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3b:T4af,To simplify the expression, we can factor out $\sqrt{\sqrt{7}-2}$ from the terms: First, rewrite the terms: - $2(\sqrt{7}-2)^{3/2} = 2(\sqrt{7}-2)\sqrt{\sqrt{7}-2}$ - To combine the fraction $\frac{18}{\sqrt{\sqrt{7}-2}}$, we can rationalize its denominator by multiplying the numerator and denominator by $\sqrt{\sqrt{7}+2}$ or express it in terms of the common factor. Since $x^2 = \sqrt{7}-2$, we have: $$\frac{1}{x} = \frac{x}{x^2} = \frac{x}{\sqrt{7}-2}$$ Rationalizing the denominator of $\frac{1}{\sqrt{7}-2}$: $$\frac{1}{\sqrt{7}-2} = \frac{\sqrt{7}+2}{(\sqrt{7}-2)(\sqrt{7}+2)} = \frac{\sqrt{7}+2}{7-4} = \frac{\sqrt{7}+2}{3}$$ Therefore, we can rewrite the fraction term: $$\frac{18}{x} = 18 \cdot \frac{\sqrt{7}+2}{3} x = 6(\sqrt{7}+2)x$$ Now, substitute this back into the expression for $k$: $$\begin{aligned} k &= 2x^3 + \frac{18}{x} + 24x + 5 \\ &= 2x(x^2) + 6(\sqrt{7}+2)x + 24x + 5 \\ &= 2x(\sqrt{7}-2) + (6\sqrt{7}+12)x + 24x + 5 \\ &= (2\sqrt{7} - 4 + 6\sqrt{7} + 12 + 24)x + 5 \\ &= (8\sqrt{7} + 32)x + 5 \\ &= 8(\sqrt{7} + 4)x + 5 \end{aligned}$$ Substituting $x = \sqrt{\sqrt{7}-2}$ back in gives our final exact value: $$k = 8(\sqrt{7} + 4)\sqrt{\sqrt{7} - 2} + 5$$3c:T4af,To simplify the expression, we can factor out $\sqrt{\sqrt{7}-2}$ from the terms: First, rewrite the terms: - $2(\sqrt{7}-2)^{3/2} = 2(\sqrt{7}-2)\sqrt{\sqrt{7}-2}$ - To combine the fraction $\frac{18}{\sqrt{\sqrt{7}-2}}$, we can rationalize its denominator by multiplying the numerator and denominator by $\sqrt{\sqrt{7}+2}$ or express it in terms of the common factor. Since $x^2 = \sqrt{7}-2$, we have: $$\frac{1}{x} = \frac{x}{x^2} = \frac{x}{\sqrt{7}-2}$$ Rationalizing the denominator of $\frac{1}{\sqrt{7}-2}$: $$\frac{1}{\sqrt{7}-2} = \frac{\sqrt{7}+2}{(\sqrt{7}-2)(\sqrt{7}+2)} = \frac{\sqrt{7}+2}{7-4} = \frac{\sqrt{7}+2}{3}$$ Therefore, we can rewrite the fraction term: $$\frac{18}{x} = 18 \cdot \frac{\sqrt{7}+2}{3} x = 6(\sqrt{7}+2)x$$ Now, substitute this back into the expression for $k$: $$\begin{aligned} k &= 2x^3 + \frac{18}{x} + 24x + 5 \\ &= 2x(x^2) + 6(\sqrt{7}+2)x + 24x + 5 \\ &= 2x(\sqrt{7}-2) + (6\sqrt{7}+12)x + 24x + 5 \\ &= (2\sqrt{7} - 4 + 6\sqrt{7} + 12 + 24)x + 5 \\ &= (8\sqrt{7} + 32)x + 5 \\ &= 8(\sqrt{7} + 4)x + 5 \end{aligned}$$ Substituting $x = \sqrt{\sqrt{7}-2}$ back in gives our final exact value: $$k = 8(\sqrt{7} + 4)\sqrt{\sqrt{7} - 2} + 5$$25:["$","$L35",null,{"bundle":{"topicIds":["141a09f3-b92b-4a52-91f4-36d5bb7f7ab0"],"paperIds":[],"stageIds":[],"stageTopicIds":[],"stageAssessmentIds":[],"topicNoteIds":[]},"children":[["$","$L36",null,{"feature":"question-bank"}],"$L37",["$","$L38",null,{"subjectCode":"a_level_maths","subjectName":"Maths","topicSlug":"a-level-pure-mathematics-1-2-algebraic-fractions","topicTitle":"1.2 Algebraic Fractions","sectionTitle":null,"activity":"worksheet","resourceTitle":"1.2 Algebraic Fractions Questions","resourceHref":"/a-level/maths/edexcel/a-level-pure-mathematics-1-2-algebraic-fractions/questions/q1","saveTrail":false}],["$","$L39",null,{"href":"/a-level/maths/edexcel/a-level-pure-mathematics-1-2-algebraic-fractions/questions/q1","label":"1.2 Algebraic Fractions","subjectCode":"a_level_maths","surface":"questions"}],["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"LearningResource\",\"name\":\"Question 5 – 1.2 Algebraic Fractions - Maths Questions\",\"description\":\"Practise 1.2 Algebraic Fractions A Level Maths exam-style questions with worked solutions on MathsGenie.\",\"learningResourceType\":\"Practice questions\",\"educationalLevel\":\"A Level\",\"about\":{\"@type\":\"Thing\",\"name\":\"1.2 Algebraic Fractions\"},\"url\":\"https://mathsgenie.co.uk/a-level/maths/edexcel/a-level-pure-mathematics-1-2-algebraic-fractions/questions/q5\",\"inLanguage\":\"en-GB\",\"provider\":{\"@id\":\"https://mathsgenie.co.uk/#organization\"}}"}}],["$","div",null,{"className":"min-h-0 w-full flex-1 overflow-hidden bg-background","children":["$","$L3a",null,{"backHref":"/a-level/maths/edexcel/questions","basePath":"/a-level/maths/edexcel/a-level-pure-mathematics-1-2-algebraic-fractions/questions","formulaUrl":"https://assets.mathsgenie.co.uk/board-resources/a-level-maths/edexcel/formula-book.pdf","initialIndex":4,"initialQuestionWindow":[{"id":"2206b2d5-983d-4e36-a279-e4872f684e17","revisionId":"7c4854ac-7e26-46f9-ac3b-a2b1089895df","questionNumber":null,"questionType":"LA","specification":"The difference in signal attenuation, \$ A(k) \$, between two experimental microwave waveguides depends on a positive integer parameter \$ k \$ according to the formula:\n\n\\[ A(k) = \\frac{8}{3+7\\sqrt{k}} - \\frac{8}{7\\sqrt{k}-3} \\]","difficulty":4,"difficultyBand":"medium","totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"e556689d-a3b6-430a-a4b9-fae36548fa54","label":"a","content":"Express \$ A(k) \$ as a single fraction that does not involve surds.","markscheme":"- Combines the terms using a common denominator: $\\frac{8(7\\sqrt{k}-3) - 8(3+7\\sqrt{k})}{(3+7\\sqrt{k})(7\\sqrt{k}-3)}$ M1\n- Expands the denominator using the difference of two squares: $(7\\sqrt{k})^2 - 3^2 = 49k - 9$ M1\n- Simplifies the numerator: $56\\sqrt{k} - 24 - (24 + 56\\sqrt{k}) = -48$ A1\n- States the final fraction: $\\frac{-48}{49k - 9}$ or $-\\frac{48}{49k - 9}$ A1","marks":4,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"common_denominator","label":"Common Denominator Method","steps":[{"type":"text","order":0,"title":"Identify the goal","content":"We need to express the difference in signal attenuation, $A(k)$, as a single fraction with no surds in either the numerator or the denominator:\n\n$$A(k) = \\frac{8}{3+7\\sqrt{k}} - \\frac{8}{7\\sqrt{k}-3}$$\n\nTo do this, we will find a common denominator for the two algebraic fractions and combine them."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$A(k) = \\frac{8}{\\htmlData{anchor=denom1}{\\color{@sky}{3+7\\sqrt{k}}}} - \\frac{8}{\\htmlData{anchor=denom2}{\\color{@amber}{7\\sqrt{k}-3}}}$"},{"latex":"$$A(k) = \\frac{8(\\htmlData{anchor=num1}{\\color{@amber}{7\\sqrt{k}-3}}) - 8(\\htmlData{anchor=num2}{\\color{@sky}{3+7\\sqrt{k}}})}{\\htmlData{anchor=com-denom}{\\color{@purple}{(3+7\\sqrt{k})(7\\sqrt{k}-3)}}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"denom1"},{"color":"amber","style":"box","anchor":"denom2"}],"connections":[{"to":"num1","from":"denom2","color":"amber","label":"multiply","style":"solid"},{"to":"num2","from":"denom1","color":"sky","label":"multiply","style":"solid"}]},"order":1,"title":"Find a common denominator","description":"To combine the two fractions, we multiply the numerator of each fraction by the denominator of the other, putting the entire expression over the product of the two denominators."},{"type":"text","order":2,"title":"Simplify the denominator","content":"Let's focus on simplifying the denominator using the difference of two squares identity, $(a+b)(a-b) = a^2 - b^2$.\n\nNotice that the terms in our denominator can be rearranged:\n$$(3+7\\sqrt{k})(7\\sqrt{k}-3) = (7\\sqrt{k}+3)(7\\sqrt{k}-3)$$\n\nApplying the identity with $a = 7\\sqrt{k}$ and $b = 3$:\n$$\\begin{aligned} (7\\sqrt{k})^2 - 3^2 &= (7^2)(\\sqrt{k})^2 - 9 \\\\ &= 49k - 9 \\end{aligned}$$\n\nThis successfully removes the surd from our denominator."},{"type":"text","order":3,"title":"Simplify the numerator","content":"Now we expand and simplify the terms in the numerator:\n\n$$\\begin{aligned} \\text{Numerator} &= 8(7\\sqrt{k}-3) - 8(3+7\\sqrt{k}) \\\\ &= 56\\sqrt{k} - 24 - (24 + 56\\sqrt{k}) \\\\ &= 56\\sqrt{k} - 24 - 24 - 56\\sqrt{k} \\\\ &= -48 \\end{aligned}$$\n\nNotice that the terms involving $\\sqrt{k}$ cancel out completely, leaving us with a simple constant numerator."},{"type":"text","order":4,"title":"State the final fraction","content":"Putting the simplified numerator and denominator back together, we obtain:\n\n$$A(k) = \\frac{-48}{49k - 9} \\quad \\text{or} \\quad -\\frac{48}{49k - 9}$$\n\nThis single fraction no longer contains any surds, which completes Part 1."},{"type":"text","order":5,"title":"Deduce rationality","content":"To deduce that $A(k)$ is a rational number for all positive integer values of $k$, we use the definition of a rational number:\n\n$$A(k) = \\frac{-48}{49k - 9}$$\n\nSince $k$ is a positive integer ($k \\in \\mathbb{Z}^+$):\n\n1. The numerator $-48$ is an integer.\n2. The denominator $49k - 9$ is also an integer.\n3. The denominator is non-zero because $49k - 9 = 0 \\implies k = \\frac{9}{49}$, which is not an integer.\n\nSince $A(k)$ can be written as the ratio of two integers with a non-zero denominator, it is a rational number for all $k \\in \\mathbb{Z}^+$."}]},{"id":"rationalizing_first","label":"Rationalizing Denominators First","steps":[{"type":"text","order":0,"title":"Understand the goal","content":"We need to express the difference in signal attenuation, $A(k)$:\n\n$$A(k) = \\frac{8}{3+7\\sqrt{k}} - \\frac{8}{7\\sqrt{k}-3}$$\n\nas a single fraction with no surds in either the numerator or the denominator. \n\nWe will use the **Rationalizing Denominators First** approach. This means we will rationalize each fraction individually by multiplying their respective numerators and denominators by their conjugates, and then we will subtract the simplified terms."