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Composite functions (A-level only) - Questions\",\"description\":\"Practise 1.3.4 Composite functions (A-level only) A Level Maths exam-style questions with worked solutions on MathsGenie.\",\"learningResourceType\":\"Practice questions\",\"educationalLevel\":\"A Level\",\"about\":{\"@type\":\"Thing\",\"name\":\"1.3.4 Composite functions (A-level only)\"},\"url\":\"https://mathsgenie.co.uk/a-level/maths/ocr/mei/composite-functions/questions/q7\",\"inLanguage\":\"en-GB\",\"provider\":{\"@id\":\"https://mathsgenie.co.uk/#organization\"}}"}}],["$","div",null,{"className":"min-h-0 w-full flex-1 overflow-hidden bg-background","children":["$","$L3b",null,{"backHref":"/a-level/maths/ocr/mei/questions","basePath":"/a-level/maths/ocr/mei/composite-functions/questions","formulaUrl":null,"initialIndex":6,"initialQuestionWindow":[{"id":"2e72a13d-0133-49dd-bb77-e5bdec80bf1b","revisionId":"6d13fda7-5e5b-40ec-af27-f0b50aef3e2e","questionNumber":null,"questionType":"LA","specification":"The functions $f$ and $g$ are defined by\n\n$$f: x \\rightarrow e^{2x+1}, \\quad x \\in \\mathbb{R}$$\n\n$$g: x \\rightarrow 2x + 3\\ln x, \\quad x \\in \\mathbb{R}, x > 0$$","difficulty":1,"difficultyBand":"medium","tier":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"663c36e5-9582-49e5-891d-073583f53e7f","label":"a","content":"Write down the range of $f$.","markscheme":"- $f(x) > 0$ B1","marks":1,"order":0,"markingRubric":{"totalMarks":1,"points":[{"id":"p1","marks":1,"description":"Correct range for exponential function.","code":"B"}],"finalAnswers":["f(x) > 0"],"generatedAt":"2026-05-06T14:05:07.865Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"analytical","label":"Using Exponential Properties","steps":[{"type":"text","order":0,"title":"Understand the Goal","content":"The question asks us to find the **range** of the function $f(x) = e^{2x+1}$ for the domain $x \\in \\mathbb{R}$.\n\nThe range represents all the possible output values (or $y$-values) that the function can produce."},{"type":"hyper_latex","block":{"nodes":[{"text":"For any real exponent $u$, the exponential function is always strictly positive:"},{"latex":"$$\\htmlData{anchor=exp-prop}{\\color{@sky}{e^u}} > \\htmlData{anchor=zero-limit}{\\color{@purple}{0}}$"},{"text":"Since $x \\in \\mathbb{R}$, the exponent can take any real value:"},{"latex":"$$\\htmlData{anchor=input}{\\color{@sky}{2x+1}} \\in \\mathbb{R}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"exp-prop"},{"color":"purple","style":"box","anchor":"zero-limit"}],"connections":[{"to":"input","from":"exp-prop","color":"sky","label":"same base e structure","style":"dashed"}]},"order":1,"title":"Analyze Exponential Properties","description":"We can determine the range by looking at the core behavior of the natural exponential function."},{"type":"text","order":2,"title":"State the Range","content":"Because the exponent $2x+1$ can take any real value from $-\\infty$ to $+\\infty$, the function $f(x) = e^{2x+1}$ behaves exactly like the standard exponential curve $y = e^u$.\n\nAs $x \\to -\\infty$, the exponent $(2x+1) \\to -\\infty$, which means $e^{2x+1} \\to 0$ (from above). As $x \\to +\\infty$, $e^{2x+1} \\to +\\infty$.\n\nTherefore, the function values are always strictly greater than $0$. We write this as:\n\n$$f(x) > 0$$"}]},{"id":"graphical","label":"Graphical Analysis","steps":[{"type":"text","order":0,"title":"Recall the base graph","content":"To find the range of $f(x) = e^{2x+1}$, let's analyze its graph. We start with the basic exponential function:\n\n$$y = e^x$$\n\nThe graph of $y = e^x$ is always positive, meaning it lies entirely above the $x$-axis. It has a horizontal asymptote at $y = 0$ as $x \\to -\\infty$."},{"type":"text","order":1,"title":"Analyze the transformations","content":"The function $f(x) = e^{2x+1}$ is a transformed version of $y = e^x$:\n\n* The $+1$ in the exponent corresponds to a horizontal translation of $1$ unit to the left.\n* The coefficient of $2$ in front of $x$ corresponds to a horizontal stretch by a scale factor of $\\frac{1}{2}$ parallel to the $x$-axis.\n\nNeither of these horizontal transformations affects the vertical position of the graph. There is no vertical shift or vertical scaling, so the horizontal asymptote remains at $y = 0$ and the graph is still restricted to the region above the $x$-axis."},{"type":"text","order":2,"title":"Determine the range","content":"Since the graph of $y = e^{2x+1}$ lies entirely above the $x$-axis and never touches or goes below it, the range of the function is all values greater than $0$.\n\nTherefore, the range of $f$ is:\n\n$$f(x) > 0$$","calculator":[{"gdc":{"expression":"e^(2*x+1)","instructions":"Go to the Graphing menu, enter the function e^(2x+1), and draw the graph to visually confirm the range."},"ti84":{"expression":"e^(2*X+1)","instructions":"Press [Y=], input Y1 = e^(2X+1) using [2nd] [LN] for the exponential function, then press [GRAPH] to view the curve."},"desmos":{"expression":"y = e^{2x+1}","instructions":"Type the equation into the input bar to see the exponential curve and its asymptote at the x-axis."},"description":"You can verify this by graphing the function on your calculator to see its shape and horizontal asymptote."}]}]}],"generatedAt":"2026-06-09T04:15:30.824Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"b1daeab3-5c0a-46ce-824b-28471b2de7ed","label":"b","content":"Find an expression for the composite function $gf$.","markscheme":"- $gf(x) = 2(e^{2x+1}) + 3\\ln(e^{2x+1})$ M1\n- $gf(x) = 2e^{2x+1} + 6x + 3$ A1","marks":2,"order":1,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Full substitution of $f$ into $g$.","code":"M"},{"id":"p2","marks":1,"description":"Correct simplification using $\\ln(e^u) = u$.","code":"A"}],"finalAnswers":["gf(x) = 2e^{2x+1} + 6x + 3"],"generatedAt":"2026-05-06T14:05:07.865Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"algebraic_substitution","label":"Direct Algebraic Substitution","steps":[{"type":"text","order":0,"title":"Understand the composite function","content":"To find an expression for the composite function $gf(x)$, we need to evaluate $g(f(x))$. This means we substitute the entire function $f(x)$ in place of $x$ in the function $g(x)$:\n\n$$gf(x) = g(f(x))$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f(x) = \\htmlData{anchor=f-expr}{\\color{@sky}{e^{2x+1}}}$"},{"latex":"$$g(x) = 2\\htmlData{anchor=x-var1}{x} + 3\\ln(\\htmlData{anchor=x-var2}{x})$"},{"latex":"$$gf(x) = 2(\\htmlData{anchor=sub1}{\\color{@sky}{e^{2x+1}}}) + 3\\ln(\\htmlData{anchor=sub2}{\\color{@sky}{e^{2x+1}}})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"f-expr"},{"color":"sky","style":"text-color","anchor":"sub1"},{"color":"sky","style":"text-color","anchor":"sub2"}],"connections":[{"to":"sub1","from":"f-expr","color":"sky","label":"substitute","style":"solid"},{"to":"sub2","from":"f-expr","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute f(x) into g(x)","description":"We replace each occurrence of $x$ in $g(x)$ with the expression for $f(x)$."},{"type":"text","order":2,"title":"Simplify using logarithm laws","content":"Now, we simplify the logarithmic term using the identity $\\ln(e^k) = k$ because the natural logarithm and the exponential function are inverses of each other:\n\n$$\\ln(e^{2x+1}) = 2x + 1$$\n\nSubstituting this back into our expression for $gf(x)$ gives:\n\n$$gf(x) = 2e^{2x+1} + 3(2x + 1)$$"},{"type":"text","order":3,"title":"Expand and write final answer","content":"Expand the brackets to obtain the final simplified expression:\n\n$$gf(x) = 2e^{2x+1} + 3(2x) + 3(1)$$\n\n$$gf(x) = 2e^{2x+1} + 6x + 3$$\n\nThis fully simplified expression matches the expectation for the final accuracy mark."