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bg-background","children":["$","$L3a",null,{"backHref":"/a-level/maths/edexcel/questions","basePath":"/a-level/maths/edexcel/a-level-pure-mathematics-1-3-partial-fractions/questions","formulaUrl":"https://assets.mathsgenie.co.uk/board-resources/a-level-maths/edexcel/formula-book.pdf","initialIndex":8,"initialQuestionWindow":[{"id":"d877c2ca-f576-439d-9a69-de51fbdd6d47","revisionId":"155f838b-750f-44aa-9740-a0d859ea2012","questionNumber":null,"questionType":"LA","specification":"","difficulty":1,"difficultyBand":"medium","totalMarks":3,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"9f307832-ffe3-4a13-bc46-654bd2c89dfe","label":"a","content":"Express $\\frac{5x+2}{(x-2)(x+2)}$ in the form $\\frac{A}{(x-2)} + \\frac{B}{(x+2)}$","markscheme":"- Sets up the identity $5x+2 = A(x+2) + B(x-2)$ M1\n- Finds $A = 3$ and $B = 2$ by substituting $x = 2$ and $x = -2$ A1\n- Final answer: $\\frac{3}{(x-2)} + \\frac{2}{(x+2)}$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Establishes the identity $5x+2 = A(x+2) + B(x-2)$.","code":"M"},{"id":"p2","marks":1,"description":"Finds the correct constants $A = 3$ and $B = 2$.","code":"A"},{"id":"p3","marks":1,"description":"Writes the final expression in the form $\\frac{3}{(x-2)} + \\frac{2}{(x+2)}$.","code":"A"}],"finalAnswers":["\\frac{3}{(x-2)} + \\frac{2}{(x+2)}"],"generatedAt":"2026-05-06T14:03:32.967Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"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 begin by setting up an identity. We equate the original fraction to the sum of the partial fractions:\n\n$$\\frac{5x+2}{(x-2)(x+2)} \\equiv \\frac{A}{x-2} + \\frac{B}{x+2}$$\n\nTo clear the denominators, we multiply both sides of the identity by the common denominator $(x-2)(x+2)$:\n\n$$5x+2 \\equiv A(x+2) + B(x-2)$$\n\nSetting up this correct identity earns the first method mark (M1)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$5\\htmlData{anchor=x-val-2}{\\color{@sky}{x}} + 2 \\equiv A(\\htmlData{anchor=x-sub-2-a}{\\color{@sky}{x}} + 2) + B(\\htmlData{anchor=x-sub-2-b}{\\color{@sky}{x}} - 2)$"},{"latex":"$$\\text{Substitute } \\htmlData{anchor=sub-val-2}{\\color{@sky}{x = 2}}:$"},{"latex":"$$5(\\color{@sky}{2}) + 2 = A(\\color{@sky}{2} + 2) + B(\\color{@sky}{2} - 2)$"},{"latex":"$$12 = 4A + 0 \\implies \\htmlData{anchor=val-a}{\\color{@sky}{A = 3}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sub-val-2"},{"color":"sky","style":"box","anchor":"val-a"}],"connections":[{"to":"x-val-2","from":"sub-val-2","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Find A using substitution","description":"We can choose any value of $x$ to substitute into our identity. Choosing $x = 2$ is strategic because it makes the $(x - 2)$ term zero, allowing us to isolate and solve for $A$ directly."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$5\\htmlData{anchor=x-val-neg2}{\\color{@amber}{x}} + 2 \\equiv A(\\htmlData{anchor=x-sub-neg2-a}{\\color{@amber}{x}} + 2) + B(\\htmlData{anchor=x-sub-neg2-b}{\\color{@amber}{x}} - 2)$"},{"latex":"$$\\text{Substitute } \\htmlData{anchor=sub-val-neg2}{\\color{@amber}{x = -2}}:$"},{"latex":"$$5(\\color{@amber}{-2}) + 2 = A(\\color{@amber}{-2} + 2) + B(\\color{@amber}{-2} - 2)$"},{"latex":"$$-8 = 0 - 4B \\implies \\htmlData{anchor=val-b}{\\color{@amber}{B = 2}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"sub-val-neg2"},{"color":"amber","style":"box","anchor":"val-b"}],"connections":[{"to":"x-val-neg2","from":"sub-val-neg2","color":"amber","label":null,"style":"solid"}]},"order":2,"title":"Find B using substitution","description":"Next, we substitute $x = -2$ to make the $(x + 2)$ term zero, which eliminates the $A$ term and lets us solve for $B$ directly. Correctly finding both values of $A$ and $B$ earns the second mark (A1)."},{"type":"text","order":3,"title":"State the final expression","content":"Now we substitute our values $A = 3$ and $B = 2$ back into the partial fraction template:\n\n$$\\frac{3}{x-2} + \\frac{2}{x+2}$$\n\nThis final answer completes the expression and earns the third mark (A1)."