STEP2 2004 -- Pure Mathematics

Exam: STEP2  |  Year: 2004  |  Questions: Q1—Q8  |  Total marks per question: 20

All questions are pure mathematics. Solutions and examiner commentary are included below.

Overview

Q	Topic	Difficulty	Key Techniques
1	代数/方程求解 Algebra/Equation Solving	Routine	平方去根号,因式分解,验根
2	代数/不等式 Algebra/Inequalities	Standard	配方,绝对值处理,韦达定理
3	微积分/曲线分析 Calculus/Curve Analysis	Standard	乘积求导法则,二阶导数判别法,曲线作图
4	微积分/最优化 Calculus/Optimisation	Challenging	三角函数求导,约束优化,消元法
5	积分/函数方程 Integration/Functional Equations	Standard	分部积分,三角恒等式展开,待定系数法
6	向量 Vectors	Challenging	向量投影,点积运算,正交分解
7	数值方法 Numerical Methods	Standard	二分法,函数符号判断,三角函数近似计算
8	微分方程 Differential Equations	Challenging	分离变量法,三角换元,不等式证明,图像分析

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Question 1

Topic: 代数/方程求解 Algebra/Equation Solving  |  Difficulty: Routine  |  Marks: 20

Problem

1 Find all real values of $x$ that satisfy:

(i) $\sqrt{3x^2 + 1} + \sqrt{x} - 2x - 1 = 0$;

(ii) $\sqrt{3x^2 + 1} - 2\sqrt{x} + x - 1 = 0$;

(iii) $\sqrt{3x^2 + 1} - 2\sqrt{x} - x + 1 = 0$.

Hint

Q1(i) Put the terms with radicals on one side and the terms without on the other and square. Repeat this strategy (S) and the equation x^4 - 6x^3 + 9x^2 - 4x = 0 () will be obtained. The roots of () are x = 0, 1, 4.

Squaring may introduce spurious roots, so these numbers must be checked to see that they are roots of the original equation. In fact, they are.

(ii) Application of S again leads to (*). Checking shows that x = 0, x = 1 are roots of the second equation but that x = 4 is not.

(iii) Again application of S leads to (*). Checking shows that x = 1, x = 4 are roots of the third equation but that x = 0 is not.

Model Solution

Part (i)

Rearrange the equation as

$\sqrt{3x^2 + 1} + \sqrt{x} = 2x + 1$

The domain requires $x \geq 0$ (for $\sqrt{x}$ to exist). For $x \geq 0$, both sides are non-negative, so squaring preserves equivalence.

Square both sides:

$3x^2 + 1 + 2\sqrt{x(3x^2 + 1)} + x = 4x^2 + 4x + 1$

Rearrange to isolate the remaining radical:

$2\sqrt{x(3x^2 + 1)} = 4x^2 + 4x + 1 - 3x^2 - 1 - x = x^2 + 3x$

So

$2\sqrt{x(3x^2 + 1)} = x(x + 3) \qquad \text{(*)}$

For $x \geq 0$, both sides of (*) are non-negative. Square again:

$4x(3x^2 + 1) = x^2(x + 3)^2$

If $x = 0$: both sides equal $0$, so $x = 0$ satisfies this equation. Check in the original: $\sqrt{1} + 0 - 0 - 1 = 0$ ✓.

If $x > 0$, divide both sides by $x$:

$4(3x^2 + 1) = x(x + 3)^2$

$12x^2 + 4 = x^3 + 6x^2 + 9x$

$x^3 - 6x^2 + 9x - 4 = 0$

Test $x = 1$: $1 - 6 + 9 - 4 = 0$ ✓. Factor out $(x - 1)$:

$x^3 - 6x^2 + 9x - 4 = (x - 1)(x^2 - 5x + 4) = (x - 1)(x - 1)(x - 4) = (x - 1)^2(x - 4)$

So the candidates are $x = 1$ and $x = 4$.

Check in the original equation $\sqrt{3x^2 + 1} + \sqrt{x} - 2x - 1 = 0$:

• $x = 1$: $\sqrt{4} + 1 - 2 - 1 = 2 + 1 - 3 = 0$ ✓
• $x = 4$: $\sqrt{49} + 2 - 8 - 1 = 7 + 2 - 9 = 0$ ✓

Solutions: $x \in \{0,\, 1,\, 4\}$.

Part (ii)

Rearrange as

$\sqrt{3x^2 + 1} - 2\sqrt{x} = 1 - x$

Square both sides:

$3x^2 + 1 - 4\sqrt{x(3x^2 + 1)} + 4x = 1 - 2x + x^2$

Rearrange:

$-4\sqrt{x(3x^2 + 1)} = 1 - 2x + x^2 - 3x^2 - 1 - 4x = -2x^2 - 6x$

$4\sqrt{x(3x^2 + 1)} = 2x^2 + 6x = 2x(x + 3)$

$2\sqrt{x(3x^2 + 1)} = x(x + 3)$

This is identical to equation (*) from part (i). The same polynomial $x^3 - 6x^2 + 9x - 4 = (x - 1)^2(x - 4) = 0$ is obtained, giving candidates $x = 0, 1, 4$.

Check in the original equation $\sqrt{3x^2 + 1} - 2\sqrt{x} + x - 1 = 0$:

• $x = 0$: $1 - 0 + 0 - 1 = 0$ ✓
• $x = 1$: $2 - 2 + 1 - 1 = 0$ ✓
• $x = 4$: $7 - 4 + 4 - 1 = 6 \neq 0$ ✗

Solutions: $x \in \{0,\, 1\}$.

Part (iii)

Rearrange as

$\sqrt{3x^2 + 1} - 2\sqrt{x} = x - 1$

Square both sides:

$3x^2 + 1 - 4\sqrt{x(3x^2 + 1)} + 4x = x^2 - 2x + 1$

Rearrange:

$-4\sqrt{x(3x^2 + 1)} = x^2 - 2x + 1 - 3x^2 - 1 - 4x = -2x^2 - 6x$

$2\sqrt{x(3x^2 + 1)} = x(x + 3)$

Again the same equation (*). Same candidates $x = 0, 1, 4$.

Check in the original equation $\sqrt{3x^2 + 1} - 2\sqrt{x} - x + 1 = 0$:

• $x = 0$: $1 - 0 - 0 + 1 = 2 \neq 0$ ✗
• $x = 1$: $2 - 2 - 1 + 1 = 0$ ✓
• $x = 4$: $7 - 4 - 4 + 1 = 0$ ✓

Solutions: $x \in \{1,\, 4\}$.

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Question 2

Topic: 代数/不等式 Algebra/Inequalities  |  Difficulty: Standard  |  Marks: 20

Problem

2 Prove that, if $|\alpha| < 2\sqrt{2}$, then there is no value of $x$ for which

$x^2 - \alpha|x| + 2 < 0 . \qquad \text{(*)}$

Find the solution set of (*) for $\alpha = 3$.

For $\alpha > 2\sqrt{2}$, the sum of the lengths of the intervals in which $x$ satisfies (*) is denoted by $S$. Find $S$ in terms of $\alpha$ and deduce that $S < 2\alpha$.

Sketch the graph of $S$ against $\alpha$.

Hint

Q2 Write Q = x^2 - α|x| + 2 = [|x| - α/2]^2 + 2 - α^2/4.

Thus α < 2√2 => 2 - α^2/4 > 0 => Q > 0 for all x.

It is therefore unnecessary to consider x > 0 and x < 0 separately and even more unnecessary to use calculus methods.