}]}],"generatedAt":"2026-06-09T04:16:47.522Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"bcec4874-6988-4495-ad1e-62a016fc3907","label":"c","content":"Write down the range of $gf$.","markscheme":"- $gf(x) \\in \\mathbb{R}$ B1","marks":1,"order":2,"markingRubric":{"totalMarks":1,"points":[{"id":"p1","marks":1,"description":"Identifies that range is the set of real numbers.","code":"B"}],"finalAnswers":["gf(x) \\in \\mathbb{R}"],"generatedAt":"2026-05-06T14:05:07.865Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"composite_algebraic","label":"Algebraic Composition and Limits","steps":[{"type":"text","order":0,"title":"Define the composite function","content":"To find the range of the composite function $gf(x)$, we first form the algebraic expression for $gf(x) = g(f(x))$. \n\nWe substitute the function $f(x) = e^{2x+1}$ into the function $g(x) = 2x + 3\\ln x$:\n\n$$gf(x) = 2(e^{2x+1}) + 3\\ln(e^{2x+1})$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$gf(x) = 2e^{2x+1} + 3\\htmlData{anchor=ln-term}{\\color{@sky}{\\ln\\left(e^{2x+1}\\right)}}$"},{"latex":"$$gf(x) = 2e^{2x+1} + 3\\htmlData{anchor=simplified-term}{\\color{@sky}{(2x + 1)}}$"},{"latex":"$$gf(x) = 2e^{2x+1} + 6x + 3$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"ln-term"},{"color":"sky","style":"text-color","anchor":"simplified-term"}],"connections":[{"to":"simplified-term","from":"ln-term","color":"sky","label":"using ln(e^A) = A","style":"solid"}]},"order":1,"title":"Simplify the expression","description":"We can simplify the natural logarithm term because the logarithmic and exponential functions are inverses of each other, i.e., $\\ln(e^A) = A$."},{"type":"text","order":2,"title":"Analyze the positive limit","content":"Now we determine the range by investigating the behavior of $gf(x)$ as $x$ approaches positive and negative infinity.\n\nFirst, let's look at the limit as $x \\to \\infty$:\n\n$$\\lim_{x \\to \\infty} gf(x) = \\lim_{x \\to \\infty} \\left( 2e^{2x+1} + 6x + 3 \\right)$$\n\nAs $x \\to \\infty$, both $2e^{2x+1} \\to \\infty$ and $6x + 3 \\to \\infty$. Therefore:\n\n$$gf(x) \\to \\infty$$"},{"type":"text","order":3,"title":"Analyze the negative limit","content":"Next, let's examine the limit as $x \\to -\\infty$:\n\n$$\\lim_{x \\to -\\infty} gf(x) = \\lim_{x \\to -\\infty} \\left( 2e^{2x+1} + 6x + 3 \\right)$$\n\nAs $x \\to -\\infty$, the exponential term decays to zero:\n\n$$2e^{2x+1} \\to 0$$\n\nHowever, the linear term decreases without bound:\n\n$$6x + 3 \\to -\\infty$$\n\nCombining these, we get:\n\n$$gf(x) \\to -\\infty$$"},{"type":"text","order":4,"title":"State the range","content":"Since $gf(x) = 2e^{2x+1} + 6x + 3$ is a continuous function defined for all real numbers $x \\in \\mathbb{R}$, and it takes values from $-\\infty$ to $\\infty$, its range is the set of all real numbers.\n\nTherefore, the range is:\n\n$$gf(x) \\in \\mathbb{R}$$"}]},{"id":"range_mapping","label":"Domain and Range Mapping","steps":[{"type":"text","order":0,"title":"Identify the inner function","content":"To find the range of the composite function $gf(x)$, we can map the range of the inner function $f(x)$ directly into the outer function $g(y)$.\n\nThe inner function is:\n$$f(x) = e^{2x+1}, \\quad x \\in \\mathbb{R}$$"},{"type":"text","order":1,"title":"Determine the range of f(x)","content":"Since $x \\in \\mathbb{R}$, the exponent $2x+1$ can take any real value, from $-\\infty$ to $\\infty$. \n\nWe know that exponential functions of the form $e^u$ are always strictly positive for any real power $u$. Therefore, as $x \\to -\\infty$, $f(x) \\to 0$, and as $x \\to \\infty$, $f(x) \\to \\infty$.\n\nThus, the range of the inner function $f(x)$ is:\n$$(0, \\infty)$$\n\nThis range now becomes the domain of our outer function $g(y)$, where we let $y = f(x) > 0$."},{"type":"text","order":2,"title":"Analyze the outer function","content":"Now we analyze the behavior of the outer function $g(y)$ over its domain $y \\in (0, \\infty)$:\n$$g(y) = 2y + 3\\ln y$$\n\nSince both $2y$ and $3\\ln y$ are continuous and strictly increasing functions for all $y > 0$, their sum $g(y)$ is also continuous and strictly increasing over $(0, \\infty)$.\n\nTo find the range of $g(y)$, let's evaluate the limits at the boundaries of its domain:\n\n* **As $y \\to 0^+$:**\n $$2y \\to 0 \\quad \\text{and} \\quad 3\\ln y \\to -\\infty \\implies g(y) \\to -\\infty$$\n\n* **As $y \\to \\infty$:**\n $$2y \\to \\infty \\quad \\text{and} \\quad 3\\ln y \\to \\infty \\implies g(y) \\to \\infty$$"},{"type":"text","order":3,"title":"State the final range","content":"Because $g(y)$ is continuous and increases monotonically from $-\\infty$ to $\\infty$ over the interval $(0, \\infty)$, its output values cover all real numbers.\n\nTherefore, the range of $gf(x)$ is the set of all real numbers:\n$$gf(x) \\in \\mathbb{R}$$\n\n*(Alternatively, this can be written as $-\\infty < gf(x) < \\infty$ or in interval notation as $(-\\infty, \\infty)$.)*"}]}],"generatedAt":"2026-06-09T04:17:18.952Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"663c36e5-9582-49e5-891d-073583f53e7f:analytical","label":"(a) Using Exponential Properties","steps":[{"type":"text","order":0,"title":"Understand the Goal","content":"The question asks us to find the **range** of the function $f(x) = e^{2x+1}$ for the domain $x \\in \\mathbb{R}$.\n\nThe range represents all the possible output values (or $y$-values) that the function can produce."},{"type":"hyper_latex","block":{"nodes":[{"text":"For any real exponent $u$, the exponential function is always strictly positive:"},{"latex":"$$\\htmlData{anchor=exp-prop}{\\color{@sky}{e^u}} > \\htmlData{anchor=zero-limit}{\\color{@purple}{0}}$"},{"text":"Since $x \\in \\mathbb{R}$, the exponent can take any real value:"},{"latex":"$$\\htmlData{anchor=input}{\\color{@sky}{2x+1}} \\in \\mathbb{R}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"exp-prop"},{"color":"purple","style":"box","anchor":"zero-limit"}],"connections":[{"to":"input","from":"exp-prop","color":"sky","label":"same base e structure","style":"dashed"}]},"order":1,"title":"Analyze Exponential Properties","description":"We can determine the range by looking at the core behavior of the natural exponential function."},{"type":"text","order":2,"title":"State the Range","content":"Because the exponent $2x+1$ can take any real value from $-\\infty$ to $+\\infty$, the function $f(x) = e^{2x+1}$ behaves exactly like the standard exponential curve $y = e^u$.\n\nAs $x \\to -\\infty$, the exponent $(2x+1) \\to -\\infty$, which means $e^{2x+1} \\to 0$ (from above). As $x \\to +\\infty$, $e^{2x+1} \\to +\\infty$.\n\nTherefore, the function values are always strictly greater than $0$. We write this as:\n\n$$f(x) > 0$$"}]},{"id":"663c36e5-9582-49e5-891d-073583f53e7f:graphical","label":"(a) Graphical Analysis","steps":[{"type":"text","order":0,"title":"Recall the base graph","content":"To find the range of $f(x) = e^{2x+1}$, let's analyze its graph. We start with the basic exponential function:\n\n$$y = e^x$$\n\nThe graph of $y = e^x$ is always positive, meaning it lies entirely above the $x$-axis. It has a horizontal asymptote at $y = 0$ as $x \\to -\\infty$."},{"type":"text","order":1,"title":"Analyze the transformations","content":"The function $f(x) = e^{2x+1}$ is a transformed version of $y = e^x$:\n\n* The $+1$ in the exponent corresponds to a horizontal translation of $1$ unit to the left.\n* The coefficient of $2$ in front of $x$ corresponds to a horizontal stretch by a scale factor of $\\frac{1}{2}$ parallel to the $x$-axis.\n\nNeither of these horizontal transformations affects the vertical position of the graph. There is no vertical shift or vertical scaling, so the horizontal asymptote remains at $y = 0$ and the graph is still restricted to the region above the $x$-axis."},{"type":"text","order":2,"title":"Determine the range","content":"Since the graph of $y = e^{2x+1}$ lies entirely above the $x$-axis and never touches or goes below it, the range of the function is all values greater than $0$.