}]},{"id":"equating_coefficients","label":"Equating Coefficients","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 with the unknown constants $A$ and $B$:\n\n$$\\frac{5x+2}{(x-2)(x+2)} \\equiv \\frac{A}{x-2} + \\frac{B}{x+2}$$\n\nTo clear the denominators, we multiply both sides of the identity by the common denominator, $(x-2)(x+2)$:\n\n$$5x+2 \\equiv A(x+2) + B(x-2)$$\n\nThis gives us our starting equation for finding the values of $A$ and $B$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$5x+2 \\equiv \\htmlData{anchor=expand1}{\\color{@sky}{A(x+2)}} + \\htmlData{anchor=expand2}{\\color{@amber}{B(x-2)}}$"},{"latex":"$$5x+2 \\equiv \\color{@sky}{Ax} + \\color{@sky}{2A} + \\color{@amber}{Bx} - \\color{@amber}{2B}$"},{"latex":"$$5x+2 \\equiv \\htmlData{anchor=grouped}{\\color{@purple}{(A+B)x + (2A-2B)}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"expand1"},{"color":"amber","style":"text-color","anchor":"expand2"},{"color":"purple","style":"box","anchor":"grouped"}],"connections":[{"to":"grouped","from":"expand1","color":"purple","label":"Group coefficients","style":"solid"}]},"order":1,"title":"Expand and group terms","description":"Now, we expand the brackets on the right-hand side and collect like terms to group the coefficients of $x$ and the constant terms."},{"type":"text","order":2,"title":"Equate the coefficients","content":"Since this is an identity, the coefficients on both sides of the equation must be equal. We can equate the coefficients of $x$ and the constant terms:\n\n* **For the coefficients of $x$:**\n $$A+B = 5$$ (Equation 1)\n\n* **For the constant terms:**\n $$2A-2B = 2$$ \n\nWe can simplify the constant term equation by dividing both sides by $2$:\n $$A-B = 1$$ (Equation 2)"},{"type":"text","order":3,"title":"Solve the system","content":"Now we can solve this system of linear equations.\n\nFirst, we add Equation 1 and Equation 2 together:\n$$(A+B) + (A-B) = 5 + 1$$\n\n$$2A = 6$$\n\n$$A = 3$$\n\nNext, we substitute $A = 3$ back into Equation 1 to find $B$:\n\n$$3+B = 5$$\n\n$$B = 2$$\n\nSo, our constants are $A = 3$ and $B = 2$."},{"type":"text","order":4,"title":"State final answer","content":"Substitute the values of $A$ and $B$ back into our original partial fractions structure to write down our final answer:\n\n$$\\frac{5x+2}{(x-2)(x+2)} \\equiv \\frac{3}{x-2} + \\frac{2}{x+2}$$"}]}],"generatedAt":"2026-06-09T09:18:51.568Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"9f307832-ffe3-4a13-bc46-654bd2c89dfe:substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the algebraic fraction as partial fractions, we begin by setting up an identity. We equate the original fraction to the sum of the partial fractions:\n\n$$\\frac{5x+2}{(x-2)(x+2)} \\equiv \\frac{A}{x-2} + \\frac{B}{x+2}$$\n\nTo clear the denominators, we multiply both sides of the identity by the common denominator $(x-2)(x+2)$:\n\n$$5x+2 \\equiv A(x+2) + B(x-2)$$\n\nSetting up this correct identity earns the first method mark (M1)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$5\\htmlData{anchor=x-val-2}{\\color{@sky}{x}} + 2 \\equiv A(\\htmlData{anchor=x-sub-2-a}{\\color{@sky}{x}} + 2) + B(\\htmlData{anchor=x-sub-2-b}{\\color{@sky}{x}} - 2)$"},{"latex":"$$\\text{Substitute } \\htmlData{anchor=sub-val-2}{\\color{@sky}{x = 2}}:$"},{"latex":"$$5(\\color{@sky}{2}) + 2 = A(\\color{@sky}{2} + 2) + B(\\color{@sky}{2} - 2)$"},{"latex":"$$12 = 4A + 0 \\implies \\htmlData{anchor=val-a}{\\color{@sky}{A = 3}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sub-val-2"},{"color":"sky","style":"box","anchor":"val-a"}],"connections":[{"to":"x-val-2","from":"sub-val-2","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Find A using substitution","description":"We can choose any value of $x$ to substitute into our identity. Choosing $x = 2$ is strategic because it makes the $(x - 2)$ term zero, allowing us to isolate and solve for $A$ directly."