Miscellaneous Exercises 6 | 综合练习6

📚 Miscellaneous Exercises 6 | 综合练习6

This article walks through a set of miscellaneous exercises typical of an A-Level Mathematics revision block. Each problem is chosen to reinforce key skills from pure mathematics, statistics, and mechanics. Work through each one carefully, covering up the solution until you have attempted it yourself.

本文带你完成一整套 A-Level 数学复习阶段常见的混合练习题。每道题都经过挑选，以巩固纯数学、统计力学等关键技能。请仔细思考每一题，先遮住答案自己尝试，然后再对照解法。

1. Quadratic Inequalities | 一元二次不等式

Solve the inequality \( x^2 – 5x + 6 < 0 \). Begin by factorising: \( (x-2)(x-3) < 0 \). The critical values are \(x=2\) and \(x=3\). Testing intervals gives the solution \( 2 < x < 3 \).

解不等式 \( x^2 – 5x + 6 < 0 \)。首先因式分解：\( (x-2)(x-3) < 0 \)。临界值为 \(x=2\) 和 \(x=3\)。通过区间检验可得解为 \( 2 < x < 3 \)。

For a quadratic inequality, always sketch the parabola or test sign intervals. If the leading coefficient is positive and the inequality is “

对于二次不等式，务必要画出抛物线草图或检验区间符号。当首项系数为正且不等式为“＜”时，解集位于两个根之间。

2. Coordinate Geometry of Circles | 圆的坐标几何

Find the centre and radius of the circle \( x^2 + y^2 – 6x + 4y – 12 = 0 \). Complete the square: \( (x-3)^2 + (y+2)^2 = 25 \). Hence centre is \( (3, -2) \) and radius is \(5\).

求圆 \( x^2 + y^2 – 6x + 4y – 12 = 0 \) 的圆心和半径。配方得：\( (x-3)^2 + (y+2)^2 = 25 \)。因此圆心为 \( (3, -2) \)，半径为 \(5\)。

When a circle equation is given in expanded form, completing the square separately for \(x\) and \(y\) reveals the standard form. Remember the constant term moves to the right-hand side.

当圆方程以一般式给出时，分别对 \(x\) 和 \(y\) 配方即可得到标准式。注意常数项要移到等号右侧。

3. Binomial Expansion | 二项式展开

Expand \( (2 + x)^5 \) up to and including the term in \(x^3\). Using the binomial theorem:
\[
(2+x)^5 = 32 + 80x + 80x^2 + 40x^3 + \cdots
\]
Check the coefficients: \( \binom{5}{1} 2^4 = 80 \), \( \binom{5}{2} 2^3 = 80 \), \( \binom{5}{3} 2^2 = 40 \).

展开 \( (2 + x)^5 \)，直到含 \(x^3\) 的项。利用二项式定理：
\[
(2+x)^5 = 32 + 80x + 80x^2 + 40x^3 + \cdots
\]
验证系数：\( \binom{5}{1} 2^4 = 80 \)，\( \binom{5}{2} 2^3 = 80 \)，\( \binom{5}{3} 2^2 = 40 \)。

For fractional or negative powers, use the general binomial series with \(|x|<1\). But for a positive integer power, the standard finite expansion is sufficient.

对于分数次幂或负次幂，需使用一般二项式级数并限制 \(|x|<1\)。但对于正整数次幂，直接使用有限二项式展开即可。

4. Differentiation: Chain Rule | 微分：链式法则

Differentiate \( y = (3x^2 + 1)^4 \). Let \(u = 3x^2 + 1\), then \(y = u^4\).
\[
\frac{dy}{dx} = 4u^3 \cdot \frac{du}{dx} = 4(3x^2+1)^3 \cdot 6x = 24x(3x^2+1)^3.
\]

求 \( y = (3x^2 + 1)^4 \) 的导数。令 \(u = 3x^2 + 1\)，则 \(y = u^4\)。
\[
\frac{dy}{dx} = 4u^3 \cdot \frac{du}{dx} = 4(3x^2+1)^3 \cdot 6x = 24x(3x^2+1)^3.
\]

The chain rule is essential for composite functions. Always differentiate the outer function first, then multiply by the derivative of the inner function.