\n\nTherefore, the range of $f$ is:\n\n$$f(x) > 0$$","calculator":[{"gdc":{"expression":"e^(2*x+1)","instructions":"Go to the Graphing menu, enter the function e^(2x+1), and draw the graph to visually confirm the range."},"ti84":{"expression":"e^(2*X+1)","instructions":"Press [Y=], input Y1 = e^(2X+1) using [2nd] [LN] for the exponential function, then press [GRAPH] to view the curve."},"desmos":{"expression":"y = e^{2x+1}","instructions":"Type the equation into the input bar to see the exponential curve and its asymptote at the x-axis."},"description":"You can verify this by graphing the function on your calculator to see its shape and horizontal asymptote."}]}]},{"id":"b1daeab3-5c0a-46ce-824b-28471b2de7ed:algebraic_substitution","label":"(b) Direct Algebraic Substitution","steps":[{"type":"text","order":0,"title":"Understand the composite function","content":"To find an expression for the composite function $gf(x)$, we need to evaluate $g(f(x))$. This means we substitute the entire function $f(x)$ in place of $x$ in the function $g(x)$:\n\n$$gf(x) = g(f(x))$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f(x) = \\htmlData{anchor=f-expr}{\\color{@sky}{e^{2x+1}}}$"},{"latex":"$$g(x) = 2\\htmlData{anchor=x-var1}{x} + 3\\ln(\\htmlData{anchor=x-var2}{x})$"},{"latex":"$$gf(x) = 2(\\htmlData{anchor=sub1}{\\color{@sky}{e^{2x+1}}}) + 3\\ln(\\htmlData{anchor=sub2}{\\color{@sky}{e^{2x+1}}})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"f-expr"},{"color":"sky","style":"text-color","anchor":"sub1"},{"color":"sky","style":"text-color","anchor":"sub2"}],"connections":[{"to":"sub1","from":"f-expr","color":"sky","label":"substitute","style":"solid"},{"to":"sub2","from":"f-expr","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute f(x) into g(x)","description":"We replace each occurrence of $x$ in $g(x)$ with the expression for $f(x)$."},{"type":"text","order":2,"title":"Simplify using logarithm laws","content":"Now, we simplify the logarithmic term using the identity $\\ln(e^k) = k$ because the natural logarithm and the exponential function are inverses of each other:\n\n$$\\ln(e^{2x+1}) = 2x + 1$$\n\nSubstituting this back into our expression for $gf(x)$ gives:\n\n$$gf(x) = 2e^{2x+1} + 3(2x + 1)$$"},{"type":"text","order":3,"title":"Expand and write final answer","content":"Expand the brackets to obtain the final simplified expression:\n\n$$gf(x) = 2e^{2x+1} + 3(2x) + 3(1)$$\n\n$$gf(x) = 2e^{2x+1} + 6x + 3$$\n\nThis fully simplified expression matches the expectation for the final accuracy mark."}]},{"id":"bcec4874-6988-4495-ad1e-62a016fc3907:composite_algebraic","label":"(c) Algebraic Composition and Limits","steps":[{"type":"text","order":0,"title":"Define the composite function","content":"To find the range of the composite function $gf(x)$, we first form the algebraic expression for $gf(x) = g(f(x))$. \n\nWe substitute the function $f(x) = e^{2x+1}$ into the function $g(x) = 2x + 3\\ln x$:\n\n$$gf(x) = 2(e^{2x+1}) + 3\\ln(e^{2x+1})$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$gf(x) = 2e^{2x+1} + 3\\htmlData{anchor=ln-term}{\\color{@sky}{\\ln\\left(e^{2x+1}\\right)}}$"},{"latex":"$$gf(x) = 2e^{2x+1} + 3\\htmlData{anchor=simplified-term}{\\color{@sky}{(2x + 1)}}$"},{"latex":"$$gf(x) = 2e^{2x+1} + 6x + 3$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"ln-term"},{"color":"sky","style":"text-color","anchor":"simplified-term"}],"connections":[{"to":"simplified-term","from":"ln-term","color":"sky","label":"using ln(e^A) = A","style":"solid"}]},"order":1,"title":"Simplify the expression","description":"We can simplify the natural logarithm term because the logarithmic and exponential functions are inverses of each other, i.e., $\\ln(e^A) = A$."},{"type":"text","order":2,"title":"Analyze the positive limit","content":"Now we determine the range by investigating the behavior of $gf(x)$ as $x$ approaches positive and negative infinity.\n\nFirst, let's look at the limit as $x \\to \\infty$:\n\n$$\\lim_{x \\to \\infty} gf(x) = \\lim_{x \\to \\infty} \\left( 2e^{2x+1} + 6x + 3 \\right)$$\n\nAs $x \\to \\infty$, both $2e^{2x+1} \\to \\infty$ and $6x + 3 \\to \\infty$. Therefore:\n\n$$gf(x) \\to \\infty$$"},{"type":"text","order":3,"title":"Analyze the negative limit","content":"Next, let's examine the limit as $x \\to -\\infty$:\n\n$$\\lim_{x \\to -\\infty} gf(x) = \\lim_{x \\to -\\infty} \\left( 2e^{2x+1} + 6x + 3 \\right)$$\n\nAs $x \\to -\\infty$, the exponential term decays to zero:\n\n$$2e^{2x+1} \\to 0$$\n\nHowever, the linear term decreases without bound:\n\n$$6x + 3 \\to -\\infty$$\n\nCombining these, we get:\n\n$$gf(x) \\to -\\infty$$"},{"type":"text","order":4,"title":"State the range","content":"Since $gf(x) = 2e^{2x+1} + 6x + 3$ is a continuous function defined for all real numbers $x \\in \\mathbb{R}$, and it takes values from $-\\infty$ to $\\infty$, its range is the set of all real numbers.\n\nTherefore, the range is:\n\n$$gf(x) \\in \\mathbb{R}$$"}]},{"id":"bcec4874-6988-4495-ad1e-62a016fc3907:range_mapping","label":"(c) Domain and Range Mapping","steps":[{"type":"text","order":0,"title":"Identify the inner function","content":"To find the range of the composite function $gf(x)$, we can map the range of the inner function $f(x)$ directly into the outer function $g(y)$.\n\nThe inner function is:\n$$f(x) = e^{2x+1}, \\quad x \\in \\mathbb{R}$$"},{"type":"text","order":1,"title":"Determine the range of f(x)","content":"Since $x \\in \\mathbb{R}$, the exponent $2x+1$ can take any real value, from $-\\infty$ to $\\infty$. \n\nWe know that exponential functions of the form $e^u$ are always strictly positive for any real power $u$. Therefore, as $x \\to -\\infty$, $f(x) \\to 0$, and as $x \\to \\infty$, $f(x) \\to \\infty$.\n\nThus, the range of the inner function $f(x)$ is:\n$$(0, \\infty)$$\n\nThis range now becomes the domain of our outer function $g(y)$, where we let $y = f(x) > 0$."},{"type":"text","order":2,"title":"Analyze the outer function","content":"Now we analyze the behavior of the outer function $g(y)$ over its domain $y \\in (0, \\infty)$:\n$$g(y) = 2y + 3\\ln y$$\n\nSince both $2y$ and $3\\ln y$ are continuous and strictly increasing functions for all $y > 0$, their sum $g(y)$ is also continuous and strictly increasing over $(0, \\infty)$.\n\nTo find the range of $g(y)$, let's evaluate the limits at the boundaries of its domain:\n\n* **As $y \\to 0^+$:**\n $$2y \\to 0 \\quad \\text{and} \\quad 3\\ln y \\to -\\infty \\implies g(y) \\to -\\infty$$\n\n* **As $y \\to \\infty$:**\n $$2y \\to \\infty \\quad \\text{and} \\quad 3\\ln y \\to \\infty \\implies g(y) \\to \\infty$$"},{"type":"text","order":3,"title":"State the final range","content":"Because $g(y)$ is continuous and increases monotonically from $-\\infty$ to $\\infty$ over the interval $(0, \\infty)$, its output values cover all real numbers.\n\nTherefore, the range of $gf(x)$ is the set of all real numbers:\n$$gf(x) \\in \\mathbb{R}$$\n\n*(Alternatively, this can be written as $-\\infty < gf(x) < \\infty$ or in interval notation as $(-\\infty, \\infty)$.)*"}]}],"generatedAt":"2026-06-09T04:17:29.802Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"98874a9e-d554-40b4-9ecf-6a130e82838e","revisionId":"4e258c72-ee25-4e02-b0d4-f2c41a4957be","questionNumber":null,"questionType":"LA","specification":"$$$f(x) = \\frac{5x+2}{x-4} \\quad x \\geq 6$$","difficulty":1,"difficultyBand":"medium","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"8465ba0e-a8bf-4c19-9c93-cc0b7afb40a6","label":"a","content":"Find $ff(6)$ ","markscheme":"- $f(6) = \\frac{5(6)+2}{6-4} = 16$ M1\n- $f(16) = \\frac{5(16)+2}{16-4} = \\frac{82}{12} = \\frac{41}{6}$ A1","marks":2,"order":0,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Evaluate $f(6)$ correctly to get $16$.","code":"M"},{"id":"p2","marks":1,"description":"Evaluate $f(16)$ to obtain $\\frac{41}{6}$.","code":"A"}],"finalAnswers":["\\frac{41}{6}","6.83"],"generatedAt":"2026-05-06T14:05:05.825Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"direct-evaluation","label":"Step-by-Step Evaluation","steps":[{"type":"text","order":0,"title":"Understand composite notation","content":"To find $ff(6)$, we are working with a **composite function**. The notation $ff(6)$ is simply a shorter way of writing:\n\n$$f(f(6))$$\n\nTo evaluate this, we work from the **inside out**:\n1. First, find the \"inside\" value: $f(6)$.