• if α = 3 then Q = (|x| - 1)(|x| - 2), in which case the solution set of Q < 0 is {x : -2 < x < -1} ∪ {x : 1 < x < 2}.

• The solutions in x of the equation Q = 0 are of the form -x2, -x1, x1, x2, where 0 < x1 < x2, so that S = 2(x2 - x1). Use of the identity x2 - x1 = sqrt((x2 + x1)^2 - 4x1x2) will lead immediately to S = 2sqrt(α^2 - 8). Thus S < 2sqrt(α^2) = 2α.

• The graph of S as a function of α is that part of the hyperbola 4α^2 - S^2 = 32 which is in the first quadrant. A sketch of this graph should, therefore, leave the other quadrants empty. It should also show the curve starting at the point (2√2, 0) and asymptotically approaching the line S = 2α.

Model Solution

Proof that no solution exists when $|\alpha| < 2\sqrt{2}$:

Let $u = |x| \geq 0$. Since $x^2 = |x|^2 = u^2$, the inequality becomes

$u^2 - \alpha u + 2 < 0$

Complete the square:

$\left(u - \frac{\alpha}{2}\right)^2 + 2 - \frac{\alpha^2}{4} < 0$

This requires

$\left(u - \frac{\alpha}{2}\right)^2 < \frac{\alpha^2}{4} - 2$

If $|\alpha| < 2\sqrt{2}$, then $\alpha^2 < 8$, so $\frac{\alpha^2}{4} < 2$, hence $\frac{\alpha^2}{4} - 2 < 0$.

Since $\left(u - \frac{\alpha}{2}\right)^2 \geq 0$ for all real $u$, we would need a non-negative quantity to be strictly less than a negative quantity, which is impossible.

Therefore, when $|\alpha| < 2\sqrt{2}$, there is no value of $x$ for which $x^2 - \alpha|x| + 2 < 0$.

Solution set for $\alpha = 3$:

$x^2 - 3|x| + 2 = (|x| - 1)(|x| - 2)$

The inequality $(|x| - 1)(|x| - 2) < 0$ holds when one factor is positive and the other is negative, i.e., $1 < |x| < 2$.

This gives $-2 < x < -1$ or $1 < x < 2$.

Solution set: $\{x : -2 < x < -1\} \cup \{x : 1 < x < 2\}$.

Finding $S$ for $\alpha > 2\sqrt{2}$:

The equation $x^2 - \alpha|x| + 2 = 0$ is equivalent to $|x|^2 - \alpha|x| + 2 = 0$. By the quadratic formula:

$|x| = \frac{\alpha \pm \sqrt{\alpha^2 - 8}}{2}$

Since $\alpha > 2\sqrt{2}$, we have $\alpha^2 > 8$, so $\sqrt{\alpha^2 - 8}$ is real and positive. Define:

$x_1 = \frac{\alpha - \sqrt{\alpha^2 - 8}}{2}, \qquad x_2 = \frac{\alpha + \sqrt{\alpha^2 - 8}}{2}$

Since $\sqrt{\alpha^2 - 8} < \alpha$ (as $8 > 0$), both roots are positive with $0 < x_1 < x_2$.

The inequality $x^2 - \alpha|x| + 2 < 0$ holds when $x_1 < |x| < x_2$, i.e.,

$x \in (-x_2,\, -x_1) \cup (x_1,\, x_2)$

The total length of these two intervals is

$S = (x_2 - x_1) + (x_2 - x_1) = 2(x_2 - x_1)$

By Vieta’s formulas applied to $|x|^2 - \alpha|x| + 2 = 0$:

$x_1 + x_2 = \alpha, \qquad x_1 x_2 = 2$

Therefore

$(x_2 - x_1)^2 = (x_1 + x_2)^2 - 4x_1 x_2 = \alpha^2 - 8$

$x_2 - x_1 = \sqrt{\alpha^2 - 8}$

$S = 2\sqrt{\alpha^2 - 8}$

Showing $S < 2\alpha$:

Since $\alpha^2 - 8 < \alpha^2$ (as $8 > 0$), taking positive square roots:

$\sqrt{\alpha^2 - 8} < \sqrt{\alpha^2} = \alpha$

(since $\alpha > 2\sqrt{2} > 0$). Therefore

$S = 2\sqrt{\alpha^2 - 8} < 2\alpha$

Sketch of $S$ against $\alpha$:

From $S = 2\sqrt{\alpha^2 - 8}$, squaring gives $S^2 = 4\alpha^2 - 32$, i.e.,

$4\alpha^2 - S^2 = 32$

This is a hyperbola in the $(\alpha, S)$-plane with asymptotes $S = \pm 2\alpha$. For $\alpha > 2\sqrt{2}$ and $S > 0$, we take the branch in the first quadrant:

• The curve starts at the point $(2\sqrt{2},\, 0)$.
• As $\alpha \to \infty$, the curve approaches the line $S = 2\alpha$ from below.
• The curve lies entirely below the line $S = 2\alpha$ for all $\alpha > 2\sqrt{2}$, consistent with the result $S < 2\alpha$.

A sketch should show only this first-quadrant branch, with the asymptote $S = 2\alpha$ drawn as a dashed line.

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Question 3

Topic: 微积分/曲线分析 Calculus/Curve Analysis  |  Difficulty: Standard  |  Marks: 20

Problem

3 The curve $C$ has equation

$y = x(x + 1)(x - 2)^4 .$

Show that the gradient of $C$ is $(x - 2)^3 (6x^2 + x - 2)$ and find the coordinates of all the stationary points. Determine the nature of each stationary point and sketch $C$.

In separate diagrams draw sketches of the curves whose equations are:

(i) $y^2 = x(x + 1)(x - 2)^4$;

(ii) $y = x^2(x^2 + 1)(x^2 - 2)^4$.

In each case, you should pay particular attention to the points where the curve meets the $x$ axis.

Hint

Q3 The obtaining of dy/dx in the form required is a routine exercise in differentiation followed by some algebra.

Setting dy/dx = 0 shows that there are stationary points where x = -2/3, 1/2, 2. Moreover d^2y/dx^2 = (x - 2)^3(12x + 1) + a term which is necessarily zero when x = -2/3, 1/2, 2. Thus d^2y/dx^2 is positive when x = -2/3 and negative when x = 1/2, so that C has a minimum at (-2/3, -8192/729) and a maximum at (1/2, 243/64). (Note that it is unnecessary to determine a simplified version of d^2y/dx^2 before inserting values of x.)

The argument d^2y/dx^2 = 0 at x = 2 => C has a point of inflexion at (2, 0) is false. In fact, in the neighbourhood of this point, y ≈ 6(x - 2)^4, so that it is obvious that C has a minimum there.

The sketch of C must have correct overall shape, location and orientation, and also show correct forms at (0,0), (2,0) and at ∞.

(i) This sketch may be deduced from that of C. It has symmetry about the x-axis and no part of it appears in the region -1 < x < 0.

(ii) This sketch may also be deduced from that of C. It has symmetry about the y-axis and no part of it appears in the region y < 0.

Model Solution

Finding $\dfrac{dy}{dx}$:

Let $y = x(x+1)(x-2)^4$. We use the product rule with $u = x(x+1) = x^2 + x$ and $v = (x-2)^4$:

$\frac{du}{dx} = 2x + 1, \qquad \frac{dv}{dx} = 4(x-2)^3$

$\frac{dy}{dx} = (2x+1)(x-2)^4 + (x^2+x) \cdot 4(x-2)^3$

Factor out $(x-2)^3$:

$\frac{dy}{dx} = (x-2)^3 \left[ (2x+1)(x-2) + 4(x^2+x) \right]$

Expand the bracket:

$(2x+1)(x-2) + 4x^2 + 4x = 2x^2 - 4x + x - 2 + 4x^2 + 4x = 6x^2 + x - 2$

Therefore

$\frac{dy}{dx} = (x-2)^3(6x^2 + x - 2) \qquad \checkmark$

Stationary points:

Set $\dfrac{dy}{dx} = 0$: either $(x-2)^3 = 0$ giving $x = 2$, or $6x^2 + x - 2 = 0$.

