Year 10 OCR Advanced Maths: Unit Test Mock Paper Solutions | Year 10 OCR 进阶数学：单元测试模拟卷解析

📚 Year 10 OCR Advanced Maths: Unit Test Mock Paper Solutions | Year 10 OCR 进阶数学：单元测试模拟卷解析

This article provides a complete walkthrough of a Year 10 OCR Advanced Maths unit test mock paper. Each question is broken down with clear, step-by-step solutions, covering surds, quadratics, polynomials, binomial expansion and introductory calculus. Use it to consolidate your revision and build confidence for the real test.

本文对一份 Year 10 OCR 进阶数学单元测试模拟卷进行了完整解析。每一道题都配有清晰的分步解答，涵盖根式、二次方程、多项式、二项式展开和入门微积分等核心主题。通过这份解析，你可以巩固复习成果，提升考试信心。

1. Simplifying Surds | 根式化简

Problem: Simplify √75 + √12 − √48.

题目：化简 √75 + √12 − √48。

1: Look for the largest square factor in each surd. 75 = 25 × 3, 12 = 4 × 3, 48 = 16 × 3.

步骤1：找出每个根式里最大的平方因数。75 = 25 × 3，12 = 4 × 3，48 = 16 × 3。

Step 2: Write each term as a product of the square root of the square number and the remaining root. √75 = √25×3 = 5√3, √12 = √4×3 = 2√3, √48 = √16×3 = 4√3.

步骤2：将每一项写成平方数的平方根与剩余根式的乘积形式。√75 = √25×3 = 5√3，√12 = √4×3 = 2√3，√48 = √16×3 = 4√3。

Step 3: Combine the like terms. 5√3 + 2√3 − 4√3 = (5 + 2 − 4)√3 = 3√3.

步骤3：合并同类项。5√3 + 2√3 − 4√3 = (5 + 2 − 4)√3 = 3√3。

Final answer: 3√3.

最终答案：3√3。

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2. Rationalising Denominators | 分母有理化

Problem: Rationalise the denominator of (5+√3) / (2−√3).

题目：将 (5+√3) / (2−√3) 分母有理化。

Step 1: Multiply numerator and denominator by the conjugate of the denominator, which is (2+√3).

步骤1：分子和分母同时乘以分母的共轭式 (2+√3)。

(5+√3)/(2−√3) × (2+√3)/(2+√3)

Step 2: Expand the numerator. (5+√3)(2+√3) = 5×2 + 5√3 + 2√3 + √3×√3 = 10 + 7√3 + 3 = 13 + 7√3.

步骤2：展开分子。(5+√3)(2+√3) = 5×2 + 5√3 + 2√3 + √3×√3 = 10 + 7√3 + 3 = 13 + 7√3。

Step 3: Expand the denominator using the difference of squares. (2−√3)(2+√3) = 2² − (√3)² = 4 − 3 = 1.

步骤3：用平方差公式展开分母。(2−√3)(2+√3) = 2² − (√3)² = 4 − 3 = 1。

Step 4: Since the denominator is 1, the expression simplifies to 13 + 7√3.

步骤4：因为分母为 1，该式简化为 13 + 7√3。

Final answer: 13 + 7√3.

最终答案：13 + 7√3。

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3. Discriminant and Equal Roots | 判别式与等根

em: The quadratic equation x² + kx + 9 = 0 has equal roots. Find the possible values of k.

题目：二次方程 x² + kx + 9 = 0 有等根，求 k 的所有可能取值。

Step 1: For equal roots, the discriminant Δ must equal zero. Here a = 1, b = k, c = 9.

步骤1：等根条件为判别式 Δ = 0。这里 a = 1，b = k，c = 9。

Step 2: Write Δ = b² − 4ac = k² − 4×1×9 = k² − 36.

步骤2：写出 Δ = b² − 4ac = k² − 4×1×9 = k² − 36。

k² − 36 = 0

Step 3: Solve for k. k² = 36, so k = ±√36 = ±6.

步骤3：解出 k。k² = 36，所以 k = ±√36 = ±6。

Final answer: k = 6 or k = −6.