链式法则是处理复合函数的关键。务必先对外层函数求导，再乘以内层函数的导数。

5. Integration by Substitution | 换元积分法

Evaluate \( \int_0^1 2x\sqrt{x^2+1} \, dx \). Let \(u = x^2+1\), so \(du = 2x\,dx\). When \(x=0\), \(u=1\); when \(x=1\), \(u=2\). Thus the integral becomes
\[
\int_1^2 \sqrt{u} \, du = \left[ \frac{2}{3} u^{3/2} \right]_1^2 = \frac{2}{3}(2\sqrt{2} – 1).
\]

计算 \( \int_0^1 2x\sqrt{x^2+1} \, dx \)。令 \(u = x^2+1\)，则 \(du = 2x\,dx\)。当 \(x=0\) 时 \(u=1\)；当 \(x=1\) 时 \(u=2\)。因此积分化为
\[
\int_1^2 \sqrt{u} \, du = \left[ \frac{2}{3} u^{3/2} \right]_1^2 = \frac{2}{3}(2\sqrt{2} – 1).
\]

When using substitution with definite integrals, remember to change the limits at the same time as the variable. This avoids the need to substitute back again.

在定积分中使用换元法时，必须同步调整积分上下限。这样在最后就不需要再代回原变量了。

6. Trigonometric Identities and Equations | 三角恒等式与方程

Solve \( 2\sin^2 x – 3\cos x = 0 \) for \( 0 \le x \le 2\pi \). Use the identity \( \sin^2 x = 1 – \cos^2 x \):
\[
2(1-\cos^2 x) – 3\cos x = 0 \implies 2\cos^2 x + 3\cos x – 2 = 0.
\]
Factorise: \( (2\cos x – 1)(\cos x + 2) = 0 \). Since \(\cos x \neq -2\), we have \(\cos x = \frac{1}{2}\). Hence \( x = \frac{\pi}{3}, \frac{5\pi}{3} \).

在 \( 0 \le x \le 2\pi \) 范围内解方程 \( 2\sin^2 x – 3\cos x = 0 \)。利用恒等式 \( \sin^2 x = 1 – \cos^2 x \)：
\[
2(1-\cos^2 x) – 3\cos x = 0 \implies 2\cos^2 x + 3\cos x – 2 = 0.
\]
因式分解：\( (2\cos x – 1)(\cos x + 2) = 0 \)。由于 \(\cos x \neq -2\)，所以 \(\cos x = \frac{1}{2}\)。因此 \( x = \frac{\pi}{3}, \frac{5\pi}{3} \)。

To solve trig equations, first reduce everything to one trigonometric function if possible. Then solve the resulting equation and find all solutions in the given interval.

解三角方程时，要尽可能将方程转化为只含一种三角函数。然后解出该方程，并在给定区间内找出所有解。

7. Exponential and Logarithmic Functions | 指数与对数函数

Simplify \( \ln(a^2) – 2\ln(a) + \ln(e^3) \). Since \( \ln(a^2) = 2\ln a \) and \( \ln(e^3) = 3 \), the expression simplifies to \( 2\ln a – 2\ln a + 3 = 3 \).

化简 \( \ln(a^2) – 2\ln(a) + \ln(e^3) \)。因为 \( \ln(a^2) = 2\ln a \)，且 \( \ln(e^3) = 3 \)，所以原式化简为 \( 2\ln a – 2\ln a + 3 = 3 \)。

For exponential equations such as \( e^{2x} – 5e^x + 6 = 0 \), set \(u = e^x\) and solve the quadratic \(u^2 – 5u + 6 = 0\). This yields \(u=2\) or \(u=3\), so \(x=\ln 2\) or \(x=\ln 3\).

对于形如 \( e^{2x} – 5e^x + 6 = 0 \) 的指数方程，令 \(u = e^x\)，解二次方程 \(u^2 – 5u + 6 = 0\)，得到 \(u=2\) 或 \(u=3\)，故 \(x=\ln 2\) 或 \(x=\ln 3\)。

8. Numerical Methods: Newton-Raphson | 数值方法：牛顿-拉弗森法

Use the Newton-Raphson formula to find a root of \( f(x) = x^3 – 2x – 5 \) starting from \(x_0 = 2\). Since \( f'(x) = 3x^2 – 2 \), the iteration is
\[
x_{n+1} = x_n – \frac{x_n^3 – 2x_n – 5}{3x_n^2 – 2}.
\]
With \(x_0=2\), \(x_1 = 2 – \frac{8-4-5}{12-2} = 2 – \frac{-1}{10} = 2.1\). Further iterations converge to the root near \(2.0946\).