For $6x^2 + x - 2 = 0$, factor: $6x^2 + x - 2 = (2x - 1)(3x + 2) = 0$, so $x = \tfrac{1}{2}$ or $x = -\tfrac{2}{3}$.

Compute $y$ at each:

• $x = -\tfrac{2}{3}$: $y = (-\tfrac{2}{3})(\tfrac{1}{3})(-\tfrac{8}{3})^4 = (-\tfrac{2}{3})(\tfrac{1}{3})(\tfrac{4096}{81}) = -\dfrac{8192}{729}$

• $x = \tfrac{1}{2}$: $y = (\tfrac{1}{2})(\tfrac{3}{2})(-\tfrac{3}{2})^4 = (\tfrac{3}{4})(\tfrac{81}{16}) = \dfrac{243}{64}$

• $x = 2$: $y = 2 \cdot 3 \cdot 0^4 = 0$

Stationary points: $\left(-\tfrac{2}{3},\, -\tfrac{8192}{729}\right)$, $\left(\tfrac{1}{2},\, \tfrac{243}{64}\right)$, $(2,\, 0)$.

Nature of each stationary point:

Using the product rule on $\dfrac{dy}{dx} = (x-2)^3(6x^2 + x - 2)$:

$\frac{d^2y}{dx^2} = 3(x-2)^2(6x^2+x-2) + (x-2)^3(12x+1)$

At $x = -\tfrac{2}{3}$:

$\frac{d^2y}{dx^2} = 3\!\left(-\tfrac{8}{3}\right)^{\!2}\!\cdot 0 + \left(-\tfrac{8}{3}\right)^{\!3}\!\left(-7\right) = -\frac{512}{27} \times (-7) = \frac{3584}{27} > 0$

So $\left(-\tfrac{2}{3},\, -\tfrac{8192}{729}\right)$ is a minimum.

At $x = \tfrac{1}{2}$:

$\frac{d^2y}{dx^2} = 3\!\left(-\tfrac{3}{2}\right)^{\!2}\!\cdot 0 + \left(-\tfrac{3}{2}\right)^{\!3}\!\left(7\right) = -\frac{27}{8} \times 7 = -\frac{189}{8} < 0$

So $\left(\tfrac{1}{2},\, \tfrac{243}{64}\right)$ is a maximum.

At $x = 2$: both terms vanish, so $y'' = 0$. We need a different approach. Near $x = 2$:

$y = x(x+1)(x-2)^4 \approx 2 \cdot 3 \cdot (x-2)^4 = 6(x-2)^4$

Since $6(x-2)^4 \geq 0$ with equality only at $x = 2$, the point $(2, 0)$ is a minimum (a flat minimum, where the curve touches the $x$-axis).

Sketch of $C$:

Key features: $y = 0$ at $x = 0, -1, 2$. The curve has a minimum at $\left(-\tfrac{2}{3},\, -\tfrac{8192}{729}\right) \approx (-0.67,\, -11.24)$, a maximum at $\left(\tfrac{1}{2},\, \tfrac{243}{64}\right) \approx (0.5,\, 3.80)$, and a flat minimum at $(2, 0)$. As $x \to +\infty$, $y \to +\infty$; as $x \to -\infty$, $y \to +\infty$ (since $x(x+1) \to +\infty$ and $(x-2)^4 \to +\infty$).

Sketch (i): $y^2 = x(x+1)(x-2)^4$

This curve has symmetry about the $x$-axis (replacing $y$ by $-y$ leaves the equation unchanged).

It exists only where $x(x+1)(x-2)^4 \geq 0$. Since $(x-2)^4 \geq 0$ always, we need $x(x+1) \geq 0$, i.e., $x \leq -1$ or $x \geq 0$. The region $-1 < x < 0$ has no curve.

The curve meets the $x$-axis at $x = -1, 0, 2$. Away from these points, $y = \pm\sqrt{x(x+1)(x-2)^4}$, so the curve is a symmetric pair of branches above and below the $x$-axis, shaped like the original curve $C$ but reflected.

Sketch (ii): $y = x^2(x^2+1)(x^2-2)^4$

This function is even: $f(-x) = f(x)$, so the curve is symmetric about the $y$-axis.

Since $x^2 \geq 0$, $(x^2+1) > 0$, and $(x^2-2)^4 \geq 0$, we have $y \geq 0$ for all $x$. The curve lies entirely in the upper half-plane.

Zeros: $x = 0$ (from $x^2$, double root — curve touches axis and turns back) and $x = \pm\sqrt{2}$ (from $(x^2-2)^4$, even multiplicity — flat minimum touching axis).

The shape in the right half-plane ($x \geq 0$) mirrors that of $C$ with $x$ replaced by $x^2$: a local maximum between $x = 0$ and $x = \sqrt{2}$, then a flat minimum at $x = \sqrt{2}$, then $y \to \infty$ as $x \to \infty$. By symmetry, the left half-plane is the mirror image.

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Question 4

Topic: 微积分/最优化 Calculus/Optimisation  |  Difficulty: Challenging  |  Marks: 20

Problem

4

Figure 1 Figure 2

(i) An attempt is made to move a rod of length $L$ from a corridor of width $a$ into a corridor of width $b$, where $a \neq b$. The corridors meet at right angles, as shown in Figure 1 and the rod remains horizontal. Show that if the attempt is to be successful then

$L \le a \operatorname{cosec} \alpha + b \sec \alpha ,$

where $\alpha$ satisfies

$\tan^3 \alpha = \frac{a}{b} .$

(ii) An attempt is made to move a rectangular table-top, of width $w$ and length $l$, from one corridor to the other, as shown in the Figure 2. The table-top remains horizontal. Show that if the attempt is to be successful then

$l \le a \operatorname{cosec} \beta + b \sec \beta - 2w \operatorname{cosec} 2\beta,$

where $\beta$ satisfies

$w = \left( \frac{a - b \tan^3 \beta}{1 - \tan^2 \beta} \right) \cos \beta .$

Hint

Q4 It is important to realise at the outset that α is a constant defined by a and b and that β is a constant defined by a, b and w. Variable angles θ/φ are needed to define the orientation of the rod/table in the general situation.

(i) Clearly, for all θ ∈ (0, π/2), it is necessary that f(θ) ≥ L, where f(θ) = a csc θ + b sec θ. Setting f’(θ) = 0 will then lead to the required result.

(ii) Here, for all φ ∈ (0, π/2), it is necessary that y ≥ l, where y is such that b = (y-x) cos φ + w sin φ and x is such that a = x sin φ + w cos φ. (Other formulations are possible.) Elimination of x leads to y = a csc φ + b sec φ - 2w csc 2φ

Setting y’(φ) = 0 plus some further working will then produce the required result.

Model Solution

Part (i).

Let $\theta$ be the variable angle the rod makes with the corridor of width $b$. The rod extends a distance $\dfrac{a}{\sin\theta} = a\operatorname{cosec}\theta$ in the $a$-corridor and $\dfrac{b}{\cos\theta} = b\sec\theta$ in the $b$-corridor.