\n2. Then, substitute that result back into the function to find $f(\\text{result})$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f(x) = \\frac{5x+2}{x-4}$"},{"latex":"$$f(\\htmlData{anchor=val-6}{\\color{@sky}{6}}) = \\frac{5(\\htmlData{anchor=sub-6a}{\\color{@sky}{6}})+2}{\\htmlData{anchor=sub-6b}{\\color{@sky}{6}}-4}$"},{"latex":"$$f(6) = \\frac{30+2}{2} = 16$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"val-6"},{"color":"sky","style":"text-color","anchor":"sub-6a"},{"color":"sky","style":"text-color","anchor":"sub-6b"}],"connections":[{"to":"sub-6a","from":"val-6","color":"sky","label":"substitute 6","style":"solid"},{"to":"sub-6b","from":"val-6","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Evaluate the inside function","description":"We start by substituting $x = 6$ into the function $f(x)$:"},{"type":"text","order":2,"title":"Earn the first mark","content":"Correctly showing the substitution and obtaining $f(6) = 16$ earns you the first mark (**M1**) on this question."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f(x) = \\frac{5x+2}{x-4}$"},{"latex":"$$f(\\htmlData{anchor=val-16}{\\color{@amber}{16}}) = \\frac{5(\\htmlData{anchor=sub-16a}{\\color{@amber}{16}})+2}{\\htmlData{anchor=sub-16b}{\\color{@amber}{16}}-4}$"},{"latex":"$$f(16) = \\frac{80+2}{12} = \\frac{82}{12}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"val-16"},{"color":"amber","style":"text-color","anchor":"sub-16a"},{"color":"amber","style":"text-color","anchor":"sub-16b"}],"connections":[{"to":"sub-16a","from":"val-16","color":"amber","label":"substitute 16","style":"solid"},{"to":"sub-16b","from":"val-16","color":"amber","label":null,"style":"solid"}]},"order":3,"title":"Evaluate the outer function","description":"Now, we substitute the output $16$ back into the function as our new input:"},{"type":"text","order":4,"title":"Simplify to final answer","content":"Finally, we simplify the fraction to its lowest terms by dividing the numerator and denominator by their greatest common divisor, $2$:\n\n$$ff(6) = \\frac{82}{12} = \\frac{41}{6}$$\n\nThis final exact fraction completes the second mark (**A1**)."}]},{"id":"algebraic-composition","label":"Algebraic Composition","steps":[{"type":"text","order":0,"title":"Define the composition","content":"To find the composite function $ff(x) = f(f(x))$, we substitute the entire function $f(x)$ in place of $x$ in the definition of $f(x)$. \n\nWe start with our given function:\n$$f(x) = \\frac{5x+2}{x-4}$$\n\nSubstituting $f(x)$ into itself, we write:\n$$f(f(x)) = \\frac{5\\left(f(x)\\right)+2}{\\left(f(x)\\right)-4}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$f(f(x)) = \\frac{5\\left(\\htmlData{anchor=fx-num}{\\frac{5x+2}{x-4}}\\right) + 2}{\\htmlData{anchor=fx-den}{\\frac{5x+2}{x-4}} - 4}$$"}],"highlights":[{"color":"sky","style":"box","anchor":"fx-num"},{"color":"sky","style":"box","anchor":"fx-den"}],"connections":[]},"order":1,"title":"Substitute the algebraic fraction","description":"We replace each occurrence of $f(x)$ with its explicit expression: $\\frac{5x+2}{x-4}$."},{"type":"text","order":2,"title":"Simplify the numerator and denominator","content":"To simplify this complex fraction, we can multiply the numerator and the denominator of the entire expression by the common denominator, $(x-4)$:\n\n$$f(f(x)) = \\frac{5(5x+2) + 2(x-4)}{(5x+2) - 4(x-4)}$$\n\nNow, let's expand both the numerator and denominator:\n- **Numerator:**\n $$5(5x+2) + 2(x-4) = 25x + 10 + 2x - 8 = 27x + 2$$\n- **Denominator:**\n $$(5x+2) - 4(x-4) = 5x + 2 - 4x + 16 = x + 18$$"},{"type":"text","order":3,"title":"Write the general expression","content":"Combining our simplified parts, we get the general algebraic expression for the composite function:\n\n$$f(f(x)) = \\frac{27x+2}{x+18}$$"},{"type":"text","order":4,"title":"Substitute x = 6","content":"Now that we have the simplified formula for $f(f(x))$, we substitute $x = 6$ into it:\n\n$$f(f(6)) = \\frac{27(6) + 2}{6 + 18}$$\n\nLet's calculate the values:\n- **Numerator:** $27 \\times 6 + 2 = 162 + 2 = 164$\n- **Denominator:** $6 + 18 = 24$\n\nThis gives:\n$$f(f(6)) = \\frac{164}{24}$$"},{"type":"text","order":5,"title":"Simplify to final answer","content":"To get the final mark, we simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is $4$:\n\n$$\\frac{164 \\div 4}{24 \\div 4} = \\frac{41}{6}$$\n\nTherefore, we have:\n$$ff(6) = \\frac{41}{6}$$"}]}],"generatedAt":"2026-06-11T03:16:06.176Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"a981fff9-8af5-4d7a-b51c-5f9cad6c386b","label":"b","content":"State the range of $f$.","markscheme":"- $f(6) = 16$. As $x \\to \\infty, f(x) \\to 5$, so $5 < f(x) \\leq 16$ B1","marks":1,"order":1,"markingRubric":{"totalMarks":1,"points":[{"id":"p1","marks":1,"description":"Correct range identified from asymptote and boundary value.","code":"B"}],"finalAnswers":["5 < f(x) \\leq 16"],"generatedAt":"2026-05-06T14:05:05.825Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"limit_endpoint","label":"Endpoint and Limit Analysis","steps":[{"type":"text","order":0,"title":"Analyze the endpoint","content":"To find the range of the function $f(x) = \\frac{5x+2}{x-4}$ for $x \\geq 6$, we begin by evaluating the function at its boundary value (the endpoint of the domain), which is $x = 6$.\n\n$$f(6) = \\frac{5(6)+2}{6-4}$$\n\n$$f(6) = \\frac{30+2}{2} = \\frac{32}{2} = 16$$\n\nSince $x \\geq 6$ is inclusive, $f(x)$ can equal $16$. This gives us our upper bound: $f(x) \\leq 16$.","highlights":[{"text":"$$$f(x) = \\frac{5x+2}{x-4} \\quad x \\geq 6$$","location":"specification"}]},{"type":"text","order":1,"title":"Analyze the limit","content":"Next, we find the behavior of $f(x)$ as $x$ increases without bound ($x \\to \\infty$) to determine the lower bound.\n\nTo find the horizontal asymptote, we divide both the numerator and the denominator by $x$:\n\n$$f(x) = \\frac{5 + \\frac{2}{x}}{1 - \\frac{4}{x}}$$\n\nAs $x \\to \\infty$, the terms $\\frac{2}{x}$ and $\\frac{4}{x}$ both approach $0$:\n\n$$\\lim_{x \\to \\infty} f(x) = \\frac{5 + 0}{1 - 0} = 5$$\n\nSince $x$ is approaching infinity but never reaches it, the value of $f(x)$ will never actually equal $5$, but will approach it from above. This gives us our strict lower bound: $f(x) > 5$."},{"type":"hyper_latex","block":{"nodes":[{"text":"To ensure that $16$ and $5$ are indeed the global bounds for $x \\geq 6$, we confirm that the function is strictly decreasing (monotonic) on this interval."},{"latex":"$$f(x) = \\frac{5x+2}{x-4} = \\frac{5(x-4) + 22}{x-4} = \n\\htmlData{anchor=asymptote}{\\color{@sky}{5}} + \\frac{\\htmlData{anchor=numerator}{\\color{@amber}{22}}}{x-4}$"},{"latex":"\\text{For } x \\geq 6, \\text{ as } x \\text{ increases, } x-4 \\text{ increases, so } \\frac{22}{x-4} \\text{ strictly decreases.}"}],"highlights":[{"color":"sky","style":"text-color","anchor":"asymptote"},{"color":"amber","style":"text-color","anchor":"numerator"}],"connections":[{"to":"numerator","from":"asymptote","color":"sky","label":"asymptote + positive term","style":"dashed"}]},"order":2,"title":"Confirm behavior is monotonic","description":"By rewriting the algebraic fraction, we can easily see how the function behaves as $x$ increases."