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$5\\htmlData{anchor=x-val-neg2}{\\color{@amber}{x}} + 2 \\equiv A(\\htmlData{anchor=x-sub-neg2-a}{\\color{@amber}{x}} + 2) + B(\\htmlData{anchor=x-sub-neg2-b}{\\color{@amber}{x}} - 2)$"},{"latex":"$$\\text{Substitute } \\htmlData{anchor=sub-val-neg2}{\\color{@amber}{x = -2}}:$"},{"latex":"$$5(\\color{@amber}{-2}) + 2 = A(\\color{@amber}{-2} + 2) + B(\\color{@amber}{-2} - 2)$"},{"latex":"$$-8 = 0 - 4B \\implies \\htmlData{anchor=val-b}{\\color{@amber}{B = 2}}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"sub-val-neg2"},{"color":"amber","style":"box","anchor":"val-b"}],"connections":[{"to":"x-val-neg2","from":"sub-val-neg2","color":"amber","label":null,"style":"solid"}]},"order":2,"title":"Find B using substitution","description":"Next, we substitute $x = -2$ to make the $(x + 2)$ term zero, which eliminates the $A$ term and lets us solve for $B$ directly. Correctly finding both values of $A$ and $B$ earns the second mark (A1)."},{"type":"text","order":3,"title":"State the final expression","content":"Now we substitute our values $A = 3$ and $B = 2$ back into the partial fraction template:\n\n$$\\frac{3}{x-2} + \\frac{2}{x+2}$$\n\nThis final answer completes the expression and earns the third mark (A1)."}]},{"id":"9f307832-ffe3-4a13-bc46-654bd2c89dfe:equating_coefficients","label":"Equating Coefficients","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the algebraic fraction as partial fractions, we start by setting up an identity with the unknown constants $A$ and $B$:\n\n$$\\frac{5x+2}{(x-2)(x+2)} \\equiv \\frac{A}{x-2} + \\frac{B}{x+2}$$\n\nTo clear the denominators, we multiply both sides of the identity by the common denominator, $(x-2)(x+2)$:\n\n$$5x+2 \\equiv A(x+2) + B(x-2)$$\n\nThis gives us our starting equation for finding the values of $A$ and $B$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$5x+2 \\equiv \\htmlData{anchor=expand1}{\\color{@sky}{A(x+2)}} + \\htmlData{anchor=expand2}{\\color{@amber}{B(x-2)}}$"},{"latex":"$$5x+2 \\equiv \\color{@sky}{Ax} + \\color{@sky}{2A} + \\color{@amber}{Bx} - \\color{@amber}{2B}$"},{"latex":"$$5x+2 \\equiv \\htmlData{anchor=grouped}{\\color{@purple}{(A+B)x + (2A-2B)}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"expand1"},{"color":"amber","style":"text-color","anchor":"expand2"},{"color":"purple","style":"box","anchor":"grouped"}],"connections":[{"to":"grouped","from":"expand1","color":"purple","label":"Group coefficients","style":"solid"}]},"order":1,"title":"Expand and group terms","description":"Now, we expand the brackets on the right-hand side and collect like terms to group the coefficients of $x$ and the constant terms."},{"type":"text","order":2,"title":"Equate the coefficients","content":"Since this is an identity, the coefficients on both sides of the equation must be equal. We can equate the coefficients of $x$ and the constant terms:\n\n* **For the coefficients of $x$:**\n $$A+B = 5$$ (Equation 1)\n\n* **For the constant terms:**\n $$2A-2B = 2$$ \n\nWe can simplify the constant term equation by dividing both sides by $2$:\n $$A-B = 1$$ (Equation 2)"},{"type":"text","order":3,"title":"Solve the system","content":"Now we can solve this system of linear equations.\n\nFirst, we add Equation 1 and Equation 2 together:\n$$(A+B) + (A-B) = 5 + 1$$\n\n$$2A = 6$$\n\n$$A = 3$$\n\nNext, we substitute $A = 3$ back into Equation 1 to find $B$:\n\n$$3+B = 5$$\n\n$$B = 2$$\n\nSo, our constants are $A = 3$ and $B = 2$."},{"type":"text","order":4,"title":"State final answer","content":"Substitute the values of $A$ and $B$ back into our original partial fractions structure to write down our final answer:\n\n$$\\frac{5x+2}{(x-2)(x+2)} \\equiv \\frac{3}{x-2} + \\frac{2}{x+2}$$"}]}],"generatedAt":"2026-06-09T09:19:04.651Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"f9c1abed-3a80-46e0-87b8-dba96c072800","revisionId":"af966380-95e7-4273-b1d2-5a13a6cb233d","questionNumber":null,"questionType":"LA","specification":"","difficulty":1,"difficultyBand":"medium","totalMarks":3,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"ccaec150-591c-43cc-8145-6a8cf2766b8c","label":"a","content":"Express $\\frac{5x+7}{3(x-1)(2x+1)}$ in partial fractions","markscheme":"- Sets up the identity $\\frac{5x+7}{3} = A(2x+1) + B(x-1)$ M1\n- Finds $A = \\frac{4}{3}$ and $B = -1$ A1\n- Final answer: $\\frac{4}{3(x-1)} - \\frac{1}{2x+1}$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Attempts to express as partial fractions with the correct