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A-Level Maths: Pure Mathematics Practice

A-Level Maths — Pure Mathematics Practice

18 MCQ practice problems covering core A-Level Pure Mathematics content.

What These Questions Test

These problems test your algebraic fluency, trigonometric reasoning, ability to work with sequences and series, and skill in constructing mathematical arguments.

Typical question types:

  • Algebra: Factorising polynomials, solving quadratic and simultaneous equations, using the discriminant to determine the nature of roots, completing the square.
  • Trigonometry: Solving equations in given intervals, using identities to simplify expressions, applying addition and double angle formulae, sketching trigonometric graphs.
  • Sequences and Series: Finding terms and sums of arithmetic and geometric sequences. Summing geometric series to infinity. Using sigma notation.
  • Logarithms and Exponentials: Converting between logarithmic and exponential form. Using log laws to simplify and solve equations. Differentiating and integrating exponential functions.
  • Proof: Proof by deduction, exhaustion, and contradiction. Proof by induction for series, divisibility, and matrix results.
  • Differentiation and Integration: Differentiating polynomials, trig functions, exponentials, and logarithms. Integrating standard forms. Using the chain rule, product rule, and quotient rule.

Approach Strategy

  1. Simplify before calculating. Many problems become much easier if you simplify the expression first. Factorise, cancel common terms, or use identities.
  2. Check your domain. For logarithms, arguments must be positive. For trigonometric equations, solutions must lie in the specified interval. For square roots, the radicand must be non-negative.
  3. Verify with substitution. After finding a solution, substitute it back into the original equation to check it works.
  4. Use diagrams where possible. A sketch of a trigonometric graph or a coordinate geometry diagram can reveal the structure of a problem.

Intuition

Pure mathematics is about finding patterns and proving they always hold. When you solve an equation, you are finding the values that make a statement true. When you prove something, you are showing it is true for all cases, not just the ones you have checked.

Think of algebra as a language. The symbols represent quantities and relationships. Manipulating expressions is like rearranging a sentence — the meaning stays the same, but the form changes to reveal something new.

Worked Examples

Example 1: Logarithmic Equation

Problem: Solve log⁡2(x−1)+log⁡2(x+1)=3\log_2(x-1) + \log_2(x+1) = 3.

Solution: Step 1: Combine using log law: log⁡2[(x−1)(x+1)]=3\log_2[(x-1)(x+1)] = 3 Step 2: Convert to exponential form: (x−1)(x+1)=23=8(x-1)(x+1) = 2^3 = 8 Step 3: Expand: x2−1=8x^2 - 1 = 8, so x2=9x^2 = 9 Step 4: Solve: x=3x = 3 or x=−3x = -3 Step 5: Check domain: log⁡2(x−1)\log_2(x-1) requires x>1x > 1, so x=−3x = -3 is rejected Step 6: Answer: x=3x = 3

Key insight: Always check the domain of logarithmic solutions. x=−3x = -3 is extraneous because log⁡2(−4)\log_2(-4) is undefined.

Example 2: Proof by Induction

Problem: Prove that ∑r=1nr2=n(n+1)(2n+1)6\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} for all n∈Z+n \in \mathbb{Z}^+.

Solution: Step 1 (Base case): n=1n = 1: LHS =12=1= 1^2 = 1, RHS =1⋅2⋅36=1= \frac{1 \cdot 2 \cdot 3}{6} = 1. ✓

Step 2 (Inductive hypothesis): Assume true for n=kn = k: ∑r=1kr2=k(k+1)(2k+1)6\sum_{r=1}^{k} r^2 = \frac{k(k+1)(2k+1)}{6}

Step 3 (Inductive step): Show true for n=k+1n = k+1: ∑r=1k+1r2=∑r=1kr2+(k+1)2=k(k+1)(2k+1)6+(k+1)2\sum_{r=1}^{k+1} r^2 = \sum_{r=1}^{k} r^2 + (k+1)^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2 =k(k+1)(2k+1)+6(k+1)26=(k+1)[k(2k+1)+6(k+1)]6= \frac{k(k+1)(2k+1) + 6(k+1)^2}{6} = \frac{(k+1)[k(2k+1) + 6(k+1)]}{6} =(k+1)(2k2+7k+6)6=(k+1)(k+2)(2k+3)6= \frac{(k+1)(2k^2 + 7k + 6)}{6} = \frac{(k+1)(k+2)(2k+3)}{6}

This equals the formula with n=k+1n = k+1. ✓

Key insight: The algebraic manipulation in Step 3 is where most errors occur. Factor out (k+1)(k+1) first, then expand carefully.

Example 3: Trigonometric Equation

Problem: Solve 2cos⁡2x−3cos⁡x+1=02\cos^2 x - 3\cos x + 1 = 0 for 0≤x≤360°0 \leq x \leq 360°.

Solution: Step 1: Let u=cos⁡xu = \cos x: 2u2−3u+1=02u^2 - 3u + 1 = 0 Step 2: Factorise: (2u−1)(u−1)=0(2u - 1)(u - 1) = 0 Step 3: Solve: u=12u = \frac{1}{2} or u=1u = 1 Step 4: cos⁡x=12\cos x = \frac{1}{2}: x=60°x = 60° or x=300°x = 300° Step 5: cos⁡x=1\cos x = 1: x=0°x = 0° Step 6: Answer: x=0°,60°,300°x = 0°, 60°, 300°

Key insight: Substitution (u=cos⁡xu = \cos x) turns a trig equation into a quadratic, which is easier to solve. Always substitute back at the end.

Common Mistakes

  1. Losing solutions when dividing by a variable. If you divide both sides by xx to solve x2=3xx^{2} = 3x, you lose the solution x=0x = 0. Always factorise instead of dividing.
  2. Incorrect use of trig identities. sin⁡(A+B)≠sin⁡A+sin⁡B\sin(A + B) \neq \sin A + \sin B. The addition formula is sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B) = \sin A\cos B + \cos A\sin B. This is one of the most common errors.
  3. Forgetting that log⁡ax\log_{a} x and axa^{x} are inverses. log⁡a(ax)=x\log_{a}(a^{x}) = x and alog⁡ax=xa^{\log_{a} x} = x, but only when x>0x > 0 for the logarithm.
  4. Sign errors in proof by induction. When proving divisibility, ensure you correctly substitute n=k+1n = k+1 and use the inductive hypothesis. A sign error here invalidates the proof.

Cross-References