},{"type":"text","order":3,"title":"State the range","content":"Since the function is strictly decreasing from the endpoint value of $16$ towards the horizontal asymptote at $y = 5$, the range of $f$ is:\n\n$$5 < f(x) \\leq 16$$\n\nIn interval notation, this is written as $(5, 16]$."}]},{"id":"inverse_inequality","label":"Inverse Function Method","steps":[{"type":"text","order":0,"title":"Understand the method","content":"To find the range of the function $f(x) = \\frac{5x+2}{x-4}$ for $x \\geq 6$, we can use the **Inverse Function Method**.\n\nSince the range of a function $f(x)$ is the domain of its inverse function $f^{-1}(y)$, we can find the expression for the inverse $x = g(y)$ and then apply the domain restriction $x \\geq 6$ to solve for $y$. This directly gives us the range of $f(x)$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\frac{5x + 2}{x - 4}$"},{"latex":"$$y(x - 4) = 5x + 2$"},{"latex":"$$yx - 4y = 5x + 2$"},{"latex":"$$\\htmlData{anchor=x-terms}{yx - 5x} = \\htmlData{anchor=y-terms}{4y + 2}$"},{"latex":"$$x(y - 5) = 4y + 2$"},{"latex":"$$\\htmlData{anchor=inv-func}{x = \\frac{4y + 2}{y - 5}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"x-terms"},{"color":"sky","style":"box","anchor":"inv-func"}],"connections":[{"to":"inv-func","from":"x-terms","color":"sky","label":"solve for x","style":"solid"}]},"order":1,"title":"Find the inverse function","description":"We set $y = f(x)$ and rearrange the equation to make $x$ the subject, which gives the inverse function formula $x = g(y)$."},{"type":"text","order":2,"title":"Apply domain restriction","content":"The domain of our original function is restricted to $x \\geq 6$.\n\nSubstituting our expression for $x$ into this inequality gives:\n\n$$\\frac{4y + 2}{y - 5} \\geq 6$$"},{"type":"text","order":3,"title":"Solve the inequality","content":"To solve $\\frac{4y + 2}{y - 5} \\geq 6$, we first note that $y \\neq 5$ because the denominator cannot be zero.\n\nNow, let's consider the two possible cases for the denominator $y - 5$:\n\n**Case 1: $y - 5 > 0$ (i.e., $y > 5$)**\nSince the denominator is positive, multiplying both sides by $(y - 5)$ preserves the inequality sign:\n$$\\begin{aligned} 4y + 2 &\\geq 6(y - 5) \\\\ 4y + 2 &\\geq 6y - 30 \\\\ 32 &\\geq 2y \\\\ y &\\leq 16 \\end{aligned}$$\nCombining this with the case condition $y > 5$, we get:\n$$5 < y \\leq 16$$\n\n**Case 2: $y - 5 < 0$ (i.e., $y < 5$)**\nSince the denominator is negative, multiplying both sides by $(y - 5)$ reverses the inequality sign:\n$$\\begin{aligned} 4y + 2 &\\leq 6(y - 5) \\\\ 4y + 2 &\\leq 6y - 30 \\\\ 32 &\\leq 2y \\\\ y &\\geq 16 \\end{aligned}$$\nBut $y$ cannot be both $y < 5$ and $y \\geq 16$ at the same time. Thus, Case 2 yields no solutions."},{"type":"text","order":4,"title":"State the range","content":"The only valid interval for the outputs $y$ is:\n\n$$5 < f(x) \\leq 16$$\n\nThis is the complete range of the function $f(x)$."}]},{"id":"graphical","label":"Graphical Analysis","steps":[{"type":"text","order":0,"title":"Analyze the starting value","content":"To find the range of the function, let's first determine the coordinate where our domain begins. Since the domain is defined for $x \\geq 6$, we substitute $x = 6$ into the function:\n\n$$f(6) = \\frac{5(6) + 2}{6 - 4} = \\frac{32}{2} = 16$$\n\nSo, the starting point on our graph is $(6, 16)$.","highlights":[{"text":"x >= 6","location":"specification"}]},{"type":"text","order":1,"title":"Analyze the end behavior","content":"Now, let's examine the behavior of $f(x)$ as $x$ increases towards positive infinity ($x \\to \\infty$). We look for the horizontal asymptote of the rational function:\n\n$$\\lim_{x \\to \\infty} f(x) = \\lim_{x \\to \\infty} \\frac{5x + 2}{x - 4} = \\lim_{x \\to \\infty} \\frac{5 + \\frac{2}{x}}{1 - \\frac{4}{x}} = 5$$\n\nThis means that as $x$ gets extremely large, the graph approaches the horizontal asymptote $y = 5$ from above, but never actually reaches it."},{"type":"text","order":2,"title":"Visualize the graph","content":"Let's sketch or visualize the graph of $y = f(x)$ for the interval $x \\geq 6$:\n\n- The graph starts at the maximum point $(6, 16)$.\n- As $x$ increases, the curve decreases smoothly and gets closer and closer to the horizontal line $y = 5$.\n- Because the curve is strictly decreasing and never crosses $y = 5$, the values of $f(x)$ are bounded below by $5$ (not included) and above by $16$ (included).","calculator":[{"gdc":{"expression":"(5*x + 2) / (x - 4)","instructions":"1. Go to the Graph menu and enter f1(x) = (5x + 2) / (x - 4).\n2. Set the viewing window with x from 6 to 50 and y from 0 to 20.\n3. Trace the function starting at x = 6 to see the y-value transition from 16 down toward the horizontal asymptote at y = 5."},"ti84":{"expression":"(5*X + 2) / (X - 4)","instructions":"1. Press [Y=] and enter Y1 = (5X + 2) / (X - 4).\n2. Press [WINDOW] and set Xmin = 6, Xmax = 50, Ymin = 0, Ymax = 20.\n3. Press [GRAPH].\n4. Use [2nd] [TRACE] and select 'Value', entering X = 6 to find the initial point (6, 16).\n5. Observe that the curve decreases and levels off near Y = 5."},"desmos":{"expression":"y = (5x + 2) / (x - 4) \\{x \\ge 6\\}","instructions":"1. In the first input box, type: y = (5x + 2) / (x - 4) {x >= 6}.\n2. Click the wrench icon to adjust the axes: set x from 5 to 50, and y from 0 to 20.\n3. Click on the left endpoint of the curve to confirm the coordinate is (6, 16).\n4. Observe that the curve slopes downward, approaching y = 5 asymptotically."},"description":"Graph the function to visually inspect its range for $x \\geq 6$."}]},{"type":"text","order":3,"title":"State the range","content":"Combining these two behaviors, we can state the range of $f$ in inequality notation:\n\n$$5 < f(x) \\leq 16$$\n\n*(Note: We use a strict inequality for 5 because the graph never touches its asymptote, and a non-strict inequality for 16 because $x=6$ is included in the domain.)*"}]}],"generatedAt":"2026-06-11T03:17:29.091Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"bedcba08-02af-4baa-b497-d6012df2da67","label":"c","content":"Find $f^{-1}(x)$ and state its domain","markscheme":"- $y = \\frac{5x+2}{x-4} \\Rightarrow yx - 4y = 5x + 2$ M1\n- $x(y-5) = 4y + 2 \\Rightarrow x = \\frac{4y+2}{y-5}$ M1\n- $f^{-1}(x) = \\frac{4x+2}{x-5}$ with domain $5 < x \\leq 16$ A1","marks":3,"order":2,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Multiply by denominator to begin isolation of $x$.","code":"M"},{"id":"p2","marks":1,"description":"Rearrange and factorise to make $x$ the subject.","code":"M"},{"id":"p3","marks":1,"description":"State correct inverse function and domain.","code":"A"}],"finalAnswers":["f^{-1}(x) = \\frac{4x+2}{x-5}, 5 < x \\leq 16"],"generatedAt":"2026-05-06T14:05:05.825Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"range_analysis","label":"Algebraic Inversion and Range Analysis","steps":[{"type":"text","order":0,"title":"Define the inverse relationship","content":"To find the inverse function $f^{-1}(x)$, we set the original function equal to $y$ and solve for $x$ in terms of $y$:\n\n$$y = \\frac{5x+2}{x-4}$$","highlights":[{"text":"f(x) = \\frac{5x+2}{x-4}","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y(x-4) = 5x+2$"},{"latex":"$$yx - 4y = 5x + 2$"},{"latex":"$$\\htmlData{anchor=step1}{yx - 5x} = \\htmlData{anchor=step2}{4y + 2}$"},{"latex":"$$x(y-5) = 4y + 2$"},{"latex":"$$x = \\frac{4y+2}{y-5}$"}],"highlights":[{"color":"sky","style":"box","anchor":"step1"},{"color":"sky","style":"box","anchor":"step2"}],"connections":[{"to":"step2","from":"step1","color":"sky","label":"rearrange terms","style":"solid"}]},"order":1,"title":"Solve for x","description":"Rearrange the equation to isolate $x$ by moving all terms containing $x$ to one side."