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{First term: } \\frac{8}{3+7\\sqrt{k}}$"},{"latex":"$$\\text{Multiply by conjugate: } \\frac{8}{3+7\\sqrt{k}} \\times \\frac{\\htmlData{anchor=conj-1}{\\color{@sky}{3-7\\sqrt{k}}}}{\\htmlData{anchor=conj-2}{\\color{@sky}{3-7\\sqrt{k}}}}$"},{"latex":"$$= \\frac{8(3-7\\sqrt{k})}{(3)^2 - (7\\sqrt{k})^2} = \\frac{24 - 56\\sqrt{k}}{9 - 49k}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"conj-1"},{"color":"sky","style":"text-color","anchor":"conj-2"}],"connections":[{"to":"conj-2","from":"conj-1","color":"sky","label":"conjugate","style":"solid"}]},"order":1,"title":"Rationalize the first term","description":"We multiply the numerator and the denominator of the first fraction by the conjugate of its denominator, which is $3-7\\sqrt{k}$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Second term: } \\frac{8}{7\\sqrt{k}-3}$"},{"latex":"$$\\text{Multiply by conjugate: } \\frac{8}{7\\sqrt{k}-3} \\times \\frac{\\htmlData{anchor=conj-3}{\\color{@amber}{7\\sqrt{k}+3}}}{\\htmlData{anchor=conj-4}{\\color{@amber}{7\\sqrt{k}+3}}}$"},{"latex":"$$= \\frac{8(7\\sqrt{k}+3)}{(7\\sqrt{k})^2 - (3)^2} = \\frac{56\\sqrt{k} + 24}{49k - 9}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"conj-3"},{"color":"amber","style":"text-color","anchor":"conj-4"}],"connections":[{"to":"conj-4","from":"conj-3","color":"amber","label":"conjugate","style":"solid"}]},"order":2,"title":"Rationalize the second term","description":"Next, we multiply the numerator and the denominator of the second fraction by its conjugate, $7\\sqrt{k}+3$."},{"type":"text","order":3,"title":"Align denominators","content":"Let's compare the denominators of our two rationalized terms:\n- First term denominator: $9 - 49k$\n- Second term denominator: $49k - 9$\n\nNotice that $9 - 49k = -(49k - 9)$. To easily subtract the terms, we can rewrite the first term so that it shares the same common denominator, $49k - 9$:\n\n$$\\frac{24 - 56\\sqrt{k}}{9 - 49k} = \\frac{24 - 56\\sqrt{k}}{-(49k - 9)} = \\frac{-(24 - 56\\sqrt{k})}{49k - 9} = \\frac{-24 + 56\\sqrt{k}}{49k - 9}$$"},{"type":"text","order":4,"title":"Subtract the fractions","content":"Now we subtract the rationalized second term from the adjusted first term:\n\n$$A(k) = \\frac{-24 + 56\\sqrt{k}}{49k - 9} - \\frac{56\\sqrt{k} + 24}{49k - 9}$$\n\nSince they share a common denominator, we combine the numerators:\n\n$$A(k) = \\frac{(-24 + 56\\sqrt{k}) - (56\\sqrt{k} + 24)}{49k - 9}$$\n\nSimplifying the numerator:\n\n$$-24 + 56\\sqrt{k} - 56\\sqrt{k} - 24 = -48$$\n\nThus, we get:\n\n$$A(k) = \\frac{-48}{49k - 9}$$"},{"type":"text","order":5,"title":"Final answer structure","content":"The final single fraction with no surds is:\n\n$$A(k) = -\\frac{48}{49k - 9} \\quad \\text{or} \\quad \\frac{48}{9 - 49k}$$\n\n### Markscheme Breakdown\n- **M1**: Combining terms or preparing to subtract with a common denominator method.\n- **M1**: Correctly using the difference of two squares to expand denominators to obtain $49k - 9$.\n- **A1**: Correctly simplifying the numerator to $-48$.\n- **A1**: Stating the final simplified fraction without surds."},{"type":"text","order":6,"title":"Deduce rationality","content":"To show that $A(k)$ is rational for all positive integers $k$:\n\n$$A(k) = \\frac{-48}{49k - 9}$$\n\nSince $k$ is a positive integer ($k \\in \\mathbb{Z}^+$):\n\n1. The numerator $-48$ and the denominator $49k - 9$ are both integers.\n2. The denominator $49k - 9 \\neq 0$ because $k = \\frac{9}{49}$ is not an integer.\n\nThus, $A(k)$ is a ratio of integers with a non-zero denominator, which by definition means $A(k)$ is a rational number for all $k \\in \\mathbb{Z}^+$."}]}],"generatedAt":"2026-06-09T03:48:56.289Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"be3cf0ef-8089-4654-a412-588780c9ab5a","label":"b","content":"Hence, deduce that the attenuation difference \$ A(k) \$ is a rational number for all positive integer values of \$ k \$.","markscheme":"- States that for all positive integers $k$, the values $-48$ and $49k - 9$ are integers B1\n- Argues that $49k - 9 \\neq 0$ for all $k \\in \\mathbb{Z}^+$ (as $k = \\frac{9}{49}$ is the only root and is not an integer), therefore $A(k)$ is a ratio of integers and thus rational B1","marks":2,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"algebraic_simplification","label":"Algebraic Simplification","steps":[{"type":"text","order":0,"title":"Find a common denominator","content":"To simplify the expression for $A(k)$, we first find a common denominator for the two fractions.\n\n$$A(k) = \\frac{8}{3 + 7\\sqrt{k}} - \\frac{8}{7\\sqrt{k} - 3}$$\n\nNotice that the denominator of the first fraction can be written as $7\\sqrt{k} + 3$. The common denominator is the product of the two denominators, which gives a difference of two squares:\n\n$$(7\\sqrt{k} + 3)(7\\sqrt{k} - 3) = (7\\sqrt{k})^2 - 3^2 = 49k - 9$$"},{"type":"text","order":1,"title":"Combine into one fraction","content":"Now, we write both fractions over this common denominator:\n\n$$\\begin{aligned} A(k) &= \\frac{8(7\\sqrt{k} - 3) - 8(7\\sqrt{k} + 3)}{(7\\sqrt{k} + 3)(7\\sqrt{k} - 3)} \\\\ &= \\frac{56\\sqrt{k} - 24 - (56\\sqrt{k} + 24)}{49k - 9} \\\\ &= \\frac{56\\sqrt{k} - 24 - 56\\sqrt{k} - 24}{49k - 9} \\\\ &= \\frac{-48}{49k - 9} \\end{aligned}$$\n\nBy simplifying algebraically, we have eliminated the radical terms in the numerator, resulting in a single rational-looking expression."},{"type":"text","order":2,"title":"Analyze the components","content":"Recall the definition of a rational number: a number that can be expressed in the form $\\frac{p}{q}$, where $p$ and $q$ are integers and $q \\neq 0$.\n\nLet's analyze the numerator and the denominator for any positive integer $k$ ($k \\in \\mathbb{Z}^+$):\n\n* The numerator is $-48$, which is clearly an integer.\n* Since $k$ is an integer, $49k - 9$ must also be an integer because integers are closed under multiplication and subtraction."},{"type":"text","order":3,"title":"Check the denominator constraint","content":"To confirm that the denominator is non-zero for all positive integers $k$, let's check when it would equal zero:\n\n$$49k - 9 = 0 \\implies k = \\frac{9}{49}$$\n\nSince $\\frac{9}{49}$ is not an integer, there is no positive integer value of $k$ that makes the denominator equal to zero. Thus, $49k - 9 \\neq 0$ for all $k \\in \\mathbb{Z}^+$."},{"type":"text","order":4,"title":"Conclude the proof","content":"Since $A(k) = \\frac{-48}{49k - 9}$ is a ratio of two integers where the denominator is non-zero, we have shown by definition that the attenuation difference $A(k)$ is a rational number for all positive integer values of $k$."}]}],"generatedAt":"2026-06-09T03:50:02.488Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"e556689d-a3b6-430a-a4b9-fae36548fa54:common_denominator","label":"(a) Common Denominator Method","steps":[{"type":"text","order":0,"title":"Identify the goal","content":"We need to express the difference in signal attenuation, $A(k)$, as a single fraction with no surds in either the numerator or the denominator:\n\n$$A(k) = \\frac{8}{3+7\\sqrt{k}} - \\frac{8}{7\\sqrt{k}-3}$$\n\nTo do this, we will find a common denominator for the two algebraic fractions and combine them."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$A(k) = \\frac{8}{\\htmlData{anchor=denom1}{\\color{@sky}{3+7\\sqrt{k}}}} - \\frac{8}{\\htmlData{anchor=denom2}{\\color{@amber}{7\\sqrt{k}-3}}}$"},{"latex":"$$A(k) = \\frac{8(\\htmlData{anchor=num1}{\\color{@amber}{7\\sqrt{k}-3}}) - 8(\\htmlData{anchor=num2}{\\color{@sky}{3+7\\sqrt{k}}})}{\\htmlData{anchor=com-denom}{\\color{@purple}{(3+7\\sqrt{k})(7\\sqrt{k}-3)}}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"denom1"},{"color":"amber","style":"box","anchor":"denom2"}],"connections":[{"to":"num1","from":"denom2","color":"amber","label":"multiply","style":"solid"},{"to":"num2","from":"denom1","color":"sky","label":"multiply","style":"solid"}]},"order":1,"title":"Find a common denominator","description":"To combine the two fractions, we multiply the numerator of each fraction by the denominator of the other, putting the entire expression over the product of the two denominators."},{"type":"text","order":2,"title":"Simplify the denominator","content":"Let's focus on simplifying the denominator using the difference of two squares identity, $(a+b)(a-b) = a^2 - b^2$.\n\nNotice that the terms in our denominator can be rearranged:\n$$(3+7\\sqrt{k})(7\\sqrt{k}-3) = (7\\sqrt{k}+3)(7\\sqrt{k}-3)$$\n\nApplying the identity with $a = 7\\sqrt{k}$ and $b = 3$:\n$$\\begin{aligned} (7\\sqrt{k})^2 - 3^2 &= (7^2)(\\sqrt{k})^2 - 9 \\\\ &= 49k - 9 \\end{aligned}$$\n\nThis successfully removes the surd from our denominator."},{"type":"text","order":3,"title":"Simplify the numerator","content":"Now we expand and simplify the terms in the numerator:\n\n$$\\begin{aligned} \\text{Numerator} &= 8(7\\sqrt{k}-3) - 8(3+7\\sqrt{k}) \\\\ &= 56\\sqrt{k} - 24 - (24 + 56\\sqrt{k}) \\\\ &= 56\\sqrt{k} - 24 - 24 - 56\\sqrt{k} \\\\ &= -48 \\end{aligned}$$\n\nNotice that the terms involving $\\sqrt{k}$ cancel out completely, leaving us with a simple constant numerator."},{"type":"text","order":4,"title":"State the final fraction","content":"Putting the simplified numerator and denominator back together, we obtain:\n\n$$A(k) = \\frac{-48}{49k - 9} \\quad \\text{or} \\quad -\\frac{48}{49k - 9}$$\n\nThis single fraction no longer contains any surds, which completes Part 1."},{"type":"text","order":5,"title":"Deduce rationality","content":"To deduce that $A(k)$ is a rational number for all positive integer values of $k$, we use the definition of a rational number:\n\n$$A(k) = \\frac{-48}{49k - 9}$$\n\nSince $k$ is a positive integer ($k \\in \\mathbb{Z}^+$):\n\n1. The numerator $-48$ is an integer.