Jamie is attempting to write 3x2+1(x−1)(x+3)2\frac{3x^2 + 1}{(x - 1)(x + 3)^2}(x−1)(x+3)23x2+1 as partial fractions with constant numerators.
His incorrect attempt is shown below.
Step 1: 3x2+1(x−1)(x+3)2≡Ax−1+B(x+3)2\frac{3x^2 + 1}{(x - 1)(x + 3)^2} \equiv \frac{A}{x - 1} + \frac{B}{(x + 3)^2}(x−1)(x+3)23x2+1≡x−1A+(x+3)2B
Step 2: 3x2+1≡A(x+3)2+B(x−1)3x^2 + 1 \equiv A(x + 3)^2 + B(x - 1)3x2+1≡A(x+3)2+B(x−1)
Step 3: Let x=1⇒4=16A⇒A=14x = 1 \Rightarrow 4 = 16A \Rightarrow A = \frac{1}{4}x=1⇒4=16A⇒A=41 Let x=−3⇒28=−4B⇒B=−7x = -3 \Rightarrow 28 = -4B \Rightarrow B = -7x=−3⇒28=−4B⇒B=−7
Answer: 3x2+1(x−1)(x+3)2≡14(x−1)−7(x+3)2\frac{3x^2 + 1}{(x - 1)(x + 3)^2} \equiv \frac{1}{4(x - 1)} - \frac{7}{(x + 3)^2}(x−1)(x+3)23x2+1≡4(x−1)1−(x+3)27
(i) By using a counter-example, show that the answer obtained by Jamie cannot be correct.
(ii) Explain the mistake Jamie made in Step 1.
Write 3x2+1(x−1)(x+3)2\frac{3x^2 + 1}{(x - 1)(x + 3)^2}(x−1)(x+3)23x2+1 as partial fractions, with constant numerators.
532 exam-style questions on AQA A Level Maths 1.5 B: Algebra and functions, covering 1.5.1 Laws of indices, 1.5.2 Surds, 1.5.3 Quadratic functions, 1.5.4 Simultaneous equations, 1.5.5 Linear and quadratic inequalities, 1.5.6 Polynomials and rational expressions, 1.5.7 Graphs of functions and proportion, 1.5.8 Composite and inverse functions (A-level only), 1.5.9 Transformations of graphs, 1.5.10 Partial fractions (A-level only), and 1.5.11 Functions in modelling (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.