},{"type":"text","order":2,"title":"Write the inverse function","content":"Replacing $y$ with $x$, we obtain the expression for $f^{-1}(x)$:\n\n$$f^{-1}(x) = \\frac{4x+2}{x-5}$$"},{"type":"text","order":3,"title":"Determine the domain","content":"Recall that the **domain** of $f^{-1}(x)$ is exactly equal to the **range** of $f(x)$. \n\nWe are given the domain of $f(x)$ as $x \\geq 6$. Let's analyze the behavior of $f(x)$ on this interval:\n\n1. **Endpoint Value:** At $x = 6$, we have:\n $$f(6) = \\frac{5(6)+2}{6-4} = \\frac{32}{2} = 16$$\n\n2. **Asymptote / Limit:** As $x \\to \\infty$, the terms without $x$ become negligible, so:\n $$\\lim_{x \\to \\infty} f(x) = 5$$\n\nSince $f(x)$ is a continuously decreasing function for $x \\geq 6$, its values range from just above $5$ up to and including $16$.","highlights":[{"text":"x \\geq 6","location":"specification"}]},{"type":"text","order":4,"title":"State final answer","content":"Thus, the range of $f(x)$ is $5 < f(x) \\leq 16$. \n\nTherefore, the domain of $f^{-1}(x)$ is:\n\n$$5 < x \\leq 16$$"}]},{"id":"inverse_inequality","label":"Algebraic Inversion and Inverse Inequality","steps":[{"type":"text","order":0,"title":"Set up the equation","content":"To find the inverse function $f^{-1}(x)$ algebraically, we start by setting $y = f(x)$:\n\n$$y = \\frac{5x+2}{x-4}$$","highlights":[{"text":"f(x) = \\frac{5x+2}{x-4}","location":"specification"}]},{"type":"text","order":1,"title":"Rearrange to make x the subject","content":"Now, we rearrange the equation to isolate $x$ in terms of $y$:\n\nMultiply both sides by $(x-4)$:\n$$y(x-4) = 5x + 2$$\n\nExpand the brackets:\n$$yx - 4y = 5x + 2$$\n\nCollect all terms containing $x$ on one side:\n$$yx - 5x = 4y + 2$$\n\nFactorise out $x$:\n$$x(y-5) = 4y + 2$$\n\nDivide by $(y-5)$:\n$$x = \\frac{4y+2}{y-5}$$\n\nThis earns the first two method marks (**M1 M1**)."},{"type":"text","order":2,"title":"Write the inverse function","content":"Swap the variables $x$ and $y$ to write the final expression for $f^{-1}(x)$:\n\n$$f^{-1}(x) = \\frac{4x+2}{x-5}$$"},{"type":"text","order":3,"title":"Set up the domain inequality","content":"The domain of $f(x)$ is given as $x \\geq 6$. Since the range of the inverse function $f^{-1}(x)$ is equal to the domain of the original function $f(x)$, we must have:\n\n$$f^{-1}(x) \\geq 6$$\n\nSubstituting our expression for $f^{-1}(x)$ gives the inequality:\n\n$$\\frac{4x+2}{x-5} \\geq 6$$","highlights":[{"text":"x \\geq 6","location":"specification"}]},{"type":"text","order":4,"title":"Solve the inequality","content":"To solve this inequality, we can subtract $6$ from both sides and combine into a single fraction:\n\n$$\\frac{4x+2}{x-5} - 6 \\geq 0$$\n\n$$\\frac{(4x+2) - 6(x-5)}{x-5} \\geq 0$$\n\n$$\\frac{4x+2 - 6x + 30}{x-5} \\geq 0$$\n\n$$\\frac{-2x + 32}{x-5} \\geq 0$$\n\nMultiply the numerator and denominator by $-1$ and flip the inequality sign:\n\n$$\\frac{2(x-16)}{x-5} \\leq 0$$"},{"type":"text","order":5,"title":"Find the domain of the inverse","content":"For the rational inequality $\\frac{2(x-16)}{x-5} \\leq 0$ to hold, the numerator and denominator must have opposite signs:\n- The denominator cannot be zero, so $x \\neq 5$. This means $x > 5$ for the expression to be well-defined and positive in the denominator.\n- For the overall expression to be non-positive ($\\leq 0$), we require $x \\leq 16$.\n\nCombining these constraints gives the domain:\n\n$$5 < x \\leq 16$$\n\nThis completes the final accuracy mark (**A1**)."}]}],"generatedAt":"2026-06-11T03:17:49.980Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"8465ba0e-a8bf-4c19-9c93-cc0b7afb40a6:direct-evaluation","label":"(a) Step-by-Step Evaluation","steps":[{"type":"text","order":0,"title":"Understand composite notation","content":"To find $ff(6)$, we are working with a **composite function**. The notation $ff(6)$ is simply a shorter way of writing:\n\n$$f(f(6))$$\n\nTo evaluate this, we work from the **inside out**:\n1. First, find the \"inside\" value: $f(6)$.\n2. Then, substitute that result back into the function to find $f(\\text{result})$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f(x) = \\frac{5x+2}{x-4}$"},{"latex":"$$f(\\htmlData{anchor=val-6}{\\color{@sky}{6}}) = \\frac{5(\\htmlData{anchor=sub-6a}{\\color{@sky}{6}})+2}{\\htmlData{anchor=sub-6b}{\\color{@sky}{6}}-4}$"},{"latex":"$$f(6) = \\frac{30+2}{2} = 16$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"val-6"},{"color":"sky","style":"text-color","anchor":"sub-6a"},{"color":"sky","style":"text-color","anchor":"sub-6b"}],"connections":[{"to":"sub-6a","from":"val-6","color":"sky","label":"substitute 6","style":"solid"},{"to":"sub-6b","from":"val-6","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Evaluate the inside function","description":"We start by substituting $x = 6$ into the function $f(x)$:"},{"type":"text","order":2,"title":"Earn the first mark","content":"Correctly showing the substitution and obtaining $f(6) = 16$ earns you the first mark (**M1**) on this question."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f(x) = \\frac{5x+2}{x-4}$"},{"latex":"$$f(\\htmlData{anchor=val-16}{\\color{@amber}{16}}) = \\frac{5(\\htmlData{anchor=sub-16a}{\\color{@amber}{16}})+2}{\\htmlData{anchor=sub-16b}{\\color{@amber}{16}}-4}$"},{"latex":"$$f(16) = \\frac{80+2}{12} = \\frac{82}{12}$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"val-16"},{"color":"amber","style":"text-color","anchor":"sub-16a"},{"color":"amber","style":"text-color","anchor":"sub-16b"}],"connections":[{"to":"sub-16a","from":"val-16","color":"amber","label":"substitute 16","style":"solid"},{"to":"sub-16b","from":"val-16","color":"amber","label":null,"style":"solid"}]},"order":3,"title":"Evaluate the outer function","description":"Now, we substitute the output $16$ back into the function as our new input:"},{"type":"text","order":4,"title":"Simplify to final answer","content":"Finally, we simplify the fraction to its lowest terms by dividing the numerator and denominator by their greatest common divisor, $2$:\n\n$$ff(6) = \\frac{82}{12} = \\frac{41}{6}$$\n\nThis final exact fraction completes the second mark (**A1**)."}]},{"id":"8465ba0e-a8bf-4c19-9c93-cc0b7afb40a6:algebraic-composition","label":"(a) Algebraic Composition","steps":[{"type":"text","order":0,"title":"Define the composition","content":"To find the composite function $ff(x) = f(f(x))$, we substitute the entire function $f(x)$ in place of $x$ in the definition of $f(x)$. \n\nWe start with our given function:\n$$f(x) = \\frac{5x+2}{x-4}$$\n\nSubstituting $f(x)$ into itself, we write:\n$$f(f(x)) = \\frac{5\\left(f(x)\\right)+2}{\\left(f(x)\\right)-4}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$f(f(x)) = \\frac{5\\left(\\htmlData{anchor=fx-num}{\\frac{5x+2}{x-4}}\\right) + 2}{\\htmlData{anchor=fx-den}{\\frac{5x+2}{x-4}} - 4}$$"}],"highlights":[{"color":"sky","style":"box","anchor":"fx-num"},{"color":"sky","style":"box","anchor":"fx-den"}],"connections":[]},"order":1,"title":"Substitute the algebraic fraction","description":"We replace each occurrence of $f(x)$ with its explicit expression: $\\frac{5x+2}{x-4}$."},{"type":"text","order":2,"title":"Simplify the numerator and denominator","content":"To simplify this complex fraction, we can multiply the numerator and the denominator of the entire expression by the common denominator, $(x-4)$:\n\n$$f(f(x)) = \\frac{5(5x+2) + 2(x-4)}{(5x+2) - 4(x-4)}$$\n\nNow, let's expand both the numerator and denominator:\n- **Numerator:**\n $$5(5x+2) + 2(x-4) = 25x + 10 + 2x - 8 = 27x + 2$$\n- **Denominator:**\n $$(5x+2) - 4(x-4) = 5x + 2 - 4x + 16 = x + 18$$"},{"type":"text","order":3,"title":"Write the general expression","content":"Combining our simplified parts, we get the general algebraic expression for the composite function:\n\n$$f(f(x)) = \\frac{27x+2}{x+18}$$"},{"type":"text","order":4,"title":"Substitute x = 6","content":"Now that we have the simplified formula for $f(f(x))$, we substitute $x = 6$ into it:\n\n$$f(f(6)) = \\frac{27(6) + 2}{6 + 18}$$\n\nLet's calculate the values:\n- **Numerator:** $27 \\times 6 + 2 = 162 + 2 = 164$\n- **Denominator:** $6 + 18 = 24$\n\nThis gives:\n$$f(f(6)) = \\frac{164}{24}$$"},{"type":"text","order":5,"title":"Simplify to final answer","content":"To get the final mark, we simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is $4$:\n\n$$\\frac{164 \\div 4}{24 \\div 4} = \\frac{41}{6}$$\n\nTherefore, we have:\n$$ff(6) = \\frac{41}{6}$$"}]},{"id":"a981fff9-8af5-4d7a-b51c-5f9cad6c386b:limit_endpoint","label":"(b) Endpoint and Limit Analysis","steps":[{"type":"text","order":0,"title":"Analyze the endpoint","content":"To find the range of the function $f(x) = \\frac{5x+2}{x-4}$ for $x \\geq 6$, we begin by evaluating the function at its boundary value (the endpoint of the domain), which is $x = 6$.