使用牛顿-拉弗森公式求 \( f(x) = x^3 – 2x – 5 \) 的根，取初值 \(x_0 = 2\)。因为 \( f'(x) = 3x^2 – 2 \)，迭代格式为
\[
x_{n+1} = x_n – \frac{x_n^3 – 2x_n – 5}{3x_n^2 – 2}.
\]
以 \(x_0=2\) 代入，得 \(x_1 = 2 – \frac{8-4-5}{12-2} = 2 – \frac{-1}{10} = 2.1\)。继续迭代可收敛到约 \(2.0946\) 的根。

Newton-Raphson is quick but may fail if the initial guess is too far from the root or if \(f'(x_n)=0\). Always check convergence by looking at the change between iterations.

牛顿-拉弗森法收敛快，但若初值离根太远或某次导数 \(f'(x_n)=0\)，可能失效。务必通过观察相邻迭代值之差来判断收敛性。

9. Vectors in 3D | 三维向量

Find the angle between vectors \( \mathbf{a} = \mathbf{i} + 2\mathbf{j} – \mathbf{k} \) and \( \mathbf{b} = 3\mathbf{i} – \mathbf{j} + 2\mathbf{k} \). The dot product is
\[
\mathbf{a}\cdot\mathbf{b} = 1(3) + 2(-1) + (-1)(2) = -1.
\]
The magnitudes are \( |\mathbf{a}| = \sqrt{1+4+1} = \sqrt{6} \) and \( |\mathbf{b}| = \sqrt{9+1+4} = \sqrt{14} \). Therefore
\[
\cos\theta = \frac{-1}{\sqrt{6}\sqrt{14}} = -\frac{1}{\sqrt{84}},
\]
so \( \theta \approx 96.3^\circ \).

求向量 \( \mathbf{a} = \mathbf{i} + 2\mathbf{j} – \mathbf{k} \) 与 \( \mathbf{b} = 3\mathbf{i} – \mathbf{j} + 2\mathbf{k} \) 之间的夹角。点积为
\[
\mathbf{a}\cdot\mathbf{b} = 1(3) + 2(-1) + (-1)(2) = -1.
\]
模长分别为 \( |\mathbf{a}| = \sqrt{1+4+1} = \sqrt{6} \)，\( |\mathbf{b}| = \sqrt{9+1+4} = \sqrt{14} \)。因此
\[
\cos\theta = \frac{-1}{\sqrt{6}\sqrt{14}} = -\frac{1}{\sqrt{84}},
\]
所以 \( \theta \approx 96.3^\circ \)。

When finding angles between vectors, the formula \( \cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|} \) is essential. The dot product may be negative, indicating an obtuse angle.

求两向量夹角时，关键在于公式 \( \cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|} \)。点积可能为负，表示夹角为钝角。

10. Hypothesis Testing for the Mean | 均值的假设检验

A sample of size \(n=50\) has sample mean \(\bar{x} = 28.4\). Test at the 5% level whether the population mean \(\mu\) is greater than 27, given that the population standard deviation is known to be \(\sigma = 6\).

设样本容量 \(n=50\)，样本均值 \(\bar{x} = 28.4\)。已知总体标准差 \(\sigma = 6\)，在 5% 显著性水平下检验总体均值 \(\mu\) 是否大于 27。

The null hypothesis is \(H_0: \mu = 27\) and the alternative is \(H_1: \mu > 27\). The test statistic is
\[
z = \frac{\bar{x} – \mu_0}{\sigma/\sqrt{n}} = \frac{28.4 – 27}{6/\sqrt{50}} = \frac{1.4}{0.8485} \approx 1.65.
\]
The critical value at the 5% one-tailed level is \(1.6449\). Since \(1.65 > 1.6449\), we reject \(H_0\). There is sufficient evidence to support the claim that \(\mu > 27\).

原假设为 \(H_0: \mu = 27\)，备择假设为 \(H_1: \mu > 27\)。检验统计量为
\[
z = \frac{\bar{x} – \mu_0}{\sigma/\sqrt{n}} = \frac{28.4 – 27}{6/\sqrt{50}} = \frac{1.4}{0.8485} \approx 1.65.
\]
5% 单尾检验的临界值为 \(1.6449\)。由于 \(1.65 > 1.6449\)，我们拒绝 \(H_0\)。有充分证据支持 \(\mu > 27\) 的断言。

When the population standard deviation is known, use the \(z\)-test. For unknown \(\sigma\) and large samples, the sample standard deviation can approximate it.

当总体标准差已知时，使用 \(z\) 检验。若 \(\sigma\) 未知但样本量较大，可用样本标准差近似代替。

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Miscellaneous Exercises 6 | 综合练习6