks)

(b) The point \(A\) lies on \(l_1\) where \(l_1\) has parameter \(\lambda = 4\). The point \(B\) lies on \(l_2\) where \(l_2\) has parameter \(\mu = 1\). Find the cosine of the angle \(APB\). (4.2 marks)

(c) Find the exact area of the triangle \(APB\). (3 marks)
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解題

(a) Equate components of \(l_1\) and \(l_2\):
1) \(1 + 2\lambda = 4 - \mu \implies 2\lambda + \mu = 3\)
2) \(-2 + \lambda = 2 + 2\mu \implies \lambda - 2\mu = 4\)
3) \(3 - \lambda = 4 + 3\mu \implies \lambda + 3\mu = -1\)

From equation 1), \\mu = 3 - 2\lambda\). Substitute this into 2):
\[ \lambda - 2(3 - 2\lambda) = 4 \implies \lambda - 6 + 4\lambda = 4 \implies 5\lambda = 10 \implies \lambda = 2 \]
Then, \(\mu = 3 - 2(2) = -1\).
Substitute \(\lambda = 2\) and \(\mu = -1\) into equation 3) to verify consistency:
\[ \text{LHS} = 2 + 3(-1) = -1 \quad \text{RHS} = -1 \]
Since the equations are consistent, the lines intersect.
Using \(\lambda = 2\) in \(l_1\), the intersection point \(P\) is:
\[ P = \begin{pmatrix} 1 + 2(2) \\ -2 + 2 \\ 3 - 2 \end{pmatrix} = \begin{pmatrix} 5 \\ 0 \\ 1 \end{pmatrix} \]
So \(P\) is \((5, 0, 1)\).

(b) Find coordinates of \(A\) (where \(\lambda = 4\)):
\[ A = \begin{pmatrix} 1 + 2(4) \\ -2 + 4 \\ 3 - 4 \end{pmatrix} = \begin{pmatrix} 9 \\ 2 \\ -1 \end{pmatrix} \]
Find coordinates of \(B\) (where \(\mu = 1\)):
\[ B = \begin{pmatrix} 4 - 1 \\ 2 + 2(1) \\ 4 + 3(1) \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \\ 7 \end{pmatrix} \]
Now, find vectors \(\vec{PA}\) and \(\vec{PB}\):
\[ \vec{PA} = A - P = \begin{pmatrix} 9 - 5 \\ 2 - 0 \\ -1 - 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \\ -2 \end{pmatrix} \]
\[ \vec{PB} = B - P = \begin{pmatrix} 3 - 5 \\ 4 - 0 \\ 7 - 1 \end{pmatrix} = \begin{pmatrix} -2 \\ 4 \\ 6 \end{pmatrix} \]
Find magnitudes and dot product:
\[ |\vec{PA}| = \sqrt{4^2 + 2^2 + (-2)^2} = \sqrt{24} = 2\sqrt{6} \]
\[ |\vec{PB}| = \sqrt{(-2)^2 + 4^2 + 6^2} = \sqrt{56} = 2\sqrt{14} \]
\[ \vec{PA} \cdot \vec{PB} = 4(-2) + 2(4) + (-2)(6) = -8 + 8 - 12 = -12 \]
Using the scalar product:
\[ \cos(\angle APB) = \frac{\vec{PA} \cdot \vec{PB}}{|\vec{PA}| |\vec{PB}|} = \frac{-12}{2\sqrt{6} \cdot 2\sqrt{14}} = \frac{-12}{4\sqrt{84}} = -\frac{3}{2\sqrt{21}} = -\frac{\sqrt{21}}{14} \]

(c) First find \\sin(\angle APB)\):
\[ \sin^2(\angle APB) = 1 - \cos^2(\angle APB) = 1 - \frac{21}{196} = \frac{175}{196} \implies \sin(\angle APB) = \frac{5\sqrt{7}}{14} \]
Now, compute area:
\[ \text{Area} = \frac{1}{2} |\vec{PA}| |\vec{PB}| \sin(\angle APB) = \frac{1}{2} (2\sqrt{6})(2\sqrt{14}) \left(\frac{5\sqrt{7}}{14}\right) \]
\[ = 2\sqrt{84} \cdot \frac{5\sqrt{7}}{14} = 4\sqrt{21} \cdot \frac{5\sqrt{7}}{14} = \frac{20\sqrt{147}}{14} = \frac{140\sqrt{3}}{14} = 10\sqrt{3} \]

評分準則

(a)
M1: Equates components of both lines to form a system of equations in terms of \(\lambda\) and \(\mu\).
M1: Solves a pair of equations to find values for \(\lambda\) and \(\mu\).
A1: Obtains \(\lambda = 2\) and \(\mu = -1\).
M1: Checks values in the third equation to confirm intersection.
A1: Correct coordinates of the intersection point \(P(5, 0, 1)\).