For the rod to fit through the corner:

$L \leq a\operatorname{cosec}\theta + b\sec\theta \quad \text{for all } \theta \in (0,\, \pi/2) \qquad \text{(*)}$

Define $f(\theta) = a\operatorname{cosec}\theta + b\sec\theta$. The tightest constraint comes at the minimum of $f$.

$f'(\theta) = -a\operatorname{cosec}\theta\cot\theta + b\sec\theta\tan\theta$

Setting $f'(\theta) = 0$:

$b\sec\theta\tan\theta = a\operatorname{cosec}\theta\cot\theta$

$\frac{b\sin\theta}{\cos^2\theta} = \frac{a\cos\theta}{\sin^2\theta}$

Cross-multiply:

$b\sin^3\theta = a\cos^3\theta$

$\tan^3\theta = \frac{a}{b}$

The critical angle $\alpha$ satisfies $\tan^3\alpha = \dfrac{a}{b}$. Since $f(\theta) \to \infty$ as $\theta \to 0^+$ (because $\operatorname{cosec}\theta \to \infty$) and as $\theta \to \frac{\pi}{2}^-$ (because $\sec\theta \to \infty$), this critical point must be a minimum.

From (*), evaluated at $\theta = \alpha$:

$L \leq a\operatorname{cosec}\alpha + b\sec\alpha$

Part (ii).

Let $\phi$ be the variable angle the table-top makes with the $b$-corridor. Let $x$ be the distance from the corner along the table’s inner edge to the point where it meets the $a$-corridor wall, and let $y - x$ be the distance from the corner to the point where it meets the $b$-corridor wall.

The table’s width $w$ is perpendicular to its length. Projecting onto each corridor direction:

$a = x\sin\phi + w\cos\phi \qquad \text{(*)}$

$b = (y - x)\cos\phi + w\sin\phi \qquad \text{(**)}$

From (*): $x = \dfrac{a - w\cos\phi}{\sin\phi}$.

From (**): $y - x = \dfrac{b - w\sin\phi}{\cos\phi}$.

Adding:

$y = \frac{a - w\cos\phi}{\sin\phi} + \frac{b - w\sin\phi}{\cos\phi}$

$= \frac{a}{\sin\phi} - \frac{w\cos\phi}{\sin\phi} + \frac{b}{\cos\phi} - \frac{w\sin\phi}{\cos\phi}$

$= a\operatorname{cosec}\phi + b\sec\phi - w\left(\frac{\cos\phi}{\sin\phi} + \frac{\sin\phi}{\cos\phi}\right)$

$= a\operatorname{cosec}\phi + b\sec\phi - w \cdot \frac{\cos^2\phi + \sin^2\phi}{\sin\phi\cos\phi}$

$= a\operatorname{cosec}\phi + b\sec\phi - \frac{w}{\sin\phi\cos\phi}$

Since $\sin\phi\cos\phi = \frac{1}{2}\sin 2\phi$, we have $\dfrac{1}{\sin\phi\cos\phi} = 2\operatorname{cosec}2\phi$, so

$y = a\operatorname{cosec}\phi + b\sec\phi - 2w\operatorname{cosec}2\phi$

For the table to fit: $l \leq y$ for all $\phi \in (0,\, \pi/2)$.

The critical angle $\beta$ minimizes $y$. Setting $\dfrac{dy}{d\phi} = 0$:

$\frac{dy}{d\phi} = -a\operatorname{cosec}\phi\cot\phi + b\sec\phi\tan\phi + 4w\operatorname{cosec}2\phi\cot 2\phi = 0$

Converting to $\sin$ and $\cos$ and using $\sin^2 2\phi = 4\sin^2\phi\cos^2\phi$:

$-\frac{a\cos\phi}{\sin^2\phi} + \frac{b\sin\phi}{\cos^2\phi} + \frac{w\cos 2\phi}{\sin^2\phi\cos^2\phi} = 0$

Multiply through by $\sin^2\phi\cos^2\phi$:

$-a\cos^3\phi + b\sin^3\phi + w\cos 2\phi = 0$

Using $\cos 2\phi = \cos^2\phi - \sin^2\phi$:

$-a\cos^3\phi + b\sin^3\phi + w(\cos^2\phi - \sin^2\phi) = 0$

Divide by $\cos^3\phi$:

$-a + b\tan^3\phi + w\left(\frac{1}{\cos\phi} - \frac{\sin^2\phi}{\cos^3\phi}\right) = 0$

$-a + b\tan^3\phi + \frac{w(\cos^2\phi - \sin^2\phi)}{\cos^3\phi} = 0$

$-a + b\tan^3\phi + \frac{w(1 - \tan^2\phi)}{\cos\phi} = 0$

$\frac{w(1 - \tan^2\phi)}{\cos\phi} = a - b\tan^3\phi$

$w = \frac{(a - b\tan^3\phi)\cos\phi}{1 - \tan^2\phi}$

At $\phi = \beta$:

$w = \frac{(a - b\tan^3\beta)\cos\beta}{1 - \tan^2\beta}$

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Question 5

Topic: 积分/函数方程 Integration/Functional Equations  |  Difficulty: Standard  |  Marks: 20

Problem

5 Evaluate $\int_0^\pi x \sin x \, dx$ and $\int_0^\pi x \cos x \, dx$.

The function $f$ satisfies the equation

$f(t) = t + \int_0^\pi f(x) \sin(x + t) \, dx . \qquad \text{(*)}$

Show that

$f(t) = t + A \sin t + B \cos t , \qquad \text{(**)}$

where $A = \int_0^\pi f(x) \cos x \, dx$ and $B = \int_0^\pi f(x) \sin x \, dx$.

Use the expression (**) to find $A$ and $B$ by substituting for $f(t)$ and $f(x)$ in (*) and equating coefficients of $\sin t$ and $\cos t$.

Hint

Q5 Using the integration by parts rule it is easy to establish the results ∫₀^π x sin x dx = π and ∫₀^π x cos x dx = -2.

• Write sin(x + t) = sin x cos t + sin t cos x and the result f(t) = t + A sin t + B cos t, where A and B are as defined in the question, follows immediately.

• Hence write t + A sin t + B cos t = t + ∫₀^π (x + A sin x + B cos x) sin(x + t) dx (***) so that as

∫₀^π x sin(x + t) dx = … = π cos t - 2 sin t,

∫₀^π sin x sin(x + t) dx = … = (π/2) cos t,

∫₀^π cos x sin(x + t) dx = … = (π/2) sin t,

then, by considering the coefficients of cos t and sin t on both sides of (***), it follows that

A = -2 + (π/2)B, B = π + (π/2)A => A = -2, B = 0.

Alternatively, equations for A and B can be obtained by putting t = 0 and t = π/2 in (***).