\n\n$$f(6) = \\frac{5(6)+2}{6-4}$$\n\n$$f(6) = \\frac{30+2}{2} = \\frac{32}{2} = 16$$\n\nSince $x \\geq 6$ is inclusive, $f(x)$ can equal $16$. This gives us our upper bound: $f(x) \\leq 16$.","highlights":[{"text":"$$$f(x) = \\frac{5x+2}{x-4} \\quad x \\geq 6$$","location":"specification"}]},{"type":"text","order":1,"title":"Analyze the limit","content":"Next, we find the behavior of $f(x)$ as $x$ increases without bound ($x \\to \\infty$) to determine the lower bound.\n\nTo find the horizontal asymptote, we divide both the numerator and the denominator by $x$:\n\n$$f(x) = \\frac{5 + \\frac{2}{x}}{1 - \\frac{4}{x}}$$\n\nAs $x \\to \\infty$, the terms $\\frac{2}{x}$ and $\\frac{4}{x}$ both approach $0$:\n\n$$\\lim_{x \\to \\infty} f(x) = \\frac{5 + 0}{1 - 0} = 5$$\n\nSince $x$ is approaching infinity but never reaches it, the value of $f(x)$ will never actually equal $5$, but will approach it from above. This gives us our strict lower bound: $f(x) > 5$."},{"type":"hyper_latex","block":{"nodes":[{"text":"To ensure that $16$ and $5$ are indeed the global bounds for $x \\geq 6$, we confirm that the function is strictly decreasing (monotonic) on this interval."},{"latex":"$$f(x) = \\frac{5x+2}{x-4} = \\frac{5(x-4) + 22}{x-4} = \n\\htmlData{anchor=asymptote}{\\color{@sky}{5}} + \\frac{\\htmlData{anchor=numerator}{\\color{@amber}{22}}}{x-4}$"},{"latex":"\\text{For } x \\geq 6, \\text{ as } x \\text{ increases, } x-4 \\text{ increases, so } \\frac{22}{x-4} \\text{ strictly decreases.}"}],"highlights":[{"color":"sky","style":"text-color","anchor":"asymptote"},{"color":"amber","style":"text-color","anchor":"numerator"}],"connections":[{"to":"numerator","from":"asymptote","color":"sky","label":"asymptote + positive term","style":"dashed"}]},"order":2,"title":"Confirm behavior is monotonic","description":"By rewriting the algebraic fraction, we can easily see how the function behaves as $x$ increases."},{"type":"text","order":3,"title":"State the range","content":"Since the function is strictly decreasing from the endpoint value of $16$ towards the horizontal asymptote at $y = 5$, the range of $f$ is:\n\n$$5 < f(x) \\leq 16$$\n\nIn interval notation, this is written as $(5, 16]$."}]},{"id":"a981fff9-8af5-4d7a-b51c-5f9cad6c386b:inverse_inequality","label":"(b) Inverse Function Method","steps":[{"type":"text","order":0,"title":"Understand the method","content":"To find the range of the function $f(x) = \\frac{5x+2}{x-4}$ for $x \\geq 6$, we can use the **Inverse Function Method**.\n\nSince the range of a function $f(x)$ is the domain of its inverse function $f^{-1}(y)$, we can find the expression for the inverse $x = g(y)$ and then apply the domain restriction $x \\geq 6$ to solve for $y$. This directly gives us the range of $f(x)$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y = \\frac{5x + 2}{x - 4}$"},{"latex":"$$y(x - 4) = 5x + 2$"},{"latex":"$$yx - 4y = 5x + 2$"},{"latex":"$$\\htmlData{anchor=x-terms}{yx - 5x} = \\htmlData{anchor=y-terms}{4y + 2}$"},{"latex":"$$x(y - 5) = 4y + 2$"},{"latex":"$$\\htmlData{anchor=inv-func}{x = \\frac{4y + 2}{y - 5}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"x-terms"},{"color":"sky","style":"box","anchor":"inv-func"}],"connections":[{"to":"inv-func","from":"x-terms","color":"sky","label":"solve for x","style":"solid"}]},"order":1,"title":"Find the inverse function","description":"We set $y = f(x)$ and rearrange the equation to make $x$ the subject, which gives the inverse function formula $x = g(y)$."},{"type":"text","order":2,"title":"Apply domain restriction","content":"The domain of our original function is restricted to $x \\geq 6$.\n\nSubstituting our expression for $x$ into this inequality gives:\n\n$$\\frac{4y + 2}{y - 5} \\geq 6$$"},{"type":"text","order":3,"title":"Solve the inequality","content":"To solve $\\frac{4y + 2}{y - 5} \\geq 6$, we first note that $y \\neq 5$ because the denominator cannot be zero.\n\nNow, let's consider the two possible cases for the denominator $y - 5$:\n\n**Case 1: $y - 5 > 0$ (i.e., $y > 5$)**\nSince the denominator is positive, multiplying both sides by $(y - 5)$ preserves the inequality sign:\n$$\\begin{aligned} 4y + 2 &\\geq 6(y - 5) \\\\ 4y + 2 &\\geq 6y - 30 \\\\ 32 &\\geq 2y \\\\ y &\\leq 16 \\end{aligned}$$\nCombining this with the case condition $y > 5$, we get:\n$$5 < y \\leq 16$$\n\n**Case 2: $y - 5 < 0$ (i.e., $y < 5$)**\nSince the denominator is negative, multiplying both sides by $(y - 5)$ reverses the inequality sign:\n$$\\begin{aligned} 4y + 2 &\\leq 6(y - 5) \\\\ 4y + 2 &\\leq 6y - 30 \\\\ 32 &\\leq 2y \\\\ y &\\geq 16 \\end{aligned}$$\nBut $y$ cannot be both $y < 5$ and $y \\geq 16$ at the same time. Thus, Case 2 yields no solutions."},{"type":"text","order":4,"title":"State the range","content":"The only valid interval for the outputs $y$ is:\n\n$$5 < f(x) \\leq 16$$\n\nThis is the complete range of the function $f(x)$."}]},{"id":"a981fff9-8af5-4d7a-b51c-5f9cad6c386b:graphical","label":"(b) Graphical Analysis","steps":[{"type":"text","order":0,"title":"Analyze the starting value","content":"To find the range of the function, let's first determine the coordinate where our domain begins. Since the domain is defined for $x \\geq 6$, we substitute $x = 6$ into the function:\n\n$$f(6) = \\frac{5(6) + 2}{6 - 4} = \\frac{32}{2} = 16$$\n\nSo, the starting point on our graph is $(6, 16)$.","highlights":[{"text":"x >= 6","location":"specification"}]},{"type":"text","order":1,"title":"Analyze the end behavior","content":"Now, let's examine the behavior of $f(x)$ as $x$ increases towards positive infinity ($x \\to \\infty$). We look for the horizontal asymptote of the rational function:\n\n$$\\lim_{x \\to \\infty} f(x) = \\lim_{x \\to \\infty} \\frac{5x + 2}{x - 4} = \\lim_{x \\to \\infty} \\frac{5 + \\frac{2}{x}}{1 - \\frac{4}{x}} = 5$$\n\nThis means that as $x$ gets extremely large, the graph approaches the horizontal asymptote $y = 5$ from above, but never actually reaches it."},{"type":"text","order":2,"title":"Visualize the graph","content":"Let's sketch or visualize the graph of $y = f(x)$ for the interval $x \\geq 6$:\n\n- The graph starts at the maximum point $(6, 16)$.\n- As $x$ increases, the curve decreases smoothly and gets closer and closer to the horizontal line $y = 5$.\n- Because the curve is strictly decreasing and never crosses $y = 5$, the values of $f(x)$ are bounded below by $5$ (not included) and above by $16$ (included).","calculator":[{"gdc":{"expression":"(5*x + 2) / (x - 4)","instructions":"1. Go to the Graph menu and enter f1(x) = (5x + 2) / (x - 4).\n2. Set the viewing window with x from 6 to 50 and y from 0 to 20.\n3. Trace the function starting at x = 6 to see the y-value transition from 16 down toward the horizontal asymptote at y = 5."},"ti84":{"expression":"(5*X + 2) / (X - 4)","instructions":"1. Press [Y=] and enter Y1 = (5X + 2) / (X - 4).\n2. Press [WINDOW] and set Xmin = 6, Xmax = 50, Ymin = 0, Ymax = 20.\n3. Press [GRAPH].\n4. Use [2nd] [TRACE] and select 'Value', entering X = 6 to find the initial point (6, 16).