\n2. The denominator $49k - 9$ is also an integer.\n3. The denominator is non-zero because $49k - 9 = 0 \\implies k = \\frac{9}{49}$, which is not an integer.\n\nSince $A(k)$ can be written as the ratio of two integers with a non-zero denominator, it is a rational number for all $k \\in \\mathbb{Z}^+$."}]},{"id":"e556689d-a3b6-430a-a4b9-fae36548fa54:rationalizing_first","label":"(a) Rationalizing Denominators First","steps":[{"type":"text","order":0,"title":"Understand the goal","content":"We need to express the difference in signal attenuation, $A(k)$:\n\n$$A(k) = \\frac{8}{3+7\\sqrt{k}} - \\frac{8}{7\\sqrt{k}-3}$$\n\nas a single fraction with no surds in either the numerator or the denominator. \n\nWe will use the **Rationalizing Denominators First** approach. This means we will rationalize each fraction individually by multiplying their respective numerators and denominators by their conjugates, and then we will subtract the simplified terms."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{First term: } \\frac{8}{3+7\\sqrt{k}}$"},{"latex":"$$\\text{Multiply by conjugate: } \\frac{8}{3+7\\sqrt{k}} \\times \\frac{\\htmlData{anchor=conj-1}{\\color{@sky}{3-7\\sqrt{k}}}}{\\htmlData{anchor=conj-2}{\\color{@sky}{3-7\\sqrt{k}}}}$"},{"latex":"$$= \\frac{8(3-7\\sqrt{k})}{(3)^2 - (7\\sqrt{k})^2} = \\frac{24 - 56\\sqrt{k}}{9 - 49k}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"conj-1"},{"color":"sky","style":"text-color","anchor":"conj-2"}],"connections":[{"to":"conj-2","from":"conj-1","color":"sky","label":"conjugate","style":"solid"}]},"order":1,"title":"Rationalize the first term","description":"We multiply the numerator and the denominator of the first fraction by the conjugate of its denominator, which is $3-7\\sqrt{k}$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Second term: } \\frac{8}{7\\sqrt{k}-3}$"},{"latex":"$$\\text{Multiply by conjugate: } \\frac{8}{7\\sqrt{k}-3} \\times \\frac{\\htmlData{anchor=conj-3}{\\color{@amber}{7\\sqrt{k}+3}}}{\\htmlData{anchor=conj-4}{\\color{@amber}{7\\sqrt{k}+3}}}$"},{"latex":"$$= \\frac{8(7\\sqrt{k}+3)}{(7\\sqrt{k})^2 - (3)^2} = \\frac{56\\sqrt{k} + 24}{49k - 9}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"conj-3"},{"color":"amber","style":"text-color","anchor":"conj-4"}],"connections":[{"to":"conj-4","from":"conj-3","color":"amber","label":"conjugate","style":"solid"}]},"order":2,"title":"Rationalize the second term","description":"Next, we multiply the numerator and the denominator of the second fraction by its conjugate, $7\\sqrt{k}+3$."},{"type":"text","order":3,"title":"Align denominators","content":"Let's compare the denominators of our two rationalized terms:\n- First term denominator: $9 - 49k$\n- Second term denominator: $49k - 9$\n\nNotice that $9 - 49k = -(49k - 9)$. To easily subtract the terms, we can rewrite the first term so that it shares the same common denominator, $49k - 9$:\n\n$$\\frac{24 - 56\\sqrt{k}}{9 - 49k} = \\frac{24 - 56\\sqrt{k}}{-(49k - 9)} = \\frac{-(24 - 56\\sqrt{k})}{49k - 9} = \\frac{-24 + 56\\sqrt{k}}{49k - 9}$$"},{"type":"text","order":4,"title":"Subtract the fractions","content":"Now we subtract the rationalized second term from the adjusted first term:\n\n$$A(k) = \\frac{-24 + 56\\sqrt{k}}{49k - 9} - \\frac{56\\sqrt{k} + 24}{49k - 9}$$\n\nSince they share a common denominator, we combine the numerators:\n\n$$A(k) = \\frac{(-24 + 56\\sqrt{k}) - (56\\sqrt{k} + 24)}{49k - 9}$$\n\nSimplifying the numerator:\n\n$$-24 + 56\\sqrt{k} - 56\\sqrt{k} - 24 = -48$$\n\nThus, we get:\n\n$$A(k) = \\frac{-48}{49k - 9}$$"},{"type":"text","order":5,"title":"Final answer structure","content":"The final single fraction with no surds is:\n\n$$A(k) = -\\frac{48}{49k - 9} \\quad \\text{or} \\quad \\frac{48}{9 - 49k}$$\n\n### Markscheme Breakdown\n- **M1**: Combining terms or preparing to subtract with a common denominator method.\n- **M1**: Correctly using the difference of two squares to expand denominators to obtain $49k - 9$.\n- **A1**: Correctly simplifying the numerator to $-48$.\n- **A1**: Stating the final simplified fraction without surds."},{"type":"text","order":6,"title":"Deduce rationality","content":"To show that $A(k)$ is rational for all positive integers $k$:\n\n$$A(k) = \\frac{-48}{49k - 9}$$\n\nSince $k$ is a positive integer ($k \\in \\mathbb{Z}^+$):\n\n1. The numerator $-48$ and the denominator $49k - 9$ are both integers.\n2. The denominator $49k - 9 \\neq 0$ because $k = \\frac{9}{49}$ is not an integer.\n\nThus, $A(k)$ is a ratio of integers with a non-zero denominator, which by definition means $A(k)$ is a rational number for all $k \\in \\mathbb{Z}^+$."}]},{"id":"be3cf0ef-8089-4654-a412-588780c9ab5a:algebraic_simplification","label":"(b) Algebraic Simplification","steps":[{"type":"text","order":0,"title":"Find a common denominator","content":"To simplify the expression for $A(k)$, we first find a common denominator for the two fractions.\n\n$$A(k) = \\frac{8}{3 + 7\\sqrt{k}} - \\frac{8}{7\\sqrt{k} - 3}$$\n\nNotice that the denominator of the first fraction can be written as $7\\sqrt{k} + 3$. The common denominator is the product of the two denominators, which gives a difference of two squares:\n\n$$(7\\sqrt{k} + 3)(7\\sqrt{k} - 3) = (7\\sqrt{k})^2 - 3^2 = 49k - 9$$"},{"type":"text","order":1,"title":"Combine into one fraction","content":"Now, we write both fractions over this common denominator:\n\n$$\\begin{aligned} A(k) &= \\frac{8(7\\sqrt{k} - 3) - 8(7\\sqrt{k} + 3)}{(7\\sqrt{k} + 3)(7\\sqrt{k} - 3)} \\\\ &= \\frac{56\\sqrt{k} - 24 - (56\\sqrt{k} + 24)}{49k - 9} \\\\ &= \\frac{56\\sqrt{k} - 24 - 56\\sqrt{k} - 24}{49k - 9} \\\\ &= \\frac{-48}{49k - 9} \\end{aligned}$$\n\nBy simplifying algebraically, we have eliminated the radical terms in the numerator, resulting in a single rational-looking expression."},{"type":"text","order":2,"title":"Analyze the components","content":"Recall the definition of a rational number: a number that can be expressed in the form $\\frac{p}{q}$, where $p$ and $q$ are integers and $q \\neq 0$.\n\nLet's analyze the numerator and the denominator for any positive integer $k$ ($k \\in \\mathbb{Z}^+$):\n\n* The numerator is $-48$, which is clearly an integer.\n* Since $k$ is an integer, $49k - 9$ must also be an integer because integers are closed under multiplication and subtraction."},{"type":"text","order":3,"title":"Check the denominator constraint","content":"To confirm that the denominator is non-zero for all positive integers $k$, let's check when it would equal zero:\n\n$$49k - 9 = 0 \\implies k = \\frac{9}{49}$$\n\nSince $\\frac{9}{49}$ is not an integer, there is no positive integer value of $k$ that makes the denominator equal to zero. Thus, $49k - 9 \\neq 0$ for all $k \\in \\mathbb{Z}^+$."},{"type":"text","order":4,"title":"Conclude the proof","content":"Since $A(k) = \\frac{-48}{49k - 9}$ is a ratio of two integers where the denominator is non-zero, we have shown by definition that the attenuation difference $A(k)$ is a rational number for all positive integer values of $k$."}]}],"generatedAt":"2026-06-09T03:50:37.822Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"3a75af72-7480-42d5-bae1-d9e615ac3f18","revisionId":"d3729183-e704-4f64-818c-30966cd1c805","questionNumber":null,"questionType":"LA","specification":"The equation\n\n\\[\\frac{2}{x} - 4 = -3x + k\\]\n\nwhere \$k\$ is a constant, has no real roots.","difficulty":4,"difficultyBand":"medium","totalMarks":5,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"7ba6a843-acce-4b63-a5fc-ef7ba993f7ad","label":null,"content":"Find the range of possible values for \$k\$.","markscheme":"- Multiplies the equation by $x$ and rearranges to form a three-term quadratic $3x^2 - (4 + k)x + 2 = 0$ M1\n- Recognises that for the equation to have no real roots, the discriminant $b^2 - 4ac < 0$ M1\n- Substitutes $a = 3$, $b = -(4 + k)$, and $c = 2$ into the discriminant to obtain $(4 + k)^2 - 24 < 0$ or equivalent M1\n- Finds the critical values $k = -4 \\pm 2\\sqrt{6}$ (or $k = -4 \\pm \\sqrt{24}$) A1\n- Correctly identifies the range for $k$ as $-4 - 2\\sqrt{6} < k < -4 + 2\\sqrt{6}$ A1","marks":5,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"algebraic_discriminant","label":"Algebraic Approach via Discriminant","steps":[{"type":"text","order":0,"title":"Form a quadratic equation","content":"To solve this algebraically, we first want to eliminate the fraction by multiplying the entire equation by $x$:\n\n$$\\left(\\frac{2}{x} - 4\\right)x = (-3x + k)x$$\n\nThis expands to:\n\n$$2 - 4x = -3x^2 + kx$$\n\nNow, rearrange all terms to one side to form a standard quadratic equation in the form $ax^2 + bx + c = 0$:\n\n$$3x^2 - 4x - kx + 2 = 0$$\n\nFactorising out $x$ from the middle terms gives:\n\n$$3x^2 - (4 + k)x + 2 = 0$$\n\nThis earns the first mark M1."},{"type":"text","order":1,"title":"Identify coefficients and discriminant","content":"For a quadratic equation $ax^2 + bx + c = 0$ to have no real roots, its discriminant must be strictly less than zero:\n\n$$\\Delta = b^2 - 4ac < 0$$\n\nThis principle earns a method mark M1.\n\nFrom our quadratic equation, we identify the coefficients:\n- $a = 3$\n- $b = -(4 + k)$\n- $c = 2$"},{"type":"text","order":2,"title":"Substitute into the discriminant","content":"Substitute these values into the discriminant inequality:\n\n$$\\Delta = (-(4+k))^2 - 4(3)(2) < 0$$\n\nSince squaring a negative gives a positive, this simplifies to:\n\n$$(4 + k)^2 - 24 < 0$$\n\nThis substitution earns a mark M1."