Model Solution

Part 1: Evaluating the integrals

Integral 1: $\displaystyle I_1 = \int_0^\pi x \sin x \, dx$

Integration by parts with $u = x$, $dv = \sin x \, dx$ (so $du = dx$, $v = -\cos x$):

$I_1 = \left[-x\cos x\right]_0^\pi + \int_0^\pi \cos x \, dx = -\pi\cos\pi + 0 + \left[\sin x\right]_0^\pi = \pi + 0 = \pi$

$\int_0^\pi x\sin x \, dx = \pi \qquad \qquad \text{(I)}$

Integral 2: $\displaystyle I_2 = \int_0^\pi x \cos x \, dx$

Integration by parts with $u = x$, $dv = \cos x \, dx$ (so $du = dx$, $v = \sin x$):

$I_2 = \left[x\sin x\right]_0^\pi - \int_0^\pi \sin x \, dx = 0 - \left[-\cos x\right]_0^\pi = \cos\pi - \cos 0 = -2$

$\int_0^\pi x\cos x \, dx = -2 \qquad \qquad \text{(II)}$

Part 2: Showing $f(t) = t + A\sin t + B\cos t$

Expand $\sin(x + t) = \sin x \cos t + \cos x \sin t$ in equation (**):

$f(t) = t + \int_0^\pi f(x)(\sin x\cos t + \cos x\sin t) \, dx$

$= t + \cos t \int_0^\pi f(x)\sin x \, dx + \sin t \int_0^\pi f(x)\cos x \, dx$

$= t + B\cos t + A\sin t \qquad \checkmark$

Part 3: Finding $A$ and $B$

Substitute $f(x) = x + A\sin x + B\cos x$ into $A = \int_0^\pi f(x)\cos x \, dx$:

$A = \int_0^\pi (x + A\sin x + B\cos x)\cos x \, dx$

$= \int_0^\pi x\cos x \, dx + A\int_0^\pi \sin x\cos x \, dx + B\int_0^\pi \cos^2 x \, dx$

Evaluate each piece. We already know $\int_0^\pi x\cos x \, dx = -2$ from (II).

For the second integral: $\int_0^\pi \sin x\cos x \, dx = \tfrac{1}{2}\int_0^\pi \sin 2x \, dx = \tfrac{1}{2}\left[-\tfrac{1}{2}\cos 2x\right]_0^\pi = -\tfrac{1}{4}(\cos 2\pi - \cos 0) = 0$.

For the third integral: $\int_0^\pi \cos^2 x \, dx = \tfrac{1}{2}\int_0^\pi (1 + \cos 2x) \, dx = \tfrac{1}{2}\left[x + \tfrac{1}{2}\sin 2x\right]_0^\pi = \tfrac{\pi}{2}$.

Therefore

$A = -2 + 0 + \frac{\pi}{2}B \qquad \qquad \text{(III)}$

Substitute into $B = \int_0^\pi f(x)\sin x \, dx$:

$B = \int_0^\pi (x + A\sin x + B\cos x)\sin x \, dx$

$= \int_0^\pi x\sin x \, dx + A\int_0^\pi \sin^2 x \, dx + B\int_0^\pi \cos x\sin x \, dx$

We know $\int_0^\pi x\sin x \, dx = \pi$ from (I), $\int_0^\pi \sin^2 x \, dx = \tfrac{\pi}{2}$ (by symmetry with $\cos^2$), and $\int_0^\pi \sin x\cos x \, dx = 0$.

Therefore

$B = \pi + \frac{\pi}{2}A + 0 \qquad \qquad \text{(IV)}$

From (III): $A = -2 + \frac{\pi}{2}B$.

Substitute into (IV):

$B = \pi + \frac{\pi}{2}\left(-2 + \frac{\pi}{2}B\right) = \pi - \pi + \frac{\pi^2}{4}B = \frac{\pi^2}{4}B$

$B - \frac{\pi^2}{4}B = 0 \qquad \Longrightarrow \qquad B\left(1 - \frac{\pi^2}{4}\right) = 0$

Since $\pi^2 \neq 4$, we must have $B = 0$.

Then from (III): $A = -2 + 0 = -2$.

$\boxed{A = -2, \quad B = 0}$

so $f(t) = t - 2\sin t$.

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Question 6

Topic: 向量 Vectors  |  Difficulty: Challenging  |  Marks: 20

Problem

6 The vectors a and b lie in the plane $\Pi$. Given that $|\mathbf{a}| = 1$ and $\mathbf{a} \cdot \mathbf{b} = 3$, find, in terms of a and b, a vector p parallel to a and a vector q perpendicular to a, both lying in the plane $\Pi$, such that

$\mathbf{p} + \mathbf{q} = \mathbf{a} + \mathbf{b} .$

The vector c is not parallel to the plane $\Pi$ and is such that $\mathbf{a} \cdot \mathbf{c} = -2$ and $\mathbf{b} \cdot \mathbf{c} = 2$. Given that $|\mathbf{b}| = 5$, find, in terms of a, b and c, vectors P, Q and R such that P and Q are parallel to p and q, respectively, R is perpendicular to the plane $\Pi$ and

$\mathbf{P} + \mathbf{Q} + \mathbf{R} = \mathbf{a} + \mathbf{b} + \mathbf{c} .$

Hint

From the data it follows that the component of b in the direction of a is $3\mathbf{a}$.

Hence $\mathbf{p} = 4\mathbf{a}$ and $\mathbf{q} = \mathbf{b} - 3\mathbf{a}$.

• Again from the data, $(\mathbf{c}\cdot\mathbf{a})\mathbf{a} = -2\mathbf{a}$ and

$|\mathbf{q}|^2 = \mathbf{b}\cdot\mathbf{b} - 6\mathbf{a}\cdot\mathbf{b} + 9\mathbf{a}\cdot\mathbf{a} = 25 - 18 + 9 = 16,$

so $|\mathbf{q}| = 4$. Also,

$\left[\frac{\mathbf{c}\cdot\mathbf{q}}{|\mathbf{q}|^2}\right]\mathbf{q} = \frac12\mathbf{b} - \frac32\mathbf{a}.$

Thus $\mathbf{P} = 2\mathbf{a}$, $\mathbf{Q} = -\frac92\mathbf{a} + \frac32\mathbf{b}$, and $\mathbf{R} = \frac72\mathbf{a} - \frac12\mathbf{b} + \mathbf{c}$.

Model Solution

Part 1: Finding $\mathbf{p}$ and $\mathbf{q}$

We need $\mathbf{p}$ parallel to $\mathbf{a}$ and $\mathbf{q}$ perpendicular to $\mathbf{a}$, both in $\Pi$, with $\mathbf{p} + \mathbf{q} = \mathbf{a} + \mathbf{b}$.

Since $\mathbf{p}$ is parallel to $\mathbf{a}$, write $\mathbf{p} = \lambda\mathbf{a}$ for some scalar $\lambda$. Then $\mathbf{q} = \mathbf{a} + \mathbf{b} - \lambda\mathbf{a}$.

The condition $\mathbf{q} \perp \mathbf{a}$ gives $\mathbf{q} \cdot \mathbf{a} = 0$:

$(\mathbf{a} + \mathbf{b} - \lambda\mathbf{a}) \cdot \mathbf{a} = 0$

$\mathbf{a} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{a} - \lambda\,\mathbf{a} \cdot \mathbf{a} = 0$

$1 + 3 - \lambda \cdot 1 = 0 \qquad \Longrightarrow \qquad \lambda = 4$

Therefore

$\mathbf{p} = 4\mathbf{a}, \qquad \mathbf{q} = \mathbf{b} - 3\mathbf{a}$

Part 2: Finding $\mathbf{P}$, $\mathbf{Q}$ and $\mathbf{R}$

We need $\mathbf{P}$ parallel to $\mathbf{p}$ (i.e., parallel to $\mathbf{a}$), $\mathbf{Q}$ parallel to $\mathbf{q}$, $\mathbf{R}$ perpendicular to $\Pi$, and $\mathbf{P} + \mathbf{Q} + \mathbf{R} = \mathbf{a} + \mathbf{b} + \mathbf{c}$.

Write $\mathbf{P} = \mu\mathbf{a}$ and $\mathbf{Q} = \nu\mathbf{q} = \nu(\mathbf{b} - 3\mathbf{a})$ for scalars $\mu, \nu$.