structure.","code":"M"},{"id":"p2","marks":1,"description":"Correct calculation of constants $A=4/3$ and $B=-1$ (or vice versa depending on setup).","code":"A"},{"id":"p3","marks":1,"description":"Final result written as $\\frac{4}{3(x-1)} - \\frac{1}{2x+1}$.","code":"A"}],"finalAnswers":["\\frac{4}{3(x-1)} - \\frac{1}{2x+1}","\\frac{4}{3(x-1)} + \\frac{-1}{2x+1}"],"generatedAt":"2026-05-06T14:03:48.467Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the algebraic fraction in partial fractions, we begin by setting up an identity with unknown constants $A$ and $B$ for each of the linear factors in the denominator:\n\n$$\\frac{5x+7}{3(x-1)(2x+1)} \\equiv \\frac{A}{x-1} + \\frac{B}{2x+1}$$\n\nTo find the values of $A$ and $B$, we multiply both sides by the common denominator $(x-1)(2x+1)$ to clear the fractions:\n\n$$\\frac{5x+7}{3} \\equiv A(2x+1) + B(x-1)$$\n\nThis is a crucial first step that earns you **1 mark** (M1) on the exam."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{5(\\htmlData{anchor=x1}{\\color{@sky}{1}}) + 7}{3} = A(2(\\htmlData{anchor=x1-sub1}{\\color{@sky}{1}}) + 1) + B(\\htmlData{anchor=x1-sub2}{\\color{@sky}{1}} - 1)$"},{"latex":"$$\\frac{12}{3} = A(3) + B(0)$"},{"latex":"$$4 = 3A \\implies A = \\htmlData{anchor=a-val}{\\color{@amber}{\\frac{4}{3}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"x1"},{"color":"amber","style":"text-color","anchor":"a-val"}],"connections":[{"to":"x1-sub1","from":"x1","color":"sky","label":"substitute x=1","style":"solid"},{"to":"x1-sub2","from":"x1","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Substitute $x = 1$","description":"We use the **Substitution Method** to find $A$ by choosing a value of $x$ that eliminates the term containing $B$. Setting $x = 1$ makes the term $(x-1) = 0$:"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{5(\\htmlData{anchor=x2}{\\color{@purple}{-\\frac{1}{2}}}) + 7}{3} = A(2(\\htmlData{anchor=x2-sub1}{\\color{@purple}{-\\frac{1}{2}}}) + 1) + B(\\htmlData{anchor=x2-sub2}{\\color{@purple}{-\\frac{1}{2}}} - 1)$"},{"latex":"$$\\frac{4.5}{3} = A(0) + B\\left(-\\frac{3}{2}\\right)$"},{"latex":"$$1.5 = -1.5B \\implies B = \\htmlData{anchor=b-val}{\\color{@teal}{-1}}$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"x2"},{"color":"teal","style":"text-color","anchor":"b-val"}],"connections":[{"to":"x2-sub1","from":"x2","color":"purple","label":"substitute x=-1/2","style":"solid"},{"to":"x2-sub2","from":"x2","color":"purple","label":null,"style":"solid"}]},"order":2,"title":"Substitute $x = -1/2$","description":"Now, we eliminate the $A$ term by substituting the root of $(2x+1)$, which is $x = -\\frac{1}{2}$:"},{"type":"text","order":3,"title":"State the final answer","content":"Finding the correct constants $A = \\frac{4}{3}$ and $B = -1$ earns the second mark (**1 mark** - A1).\n\nFinally, we substitute these values back into our original expression to get:\n\n$$\\frac{\\frac{4}{3}}{x-1} + \\frac{-1}{2x+1}$$\n\nSimplifying this gives the final answer:\n\n$$\\frac{4}{3(x-1)} - \\frac{1}{2x+1}$$\n\nThis correct final simplified expression earns the final mark (**1 mark** - A1)."}]},{"id":"equating_coefficients","label":"Equating Coefficients Method","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the algebraic fraction in partial fractions, we first write it as a sum of two simpler fractions with unknown constants $A$ and $B$ as numerators:\n\n$$\\frac{5x+7}{3(x-1)(2x+1)} = \\frac{A}{x-1} + \\frac{B}{2x+1}$$\n\nNext, we multiply both sides of the equation by $(x-1)(2x+1)$ to clear these denominators:\n\n$$\\frac{5x+7}{3} = A(2x+1) + B(x-1)$$\n\nThis is an identity that holds true for all values of $x$."},{"type":"text","order":1,"title":"Expand and group terms","content":"We now expand the right-hand side of our identity to group the terms by powers of $x$:\n\n$$\\frac{5}{3}x + \\frac{7}{3} = A(2x+1) + B(x-1)$$\n\n$$\\frac{5}{3}x + \\frac{7}{3} = 2Ax + A + Bx - B$$\n\nCombine the $x$ terms and the constant terms on the right-hand side:\n\n$$\\frac{5}{3}x + \\frac{7}{3} = (2A + B)x + (A - B)$$\n\nThis puts the equation in a perfect form to compare both sides directly."