},{"type":"text","order":3,"title":"Find the critical values","content":"To solve the inequality, we first find the critical values by setting the discriminant equal to zero:\n\n$$(4 + k)^2 - 24 = 0$$\n\n$$(4 + k)^2 = 24$$\n\nTaking the square root of both sides:\n\n$$4 + k = \\pm \\sqrt{24}$$\n\nSimplify the radical $\\sqrt{24} = \\sqrt{4 \\times 6} = 2\\sqrt{6}$:\n\n$$k = -4 \\pm 2\\sqrt{6}$$\n\nFinding these critical values earns an accuracy mark A1."},{"type":"text","order":4,"title":"Determine the range for k","content":"The inequality we are solving is a quadratic inequality of the form $(4+k)^2 - 24 < 0$, which represents an upward-opening parabola.\n\nSince we want the region where this expression is **less than zero** ($< 0$), the solution lies **between** our two critical values:\n\n$$-4 - 2\\sqrt{6} < k < -4 + 2\\sqrt{6}$$\n\nThis correct final inequality earns the accuracy mark A1.","calculator":[{"gdc":{"expression":"(x+4)^2 - 24","instructions":"1. Go to the Graph application.\n2. Sketch \$f1(x) = (4 + x)^2 - 24\$.\n3. Use the 'Analyze Graph' tool to find the roots (zeros).\n4. Observe that the graph is below the x-axis between the two zeros, which represents the interval where the inequality is satisfied."},"ti84":{"expression":"(X+4)^2 - 24","instructions":"1. Press [Y=] and enter \$Y_1 = (4 + X)^2 - 24\$.\n2. Press [WINDOW] and set Xmin = -12, Xmax = 4, Ymin = -30, Ymax = 10.\n3. Press [GRAPH] to view the parabola.\n4. Use [2nd] [CALC] [2:zero] to find the two x-intercepts, which correspond to the critical values of \$k\$ (approximately -8.90 and 0.90).\n5. Since we require \$Y_1 < 0\$, the solution is the interval between these two intercepts."},"desmos":{"expression":"y = (4 + x)^2 - 24","instructions":"1. Type `y = (4 + x)^2 - 24` in the expression list.\n2. Click on the x-intercepts to reveal their coordinates: \$(-4-2\\sqrt{6}, 0) \\approx (-8.899, 0)\$ and \$(-4+2\\sqrt{6}, 0) \\approx (0.899, 0)\$.\n3. The inequality is satisfied for values of \$x\$ (representing \$k\$) strictly between these two values."},"description":"You can verify the range of the inequality graphically on your graphing calculator."}]}]},{"id":"graphical_calculus","label":"Graphical Approach via Calculus","steps":[{"type":"text","order":0,"title":"Isolate the constant k","content":"To solve this problem using a graphical approach, we want to separate the constant $k$ from the rest of the equation. This allows us to think of the problem as finding the intersections between a curve $y = f(x)$ and a horizontal line $y = k$.\n\nStarting with the given equation:\n$$\\frac{2}{x} - 4 = -3x + k$$\n\nWe add $3x$ to both sides to isolate $k$:\n$$k = \\frac{2}{x} + 3x - 4$$\n\nLet's define our function as:\n$$f(x) = \\frac{2}{x} + 3x - 4 \\quad \\text{for } x \\neq 0$$"},{"type":"text","order":1,"title":"Differentiate the function","content":"To understand the behavior of the curve $y = f(x)$ and locate its turning points, we find its derivative with respect to $x$.\n\nFirst, rewrite $f(x)$ using a negative index:\n$$f(x) = 2x^{-1} + 3x - 4$$\n\nNow, apply the power rule to find $f'(x)$:\n$$f'(x) = -2x^{-2} + 3 = -\\frac{2}{x^2} + 3$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$-\\frac{2}{\\htmlData{anchor=x-sq}{x^2}} + 3 = 0$"},{"latex":"$$x^2 = \\htmlData{anchor=x-val}{\\frac{2}{3}}$"},{"latex":"$$x = \\pm \\sqrt{\\frac{2}{3}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"x-sq"},{"color":"sky","style":"box","anchor":"x-val"}],"connections":[{"to":"x-val","from":"x-sq","color":"sky","label":"solve for x^2","style":"solid"}]},"order":2,"title":"Find the stationary points","description":"Stationary points occur where the gradient is zero, so we set $f'(x) = 0$ and solve for $x$."},{"type":"text","order":3,"title":"Calculate stationary values","content":"Now, we find the corresponding $y$-values for each stationary point to determine the local maximum and minimum values.\n\n**Case 1: $x = \\sqrt{\\frac{2}{3}}$**\n\nWe substitute this value into our function $f(x) = \\frac{2}{x} + 3x - 4$:\n\n$$f\\left(\\sqrt{\\frac{2}{3}}\\right) = \\frac{2}{\\sqrt{2/3}} + 3\\sqrt{\\frac{2}{3}} - 4$$\n\nSince $\\frac{2}{\\sqrt{2/3}} = \\sqrt{6}$ and $3\\sqrt{\\frac{2}{3}} = \\sqrt{6}$, this simplifies to:\n\n$$f\\left(\\sqrt{\\frac{2}{3}}\\right) = \\sqrt{6} + \\sqrt{6} - 4 = 2\\sqrt{6} - 4$$\n\n**Case 2: $x = -\\sqrt{\\frac{2}{3}}$**\n\nSimilarly, substituting the negative value:\n\n$$f\\left(-\\sqrt{\\frac{2}{3}}\\right) = -\\sqrt{6} - \\sqrt{6} - 4 = -2\\sqrt{6} - 4$$"},{"type":"text","order":4,"title":"Use a calculator to visualize","content":"Let's check our work and visualize the curve $y = \\frac{2}{x} + 3x - 4$ and the horizontal lines $y = k$ on a graphing utility.\n\n- The local minimum is at $\\left(\\sqrt{\\frac{2}{3}}, 2\\sqrt{6}-4\\right) \\approx (0.816, 0.899)$.\n- The local maximum is at $\\left(-\\sqrt{\\frac{2}{3}}, -2\\sqrt{6}-4\\right) \\approx (-0.816, -8.899)$.","calculator":[{"gdc":{"expression":"y = 2/x + 3*x - 4","instructions":"1. Go to the Graph menu.\n2. Enter the function y = 2/x + 3x - 4.\n3. Use G-Solve (F5) then select MIN (F3) and MAX (F2) to find the local extrema."},"ti84":{"expression":"Y1 = 2/X + 3*X - 4","instructions":"1. Press [Y=] and enter Y1 = 2/X + 3X - 4.\n2. Press [GRAPH] (adjust WINDOW if needed).\n3. Use [2nd] [TRACE] and select 'minimum' or 'maximum' to verify the coordinates of the turning points."},"desmos":{"expression":"y = \\frac{2}{x} + 3x - 4","instructions":"1. Type y = 2/x + 3x - 4 into the first expression line.\n2. Click on the gray dots at the peaks and valleys to display the local maximum and minimum coordinates."},"description":"Graph the function to see where a horizontal line $y=k$ does not touch the curve."}]},{"type":"text","order":5,"title":"Determine range with no roots","content":"For the equation to have **no real roots**, the horizontal line $y = k$ must not intersect the curve $y = f(x)$ at any point.\n\nLooking at the graph of $f(x)$:\n- For $x > 0$, the curve has a local minimum at $y = -4 + 2\\sqrt{6}$. Thus, $f(x) \\ge -4 + 2\\sqrt{6}$ for all positive $x$.\n- For $x < 0$, the curve has a local maximum at $y = -4 - 2\\sqrt{6}$. Thus, $f(x) \\le -4 - 2\\sqrt{6}$ for all negative $x$.\n\nTherefore, the \"gap\" where no part of the curve exists is between the local maximum and the local minimum:\n\n$$-4 - 2\\sqrt{6} < k < -4 + 2\\sqrt{6}$$"}]}],"generatedAt":"2026-06-09T02:34:08.816Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"7ba6a843-acce-4b63-a5fc-ef7ba993f7ad:algebraic_discriminant","label":"Algebraic Approach via Discriminant","steps":[{"type":"text","order":0,"title":"Form a quadratic equation","content":"To solve this algebraically, we first want to eliminate the fraction by multiplying the entire equation by $x$:\n\n$$\\left(\\frac{2}{x} - 4\\right)x = (-3x + k)x$$\n\nThis expands to:\n\n$$2 - 4x = -3x^2 + kx$$\n\nNow, rearrange all terms to one side to form a standard quadratic equation in the form $ax^2 + bx + c = 0$:\n\n$$3x^2 - 4x - kx + 2 = 0$$\n\nFactorising out $x$ from the middle terms gives:\n\n$$3x^2 - (4 + k)x + 2 = 0$$\n\nThis earns the first mark M1."},{"type":"text","order":1,"title":"Identify coefficients and discriminant","content":"For a quadratic equation $ax^2 + bx + c = 0$ to have no real roots, its discriminant must be strictly less than zero:\n\n$$\\Delta = b^2 - 4ac < 0$$\n\nThis principle earns a method mark M1.\n\nFrom our quadratic equation, we identify the coefficients:\n- $a = 3$\n- $b = -(4 + k)$\n- $c = 2$"},{"type":"text","order":2,"title":"Substitute into the discriminant","content":"Substitute these values into the discriminant inequality:\n\n$$\\Delta = (-(4+k))^2 - 4(3)(2) < 0$$\n\nSince squaring a negative gives a positive, this simplifies to:\n\n$$(4 + k)^2 - 24 < 0$$\n\nThis substitution earns a mark M1."},{"type":"text","order":3,"title":"Find the critical values","content":"To solve the inequality, we first find the critical values by setting the discriminant equal to zero:\n\n$$(4 + k)^2 - 24 = 0$$\n\n$$(4 + k)^2 = 24$$\n\nTaking the square root of both sides:\n\n$$4 + k = \\pm \\sqrt{24}$$\n\nSimplify the radical $\\sqrt{24} = \\sqrt{4 \\times 6} = 2\\sqrt{6}$:\n\n$$k = -4 \\pm 2\\sqrt{6}$$\n\nFinding these critical values earns an accuracy mark A1."},{"type":"text","order":4,"title":"Determine the range for k","content":"The inequality we are solving is a quadratic inequality of the form $(4+k)^2 - 24 < 0$, which represents an upward-opening parabola.\n\nSince we want the region where this expression is **less than zero** ($< 0$), the solution lies **between** our two critical values:\n\n$$-4 - 2\\sqrt{6} < k < -4 + 2\\sqrt{6}$$\n\nThis correct final inequality earns the accuracy mark A1.","calculator":[{"gdc":{"expression":"(x+4)^2 - 24","instructions":"1. Go to the Graph application.\n2. Sketch \$f1(x) = (4 + x)^2 - 24\$.\n3. Use the 'Analyze Graph' tool to find the roots (zeros).\n4. Observe that the graph is below the x-axis between the two zeros, which represents the interval where the inequality is satisfied."},"ti84":{"expression":"(X+4)^2 - 24","instructions":"1. Press [Y=] and enter \$Y_1 = (4 + X)^2 - 24\$.\n2. Press [WINDOW] and set Xmin = -12, Xmax = 4, Ymin = -30, Ymax = 10.\n3. Press [GRAPH] to view the parabola.