Then $\mathbf{R} = \mathbf{a} + \mathbf{b} + \mathbf{c} - \mu\mathbf{a} - \nu(\mathbf{b} - 3\mathbf{a})$.

Since $\mathbf{R} \perp \Pi$ and $\mathbf{a}, \mathbf{b} \in \Pi$, we need $\mathbf{R} \cdot \mathbf{a} = 0$ and $\mathbf{R} \cdot \mathbf{q} = 0$ (where $\mathbf{q}$ also lies in $\Pi$ and is perpendicular to $\mathbf{a}$, so $\{\mathbf{a}, \mathbf{q}\}$ spans $\Pi$).

Condition $\mathbf{R} \cdot \mathbf{a} = 0$:

$\mathbf{R} \cdot \mathbf{a} = (\mathbf{a} + \mathbf{b} + \mathbf{c} - \mu\mathbf{a} - \nu\mathbf{b} + 3\nu\mathbf{a}) \cdot \mathbf{a} = 0$

$= \mathbf{a}\cdot\mathbf{a} + \mathbf{b}\cdot\mathbf{a} + \mathbf{c}\cdot\mathbf{a} - \mu\,\mathbf{a}\cdot\mathbf{a} - \nu\,\mathbf{b}\cdot\mathbf{a} + 3\nu\,\mathbf{a}\cdot\mathbf{a} = 0$

$= 1 + 3 + (-2) - \mu - 3\nu + 3\nu = 0$

$2 - \mu = 0 \qquad \Longrightarrow \qquad \mu = 2$

So $\mathbf{P} = 2\mathbf{a}$.

Condition $\mathbf{R} \cdot \mathbf{q} = 0$:

First compute $\mathbf{q} \cdot \mathbf{q} = |\mathbf{q}|^2$:

$|\mathbf{q}|^2 = (\mathbf{b} - 3\mathbf{a}) \cdot (\mathbf{b} - 3\mathbf{a}) = |\mathbf{b}|^2 - 6\,\mathbf{a}\cdot\mathbf{b} + 9|\mathbf{a}|^2 = 25 - 18 + 9 = 16$

So $|\mathbf{q}| = 4$.

Now compute $\mathbf{c} \cdot \mathbf{q} = \mathbf{c} \cdot (\mathbf{b} - 3\mathbf{a}) = \mathbf{b}\cdot\mathbf{c} - 3\,\mathbf{a}\cdot\mathbf{c} = 2 - 3(-2) = 8$.

Also $\mathbf{a} \cdot \mathbf{q} = \mathbf{a} \cdot (\mathbf{b} - 3\mathbf{a}) = 3 - 3 = 0$ (as expected, since $\mathbf{q} \perp \mathbf{a}$).

Now $\mathbf{R} \cdot \mathbf{q} = 0$:

$\mathbf{R} \cdot \mathbf{q} = (\mathbf{a} + \mathbf{b} + \mathbf{c} - 2\mathbf{a} - \nu\mathbf{b} + 3\nu\mathbf{a}) \cdot \mathbf{q} = 0$

$= (-\mathbf{a} + (1-\nu)\mathbf{b} + \mathbf{c} + 3\nu\mathbf{a}) \cdot \mathbf{q} = 0$

$= ((-1+3\nu)\mathbf{a} + (1-\nu)\mathbf{b} + \mathbf{c}) \cdot \mathbf{q} = 0$

Since $\mathbf{a} \cdot \mathbf{q} = 0$:

$(1-\nu)\,\mathbf{b}\cdot\mathbf{q} + \mathbf{c}\cdot\mathbf{q} = 0$

We have $\mathbf{b}\cdot\mathbf{q} = \mathbf{b}\cdot(\mathbf{b}-3\mathbf{a}) = 25 - 9 = 16$.

So: $(1-\nu)(16) + 8 = 0$, giving $1 - \nu = -\tfrac{1}{2}$, hence $\nu = \tfrac{3}{2}$.

Therefore

$\mathbf{Q} = \tfrac{3}{2}(\mathbf{b} - 3\mathbf{a}) = \tfrac{3}{2}\mathbf{b} - \tfrac{9}{2}\mathbf{a}$

Finding $\mathbf{R}$:

$\mathbf{R} = \mathbf{a} + \mathbf{b} + \mathbf{c} - \mathbf{P} - \mathbf{Q} = \mathbf{a} + \mathbf{b} + \mathbf{c} - 2\mathbf{a} - \tfrac{3}{2}\mathbf{b} + \tfrac{9}{2}\mathbf{a}$

$= \left(1 - 2 + \tfrac{9}{2}\right)\mathbf{a} + \left(1 - \tfrac{3}{2}\right)\mathbf{b} + \mathbf{c} = \tfrac{7}{2}\mathbf{a} - \tfrac{1}{2}\mathbf{b} + \mathbf{c}$

$\boxed{\mathbf{P} = 2\mathbf{a}, \quad \mathbf{Q} = -\tfrac{9}{2}\mathbf{a} + \tfrac{3}{2}\mathbf{b}, \quad \mathbf{R} = \tfrac{7}{2}\mathbf{a} - \tfrac{1}{2}\mathbf{b} + \mathbf{c}}$

Verification: $\mathbf{P} + \mathbf{Q} + \mathbf{R} = (2 - \tfrac{9}{2} + \tfrac{7}{2})\mathbf{a} + (0 + \tfrac{3}{2} - \tfrac{1}{2})\mathbf{b} + \mathbf{c} = \mathbf{a} + \mathbf{b} + \mathbf{c}$ ✓

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Question 7

Topic: 数值方法 Numerical Methods  |  Difficulty: Standard  |  Marks: 20

Problem

7 The function $f$ is defined by

$f(x) = 2 \sin x - x .$

Show graphically that the equation $f(x) = 0$ has exactly one root in the interval $[\frac{1}{2}\pi, \pi]$. This interval is denoted $I_0$.

In order to determine the root, a sequence of intervals $I_1, I_2, \dots$ is generated in the following way. If the interval $I_n = [a_n, b_n]$, and $c_n = (a_n + b_n)/2$, then

$I_{n+1} = \begin{cases} [a_n, c_n] & \text{if } f(a_n)f(c_n) < 0 ; \\ [c_n, b_n] & \text{if } f(c_n)f(b_n) < 0 . \end{cases}$

By using the approximations $\frac{1}{\sqrt{2}} \approx 0.7$ and $\pi \approx \sqrt{10}$, show that $I_2 = [\frac{1}{2}\pi, \frac{5}{8}\pi]$ and find $I_3$.

Hint

Good sketch graphs of $y = x$ and $y = 2 \sin x$, in the same diagram and over the interval $[0, \pi]$, will readily show that the equation $f(x) = 0$ has exactly one root in the interval $[\pi/2, \pi]$.

• $f(3\pi/4) = \sqrt{2} - 3\pi/4$ has the same sign as $2 - 9\pi^2/16 \approx 2 - 45/8 = -29/8 < 0$. Hence, as $f(\pi/2) = 2 - \pi/2 > 0$ and $f(\pi) = -\pi < 0$, we have $I_1 = [\pi/2, 3\pi/4]$.

• If $x = \sin(5\pi/8)$, then $2x\sqrt{1 - x^2} = 1/\sqrt{2}$, so $8x^4 - 8x^2 + 1 = 0$ and hence $x^2 = \frac12 + \frac{1}{2\sqrt{2}} \approx 0.85$. The sign of $f(5\pi/8)$ is the same as that of $4x^2 - 25\pi^2/64 \approx 17/5 - 125/32 < 0$. Hence $I_2 = [\pi/2, 5\pi/8]$.

