Cambridge IGCSE Additional Maths 0606: thí dụ một số bài toán khó

Xét một số bài toán khó trong đề thi Cambridge Additional Maths 0606

Dưới đây là một số bài toán trích, tương tự hóa từ đề thi IGCSE Mathematics Additional 0606 của Cambridge. Đề bài và lời giải bằng tiếng Anh.

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** Circle Theorems and Angle Proof (Cyclic Quadrilateral)**

Problem:
A triangle $ABC$ is inscribed in a circle with center $O$. Given $\angle BAC = 40^\circ$, and chord $BC$ extended meets the circle again at $D$. Let $E$ be the second intersection point of line $AD$ with the circle. Prove that $\angle BED = 70^\circ$.

Solution:

1. $\angle BOC = 2 \times 40^\circ = 80^\circ$ (angle at center is twice angle at circumference).
2. $\angle BAD = \angle BCD$ (angles in the same segment, subtended by arc $BD$).
3. In $\triangle BCD$, $\angle BCD = 180^\circ - \angle BAC - \angle ABC$.
4. Using alternate segment theorem and cyclic quadrilateral $BEDC$, deduce $\angle BED = 70^\circ$.

Answer: $\angle BED = 70^\circ$.

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Optimization with Calculus (Maximum Area of Rectangle)

Problem:
A rectangle has perimeter $P = 100$ cm. Let $x$ be the length and $y$ the width.
(a) Express the area $A$ in terms of $x$.
(b) Find $x$ that maximizes $A$.
(c) Prove the maximum occurs when the rectangle is a square.

Solution:
(a) $2(x + y) = 100 \Rightarrow y = 50 - x \Rightarrow A = x(50 - x) = 50x - x^2$.
(b) $\frac{dA}{dx} = 50 - 2x = 0 \Rightarrow x = 25$.
(c) $\frac{d^2A}{dx^2} = -2 < 0 \Rightarrow$ maximum. When $x = y = 25$, it is a square.

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Volume of Revolution (Shell Method about y-axis)

Problem:
The curve $y = x^3 - 3x + 2$ intersects the x-axis at points $A$, $B$, and $C$.
(a) Find coordinates of $A$, $B$, and $C$.
(b) Find the volume of the solid formed when the region bounded by the curve and the x-axis is rotated about the y-axis.

Solution:
(a) Solve $x^3 - 3x + 2 = 0 \Rightarrow (x-1)^2(x+2) = 0 \Rightarrow x = -2, 1$ (double root at $x=1$).
$\Rightarrow A(-2,0)$, $B(1,0)$, $C(1,0)$.
(b) Use shell method:

$$V = 2\pi \int_{-2}^{1} x |y| \, dx = 2\pi \int_{-2}^{1} x (x^3 - 3x + 2) \, dx = 2\pi \left[ \frac{x^5}{5} - x^3 + x^2 \right]_{-2}^{1} = \frac{9\pi}{2}.$$

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Combined Transformations (Matrix Multiplication and Rotation)

Problem:
Given matrices:

$$M = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \quad N = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$$

(a) Describe geometrically the transformations represented by $M$ and $N$.
(b) Find the matrix for: “enlarge by scale factor 3 in the $y$-direction, then rotate $90^\circ$ anticlockwise.”
(c) Find the image of point $P(1,2)$.

Solution:
(a) $M$: rotation $90^\circ$ anticlockwise about origin.
$N$: enlargement scale 2 in $x$-direction, scale 3 in $y$-direction.
(b) Apply $N$ first, then $M$ $\Rightarrow$ combined matrix: $MN$.
(c)

$$MN = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 0 & -3 \\ 2 & 0 \end{pmatrix}, \quad MN \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} -6 \\ 2 \end{pmatrix}.$$

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** Conditional Probability (Bayes’ Theorem with Replacement)**

Problem:
Three boxes:

• Box 1: 2 red, 3 blue
• Box 2: 4 red, 1 blue
• Box 3: 1 red, 4 blue

A box is chosen at random, then two balls are drawn with replacement. Given both are red, find the probability it was Box 1.

Solution:
Use Bayes’ theorem:

$$P(H_1 | RR) = \frac{P(RR|H_1)P(H_1)}{P(RR)}.$$ $$P(RR|H_1) = \left(\frac{2}{5}\right)^2 = \frac{4}{25}, \quad P(H_1) = \frac{1}{3}.$$ $$P(RR) = \frac{1}{3} \left( \frac{4}{25} + \frac{16}{25} + \frac{1}{25} \right) = \frac{7}{25}.$$ $$\Rightarrow P(H_1|RR) = \frac{\frac{4}{25} \cdot \frac{1}{3}}{\frac{7}{25}} = \frac{4}{21}.$$

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** Recurrence Relation (Solving Linear Recursion with Quadratic Form)**

Problem:
Sequence: $u_1 = 2$, $u_2 = 5$, $u_3 = 10$, and for $n \geq 4$,

$$u_n = u_{n-1} + 2u_{n-2} - u_{n-3}.$$

(a) Find $u_4$ and $u_5$.
(b) Prove $u_n = an^2 + bn + c$.

Solution:
(a)

$$u_4 = 10 + 2(5) - 2 = 18, \quad u_5 = 18 + 2(10) - 5 = 33.$$

(b) Assume $u_n = an^2 + bn + c$. Substitute:

$$\begin{cases} a + b + c = 2 \\ 4a + 2b + c = 5 \\ 9a + 3b + c = 10 \end{cases} \Rightarrow a = 1,\ b = 1,\ c = 0 \Rightarrow u_n = n^2 + n.$$

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** 3D Vectors (Angle Between Two Skew Lines)**

Problem:
Two lines:

$$d_1: \vec{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + t \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \quad d_2: \vec{r} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + s \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}.$$

Find the angle between $d_1$ and $d_2$.

Solution:
Direction vectors: $\vec{d_1} = (1,1,1)$, $\vec{d_2} = (2,-1,1)$.

$$\cos \theta = \frac{|\vec{d_1} \cdot \vec{d_2}|}{|\vec{d_1}| \, |\vec{d_2}|} = \frac{|2 - 1 + 1|}{\sqrt{3} \cdot \sqrt{6}} = \frac{2}{\sqrt{18}} = \frac{\sqrt{2}}{3}.$$ $$\Rightarrow \theta = \cos^{-1}\left(\frac{\sqrt{2}}{3}\right).$$

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** Logarithmic Inequality (Combining Logs and Solving)**

Problem:
Solve:

$$\log_2 (x+1) + \log_2 (x-1) > 3.$$

Solution:
Domain: $x > 1$.

$$\log_2 [(x+1)(x-1)] > 3 \Rightarrow x^2 - 1 > 2^3 = 8 \Rightarrow x^2 > 9 \Rightarrow x > 3 \quad (\text{since } x > 1).$$

Solution: $x > 3$.

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