\n4. Use [2nd] [CALC] [2:zero] to find the two x-intercepts, which correspond to the critical values of \$k\$ (approximately -8.90 and 0.90).\n5. Since we require \$Y_1 < 0\$, the solution is the interval between these two intercepts."},"desmos":{"expression":"y = (4 + x)^2 - 24","instructions":"1. Type `y = (4 + x)^2 - 24` in the expression list.\n2. Click on the x-intercepts to reveal their coordinates: \$(-4-2\\sqrt{6}, 0) \\approx (-8.899, 0)\$ and \$(-4+2\\sqrt{6}, 0) \\approx (0.899, 0)\$.\n3. The inequality is satisfied for values of \$x\$ (representing \$k\$) strictly between these two values."},"description":"You can verify the range of the inequality graphically on your graphing calculator."}]}]},{"id":"7ba6a843-acce-4b63-a5fc-ef7ba993f7ad:graphical_calculus","label":"Graphical Approach via Calculus","steps":[{"type":"text","order":0,"title":"Isolate the constant k","content":"To solve this problem using a graphical approach, we want to separate the constant $k$ from the rest of the equation. This allows us to think of the problem as finding the intersections between a curve $y = f(x)$ and a horizontal line $y = k$.\n\nStarting with the given equation:\n$$\\frac{2}{x} - 4 = -3x + k$$\n\nWe add $3x$ to both sides to isolate $k$:\n$$k = \\frac{2}{x} + 3x - 4$$\n\nLet's define our function as:\n$$f(x) = \\frac{2}{x} + 3x - 4 \\quad \\text{for } x \\neq 0$$"},{"type":"text","order":1,"title":"Differentiate the function","content":"To understand the behavior of the curve $y = f(x)$ and locate its turning points, we find its derivative with respect to $x$.\n\nFirst, rewrite $f(x)$ using a negative index:\n$$f(x) = 2x^{-1} + 3x - 4$$\n\nNow, apply the power rule to find $f'(x)$:\n$$f'(x) = -2x^{-2} + 3 = -\\frac{2}{x^2} + 3$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$-\\frac{2}{\\htmlData{anchor=x-sq}{x^2}} + 3 = 0$"},{"latex":"$$x^2 = \\htmlData{anchor=x-val}{\\frac{2}{3}}$"},{"latex":"$$x = \\pm \\sqrt{\\frac{2}{3}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"x-sq"},{"color":"sky","style":"box","anchor":"x-val"}],"connections":[{"to":"x-val","from":"x-sq","color":"sky","label":"solve for x^2","style":"solid"}]},"order":2,"title":"Find the stationary points","description":"Stationary points occur where the gradient is zero, so we set $f'(x) = 0$ and solve for $x$."},{"type":"text","order":3,"title":"Calculate stationary values","content":"Now, we find the corresponding $y$-values for each stationary point to determine the local maximum and minimum values.\n\n**Case 1: $x = \\sqrt{\\frac{2}{3}}$**\n\nWe substitute this value into our function $f(x) = \\frac{2}{x} + 3x - 4$:\n\n$$f\\left(\\sqrt{\\frac{2}{3}}\\right) = \\frac{2}{\\sqrt{2/3}} + 3\\sqrt{\\frac{2}{3}} - 4$$\n\nSince $\\frac{2}{\\sqrt{2/3}} = \\sqrt{6}$ and $3\\sqrt{\\frac{2}{3}} = \\sqrt{6}$, this simplifies to:\n\n$$f\\left(\\sqrt{\\frac{2}{3}}\\right) = \\sqrt{6} + \\sqrt{6} - 4 = 2\\sqrt{6} - 4$$\n\n**Case 2: $x = -\\sqrt{\\frac{2}{3}}$**\n\nSimilarly, substituting the negative value:\n\n$$f\\left(-\\sqrt{\\frac{2}{3}}\\right) = -\\sqrt{6} - \\sqrt{6} - 4 = -2\\sqrt{6} - 4$$"},{"type":"text","order":4,"title":"Use a calculator to visualize","content":"Let's check our work and visualize the curve $y = \\frac{2}{x} + 3x - 4$ and the horizontal lines $y = k$ on a graphing utility.\n\n- The local minimum is at $\\left(\\sqrt{\\frac{2}{3}}, 2\\sqrt{6}-4\\right) \\approx (0.816, 0.899)$.\n- The local maximum is at $\\left(-\\sqrt{\\frac{2}{3}}, -2\\sqrt{6}-4\\right) \\approx (-0.816, -8.899)$.","calculator":[{"gdc":{"expression":"y = 2/x + 3*x - 4","instructions":"1. Go to the Graph menu.\n2. Enter the function y = 2/x + 3x - 4.\n3. Use G-Solve (F5) then select MIN (F3) and MAX (F2) to find the local extrema."},"ti84":{"expression":"Y1 = 2/X + 3*X - 4","instructions":"1. Press [Y=] and enter Y1 = 2/X + 3X - 4.\n2. Press [GRAPH] (adjust WINDOW if needed).\n3. Use [2nd] [TRACE] and select 'minimum' or 'maximum' to verify the coordinates of the turning points."},"desmos":{"expression":"y = \\frac{2}{x} + 3x - 4","instructions":"1. Type y = 2/x + 3x - 4 into the first expression line.\n2. Click on the gray dots at the peaks and valleys to display the local maximum and minimum coordinates."},"description":"Graph the function to see where a horizontal line $y=k$ does not touch the curve."}]},{"type":"text","order":5,"title":"Determine range with no roots","content":"For the equation to have **no real roots**, the horizontal line $y = k$ must not intersect the curve $y = f(x)$ at any point.\n\nLooking at the graph of $f(x)$:\n- For $x > 0$, the curve has a local minimum at $y = -4 + 2\\sqrt{6}$. Thus, $f(x) \\ge -4 + 2\\sqrt{6}$ for all positive $x$.\n- For $x < 0$, the curve has a local maximum at $y = -4 - 2\\sqrt{6}$. Thus, $f(x) \\le -4 - 2\\sqrt{6}$ for all negative $x$.\n\nTherefore, the \"gap\" where no part of the curve exists is between the local maximum and the local minimum:\n\n$$-4 - 2\\sqrt{6} < k < -4 + 2\\sqrt{6}$$"}]}],"generatedAt":"2026-06-09T02:34:25.384Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":58},"walkthroughVideo":null},{"id":"4d2a0a7c-e5b1-4b00-8ca5-269148915639","revisionId":"3226f683-dcb8-43fe-b4e3-f28c5dd49886","questionNumber":null,"questionType":"LA","specification":"The curve \$C\$ has equation\n\\[y = 2x^3 + \\frac{18}{x} + 5, \\quad x > 0\\]","difficulty":8,"difficultyBand":"medium","totalMarks":9,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"c0d117e6-790c-43eb-8860-c6e8fb68919f","label":"a","content":"Find \$\\frac{\\mathrm{d}y}{\\mathrm{d}x}\$, giving your answer in simplest form.","markscheme":"- Differentiates the curve equation to find $\\frac{dy}{dx} = 6x^2 - 18x^{-2}$ M1\n- Simplifies the derivative to $\\frac{dy}{dx} = 6x^2 - \\frac{18}{x^2}$ A1","marks":2,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"power_rule","label":"Power Rule with Negative Indices","steps":[{"type":"text","order":0,"title":"Rewrite using negative indices","content":"To differentiate the equation term-by-term using the standard power rule, we first need to express the fraction $\\frac{18}{x}$ in index form using a negative exponent.\n\nRecall that $\\frac{1}{x} = x^{-1}$. Therefore, we can rewrite the equation of the curve as:\n\n$$y = 2x^3 + 18x^{-1} + 5$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = 2x^{\\htmlData{anchor=pow-1}{\\color{@sky}{3}}} + 18x^{\\htmlData{anchor=pow-2}{\\color{@amber}{-1}}} + 5$"},{"latex":"$$\\frac{\\mathrm{d}y}{\\mathrm{d}x} = 2 \\cdot \\htmlData{anchor=mult-1}{\\color{@sky}{3}}x^{\\htmlData{anchor=sub-1}{\\color{@sky}{3}} - 1} + 18 \\cdot (\\htmlData{anchor=mult-2}{\\color{@amber}{-1}})x^{\\htmlData{anchor=sub-2}{\\color{@amber}{-1}} - 1} + 0$"},{"latex":"$$\\frac{\\mathrm{d}y}{\\mathrm{d}x} = 6x^2 - 18x^{-2}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"pow-1"},{"color":"amber","style":"text-color","anchor":"pow-2"}],"connections":[{"to":"mult-1","from":"pow-1","color":"sky","label":null,"style":"solid"},{"to":"mult-2","from":"pow-2","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Differentiate each term","description":"Now, we apply the power rule, which states that $\\frac{\\mathrm{d}}{\\mathrm{d}x}(x^n) = n x^{n-1}$, to each term:\n\n- For $2x^3$, we multiply by the power $3$ and reduce the power by $1$: $2(3)x^2 = 6x^2$.\n- For $18x^{-1}$, we multiply by the power $-1$ and reduce the power by $1$: $18(-1)x^{-2} = -18x^{-2}$.\n- The constant term $5$ differentiates to $0$.\n\nThis step earns you the first method mark (**M1**)."},{"type":"text","order":2,"title":"Simplify the final expression","content":"To give the answer in its simplest and most standard form, we convert the negative index $x^{-2}$ back into a fraction using the rule $x^{-a} = \\frac{1}{x^a}$:\n\n$$\\frac{\\mathrm{d}y}{\\mathrm{d}x} = 6x^2 - \\frac{18}{x^2}$$\n\nThis completes the differentiation and earns you the final accuracy mark (**A1**)."}]},{"id":"quotient_rule","label":"Using the Quotient Rule","steps":[{"type":"text","order":0,"title":"Analyze the equation","content":"We need to find the derivative $\\frac{\\mathrm{d}y}{\\mathrm{d}x}$ of the given curve equation:\n\n$$y = 2x^3 + \\frac{18}{x} + 5$$\n\nTo differentiate this equation, we can find the derivative of each term individually:\n1. For $2x^3$, we can use the **Power Rule**.\n2. For $\\frac{18}{x}$, we will apply the **Quotient Rule** directly.\n3. For the constant $+5$, the derivative is simply $0$."},{"type":"text","order":1,"title":"Differentiate first term","content":"First, let's differentiate the term $2x^3$ using the power rule $\\frac{\\mathrm{d}}{\\mathrm{d}x}(x^n) = nx^{n-1}$:\n\n$$\\frac{\\mathrm{d}}{\\mathrm{d}x}(2x^3) = 2 \\cdot 3x^2 = 6x^2$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Let } \\frac{u}{v} = \\frac{\\htmlData{anchor=u-term}{\\color{@sky}{18}}}{\\htmlData{anchor=v-term}{\\color{@amber}{x}}}$"},{"latex":"$$\\text{Here, } \\htmlData{anchor=u-val}{u = 18} \\implies \\htmlData{anchor=du-val}{\\frac{\\mathrm{d}u}{\\mathrm{d}x} = 0} \\quad \\text{and} \\quad \\htmlData{anchor=v-val}{v = x} \\implies \\htmlData{anchor=dv-val}{\\frac{\\mathrm{d}v}{\\mathrm{d}x} = 1}$"},{"latex":"$$\\frac{\\mathrm{d}}{\\mathrm{d}x}\\left(\\frac{u}{v}\\right) = \\frac{\\htmlData{anchor=du-sub}{\\frac{\\mathrm{d}u}{\\mathrm{d}x}}v - u\\htmlData{anchor=dv-sub}{\\frac{\\mathrm{d}v}{\\mathrm{d}x}}}{v^2}$"},{"latex":"$$\\frac{\\mathrm{d}}{\\mathrm{d}x}\\left(\\frac{18}{x}\\right) = \\frac{\\htmlData{anchor=num-calc}{\\color{@sky}{(0)(x) - (18)(1)}}}{\\htmlData{anchor=den-calc}{\\color{@amber}{x^2}}} = -\\frac{18}{x^2}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-term"},{"color":"amber","style":"text-color","anchor":"v-term"}],"connections":[{"to":"u-val","from":"u-term","color":"sky","label":null,"style":"solid"},{"to":"v-val","from":"v-term","color":"amber","label":null,"style":"solid"},{"to":"du-sub","from":"du-val","color":"purple","label":null,"style":"solid"},{"to":"dv-sub","from":"dv-val","color":"indigo","label":null,"style":"solid"}]},"order":2,"title":"Apply the Quotient Rule","description":"Now, we apply the quotient rule directly to differentiate the fraction term $\\frac{18}{x}$."