• A good approximation to $x = \sin 9\pi/16$ may also be obtained in a similar way. In fact, it will be found that $f(9\pi/16) > 0$ so that $I_3 = [9\pi/16, 5\pi/8]$.

Model Solution

Graphical argument for exactly one root in $\left[\tfrac{\pi}{2},\, \pi\right]$:

Consider $f(x) = 2\sin x - x$ on $[0, \pi]$. At the endpoints of the interval:

$f\!\left(\tfrac{\pi}{2}\right) = 2\sin\tfrac{\pi}{2} - \tfrac{\pi}{2} = 2 - \tfrac{\pi}{2} \approx 2 - 1.57 = 0.43 > 0$

$f(\pi) = 2\sin\pi - \pi = -\pi < 0$

Since $f$ is continuous and changes sign on $\left[\tfrac{\pi}{2},\, \pi\right]$, there is at least one root. Moreover, $f'(x) = 2\cos x - 1$, which is negative on $\left[\tfrac{\pi}{2},\, \pi\right]$ (since $\cos x \leq 0$ there), so $f$ is strictly decreasing. Hence there is exactly one root in this interval.

Bisection method:

We use the approximations $\dfrac{1}{\sqrt{2}} \approx 0.7$ and $\pi \approx \sqrt{10}$.

Step 1: $I_0 = \left[\tfrac{\pi}{2},\, \pi\right]$. Midpoint $c_0 = \tfrac{3\pi}{4}$.

Evaluate $f\!\left(\tfrac{3\pi}{4}\right) = 2\sin\tfrac{3\pi}{4} - \tfrac{3\pi}{4} = 2 \cdot \tfrac{1}{\sqrt{2}} - \tfrac{3\pi}{4} = \sqrt{2} - \tfrac{3\pi}{4}$.

To determine the sign, note $\left(\sqrt{2}\right)^2 = 2$ and $\left(\tfrac{3\pi}{4}\right)^2 = \tfrac{9\pi^2}{16} \approx \tfrac{90}{16} = \tfrac{45}{8} = 5.625$. Since $2 < 5.625$, we have $\sqrt{2} < \tfrac{3\pi}{4}$, so $f\!\left(\tfrac{3\pi}{4}\right) < 0$.

Since $f\!\left(\tfrac{\pi}{2}\right) > 0$ and $f\!\left(\tfrac{3\pi}{4}\right) < 0$, the root lies in $I_1 = \left[\tfrac{\pi}{2},\, \tfrac{3\pi}{4}\right]$.

Step 2: Midpoint $c_1 = \tfrac{1}{2}\!\left(\tfrac{\pi}{2} + \tfrac{3\pi}{4}\right) = \tfrac{5\pi}{8}$.

Evaluate $f\!\left(\tfrac{5\pi}{8}\right) = 2\sin\tfrac{5\pi}{8} - \tfrac{5\pi}{8}$.

We need $\sin\tfrac{5\pi}{8}$. Using the double angle formula with $\tfrac{5\pi}{8} = \tfrac{\pi}{2} + \tfrac{\pi}{8}$:

$\sin\tfrac{5\pi}{8} = \cos\tfrac{\pi}{8}$

By the half-angle formula: $\cos\tfrac{\pi}{8} = \sqrt{\tfrac{1+\cos(\pi/4)}{2}} = \sqrt{\tfrac{1 + 1/\sqrt{2}}{2}} = \sqrt{\tfrac{1 + 0.7}{2}} = \sqrt{0.85} \approx 0.92$.

So $f\!\left(\tfrac{5\pi}{8}\right) \approx 2(0.92) - \tfrac{5\sqrt{10}}{8} \approx 1.84 - \tfrac{5 \times 3.16}{8} \approx 1.84 - 1.98 = -0.14 < 0$.

More precisely: $f\!\left(\tfrac{5\pi}{8}\right)$ has the same sign as $\left(2\sin\tfrac{5\pi}{8}\right)^2 - \left(\tfrac{5\pi}{8}\right)^2 = 4\sin^2\tfrac{5\pi}{8} - \tfrac{25\pi^2}{64}$.

From $\sin\tfrac{5\pi}{8} = \cos\tfrac{\pi}{8}$: $4\cos^2\tfrac{\pi}{8} = 4 \cdot \tfrac{1+1/\sqrt{2}}{2} = 2 + \sqrt{2} \approx 2 + 1.41 = 3.41$.

And $\tfrac{25\pi^2}{64} \approx \tfrac{250}{64} = 3.91$. Since $3.41 < 3.91$, $f\!\left(\tfrac{5\pi}{8}\right) < 0$.

Since $f\!\left(\tfrac{\pi}{2}\right) > 0$ and $f\!\left(\tfrac{5\pi}{8}\right) < 0$:

$I_2 = \left[\tfrac{\pi}{2},\, \tfrac{5\pi}{8}\right] \qquad \checkmark$

Step 3: Midpoint $c_2 = \tfrac{1}{2}\!\left(\tfrac{\pi}{2} + \tfrac{5\pi}{8}\right) = \tfrac{9\pi}{16}$.

Evaluate $f\!\left(\tfrac{9\pi}{16}\right) = 2\sin\tfrac{9\pi}{16} - \tfrac{9\pi}{16}$.

We need $\sin\tfrac{9\pi}{16}$. Since $\tfrac{9\pi}{16} = \tfrac{\pi}{2} + \tfrac{\pi}{16}$:

$\sin\tfrac{9\pi}{16} = \cos\tfrac{\pi}{16}$

By the half-angle formula: $\cos\tfrac{\pi}{16} = \sqrt{\tfrac{1+\cos(\pi/8)}{2}}$.

We already found $\cos\tfrac{\pi}{8} \approx 0.92$, so $\cos\tfrac{\pi}{16} \approx \sqrt{\tfrac{1.92}{2}} = \sqrt{0.96} \approx 0.98$.

Then $f\!\left(\tfrac{9\pi}{16}\right) \approx 2(0.98) - \tfrac{9\sqrt{10}}{16} \approx 1.96 - \tfrac{28.46}{16} \approx 1.96 - 1.78 = 0.18 > 0$.

Check precisely: $4\cos^2\tfrac{\pi}{16} = 4 \cdot \tfrac{1 + \cos(\pi/8)}{2} = 2 + 2\cos\tfrac{\pi}{8} = 2 + 2\sqrt{\tfrac{1+1/\sqrt{2}}{2}} \approx 2 + 2(0.92) = 3.84$.

And $\left(\tfrac{9\pi}{16}\right)^2 = \tfrac{81\pi^2}{256} \approx \tfrac{810}{256} = 3.16$. Since $3.84 > 3.16$, $f\!\left(\tfrac{9\pi}{16}\right) > 0$.

Since $f\!\left(\tfrac{9\pi}{16}\right) > 0$ and $f\!\left(\tfrac{5\pi}{8}\right) < 0$:

$I_3 = \left[\tfrac{9\pi}{16},\, \tfrac{5\pi}{8}\right]$

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Question 8

Topic: 微分方程 Differential Equations  |  Difficulty: Challenging  |  Marks: 20

Problem

8 Let $x$ satisfy the differential equation

$\frac{dx}{dt} = (1 - x^n)^{1/n}$

and the condition $x = 0$ when $t = 0$.

(i) Solve the equation in the case $n = 1$ and sketch the graph of the solution for $t > 0$.

(ii) Prove that $1 - x < (1 - x^2)^{1/2}$ for $0 < x < 1$.