},{"type":"text","order":2,"title":"Equate the coefficients","content":"Since this is an identity, the coefficients on both sides of the equation must be equal.\n\nBy comparing the coefficients of $x$:\n$$2A + B = \\frac{5}{3} \\quad \\text{--- (Equation 1)}$$\n\nBy comparing the constant terms:\n$$A - B = \\frac{7}{3} \\quad \\text{--- (Equation 2)}$$"},{"type":"text","order":3,"title":"Solve the system","content":"We have a system of two simultaneous equations. We can solve them by adding Equation 1 and Equation 2 together to eliminate $B$:\n\n$$\\begin{aligned} (2A + B) + (A - B) &= \\frac{5}{3} + \\frac{7}{3} \\\\ 3A &= \\frac{12}{3} \\\\ 3A &= 4 \\\\ A &= \\frac{4}{3} \\end{aligned}$$\n\nNow substitute $A = \\frac{4}{3}$ back into Equation 2 to find $B$:\n\n$$\\begin{aligned} \\frac{4}{3} - B &= \\frac{7}{3} \\\\ -B &= \\frac{7}{3} - \\frac{4}{3} \\\\ -B &= 1 \\\\ B &= -1 \\end{aligned}$$"},{"type":"text","order":4,"title":"Write final answer","content":"Substitute the values of $A = \\frac{4}{3}$ and $B = -1$ back into our initial partial fraction setup:\n\n$$\\frac{\\frac{4}{3}}{x-1} + \\frac{-1}{2x+1}$$\n\nSimplifying this expression gives the final answer:\n\n$$\\frac{4}{3(x-1)} - \\frac{1}{2x+1}$$"}]}],"generatedAt":"2026-06-11T03:44:10.537Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"ccaec150-591c-43cc-8145-6a8cf2766b8c:substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the algebraic fraction in partial fractions, we begin by setting up an identity with unknown constants $A$ and $B$ for each of the linear factors in the denominator:\n\n$$\\frac{5x+7}{3(x-1)(2x+1)} \\equiv \\frac{A}{x-1} + \\frac{B}{2x+1}$$\n\nTo find the values of $A$ and $B$, we multiply both sides by the common denominator $(x-1)(2x+1)$ to clear the fractions:\n\n$$\\frac{5x+7}{3} \\equiv A(2x+1) + B(x-1)$$\n\nThis is a crucial first step that earns you **1 mark** (M1) on the exam."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{5(\\htmlData{anchor=x1}{\\color{@sky}{1}}) + 7}{3} = A(2(\\htmlData{anchor=x1-sub1}{\\color{@sky}{1}}) + 1) + B(\\htmlData{anchor=x1-sub2}{\\color{@sky}{1}} - 1)$"},{"latex":"$$\\frac{12}{3} = A(3) + B(0)$"},{"latex":"$$4 = 3A \\implies A = \\htmlData{anchor=a-val}{\\color{@amber}{\\frac{4}{3}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"x1"},{"color":"amber","style":"text-color","anchor":"a-val"}],"connections":[{"to":"x1-sub1","from":"x1","color":"sky","label":"substitute x=1","style":"solid"},{"to":"x1-sub2","from":"x1","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Substitute $x = 1$","description":"We use the **Substitution Method** to find $A$ by choosing a value of $x$ that eliminates the term containing $B$. Setting $x = 1$ makes the term $(x-1) = 0$:"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{5(\\htmlData{anchor=x2}{\\color{@purple}{-\\frac{1}{2}}}) + 7}{3} = A(2(\\htmlData{anchor=x2-sub1}{\\color{@purple}{-\\frac{1}{2}}}) + 1) + B(\\htmlData{anchor=x2-sub2}{\\color{@purple}{-\\frac{1}{2}}} - 1)$"},{"latex":"$$\\frac{4.5}{3} = A(0) + B\\left(-\\frac{3}{2}\\right)$"},{"latex":"$$1.5 = -1.5B \\implies B = \\htmlData{anchor=b-val}{\\color{@teal}{-1}}$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"x2"},{"color":"teal","style":"text-color","anchor":"b-val"}],"connections":[{"to":"x2-sub1","from":"x2","color":"purple","label":"substitute x=-1/2","style":"solid"},{"to":"x2-sub2","from":"x2","color":"purple","label":null,"style":"solid"}]},"order":2,"title":"Substitute $x = -1/2$","description":"Now, we eliminate the $A$ term by substituting the root of $(2x+1)$, which is $x = -\\frac{1}{2}$:"},{"type":"text","order":3,"title":"State the final answer","content":"Finding the correct constants $A = \\frac{4}{3}$ and $B = -1$ earns the second mark (**1 mark** - A1).\n\nFinally, we substitute these values back into our original expression to get:\n\n$$\\frac{\\frac{4}{3}}{x-1} + \\frac{-1}{2x+1}$$\n\nSimplifying this gives the final answer:\n\n$$\\frac{4}{3(x-1)} - \\frac{1}{2x+1}$$\n\nThis correct final simplified expression earns the final mark (**1 mark** - A1)."