},{"type":"text","order":3,"title":"Combine the terms","content":"Combine the derivatives of each term to find the overall derivative $\\frac{\\mathrm{d}y}{\\mathrm{d}x}$:\n\n$$\\frac{\\mathrm{d}y}{\\mathrm{d}x} = 6x^2 - \\frac{18}{x^2}$$\n\nThis expression is in its simplest simplified form and is equivalent to the alternative representation $6x^2 - 18x^{-2}$."}]}],"generatedAt":"2026-06-09T05:24:06.551Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"7f646ee7-1ffc-4518-bdbb-29c67f841dd3","label":"b","content":"Given that\n- the point \$P\$ lies on \$C\$\n- the line with equation \$y = k - 24x\$, where \$k\$ is a constant, is the tangent to \$C\$ at \$P\$\n\nShow that the \$x\$-coordinate of \$P\$ satisfies the equation\n\\[x^4 + 4x^2 - 3 = 0\\]","markscheme":"- Identifies the gradient of the tangent $y = k - 24x$ is $-24$ B1\n- Sets the derivative equal to the gradient: $6x^2 - \\frac{18}{x^2} = -24$ M1\n- Multiplies by $x^2$ and rearranges to show $6x^4 + 24x^2 - 18 = 0$, which simplifies to $x^4 + 4x^2 - 3 = 0$ A1","marks":3,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"calculus","label":"Using Differentiation","steps":[{"type":"text","order":0,"title":"Find the curve's derivative","content":"First, we find the derivative of the curve's equation to represent its gradient at any point. Let's write the equation of $C$ using index notation:\n\n$$y = 2x^3 + 18x^{-1} + 5$$\n\nDifferentiating with respect to $x$ using the power rule:\n\n$$\\frac{dy}{dx} = 6x^2 - 18x^{-2}$$\n\n$$\\frac{dy}{dx} = 6x^2 - \\frac{18}{x^2}$$"},{"type":"text","order":1,"title":"Determine the tangent's gradient","content":"The equation of the tangent line is given as:\n\n$$y = k - 24x$$\n\nComparing this to the standard linear form $y = mx + c$, we can see that the gradient of this line is constant and equal to $-24$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Curve's Gradient} = \\htmlData{anchor=dy-dx}{\\color{@sky}{6x^2 - \\frac{18}{x^2}}}$"},{"latex":"$$\\text{Tangent's Gradient} = \\htmlData{anchor=m-val}{\\color{@amber}{-24}}$"},{"latex":"$$\\htmlData{anchor=eq}{\\color{@sky}{6x^2 - \\frac{18}{x^2}}} = \\htmlData{anchor=eq-val}{\\color{@amber}{-24}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"dy-dx"},{"color":"amber","style":"text-color","anchor":"m-val"}],"connections":[{"to":"eq","from":"dy-dx","color":"sky","label":"equate","style":"solid"},{"to":"eq-val","from":"m-val","color":"amber","label":null,"style":"solid"}]},"order":2,"title":"Equate the gradients","description":"At the point of tangency $P$, the gradient of the curve must equal the gradient of the tangent line."},{"type":"text","order":3,"title":"Form the quartic equation","content":"To eliminate the fraction, multiply the entire equation by $x^2$ (noting that $x > 0$ so $x \\neq 0$):\n\n$$6x^4 - 18 = -24x^2$$\n\nRearrange all terms to one side to set the equation to zero:\n\n$$6x^4 + 24x^2 - 18 = 0$$"},{"type":"text","order":4,"title":"Simplify to show result","content":"Divide each term in the equation by $6$:\n\n$$x^4 + 4x^2 - 3 = 0$$\n\nThis is the required equation showing that the $x$-coordinate of $P$ satisfies $x^4 + 4x^2 - 3 = 0$."},{"type":"text","order":5,"title":"Solve the quadratic in $x^2$","content":"To find the exact value of $k$, we first need to find the coordinates of $P$. Let's solve the equation $x^4 + 4x^2 - 3 = 0$ by treating it as a quadratic in $x^2$:\n\n$$x^2 = \\frac{-4 \\pm \\sqrt{4^2 - 4(1)(-3)}}{2(1)}$$\n\n$$x^2 = \\frac{-4 \\pm \\sqrt{16 + 12}}{2} = \\frac{-4 \\pm \\sqrt{28}}{2}$$\n\n$$x^2 = \\frac{-4 \\pm 2\\sqrt{7}}{2} = -2 \\pm \\sqrt{7}$$\n\nSince $x^2$ must be positive for any real value of $x$, we reject the negative root $-2 - \\sqrt{7}$. Therefore:\n\n$$x^2 = \\sqrt{7} - 2$$\n\nSince $x > 0$, the $x$-coordinate of $P$ is:\n\n$$x = \\sqrt{\\sqrt{7} - 2}$$"},{"type":"text","order":6,"title":"Find the $y$-coordinate of $P$","content":"Now we substitute our expression for $x$ back into the curve's equation to find the corresponding $y$-coordinate of $P$:\n\n$$y = 2x^3 + \\frac{18}{x} + 5$$\n\nSince $x^2 = \\sqrt{7} - 2$, we can write $2x^3 = 2x(x^2) = 2x(\\sqrt{7}-2)$. Substituting this into the equation:\n\n$$y = 2x(\\sqrt{7}-2) + \\frac{18}{x} + 5$$"},{"type":"text","order":7,"title":"Express $k$ in terms of $x$","content":"The point $P(x, y)$ lies on both the curve and the tangent line $y = k - 24x$. Rearranging the tangent equation for $k$:\n\n$$k = y + 24x$$\n\nSubstitute our expression for $y$ into this equation:\n\n$$k = 2x(\\sqrt{7}-2) + \\frac{18}{x} + 5 + 24x$$\n\n$$k = x(2\\sqrt{7} - 4 + 24) + \\frac{18}{x} + 5$$\n\n$$k = x(2\\sqrt{7} + 20) + \\frac{18}{x} + 5$$"},{"type":"text","order":8,"title":"Calculate the exact value of $k$","content":"Let's simplify the fraction $\\frac{18}{x}$ by rationalising with $x^2 = \\sqrt{7}-2$:\n\n$$\\frac{18}{x} = \\frac{18x}{x^2} = \\frac{18x}{\\sqrt{7}-2}$$\n\nMultiply the numerator and denominator by $(\\sqrt{7}+2)$:\n\n$$\\frac{18x(\\sqrt{7}+2)}{7-4} = \\frac{18x(\\sqrt{7}+2)}{3} = 6x(\\sqrt{7}+2)$$\n\nSubstitute this back into our expression for $k$:\n\n$$k = x(2\\sqrt{7} + 20) + 6x(\\sqrt{7}+2) + 5$$\n\n$$k = x(2\\sqrt{7} + 20 + 6\\sqrt{7} + 12) + 5$$\n\n$$k = x(8\\sqrt{7} + 32) + 5$$\n\nFactor out $8$ from the parenthesis:\n\n$$k = 8(\\sqrt{7} + 4)x + 5$$\n\nFinally, substituting $x = \\sqrt{\\sqrt{7}-2}$ gives the exact value:\n\n$$k = 8(\\sqrt{7} + 4)\\sqrt{\\sqrt{7}-2} + 5$$"}]}],"generatedAt":"2026-06-09T05:25:03.938Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"bf6510a3-20ad-40a7-8339-40d2898acfd3","label":"c","content":"Hence find the exact value of \$k\$.","markscheme":"- Solves the quadratic in $x^2$ using the quadratic formula: $x^2 = \\frac{-4 \\pm \\sqrt{16 - 4(1)(-3)}}{2}$ M1\n- Obtains $x^2 = -2 + \\sqrt{7}$ (rejecting $-2 - \\sqrt{7}$ as $x^2 > 0$) and finds $x = \\sqrt{\\sqrt{7}-2}$ A1\n- Recognises that $k = y + 24x$ and substitutes the expressions for $y$ and $x$ into this equation M1\n- Obtains the exact value $k = 2(\\sqrt{7}-2)^{3/2} + \\frac{18}{\\sqrt{\\sqrt{7}-2}} + 24\\sqrt{\\sqrt{7}-2} + 5$ or simplified equivalent $k = 8(\\sqrt{7}+4)\\sqrt{\\sqrt{7}-2} + 5$ A1","marks":4,"order":2,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"direct_substitution","label":"Direct Substitution","steps":[{"type":"text","order":0,"title":"Solve for $x^2$","content":"From the previous parts of the question, we have a quadratic equation in terms of $x^2$:\n\n$$(x^2)^2 + 4(x^2) - 3 = 0$$\n\nWe solve this using the quadratic formula for $x^2$:\n\n$$x^2 = \\frac{-4 \\pm \\sqrt{4^2 - 4(1)(-3)}}{2(1)}$$\n\n$$x^2 = \\frac{-4 \\pm \\sqrt{16 + 12}}{2} = \\frac{-4 \\pm \\sqrt{28}}{2}$$\n\nSince $\\sqrt{28} = 2\\sqrt{7}$, this simplifies to:\n\n$$x^2 = -2 \\pm \\sqrt{7}$$\n\nSince $x$ is real and $x > 0$, we must have $x^2 > 0$. Because $\\sqrt{7} \\approx 2.65$, the value $-2 - \\sqrt{7}$ is negative and must be rejected.\n\nThus, we have:\n\n$$x^2 = \\sqrt{7} - 2$$"},{"type":"text","order":1,"title":"Find the value of $x$","content":"Taking the positive square root since $x > 0$, we get:\n\n$$x = \\sqrt{\\sqrt{7} - 2}$$"},{"type":"text","order":2,"title":"Express $k$ in terms of $x$","content":"From the definition of $k$, we use the relationship:\n\n$$k = y + 24x$$\n\nSubstituting the equation for the curve $y = 2x^3 + \\frac{18}{x} + 5$ into this expression gives:\n\n$$k = 2x^3 + \\frac{18}{x} + 24x + 5$$"},{"type":"text","order":3,"title":"Directly substitute $x$","content":"Now we substitute our exact value of $x = \\sqrt{\\sqrt{7} - 2}$ directly into the expression for $k$:\n\n$$k = 2(\\sqrt{\\sqrt{7}-2})^3 + \\frac{18}{\\sqrt{\\sqrt{7}-2}} + 24\\sqrt{\\sqrt{7}-2} + 5$$\n\nUsing fractional powers, this is written as:\n\n$$k = 2(\\sqrt{7}-2)^{3/2} + \\frac{18}{\\sqrt{\\sqrt{7}-2}} + 24\\sqrt{\\sqrt{7}-2} + 5$$"},{"type":"text","order":4,"title":"Simplify the expression","content":"$3b"}]},{"id":"algebraic_reduction","label":"Algebraic Reduction","steps":[{"type":"text","order":0,"title":"Solve for x²","content":"First, we need to solve the given polynomial relation $x^4 + 4x^2 - 3 = 0$ for $x^2$. Since this equation is quadratic in terms of $x^2$, we can apply the quadratic formula:\n\n$$x^2 = \\frac{-4 \\pm \\sqrt{4^2 - 4(1)(-3)}}{2(1)}$$\n\nSimplifying inside the square root:\n\n$$x^2 = \\frac{-4 \\pm \\sqrt{16 + 12}}{2} = \\frac{-4 \\pm \\sqrt{28}}{2} = \\frac{-4 \\pm 2\\sqrt{7}}{2} = -2 \\pm \\sqrt{7}$$\n\nSince $x > 0$, we must have $x^2 > 0$. Since $-2 - \\sqrt{7} < 0$, we reject the negative root and obtain:\n\n$$x^2 = \\sqrt{7} - 2$$\n\nThus, the positive value of $x$ is:\n\n$$x = \\sqrt{\\sqrt{7} - 2}$$"},{"type":"text","order":1,"title":"Express k in x","content":"From the relationship given in the problem, we define $k$ by adding $24x$ to the $y$-coordinate of the curve $C$:\n\n$$k = y + 24x$$\n\nSubstituting the equation of the curve $y = 2x^3 + \\frac{18}{x} + 5$, we write $k$ entirely in terms of $x$:\n\n$$k = 2x^3 + \\frac{18}{x} + 24x + 5$$"},{"type":"text","order":2,"title":"Perform algebraic reduction","content":"To simplify the calculation significantly, let's find a common denominator for the variable terms in $k - 5$:\n\n$$k - 5 = \\frac{2x^4 + 24x^2 + 18}{x}$$\n\nWe can use our polynomial relation $x^4 + 4x^2 - 3 = 0$ to reduce the degree of the numerator. Rearranging this relation gives:\n\n$$x^4 = 3 - 4x^2$$\n\nNow, substitute this expression for $x^4$ into the numerator of our equation for $k - 5$:\n\n$$2x^4 + 24x^2 + 18 = 2(3 - 4x^2) + 24x^2 + 18$$\n\n$$= 6 - 8x^2 + 24x^2 + 18$$\n\n$$= 16x^2 + 24$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$k - 5 = \\frac{\\htmlData{anchor=num-reduced}{\\color{@sky}{16x^2 + 24}}}{x}$"},{"latex":"$$k = \\frac{8(\\htmlData{anchor=x2-term}{\\color{@amber}{2x^2 + 3}})}{x} + 5$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"num-reduced"},{"color":"amber","style":"text-color","anchor":"x2-term"}],"connections":[{"to":"x2-term","from":"num-reduced","color":"sky","label":"Factor out 8","style":"solid"}]},"order":3,"title":"Simplify k expression","description":"Substitute the reduced numerator back into our expression for $k - 5$ and factor out the common constant."