lash","explanationVersion":"v2"},"firstAttemptFullMarks":null},{"id":"bf5c00ea-3f06-43fb-bd0d-ec2689b1e986","revisionId":"d692a74f-e1d4-4dbd-a9ee-705c85b35411","questionNumber":null,"questionType":"LA","specification":"","difficulty":1,"difficultyBand":"easy","totalMarks":3,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"cea10272-d27e-4a87-bbb0-0be524f49467","label":"a","content":"Express $\\frac{3x+5}{(x+1)(x+2)}$ in the form $\\frac{A}{(x+1)} + \\frac{B}{(x+2)}$","markscheme":"- Sets up the identity $3x+5 = A(x+2) + B(x+1)$ M1\n- Finds $A = 2$ and $B = 1$ by substituting $x = -1$ and $x = -2$ A1\n- Final answer: $\\frac{2}{(x+1)} + \\frac{1}{(x+2)}$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Sets up the correct partial fraction identity $3x+5 = A(x+2) + B(x+1)$.","code":"M"},{"id":"p2","marks":1,"description":"Solves for $A = 2$ and $B = 1$ correctly.","code":"A"},{"id":"p3","marks":1,"description":"Expresses the partial fractions correctly as $\\frac{2}{(x+1)} + \\frac{1}{(x+2)}$.","code":"A"}],"finalAnswers":["\\frac{2}{(x+1)} + \\frac{1}{(x+2)}"],"generatedAt":"2026-05-06T14:03:31.168Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the algebraic fraction as partial fractions, we start by setting up an identity. We equate the original fraction to the sum of two simpler fractions with unknown constants $A$ and $B$:\n\n$$\\frac{3x+5}{(x+1)(x+2)} \\equiv \\frac{A}{x+1} + \\frac{B}{x+2}$$\n\nTo clear the denominators, multiply both sides of the identity by the common denominator $(x+1)(x+2)$:\n\n$$3x+5 \\equiv A(x+2) + B(x+1)$$\n\nThis is a crucial step that earns the first method mark M1.","highlights":[{"text":"Express $\\frac{3x+5}{(x+1)(x+2)}$ in the form $\\frac{A}{(x+1)} + \\frac{B}{(x+2)}$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$3(\\htmlData{anchor=x-sub-1}{\\color{@sky}{-1}}) + 5 = A(\\htmlData{anchor=x-term-1}{\\color{@sky}{-1}} + 2) + B(\\htmlData{anchor=zero-term-1}{\\color{@sky}{-1}} + 1)$"},{"latex":"$$2 = A(1) + B(0)$"},{"latex":"$$\\htmlData{anchor=a-val}{\\color{@purple}{A = 2}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"x-sub-1"},{"color":"purple","style":"text-color","anchor":"a-val"}],"connections":[{"to":"x-term-1","from":"x-sub-1","color":"sky","label":"substitute x = -1","style":"solid"}]},"order":1,"title":"Find A by substitution","description":"To eliminate the $B$ term and find $A$, we choose a value of $x$ that makes the coefficient of $B$ equal to zero. This is the root of the factor $(x+1)$, which is $x = -1$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$3(\\htmlData{anchor=x-sub-2}{\\color{@amber}{-2}}) + 5 = A(\\htmlData{anchor=zero-term-2}{\\color{@amber}{-2}} + 2) + B(\\htmlData{anchor=x-term-2}{\\color{@amber}{-2}} + 1)$"},{"latex":"$$-6 + 5 = A(0) + B(-1)$"},{"latex":"$$-1 = -B \\implies \\htmlData{anchor=b-val}{\\color{@purple}{B = 1}}$"}],"highlights":[{"color":"amber","style":"box","anchor":"x-sub-2"},{"color":"purple","style":"text-color","anchor":"b-val"}],"connections":[{"to":"x-term-2","from":"x-sub-2","color":"amber","label":"substitute x = -2","style":"solid"}]},"order":2,"title":"Find B by substitution","description":"Next, to eliminate the $A$ term and find $B$, we choose the value of $x$ that makes the coefficient of $A$ equal to zero. This is the root of the factor $(x+2)$, which is $x = -2$. Finding both values correctly earns the second mark A1."},{"type":"text","order":3,"title":"Write the final expression","content":"Now substitute $A = 2$ and $B = 1$ back into our initial partial fraction form to obtain the final answer A1:\n\n$$\\frac{2}{x+1} + \\frac{1}{x+2}$$"}]},{"id":"equating_coefficients","label":"Equating Coefficients","steps":[{"type":"text","order":0,"title":"Set up the identity","conte

1.2.25 Partial fractions (A-level only)

Question 6

1.2.25 Partial fractions (A-level only) Questions

1.2.25 Partial fractions (A-level only)

Question 6

1.2.25 Partial fractions (A-level only) Questions