(b)
M1: Finds coordinates of \(A\) and \(B\) and forms vectors \(\vec{PA}\) and \(\vec{PB}\).
M1: Finds the dot product \(\vec{PA} \cdot \vec{PB}\) and magnitudes of both vectors.
M1: Applies the cosine formula correctly.
A1.2: Correct simplified exact value of \(\cos(\angle APB) = -\frac{\sqrt{21}}{14}\).

(c)
M1: Finds \(\sin(\angle APB)\) using trigonometric identity \(\sin^2\theta + \cos^2\theta = 1\).
M1: Applies the triangle area formula \(\frac{1}{2} ab \sin C\).
A1: Correct exact area of \(10\sqrt{3}\).
題目 8 · Structured
12.2 分
Answer all questions. Show sufficient working. Calculators allowed (no symbolic algebra).

A curve \(C\) has parametric equations
\[x = \ln(t + 2), \quad y = \frac{t}{t+1}, \quad t > -1\]
The region \(R\) is bounded by the curve \(C\), the x-axis, and the lines \(x = \ln(2)\) and \(x = \ln(3)\).

(a) Show that the area of \(R\) is given by \(\int_0^1 \frac{t}{(t+1)(t+2)}\,\mathrm{d}t\). (4 marks)

(b) Using partial fractions, find the exact area of \(R\), giving your answer in the form \(\ln(a)\) where \(a\) is a rational number to be found. (6.2 marks)

(c) Find a cartesian equation of the curve \(C\) in the form \(y = f(x)\). (2 marks)
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解題

(a) The area of the region \(R\) is given by:
\[ \text{Area} = \int y\,\mathrm{d}x \]
Since \(x = \ln(t + 2)\), we have:
\[ \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{1}{t+2} \implies \mathrm{d}x = \frac{1}{t+2}\,\mathrm{d}t \]
Now find the limits of integration in terms of \(t\):
When \(x = \ln(2)\):
\[ \ln(t+2) = \ln(2) \implies t + 2 = 2 \implies t = 0 \]
When \(x = \ln(3)\):
\[ \ln(t+2) = \ln(3) \implies t + 2 = 3 \implies t = 1 \]
Substitute \(y\), \(\mathrm{d}x\), and limits into the integral:
\[ \text{Area} = \int_0^1 \frac{t}{t+1} \cdot \frac{1}{t+2}\,\mathrm{d}t = \int_0^1 \frac{t}{(t+1)(t+2)}\,\mathrm{d}t \]

(b) Express the integrand in partial fractions:
\[ \frac{t}{(t+1)(t+2)} = \frac{A}{t+1} + \frac{B}{t+2} \]
\[ t = A(t+2) + B(t+1) \]
Let \(t = -1\):
\[ -1 = A(1) \implies A = -1 \]
Let \(t = -2\):
\[ -2 = B(-1) \implies B = 2 \]
Thus:
\[ \int_0^1 \left( \frac{2}{t+2} - \frac{1}{t+1} \right)\,\mathrm{d}t = \Big[ 2\ln(t+2) - \ln(t+1) \Big]_0^1 \]
Evaluate at the limits:
\[ = \Big( 2\ln(3) - \ln(2) \Big) - \Big( 2\ln(2) - \ln(1) \Big) \]
\[ = 2\ln(3) - 3\ln(2) \]
\[ = \ln(3^2) - \ln(2^3) = \ln(9) - \ln(8) = \ln\left(\frac{9}{8}\right) \]
So \(a = \frac{9}{8}\).

(c) We have:
\[ x = \ln(t+2) \implies \mathrm{e}^x = t+2 \implies t = \mathrm{e}^x - 2 \]
Substitute into the equation for \(y\):
\[ y = \frac{\mathrm{e}^x - 2}{(\mathrm{e}^x - 2) + 1} = \frac{\mathrm{e}^x - 2}{\mathrm{e}^x - 1} \]

評分準則

(a)
M1: Correct expression for \(\frac{\mathrm{d}x}{\mathrm{d}t}\).
M1: Finding correct \(t\) limits corresponding to \(x = \ln(2)\) and \(x = \ln(3)\).
M1: Integrates using \(\int y \frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t\).
A1: Correctly derives the given integral with clear working.

(b)
M1: Correct form for partial fractions and finds values for \(A\) and \(B\).
A1: Correct partial fractions: \(\frac{2}{t+2} - \frac{1}{t+1}\).
M1: Integrates partial fractions to logarithmic terms.
A1: Correct integration: \(2\ln(t+2) - \ln(t+1)\).
M1: Substitutes limits 0 and 1, applying logarithmic subtraction/addition laws.
A1.2: Finds \(a = \frac{9}{8}\) in the final form \(\ln\left(\frac{9}{8}\right)\).

(c)
M1: Rearranges \(x = \ln(t+2)\) to make \(t\) the subject.
A1: Correct Cartesian equation: \(y = \frac{\mathrm{e}^x - 2}{\mathrm{e}^x - 1}\).

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