},{"type":"text","order":4,"title":"Substitute the values","content":"Now we substitute the values of $x^2 = \\sqrt{7} - 2$ and $x = \\sqrt{\\sqrt{7} - 2}$ into our simplified expression:\n\n$$2x^2 + 3 = 2(\\sqrt{7} - 2) + 3 = 2\\sqrt{7} - 4 + 3 = 2\\sqrt{7} - 1$$\n\nSubstituting this back into $k$:\n\n$$k = \\frac{8(2\\sqrt{7} - 1)}{\\sqrt{\\sqrt{7} - 2}} + 5$$\n\nTo rationalize the denominator, we can multiply the numerator and denominator by $\\sqrt{\\sqrt{7} + 2}$:\n\n$$k = \\frac{8(2\\sqrt{7} - 1)\\sqrt{\\sqrt{7} + 2}}{\\sqrt{(\\sqrt{7} - 2)(\\sqrt{7} + 2)}} + 5$$\n\nSince $(\\sqrt{7} - 2)(\\sqrt{7} + 2) = 7 - 4 = 3$, the denominator becomes $\\sqrt{3}$:\n\n$$k = \\frac{8(2\\sqrt{7} - 1)\\sqrt{\\sqrt{7} + 2}}{\\sqrt{3}} + 5$$"},{"type":"text","order":5,"title":"Alternative exact form","content":"Alternatively, we can express the final answer in terms of $x = \\sqrt{\\sqrt{7}-2}$ using the equivalence $(\\sqrt{7} - 2)^{-1/2} = \\frac{\\sqrt{\\sqrt{7}+2}}{\\sqrt{3}}$. Both representations are mathematically correct.\n\nThus, the exact value of $k$ is:\n\n$$k = 8(\\sqrt{7}+4)\\sqrt{\\sqrt{7}-2} + 5$$"}]}],"generatedAt":"2026-06-09T05:34:10.801Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"c0d117e6-790c-43eb-8860-c6e8fb68919f:power_rule","label":"(a) Power Rule with Negative Indices","steps":[{"type":"text","order":0,"title":"Rewrite using negative indices","content":"To differentiate the equation term-by-term using the standard power rule, we first need to express the fraction $\\frac{18}{x}$ in index form using a negative exponent.\n\nRecall that $\\frac{1}{x} = x^{-1}$. Therefore, we can rewrite the equation of the curve as:\n\n$$y = 2x^3 + 18x^{-1} + 5$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = 2x^{\\htmlData{anchor=pow-1}{\\color{@sky}{3}}} + 18x^{\\htmlData{anchor=pow-2}{\\color{@amber}{-1}}} + 5$"},{"latex":"$$\\frac{\\mathrm{d}y}{\\mathrm{d}x} = 2 \\cdot \\htmlData{anchor=mult-1}{\\color{@sky}{3}}x^{\\htmlData{anchor=sub-1}{\\color{@sky}{3}} - 1} + 18 \\cdot (\\htmlData{anchor=mult-2}{\\color{@amber}{-1}})x^{\\htmlData{anchor=sub-2}{\\color{@amber}{-1}} - 1} + 0$"},{"latex":"$$\\frac{\\mathrm{d}y}{\\mathrm{d}x} = 6x^2 - 18x^{-2}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"pow-1"},{"color":"amber","style":"text-color","anchor":"pow-2"}],"connections":[{"to":"mult-1","from":"pow-1","color":"sky","label":null,"style":"solid"},{"to":"mult-2","from":"pow-2","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Differentiate each term","description":"Now, we apply the power rule, which states that $\\frac{\\mathrm{d}}{\\mathrm{d}x}(x^n) = n x^{n-1}$, to each term:\n\n- For $2x^3$, we multiply by the power $3$ and reduce the power by $1$: $2(3)x^2 = 6x^2$.\n- For $18x^{-1}$, we multiply by the power $-1$ and reduce the power by $1$: $18(-1)x^{-2} = -18x^{-2}$.\n- The constant term $5$ differentiates to $0$.\n\nThis step earns you the first method mark (**M1**)."},{"type":"text","order":2,"title":"Simplify the final expression","content":"To give the answer in its simplest and most standard form, we convert the negative index $x^{-2}$ back into a fraction using the rule $x^{-a} = \\frac{1}{x^a}$:\n\n$$\\frac{\\mathrm{d}y}{\\mathrm{d}x} = 6x^2 - \\frac{18}{x^2}$$\n\nThis completes the differentiation and earns you the final accuracy mark (**A1**)."}]},{"id":"c0d117e6-790c-43eb-8860-c6e8fb68919f:quotient_rule","label":"(a) Using the Quotient Rule","steps":[{"type":"text","order":0,"title":"Analyze the equation","content":"We need to find the derivative $\\frac{\\mathrm{d}y}{\\mathrm{d}x}$ of the given curve equation:\n\n$$y = 2x^3 + \\frac{18}{x} + 5$$\n\nTo differentiate this equation, we can find the derivative of each term individually:\n1. For $2x^3$, we can use the **Power Rule**.\n2. For $\\frac{18}{x}$, we will apply the **Quotient Rule** directly.\n3. For the constant $+5$, the derivative is simply $0$."},{"type":"text","order":1,"title":"Differentiate first term","content":"First, let's differentiate the term $2x^3$ using the power rule $\\frac{\\mathrm{d}}{\\mathrm{d}x}(x^n) = nx^{n-1}$:\n\n$$\\frac{\\mathrm{d}}{\\mathrm{d}x}(2x^3) = 2 \\cdot 3x^2 = 6x^2$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Let } \\frac{u}{v} = \\frac{\\htmlData{anchor=u-term}{\\color{@sky}{18}}}{\\htmlData{anchor=v-term}{\\color{@amber}{x}}}$"},{"latex":"$$\\text{Here, } \\htmlData{anchor=u-val}{u = 18} \\implies \\htmlData{anchor=du-val}{\\frac{\\mathrm{d}u}{\\mathrm{d}x} = 0} \\quad \\text{and} \\quad \\htmlData{anchor=v-val}{v = x} \\implies \\htmlData{anchor=dv-val}{\\frac{\\mathrm{d}v}{\\mathrm{d}x} = 1}$"},{"latex":"$$\\frac{\\mathrm{d}}{\\mathrm{d}x}\\left(\\frac{u}{v}\\right) = \\frac{\\htmlData{anchor=du-sub}{\\frac{\\mathrm{d}u}{\\mathrm{d}x}}v - u\\htmlData{anchor=dv-sub}{\\frac{\\mathrm{d}v}{\\mathrm{d}x}}}{v^2}$"},{"latex":"$$\\frac{\\mathrm{d}}{\\mathrm{d}x}\\left(\\frac{18}{x}\\right) = \\frac{\\htmlData{anchor=num-calc}{\\color{@sky}{(0)(x) - (18)(1)}}}{\\htmlData{anchor=den-calc}{\\color{@amber}{x^2}}} = -\\frac{18}{x^2}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-term"},{"color":"amber","style":"text-color","anchor":"v-term"}],"connections":[{"to":"u-val","from":"u-term","color":"sky","label":null,"style":"solid"},{"to":"v-val","from":"v-term","color":"amber","label":null,"style":"solid"},{"to":"du-sub","from":"du-val","color":"purple","label":null,"style":"solid"},{"to":"dv-sub","from":"dv-val","color":"indigo","label":null,"style":"solid"}]},"order":2,"title":"Apply the Quotient Rule","description":"Now, we apply the quotient rule directly to differentiate the fraction term $\\frac{18}{x}$."},{"type":"text","order":3,"title":"Combine the terms","content":"Combine the derivatives of each term to find the overall derivative $\\frac{\\mathrm{d}y}{\\mathrm{d}x}$:\n\n$$\\frac{\\mathrm{d}y}{\\mathrm{d}x} = 6x^2 - \\frac{18}{x^2}$$\n\nThis expression is in its simplest simplified form and is equivalent to the alternative representation $6x^2 - 18x^{-2}$."}]},{"id":"7f646ee7-1ffc-4518-bdbb-29c67f841dd3:calculus","label":"(b) Using Differentiation","steps":[{"type":"text","order":0,"title":"Find the curve's derivative","content":"First, we find the derivative of the curve's equation to represent its gradient at any point. Let's write the equation of $C$ using index notation:\n\n$$y = 2x^3 + 18x^{-1} + 5$$\n\nDifferentiating with respect to $x$ using the power rule:\n\n$$\\frac{dy}{dx} = 6x^2 - 18x^{-2}$$\n\n$$\\frac{dy}{dx} = 6x^2 - \\frac{18}{x^2}$$"},{"type":"text","order":1,"title":"Determine the tangent's gradient","content":"The equation of the tangent line is given as:\n\n$$y = k - 24x$$\n\nComparing this to the standard linear form $y = mx + c$, we can see that the gradient of this line is constant and equal to $-24$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Curve's Gradient} = \\htmlData{anchor=dy-dx}{\\color{@sky}{6x^2 - \\frac{18}{x^2}}}$"},{"latex":"$$\\text{Tangent's Gradient} = \\htmlData{anchor=m-val}{\\color{@amber}{-24}}$"},{"latex":"$$\\htmlData{anchor=eq}{\\color{@sky}{6x^2 - \\frac{18}{x^2}}} = \\htmlData{anchor=eq-val}{\\color{@amber}{-24}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"dy-dx"},{"color":"amber","style":"text-color","anchor":"m-val"}],"connections":[{"to":"eq","from":"dy-dx","color":"sky","label":"equate","style":"solid"},{"to":"eq-val","from":"m-val","color":"amber","label":null,"style":"solid"}]},"order":2,"title":"Equate the gradients","description":"At the point of tangency $P$, the gradient of the curve must equal the gradient of the tangent line."},{"type":"text","order":3,"title":"Form the quartic equation","content":"To eliminate the fraction, multiply the entire equation by $x^2$ (noting that $x > 0$ so $x \\neq 0$):\n\n$$6x^4 - 18 = -24x^2$$\n\nRearrange all terms to one side to set the equation to zero:\n\n$$6x^4 + 24x^2 - 18 = 0$$"},{"type":"text","order":4,"title":"Simplify to show result","content":"Divide each term in the equation by $6$:\n\n$$x^4 + 4x^2 - 3 = 0$$\n\nThis is the required equation showing that the $x$-coordinate of $P$ satisfies $x^4 + 4x^2 - 3 = 0$."