}]},{"id":"ccaec150-591c-43cc-8145-6a8cf2766b8c:equating_coefficients","label":"Equating Coefficients Method","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the algebraic fraction in partial fractions, we first write it as a sum of two simpler fractions with unknown constants $A$ and $B$ as numerators:\n\n$$\\frac{5x+7}{3(x-1)(2x+1)} = \\frac{A}{x-1} + \\frac{B}{2x+1}$$\n\nNext, we multiply both sides of the equation by $(x-1)(2x+1)$ to clear these denominators:\n\n$$\\frac{5x+7}{3} = A(2x+1) + B(x-1)$$\n\nThis is an identity that holds true for all values of $x$."},{"type":"text","order":1,"title":"Expand and group terms","content":"We now expand the right-hand side of our identity to group the terms by powers of $x$:\n\n$$\\frac{5}{3}x + \\frac{7}{3} = A(2x+1) + B(x-1)$$\n\n$$\\frac{5}{3}x + \\frac{7}{3} = 2Ax + A + Bx - B$$\n\nCombine the $x$ terms and the constant terms on the right-hand side:\n\n$$\\frac{5}{3}x + \\frac{7}{3} = (2A + B)x + (A - B)$$\n\nThis puts the equation in a perfect form to compare both sides directly."},{"type":"text","order":2,"title":"Equate the coefficients","content":"Since this is an identity, the coefficients on both sides of the equation must be equal.\n\nBy comparing the coefficients of $x$:\n$$2A + B = \\frac{5}{3} \\quad \\text{--- (Equation 1)}$$\n\nBy comparing the constant terms:\n$$A - B = \\frac{7}{3} \\quad \\text{--- (Equation 2)}$$"},{"type":"text","order":3,"title":"Solve the system","content":"We have a system of two simultaneous equations. We can solve them by adding Equation 1 and Equation 2 together to eliminate $B$:\n\n$$\\begin{aligned} (2A + B) + (A - B) &= \\frac{5}{3} + \\frac{7}{3} \\\\ 3A &= \\frac{12}{3} \\\\ 3A &= 4 \\\\ A &= \\frac{4}{3} \\end{aligned}$$\n\nNow substitute $A = \\frac{4}{3}$ back into Equation 2 to find $B$:\n\n$$\\begin{aligned} \\frac{4}{3} - B &= \\frac{7}{3} \\\\ -B &= \\frac{7}{3} - \\frac{4}{3} \\\\ -B &= 1 \\\\ B &= -1 \\end{aligned}$$"},{"type":"text","order":4,"title":"Write final answer","content":"Substitute the values of $A = \\frac{4}{3}$ and $B = -1$ back into our initial partial fraction setup:\n\n$$\\frac{\\frac{4}{3}}{x-1} + \\frac{-1}{2x+1}$$\n\nSimplifying this expression gives the final answer:\n\n$$\\frac{4}{3(x-1)} - \\frac{1}{2x+1}$$"}]}],"generatedAt":"2026-06-11T03:44:10.589Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":86},"walkthroughVideo":null},{"id":"1fa2f970-a7dc-494e-8df0-cb6f72019df4","revisionId":"87991e3d-1141-4d63-8ed3-7477a2e5e482","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":"medium","totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"d9130c81-5ce7-4919-a38e-268012c05efa","label":null,"content":"In a high-frequency electronics design, the transfer function \$ H(s) \$ of a specific signal filter is given by:\n\n\\[ H(s) = \\frac{16s + 11}{(2s - 3)(4s + 1)} \\]\n\nDetermine the values of the constants \$ A \$ and \$ B \$ such that \$ H(s) \$ can be expressed as:\n\n\\[ \\frac{A}{2s - 3} + \\frac{B}{4s + 1} \\]\n\nwhere \$ A \$ and \$ B \$ are integers.","markscheme":"- Sets up the identity $16s + 11 \\equiv A(4s + 1) + B(2s - 3)$ M1\n- Uses a valid method to find a constant, such as substituting $s = 1.5$ or $s = -0.25$ into their identity M1\n- Obtains $A = 5$ A1\n- Obtains $B = -2$ A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the transfer function $H(s) = \\frac{16s + 11}{(2s - 3)(4s + 1)}$ in partial fractions, we set up the following identity:\n\n$$\\frac{16s + 11}{(2s - 3)(4s + 1)} \\equiv \\frac{A}{2s - 3} + \\frac{B}{4s + 1}$$\n\nMultiplying both sides by the denominator $(2s - 3)(4s + 1)$ gives the identity:\n\n$$16s + 11 \\equiv A(4s + 1) + B(2s - 3)$$\n\nThis must hold for all values of $s$."