最终答案：k = 6 或 k = −6。

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4. Completing the Square and Minimum Value | 配方法与最小值

Express f(x) = 2x² − 8x + 5 in the form a(x−p)² + q, and state the minimum value of f(x).

题目：将 f(x) = 2x² − 8x + 5 写成 a(x−p)² + q 的形式，并指出 f(x) 的最小值。

Step 1: Factor out the coefficient of x² from the first two terms. f(x) = 2(x² − 4x) + 5.

步骤1：从含 x² 和 x 的项中提出系数 2。f(x) = 2(x² − 4x) + 5。

Step 2: Complete the square inside the bracket. Half the coefficient of x is −2, square it to get 4. Add and subtract 4 inside the bracket.

步骤2：在括号内配方。x 系数的一半是 −2，平方得 4。在括号内加上并减去 4。

f(x) = 2[(x² − 4x + 4) − 4] + 5

Step 3: Rewrite the perfect square trinomial as (x−2)². f(x) = 2[(x−2)² − 4] + 5.

步骤3：将完全平方式写成 (x−2)²。f(x) = 2[(x−2)² − 4] + 5。

Step 4: Expand and simplify. f(x) = 2(x−2)² − 8 + 5 = 2(x−2)² − 3.

步骤4：展开并化简。f(x) = 2(x−2)² − 8 + 5 = 2(x−2)² − 3。

f(x) = 2(x−2)² − 3, so p = 2, q = −3.

erm is zero. Since a=2 > 0, the parabola opens upwards, so the minimum is q = −3 at x = 2.

步骤5：当平方项为零时取得最小值。因为 a=2 > 0，抛物线开口向上，所以最小值为 q = −3，此时 x = 2。

Final answer: f(x) = 2(x−2)² − 3, minimum value = −3.

最终答案：f(x) = 2(x−2)² − 3，最小值为 −3。

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5. Factor Theorem and Polynomial Division | 因式定理与多项式除法

= x³ − 4x² + x + 6. (a) Show that (x−2) is a factor. (b) Hence factorise f(x) completely.

题目：f(x) = x³ − 4x² + x + 6。(a) 证明 (x−2) 是一个因式。(b) 由此将 f(x) 完全分解。

problem (a): Substitute x = 2 into f(x). f(2) = (2)³ − 4(2)² + 2 + 6 = 8 − 16 + 2 + 6 = 0.

小题 (a)：将 x = 2 代入 f(x)。f(2) = 2³ − 4×2² + 2 + 6 = 8 − 16 + 2 + 6 = 0。

Since f(2) = 0, by the Factor Theorem, (x−2) is a factor.

因为 f(2)=0，根据因式定理，(x−2) 是一个因式。

Sub-problem (b): Divide f(x) by (x−2). Use synthetic or long division. The quotient is x² − 2x − 3.

小题 (b)：用 (x−2) 除 f(x)。使用综合除法或长除法，商为 x² − 2x − 3。

4x; subtract: (x)−(4x)=−3x; bring down +6; −3x ÷ x = −3; multiply: −3(x−2)=−3x+6; remainder 0. Quotient is x² − 2x − 3.

长除法步骤：x³ ÷ x = x²；乘：x²(x−2)=x³−2x²；减：(−4x²)−(−2x²)=−2x²；落下 +x；−2x² ÷ x = −2x；乘：−2x(x−2)=−2x²+4x；减：(x)−(4x)=−3x；落下 +6；−3x ÷ x = −3；乘：−3(x−2)=−3x+6；余式为 0。商为 x² − 2x − 3。

Factorise the quadratic: x² − 2x − 3 = (x−3)(x+1).

将二次式分解：x² − 2x − 3 = (x−3)(x+1)。

Complete factorisation: f(x) = (x−2)(x−3)(x+1).

完全分解为：f(x) = (x−2)(x−3)(x+1)。

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6. Solving the Cubic Equation | 解三次方程

Problem: Solve f(x) = 0 using the factorisation from the previous section.

题目：利用上一节得到的因式分解，解方程 f(x) = 0。

Set each factor equal to zero: (x−2)=0 ⇒ x=2; (x−3)=0 ⇒ x=3; (x+1)=0 ⇒ x=−1.