},{"type":"text","order":5,"title":"Solve the quadratic in $x^2$","content":"To find the exact value of $k$, we first need to find the coordinates of $P$. Let's solve the equation $x^4 + 4x^2 - 3 = 0$ by treating it as a quadratic in $x^2$:\n\n$$x^2 = \\frac{-4 \\pm \\sqrt{4^2 - 4(1)(-3)}}{2(1)}$$\n\n$$x^2 = \\frac{-4 \\pm \\sqrt{16 + 12}}{2} = \\frac{-4 \\pm \\sqrt{28}}{2}$$\n\n$$x^2 = \\frac{-4 \\pm 2\\sqrt{7}}{2} = -2 \\pm \\sqrt{7}$$\n\nSince $x^2$ must be positive for any real value of $x$, we reject the negative root $-2 - \\sqrt{7}$. Therefore:\n\n$$x^2 = \\sqrt{7} - 2$$\n\nSince $x > 0$, the $x$-coordinate of $P$ is:\n\n$$x = \\sqrt{\\sqrt{7} - 2}$$"},{"type":"text","order":6,"title":"Find the $y$-coordinate of $P$","content":"Now we substitute our expression for $x$ back into the curve's equation to find the corresponding $y$-coordinate of $P$:\n\n$$y = 2x^3 + \\frac{18}{x} + 5$$\n\nSince $x^2 = \\sqrt{7} - 2$, we can write $2x^3 = 2x(x^2) = 2x(\\sqrt{7}-2)$. Substituting this into the equation:\n\n$$y = 2x(\\sqrt{7}-2) + \\frac{18}{x} + 5$$"},{"type":"text","order":7,"title":"Express $k$ in terms of $x$","content":"The point $P(x, y)$ lies on both the curve and the tangent line $y = k - 24x$. Rearranging the tangent equation for $k$:\n\n$$k = y + 24x$$\n\nSubstitute our expression for $y$ into this equation:\n\n$$k = 2x(\\sqrt{7}-2) + \\frac{18}{x} + 5 + 24x$$\n\n$$k = x(2\\sqrt{7} - 4 + 24) + \\frac{18}{x} + 5$$\n\n$$k = x(2\\sqrt{7} + 20) + \\frac{18}{x} + 5$$"},{"type":"text","order":8,"title":"Calculate the exact value of $k$","content":"Let's simplify the fraction $\\frac{18}{x}$ by rationalising with $x^2 = \\sqrt{7}-2$:\n\n$$\\frac{18}{x} = \\frac{18x}{x^2} = \\frac{18x}{\\sqrt{7}-2}$$\n\nMultiply the numerator and denominator by $(\\sqrt{7}+2)$:\n\n$$\\frac{18x(\\sqrt{7}+2)}{7-4} = \\frac{18x(\\sqrt{7}+2)}{3} = 6x(\\sqrt{7}+2)$$\n\nSubstitute this back into our expression for $k$:\n\n$$k = x(2\\sqrt{7} + 20) + 6x(\\sqrt{7}+2) + 5$$\n\n$$k = x(2\\sqrt{7} + 20 + 6\\sqrt{7} + 12) + 5$$\n\n$$k = x(8\\sqrt{7} + 32) + 5$$\n\nFactor out $8$ from the parenthesis:\n\n$$k = 8(\\sqrt{7} + 4)x + 5$$\n\nFinally, substituting $x = \\sqrt{\\sqrt{7}-2}$ gives the exact value:\n\n$$k = 8(\\sqrt{7} + 4)\\sqrt{\\sqrt{7}-2} + 5$$"}]},{"id":"bf6510a3-20ad-40a7-8339-40d2898acfd3:direct_substitution","label":"(c) Direct Substitution","steps":[{"type":"text","order":0,"title":"Solve for $x^2$","content":"From the previous parts of the question, we have a quadratic equation in terms of $x^2$:\n\n$$(x^2)^2 + 4(x^2) - 3 = 0$$\n\nWe solve this using the quadratic formula for $x^2$:\n\n$$x^2 = \\frac{-4 \\pm \\sqrt{4^2 - 4(1)(-3)}}{2(1)}$$\n\n$$x^2 = \\frac{-4 \\pm \\sqrt{16 + 12}}{2} = \\frac{-4 \\pm \\sqrt{28}}{2}$$\n\nSince $\\sqrt{28} = 2\\sqrt{7}$, this simplifies to:\n\n$$x^2 = -2 \\pm \\sqrt{7}$$\n\nSince $x$ is real and $x > 0$, we must have $x^2 > 0$. Because $\\sqrt{7} \\approx 2.65$, the value $-2 - \\sqrt{7}$ is negative and must be rejected.\n\nThus, we have:\n\n$$x^2 = \\sqrt{7} - 2$$"},{"type":"text","order":1,"title":"Find the value of $x$","content":"Taking the positive square root since $x > 0$, we get:\n\n$$x = \\sqrt{\\sqrt{7} - 2}$$"},{"type":"text","order":2,"title":"Express $k$ in terms of $x$","content":"From the definition of $k$, we use the relationship:\n\n$$k = y + 24x$$\n\nSubstituting the equation for the curve $y = 2x^3 + \\frac{18}{x} + 5$ into this expression gives:\n\n$$k = 2x^3 + \\frac{18}{x} + 24x + 5$$"},{"type":"text","order":3,"title":"Directly substitute $x$","content":"Now we substitute our exact value of $x = \\sqrt{\\sqrt{7} - 2}$ directly into the expression for $k$:\n\n$$k = 2(\\sqrt{\\sqrt{7}-2})^3 + \\frac{18}{\\sqrt{\\sqrt{7}-2}} + 24\\sqrt{\\sqrt{7}-2} + 5$$\n\nUsing fractional powers, this is written as:\n\n$$k = 2(\\sqrt{7}-2)^{3/2} + \\frac{18}{\\sqrt{\\sqrt{7}-2}} + 24\\sqrt{\\sqrt{7}-2} + 5$$"},{"type":"text","order":4,"title":"Simplify the expression","content":"$3c"}]},{"id":"bf6510a3-20ad-40a7-8339-40d2898acfd3:algebraic_reduction","label":"(c) Algebraic Reduction","steps":[{"type":"text","order":0,"title":"Solve for x²","content":"First, we need to solve the given polynomial relation $x^4 + 4x^2 - 3 = 0$ for $x^2$. Since this equation is quadratic in terms of $x^2$, we can apply the quadratic formula:\n\n$$x^2 = \\frac{-4 \\pm \\sqrt{4^2 - 4(1)(-3)}}{2(1)}$$\n\nSimplifying inside the square root:\n\n$$x^2 = \\frac{-4 \\pm \\sqrt{16 + 12}}{2} = \\frac{-4 \\pm \\sqrt{28}}{2} = \\frac{-4 \\pm 2\\sqrt{7}}{2} = -2 \\pm \\sqrt{7}$$\n\nSince $x > 0$, we must have $x^2 > 0$. Since $-2 - \\sqrt{7} < 0$, we reject the negative root and obtain:\n\n$$x^2 = \\sqrt{7} - 2$$\n\nThus, the positive value of $x$ is:\n\n$$x = \\sqrt{\\sqrt{7} - 2}$$"},{"type":"text","order":1,"title":"Express k in x","content":"From the relationship given in the problem, we define $k$ by adding $24x$ to the $y$-coordinate of the curve $C$:\n\n$$k = y + 24x$$\n\nSubstituting the equation of the curve $y = 2x^3 + \\frac{18}{x} + 5$, we write $k$ entirely in terms of $x$:\n\n$$k = 2x^3 + \\frac{18}{x} + 24x + 5$$"},{"type":"text","order":2,"title":"Perform algebraic reduction","content":"To simplify the calculation significantly, let's find a common denominator for the variable terms in $k - 5$:\n\n$$k - 5 = \\frac{2x^4 + 24x^2 + 18}{x}$$\n\nWe can use our polynomial relation $x^4 + 4x^2 - 3 = 0$ to reduce the degree of the numerator. Rearranging this relation gives:\n\n$$x^4 = 3 - 4x^2$$\n\nNow, substitute this expression for $x^4$ into the numerator of our equation for $k - 5$:\n\n$$2x^4 + 24x^2 + 18 = 2(3 - 4x^2) + 24x^2 + 18$$\n\n$$= 6 - 8x^2 + 24x^2 + 18$$\n\n$$= 16x^2 + 24$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$k - 5 = \\frac{\\htmlData{anchor=num-reduced}{\\color{@sky}{16x^2 + 24}}}{x}$"},{"latex":"$$k = \\frac{8(\\htmlData{anchor=x2-term}{\\color{@amber}{2x^2 + 3}})}{x} + 5$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"num-reduced"},{"color":"amber","style":"text-color","anchor":"x2-term"}],"connections":[{"to":"x2-term","from":"num-reduced","color":"sky","label":"Factor out 8","style":"solid"}]},"order":3,"title":"Simplify k expression","description":"Substitute the reduced numerator back into our expression for $k - 5$ and factor out the common constant."},{"type":"text","order":4,"title":"Substitute the values","content":"Now we substitute the values of $x^2 = \\sqrt{7} - 2$ and $x = \\sqrt{\\sqrt{7} - 2}$ into our simplified expression:\n\n$$2x^2 + 3 = 2(\\sqrt{7} - 2) + 3 = 2\\sqrt{7} - 4 + 3 = 2\\sqrt{7} - 1$$\n\nSubstituting this back into $k$:\n\n$$k = \\frac{8(2\\sqrt{7} - 1)}{\\sqrt{\\sqrt{7} - 2}} + 5$$\n\nTo rationalize the denominator, we can multiply the numerator and denominator by $\\sqrt{\\sqrt{7} + 2}$:\n\n$$k = \\frac{8(2\\sqrt{7} - 1)\\sqrt{\\sqrt{7} + 2}}{\\sqrt{(\\sqrt{7} - 2)(\\sqrt{7} + 2)}} + 5$$\n\nSince $(\\sqrt{7} - 2)(\\sqrt{7} + 2) = 7 - 4 = 3$, the denominator becomes $\\sqrt{3}$:\n\n$$k = \\frac{8(2\\sqrt{7} - 1)\\sqrt{\\sqrt{7} + 2}}{\\sqrt{3}} + 5$$"},{"type":"text","order":5,"title":"Alternative exact form","content":"Alternatively, we can express the final answer in terms of $x = \\sqrt{\\sqrt{7}-2}$ using the equivalence $(\\sqrt{7} - 2)^{-1/2} = \\frac{\\sqrt{\\sqrt{7}+2}}{\\sqrt{3}}$. Both representations are mathematically correct.\n\nThus, the exact value of $k$ is:\n\n$$k = 8(\\sqrt{7}+4)\\sqrt{\\sqrt{7}-2} + 5$$"}]}],"generatedAt":"2026-06-09T05:34:29.459Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths","summaries":[{"id":"0f02016a-b3fe-4bb8-a3d8-46426609ae5a","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"1897fb1a-feca-4226-b37c-d34dc8dba7f3","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"1eb6b943-4be5-461b-a944-fbcaa6e7688a","difficultyBand":"medium","calculatorUse":null,"totalMarks":13,"questionType":"LA","isCalculation":true},{"id":"2206b2d5-983d-4e36-a279-e4872f684e17","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"3a75af72-7480-42d5-bae1-d9e615ac3f18","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"4d2a0a7c-e5b1-4b00-8ca5-269148915639","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"586574aa-4a9c-497d-bc36-11f45cf4ba6b","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"6bd4d4b0-e670-4aa5-818e-85382ca84d2e","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"84836652-f175-4bf3-90b4-afbac37d5ab8","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"85c13032-2675-479b-a892-7bfbb2007bc4","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"9f56b6ac-56a2-440a-a3f4-1df3a47e5f65","difficultyBand":"medium","calculatorUse":null,"totalMarks":13,"questionType":"LA","isCalculation":true},{"id":"a0094748-7018-47e2-993b-4fe39a19ae04","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"a616a3c5-c03e-42e4-821b-e50b1ff8051b","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"d9f7ed48-e7ce-47f8-8af9-943dcc528fdc","difficultyBand":"medium","calculatorUse":null,"totalMarks":13,"questionType":"LA","isCalculation":true},{"id":"eb9a3297-879b-43de-88d5-7de1fdba621a","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"f9ce4a52-4b0d-49fa-a5dd-8e5bb1355e95","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true}],"title":"1.2 Algebraic Fractions","topicId":"141a09f3-b92b-4a52-91f4-36d5bb7f7ab0","topicSidebar":"$L3d","topicSlug":"a-level-pure-mathematics-1-2-algebraic-fractions","topicTitle":"1.2 Algebraic Fractions"}]}],"$L3e","$L3f"]}]