},{"type":"text","order":1,"title":"Solve for $A$","content":"To solve for $A$, we can eliminate the term involving $B$ by substituting the root of $2s - 3 = 0$, which is $s = 1.5$ (or $s = \\frac{3}{2}$), into the identity:\n\n$$16(1.5) + 11 = A(4(1.5) + 1) + B(2(1.5) - 3)$$\n\nSimplifying this equation yields:\n\n$$24 + 11 = A(6 + 1) + B(0)$$\n\n$$35 = 7A$$\n\n$$A = 5$$"},{"type":"text","order":2,"title":"Solve for $B$","content":"To solve for $B$, we eliminate the term involving $A$ by substituting the root of $4s + 1 = 0$, which is $s = -0.25$ (or $s = -\\frac{1}{4}$), into the identity:\n\n$$16(-0.25) + 11 = A(4(-0.25) + 1) + B(2(-0.25) - 3)$$\n\nSimplifying this equation yields:\n\n$$-4 + 11 = A(0) + B(-0.5 - 3)$$\n\n$$7 = -3.5B$$\n\n$$B = -2$$"},{"type":"text","order":3,"title":"State the final values","content":"Thus, the solved integer values for the constants are:\n\n$$A = 5$$\n\n$$B = -2$$\n\nThe completed partial fraction expansion of the filter's transfer function is:\n\n$$H(s) = \\frac{5}{2s - 3} - \\frac{2}{4s + 1}$$"}]},{"id":"equating_coefficients","label":"Equating Coefficients","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the transfer function $H(s)$ in partial fractions, we start by setting up an identity:\n\n$$\\frac{16s + 11}{(2s - 3)(4s + 1)} \\equiv \\frac{A}{2s - 3} + \\frac{B}{4s + 1}$$\n\nMultiplying both sides by the common denominator $(2s - 3)(4s + 1)$ gives:\n\n$$16s + 11 \\equiv A(4s + 1) + B(2s - 3)$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$A(4s + 1) + B(2s - 3) = \\htmlData{anchor=expand1}{\\color{@sky}{4As + A}} + \\htmlData{anchor=expand2}{\\color{@amber}{2Bs - 3B}}$"},{"latex":"$$= \\htmlData{anchor=group-s}{(\\color{@sky}{4A} + \\color{@amber}{2B})s} + \\htmlData{anchor=group-const}{(\\color{@sky}{A} - \\color{@amber}{3B})}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"expand1"},{"color":"amber","style":"text-color","anchor":"expand2"}],"connections":[{"to":"group-s","from":"expand1","color":"sky","label":null,"style":"solid"},{"to":"group-const","from":"expand2","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Expand and group terms","description":"Next, expand the right-hand side of the identity and group the terms containing $s$ together, and the constant terms together."},{"type":"text","order":2,"title":"Equate the coefficients","content":"Since this is an identity that must hold for all values of $s$, we equate the coefficients of the terms on both sides.\n\nEquating the coefficients of $s$:\n\n$$4A + 2B = 16$$\n\nEquating the constant terms:\n\n$$A - 3B = 11$$"},{"type":"text","order":3,"title":"Solve the system","content":"Now we solve this system of linear equations. Let's simplify the first equation by dividing both sides by $2$:\n\n$$2A + B = 8$$\n\n$$B = 8 - 2A$$\n\nSubstitute this expression for $B$ into the second equation and solve for $A$:\n\n$$A - 3(8 - 2A) = 11$$\n\n$$A - 24 + 6A = 11$$\n\n$$7A - 24 = 11$$\n\n$$7A = 35$$\n\n$$A = 5$$\n\nNow, substitute $A = 5$ back to find $B$:\n\n$$B = 8 - 2(5)$$\n\n$$B = -2$$"},{"type":"text","order":4,"title":"State final answer","content":"The required constants are:\n\n$$A = 5$$\n\n$$B = -2$$\n\nThis means our partial fraction decomposition is:\n\n$$H(s) = \\frac{5}{2s - 3} - \\frac{2}{4s + 1}$$"}]}],"generatedAt":"2026-06-09T03:52:57.722Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"d9130c81-5ce7-4919-a38e-268012c05efa:substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the transfer function $H(s) = \\frac{16s + 11}{(2s - 3)(4s + 1)}$ in partial fractions, we set up the following identity:\n\n$$\\frac{16s + 11}{(2s - 3)(4s + 1)} \\equiv \\frac{A}{2s - 3} + \\frac{B}{4s + 1}$$\n\nMultiplying both sides by the denominator $(2s - 3)(4s + 1)$ gives the identity:\n\n$$16s + 11 \\equiv A(4s + 1) + B(2s - 3)$$\n\nThis must hold for all values of $s$."},{"type":"text","order":1,"title":"Solve for $A$","content":"To solve for $A$, we can eliminate the term involving $B$ by substituting the root of $2s - 3 = 0$, which is $s = 1.5$ (or $s = \\frac{3}{2}$), into the identity:\n\n$$16(1.5) + 11 = A(4(1.5) + 1) + B(2(1.5) - 3)$$\n\nSimplifying this equation yields:\n\n$$24 + 11 = A(6 + 1) + B(0)$$\n\n$$35 = 7A$$\n\n$$A = 5$$"},{"type":"text","order":2,"title":"Solve for $B$","content":"To solve for $B$, we eliminate the term involving $A$ by substituting the root of $4s + 1 = 0$, which is $s = -0.25$ (or $s = -\\frac{1}{4}$), into the identity:\n\n$$16(-0.25) + 11 = A(4(-0.25) + 1) + B(2(-0.25) - 3)$$\n\nSimplifying this equation yields:\n\n$$-4 + 11 = A(0) + B(-0.5 - 3)$$\n\n$$7 = -3.5B$$\n\n$$B = -2$$"},{"type":"text","order":3,"title":"State the final values","content":"Thus, the solved integer values for the constants are:\n\n$$A = 5$$\n\n$$B = -2$$\n\nThe completed partial fraction expansion