Use this result to sketch the graph of the solution in the case $n = 2$ for $0 < t < \frac{1}{2}\pi$, using the same axes as your previous sketch.

By setting $x = \sin y$, solve the equation in this case.

(iii) Use the result (which you need not prove)

$(1 - x^2)^{1/2} < (1 - x^3)^{1/3} \quad \text{for} \quad 0 < x < 1 ,$

to sketch, without solving the equation, the graph of the solution of the equation in the case $n = 3$ using the same axes as your previous sketches. Use your sketch to show that $x = 1$ at a value of $t$ less than $\frac{1}{2}\pi$.

Hint

Q8(i) Integration leads to the general solution $t = A - \ln(1 - x)$ and $x(0) = 0 \Rightarrow A = 0$. Thus $x = 1 - e^{-t}$.

(ii) Obviously, $(1 - x)^{1/2} < (1 + x)^{1/2}$ for all $x \in (0, 1]$. Hence multiplying this inequality through by $(1 - x)^{1/2}$ leads immediately to the required result.

Arguments which go in the wrong direction, e.g., $1 - x < (1 - x^2)^{1/2} \Rightarrow \dots \Rightarrow x - x^2 > 0$, etc., are invalid. It may be possible to salvage them by replacing ’$\Rightarrow'$ by ’$\Leftarrow'$.

In the case $n = 2$, the substitution $x = \sin y$ will lead to $t = y + B$ and hence to $t = \sin^{-1}(x) + B$ as the general solution. In particular, $x(0) = 0 \Rightarrow B = 0 \Rightarrow x = \sin t$.

Note that the question does not allow the use of the standard form $\int (1 - x^2)^{-1/2} dx = \sin^{-1}(x) +$ an arbitrary constant, without proof.

(iii) If $G_n$ is the graph of $x$ for $0 \le x \le 1$, then the given inequality shows that the gradient of $G_3$ is greater than the gradient of $G_2$ for each $x$ in this interval. (The inequality of (ii) shows that the same is true of $G_2$ in relation to $G_1$.) These considerations will help to clarify ideas when drawing sketches of $G_n$ for $n = 1, 2, 3$ in the same diagram. In particular, the sketch of $G_3$ should make it clear that once $x$ reaches the value 1 it remains there.

Model Solution

Part (i): $n = 1$

The equation becomes $\dfrac{dx}{dt} = 1 - x$, with $x(0) = 0$.

Separate variables:

$\frac{dx}{1 - x} = dt \qquad (x \neq 1)$

$-\ln|1 - x| = t + C$

At $t = 0$, $x = 0$: $-\ln 1 = C$, so $C = 0$.

$-\ln(1 - x) = t \qquad \Longrightarrow \qquad 1 - x = e^{-t} \qquad \Longrightarrow \qquad x = 1 - e^{-t}$

(valid for $x < 1$, which holds for all finite $t > 0$).

Sketch for $n = 1$: The curve starts at the origin, rises steeply at first (gradient 1 at $t = 0$), and approaches $x = 1$ asymptotically as $t \to \infty$. It never actually reaches $x = 1$.

Part (ii): Proving $1 - x < (1 - x^2)^{1/2}$ for $0 < x < 1$

Both sides are positive for $0 < x < 1$, so squaring preserves the inequality. We need to show:

$(1 - x)^2 < 1 - x^2$

Expand the left side: $1 - 2x + x^2$.

So we need: $1 - 2x + x^2 < 1 - x^2$, i.e., $2x^2 - 2x < 0$, i.e., $2x(x - 1) < 0$.

Since $0 < x < 1$, we have $x > 0$ and $x - 1 < 0$, so $2x(x-1) < 0$ ✓.

This is a valid proof: we started from $2x(x-1) < 0$ (true for $0 < x < 1$) and derived $(1-x)^2 < 1 - x^2$ by reversible algebraic steps, from which $1 - x < (1-x^2)^{1/2}$ follows since both sides are positive.

Solving for $n = 2$:

The equation is $\dfrac{dx}{dt} = (1 - x^2)^{1/2}$ with $x(0) = 0$.

Set $x = \sin y$, so $\dfrac{dx}{dt} = \cos y \cdot \dfrac{dy}{dt}$. Then:

$\cos y \cdot \frac{dy}{dt} = (1 - \sin^2 y)^{1/2} = \cos y$

For $0 < y < \tfrac{\pi}{2}$, $\cos y > 0$, so:

$\frac{dy}{dt} = 1 \qquad \Longrightarrow \qquad y = t + C$

At $t = 0$, $x = 0$, so $\sin y = 0$, giving $y = 0$ and $C = 0$.

Therefore $y = t$, and $x = \sin t$.

Sketch for $n = 2$: $x = \sin t$ starts at the origin, rises more steeply than $x = 1 - e^{-t}$ (since $(1-x^2)^{1/2} > 1 - x$ for $0 < x < 1$), and reaches $x = 1$ at $t = \tfrac{\pi}{2}$. For $t \in (0, \tfrac{\pi}{2})$, the curve $x = \sin t$ lies above $x = 1 - e^{-t}$.

Part (iii): Sketch for $n = 3$

We are given that $(1 - x^2)^{1/2} < (1 - x^3)^{1/3}$ for $0 < x < 1$.

This means $\dfrac{dx}{dt}\Big|_{n=3} > \dfrac{dx}{dt}\Big|_{n=2}$ for each $x \in (0, 1)$. So at any given $x$, the solution for $n = 3$ has a larger gradient than for $n = 2$.

Combined with part (ii), for $0 < x < 1$:

$\underbrace{1 - x}_{\text{gradient for } n=1} < \underbrace{(1-x^2)^{1/2}}_{\text{gradient for } n=2} < \underbrace{(1-x^3)^{1/3}}_{\text{gradient for } n=3}$

Since the gradient is larger at every $x$, the solution $x(t)$ for $n = 3$ rises more steeply than for $n = 2$, which rises more steeply than for $n = 1$. Starting from the origin, the three curves are ordered: $G_1$ (lowest), $G_2$ (middle), $G_3$ (highest) for $t > 0$.

Since $G_2$ reaches $x = 1$ at $t = \tfrac{\pi}{2}$, and $G_3$ rises more steeply, $G_3$ must reach $x = 1$ at some value $t^* < \tfrac{\pi}{2}$.

Once $x = 1$, we have $\dfrac{dx}{dt} = (1 - 1^n)^{1/n} = 0$, so $x$ stays at $1$ for all $t \geq t^*$. The graph of $G_3$ is therefore a curve from the origin, rising steeply, reaching $x = 1$ at $t = t^* < \tfrac{\pi}{2}$, and then remaining at $x = 1$ (a horizontal line) for $t \geq t^*$.

Showing $x = 1$ at a value less than $\tfrac{\pi}{2}$ more formally:

The time to reach $x = 1$ is $t^* = \int_0^1 \dfrac{dx}{(1-x^3)^{1/3}}$. We need to show this is less than $\tfrac{\pi}{2} = \int_0^1 \dfrac{dx}{(1-x^2)^{1/2}}$.

Since $(1-x^3)^{1/3} > (1-x^2)^{1/2}$ for $0 < x < 1$, we have $\dfrac{1}{(1-x^3)^{1/3}} < \dfrac{1}{(1-x^2)^{1/2}}$ on this interval, so

$t^* = \int_0^1 \frac{dx}{(1-x^3)^{1/3}} < \int_0^1 \frac{dx}{(1-x^2)^{1/2}} = \frac{\pi}{2}$

as required.