令每个因式等于零：(x−2)=0 ⇒ x=2；(x−3)=0 ⇒ x=3；(x+1)=0 ⇒ x=−1。

Therefore, the solution set is x = −1, 2, 3.

因此，解集为 x = −1, 2, 3。

lways check roots in the original equation: f(−1)=0, f(2)=0, f(3)=0, confirming the cubic.

始终将根代入原方程检验：f(−1)=0，f(2)=0，f(3)=0，验证无误。

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7. Binomial Expansion | 二项式展开

Problem: Find the first three terms in the binomial expansion of (1+2x)⁵.

题目：写出 (1+2x)⁵ 的二项展开式的前三项。

Use the binomial theorem: (a+b)ⁿ = aⁿ + nC1 aⁿ⁻¹b + nC2 aⁿ⁻²b² + … . Here a = 1, b = 2x, n = 5.

使用二项式定理：(a+b)ⁿ = aⁿ + nC1 aⁿ⁻¹b + nC2 aⁿ⁻²b² + … 。这里 a=1，b=2x，n=5。

Term 1: 1⁵ = 1.

第一项：1⁵ = 1。

Term 2: 5C1 × 1⁴ × (2x) = 5 × 2x = 10x.

第二项：5C1 × 1⁴ × (2x) = 5 × 2x = 10x。

Term 3: 5C2 × 1³ × (2x)² = 10 × 4x² = 40x².

第三项：5C2 × 1³ × (2x)² = 10 × 4x² = 40x²。

(1+2x)⁵ = 1 + 10x + 40x² + …

ote: nC1 = C(5,1) = 5, nC2 = C(5,2) = 10. The expansion continues with higher powers of x.

注：nC1 = C(5,1) = 5，nC2 = C(5,2) = 10。展开式后续还包含 x 的更高次幂。

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8. Binomial Estimation | 二项式估算

Problem: Use the expansion obtained to estimate (1.02)⁵.

题目：运用上述展开式估算 (1.02)⁵。

Observe that 1.02 = 1 + 2×0.01. So set x = 0.01 in the expansion (1+2x)⁵.

观察得 1.02 = 1 + 2×0.01。因此在展开式中令 x = 0.01。

From the first three terms: (1+2x)⁵ ≈ 1 + 10x + 40x².

根据前三项：(1+2x)⁵ ≈ 1 + 10x + 40x²。

Substitute x = 0.01: 1 + 10(0.01) + 40(0.01)² = 1 + 0.1 + 40(0.0001).

代入 x = 0.01：1 + 10(0.01) + 40(0.01)² = 1 + 0.1 + 40(0.0001)。

Calculate: 40 × 0.0001 = 0.004. So 1 + 0.1 + 0.004 = 1.104.

计算：40 × 0.0001 = 0.004。因此 1 + 0.1 + 0.004 = 1.104。

Estimated value is 1.104. (The exact value is 1.1040808…, so the approximation is accurate to three decimal places.)

估算值为 1.104。（精确值为 1.1040808…，因此该近似值精确至三位小数。）

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9. Differentiation of Polynomials | 多项式微分

Problem: Find dy/dx for y = 3x² − 5x.

题目：对 y = 3x² − 5x 求 dy/dx。

Use the power rule: d/dx (xⁿ) = nxⁿ⁻¹.

使用幂法则：d/dx (xⁿ) = nxⁿ⁻¹。

Differentiate term by term. For 3x²: bring down the power 2, multiply by 3 to get 6x.

逐项求导。对于 3x²：将指数 2 下拉，乘以 3 得 6x。

For −5x: the derivative of x¹ is 1, so the term becomes −5 × 1 = −5.

对于 −5x：x¹ 的导数为 1，因此该项变为 −5 × 1 = −5。

dy/dx = 6x − 5

Always check: derivative of a constant is zero, so no extra term appears.

检查：常数的导数为零，因此不会出现额外的项。

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10. Equation of a Tangent | 切线方程

oblem: Find the equation of the tangent to the curve y = x³ − 2x at the point where x = 1.

题目：求曲线 y = x³ − 2x 在 x = 1 处的切线方程。

Step 1: Find

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