of the filter's transfer function is:\n\n$$H(s) = \\frac{5}{2s - 3} - \\frac{2}{4s + 1}$$"}]},{"id":"d9130c81-5ce7-4919-a38e-268012c05efa:equating_coefficients","label":"Equating Coefficients","steps":[{"type":"text","order":0,"title":"Set up the identity","content":"To express the transfer function $H(s)$ in partial fractions, we start by setting up an identity:\n\n$$\\frac{16s + 11}{(2s - 3)(4s + 1)} \\equiv \\frac{A}{2s - 3} + \\frac{B}{4s + 1}$$\n\nMultiplying both sides by the common denominator $(2s - 3)(4s + 1)$ gives:\n\n$$16s + 11 \\equiv A(4s + 1) + B(2s - 3)$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$A(4s + 1) + B(2s - 3) = \\htmlData{anchor=expand1}{\\color{@sky}{4As + A}} + \\htmlData{anchor=expand2}{\\color{@amber}{2Bs - 3B}}$"},{"latex":"$$= \\htmlData{anchor=group-s}{(\\color{@sky}{4A} + \\color{@amber}{2B})s} + \\htmlData{anchor=group-const}{(\\color{@sky}{A} - \\color{@amber}{3B})}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"expand1"},{"color":"amber","style":"text-color","anchor":"expand2"}],"connections":[{"to":"group-s","from":"expand1","color":"sky","label":null,"style":"solid"},{"to":"group-const","from":"expand2","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Expand and group terms","description":"Next, expand the right-hand side of the identity and group the terms containing $s$ together, and the constant terms together."},{"type":"text","order":2,"title":"Equate the coefficients","content":"Since this is an identity that must hold for all values of $s$, we equate the coefficients of the terms on both sides.\n\nEquating the coefficients of $s$:\n\n$$4A + 2B = 16$$\n\nEquating the constant terms:\n\n$$A - 3B = 11$$"},{"type":"text","order":3,"title":"Solve the system","content":"Now we solve this system of linear equations. Let's simplify the first equation by dividing both sides by $2$:\n\n$$2A + B = 8$$\n\n$$B = 8 - 2A$$\n\nSubstitute this expression for $B$ into the second equation and solve for $A$:\n\n$$A - 3(8 - 2A) = 11$$\n\n$$A - 24 + 6A = 11$$\n\n$$7A - 24 = 11$$\n\n$$7A = 35$$\n\n$$A = 5$$\n\nNow, substitute $A = 5$ back to find $B$:\n\n$$B = 8 - 2(5)$$\n\n$$B = -2$$"},{"type":"text","order":4,"title":"State final answer","content":"The required constants are:\n\n$$A = 5$$\n\n$$B = -2$$\n\nThis means our partial fraction decomposition is:\n\n$$H(s) = \\frac{5}{2s - 3} - \\frac{2}{4s + 1}$$"}]}],"generatedAt":"2026-06-09T03:53:12.601Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":79},"walkthroughVideo":null}],"subjectCode":"a_level_maths","summaries":[{"id":"f209f2a2-5723-43ba-82a1-fc17edfc8c67","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"9d3c6e50-5295-4a06-b680-b33bd067334d","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"ccd01f36-c50a-4e45-ad97-44a918590c37","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"61bf9309-865b-4108-a293-c8fc670013b0","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"ba7e7e9f-cffd-4764-bae7-369bfddc13e7","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"bf5c00ea-3f06-43fb-bd0d-ec2689b1e986","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"edbf14ba-6206-41a7-aaa9-7fc7f936d530","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"d877c2ca-f576-439d-9a69-de51fbdd6d47","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"f9c1abed-3a80-46e0-87b8-dba96c072800","difficultyBand":"medium","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"1fa2f970-a7dc-494e-8df0-cb6f72019df4","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"5e736fe7-d52f-4446-9999-56c60bf286a7","difficultyBand":"hard","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"60f1d250-7fc0-4450-bfae-f94c3355b1db","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"faa915aa-44fb-41cd-9000-18defed8f4cf","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true}],"title":"1.3 Partial Fractions","topicId":"5d96c6a6-9bd0-49bd-9a4a-f7ce0af48c3d","topicSidebar":"$L3b","topicSlug":"a-level-pure-mathematics-1-3-partial-fractions","topicTitle":"1.3 Partial Fractions"}]}],"$L3c","$L3d"]}]

1.3 Partial Fractions

Question 7

1.3 Partial Fractions Questions

1.3 Partial Fractions

Question 7

1.3 Partial Fractions Questions