\n5. Observe that the curve decreases and levels off near Y = 5."},"desmos":{"expression":"y = (5x + 2) / (x - 4) \\{x \\ge 6\\}","instructions":"1. In the first input box, type: y = (5x + 2) / (x - 4) {x >= 6}.\n2. Click the wrench icon to adjust the axes: set x from 5 to 50, and y from 0 to 20.\n3. Click on the left endpoint of the curve to confirm the coordinate is (6, 16).\n4. Observe that the curve slopes downward, approaching y = 5 asymptotically."},"description":"Graph the function to visually inspect its range for $x \\geq 6$."}]},{"type":"text","order":3,"title":"State the range","content":"Combining these two behaviors, we can state the range of $f$ in inequality notation:\n\n$$5 < f(x) \\leq 16$$\n\n*(Note: We use a strict inequality for 5 because the graph never touches its asymptote, and a non-strict inequality for 16 because $x=6$ is included in the domain.)*"}]},{"id":"bedcba08-02af-4baa-b497-d6012df2da67:range_analysis","label":"(c) Algebraic Inversion and Range Analysis","steps":[{"type":"text","order":0,"title":"Define the inverse relationship","content":"To find the inverse function $f^{-1}(x)$, we set the original function equal to $y$ and solve for $x$ in terms of $y$:\n\n$$y = \\frac{5x+2}{x-4}$$","highlights":[{"text":"f(x) = \\frac{5x+2}{x-4}","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y(x-4) = 5x+2$"},{"latex":"$$yx - 4y = 5x + 2$"},{"latex":"$$\\htmlData{anchor=step1}{yx - 5x} = \\htmlData{anchor=step2}{4y + 2}$"},{"latex":"$$x(y-5) = 4y + 2$"},{"latex":"$$x = \\frac{4y+2}{y-5}$"}],"highlights":[{"color":"sky","style":"box","anchor":"step1"},{"color":"sky","style":"box","anchor":"step2"}],"connections":[{"to":"step2","from":"step1","color":"sky","label":"rearrange terms","style":"solid"}]},"order":1,"title":"Solve for x","description":"Rearrange the equation to isolate $x$ by moving all terms containing $x$ to one side."},{"type":"text","order":2,"title":"Write the inverse function","content":"Replacing $y$ with $x$, we obtain the expression for $f^{-1}(x)$:\n\n$$f^{-1}(x) = \\frac{4x+2}{x-5}$$"},{"type":"text","order":3,"title":"Determine the domain","content":"Recall that the **domain** of $f^{-1}(x)$ is exactly equal to the **range** of $f(x)$. \n\nWe are given the domain of $f(x)$ as $x \\geq 6$. Let's analyze the behavior of $f(x)$ on this interval:\n\n1. **Endpoint Value:** At $x = 6$, we have:\n $$f(6) = \\frac{5(6)+2}{6-4} = \\frac{32}{2} = 16$$\n\n2. **Asymptote / Limit:** As $x \\to \\infty$, the terms without $x$ become negligible, so:\n $$\\lim_{x \\to \\infty} f(x) = 5$$\n\nSince $f(x)$ is a continuously decreasing function for $x \\geq 6$, its values range from just above $5$ up to and including $16$.","highlights":[{"text":"x \\geq 6","location":"specification"}]},{"type":"text","order":4,"title":"State final answer","content":"Thus, the range of $f(x)$ is $5 < f(x) \\leq 16$. \n\nTherefore, the domain of $f^{-1}(x)$ is:\n\n$$5 < x \\leq 16$$"}]},{"id":"bedcba08-02af-4baa-b497-d6012df2da67:inverse_inequality","label":"(c) Algebraic Inversion and Inverse Inequality","steps":[{"type":"text","order":0,"title":"Set up the equation","content":"To find the inverse function $f^{-1}(x)$ algebraically, we start by setting $y = f(x)$:\n\n$$y = \\frac{5x+2}{x-4}$$","highlights":[{"text":"f(x) = \\frac{5x+2}{x-4}","location":"specification"}]},{"type":"text","order":1,"title":"Rearrange to make x the subject","content":"Now, we rearrange the equation to isolate $x$ in terms of $y$:\n\nMultiply both sides by $(x-4)$:\n$$y(x-4) = 5x + 2$$\n\nExpand the brackets:\n$$yx - 4y = 5x + 2$$\n\nCollect all terms containing $x$ on one side:\n$$yx - 5x = 4y + 2$$\n\nFactorise out $x$:\n$$x(y-5) = 4y + 2$$\n\nDivide by $(y-5)$:\n$$x = \\frac{4y+2}{y-5}$$\n\nThis earns the first two method marks (**M1 M1**)."},{"type":"text","order":2,"title":"Write the inverse function","content":"Swap the variables $x$ and $y$ to write the final expression for $f^{-1}(x)$:\n\n$$f^{-1}(x) = \\frac{4x+2}{x-5}$$"},{"type":"text","order":3,"title":"Set up the domain inequality","content":"The domain of $f(x)$ is given as $x \\geq 6$. Since the range of the inverse function $f^{-1}(x)$ is equal to the domain of the original function $f(x)$, we must have:\n\n$$f^{-1}(x) \\geq 6$$\n\nSubstituting our expression for $f^{-1}(x)$ gives the inequality:\n\n$$\\frac{4x+2}{x-5} \\geq 6$$","highlights":[{"text":"x \\geq 6","location":"specification"}]},{"type":"text","order":4,"title":"Solve the inequality","content":"To solve this inequality, we can subtract $6$ from both sides and combine into a single fraction:\n\n$$\\frac{4x+2}{x-5} - 6 \\geq 0$$\n\n$$\\frac{(4x+2) - 6(x-5)}{x-5} \\geq 0$$\n\n$$\\frac{4x+2 - 6x + 30}{x-5} \\geq 0$$\n\n$$\\frac{-2x + 32}{x-5} \\geq 0$$\n\nMultiply the numerator and denominator by $-1$ and flip the inequality sign:\n\n$$\\frac{2(x-16)}{x-5} \\leq 0$$"},{"type":"text","order":5,"title":"Find the domain of the inverse","content":"For the rational inequality $\\frac{2(x-16)}{x-5} \\leq 0$, check the signs around $x=5$ and $x=16$:\n- If $x<5$, both numerator and denominator are negative, so the fraction is positive and does not satisfy the inequality.\n- If $516$, the fraction is positive.\n\nTherefore, the domain is:\n\n$$5 < x \\leq 16$$\n\nThis completes the final accuracy mark (**A1**)."}]}],"generatedAt":"2026-06-11T03:17:50.048Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"fac15d6f-33ea-49de-8341-30b527130c30","revisionId":"f8ee2464-b874-447f-a141-6113ed5e0042","questionNumber":null,"questionType":"LA","specification":"The functions $f$ and $g$ are defined by\n\n$$f: x \\rightarrow 4x - 3, \\quad x \\in \\mathbb{R}$$\n\n$$g: x \\rightarrow \\frac{6}{x+5}, \\quad x \\in \\mathbb{R}, x \\neq -5$$","difficulty":1,"difficultyBand":"medium","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"42324282-2795-4f5e-8b28-76df8f60c575","label":"a","content":"Evaluate $fg(-2)$","markscheme":"- $g(-2) = \\frac{6}{-2+5} = 2$ M1\n- $f(2) = 4(2) - 3 = 5$ A1","marks":2,"order":0,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Finding $g(-2)$.","code":"M"},{"id":"p2","marks":1,"description":"Evaluating $f$ of the result.","code":"A"}],"finalAnswers":["5"],"generatedAt":"2026-05-06T14:04:47.436Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"inside_out_evaluation","label":"Inside-Out Evaluation","steps":[{"type":"text","order":0,"title":"Understand the composite function","content":"To evaluate the composite function $fg(-2)$ using the **Inside-Out Evaluation** approach, we start from the \"inside\" function and work our way \"out\".\n\nIn the expression $fg(-2)$, which can also be written as $f(g(-2))$:\n* The **inner function** is $g(x)$.\n* The **outer function** is $f(x)$.\n\nFirst, we will calculate the numerical value of the inner function, $g(-2)$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$g(\u0001color{@sky}{-2}) = \\frac{6}{\\htmlData{anchor=x-val}{\\color{@sky}{-2}} + 5}$"},{"latex":"$$g(-2) = \\frac{6}{\\htmlData{anchor=denom}{\\color{@amber}{3}}}$"},{"latex":"$$g(-2) = \\htmlData{anchor=inner-res}{\\color{@purple}{2}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"x-val"},{"color":"amber","style":"text-color","anchor":"denom"},{"color":"purple","style":"text-color","anchor":"inner-res"}],"connections":[{"to":"denom","from":"x-val","color":"amber","label":"simplify","style":"solid"},{"to":"inner-res","from":"denom","color":"purple","label":"divide","style":"solid"}]},"order":1,"title":"Evaluate the inner function","description":"We substitute $x = -2$ into the definition of $g(x) = \\frac{6}{x+5}$ and simplify to get our intermediate result."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f(\\htmlData{anchor=input-val}{\\color{@purple}{2}}) = 4(\\htmlData{anchor=x-sub}{\\color{@purple}{2}}) - 3$"},{"latex":"$$f(2) = \\htmlData{anchor=mult}{\\color{@orange}{8}} - 3$"},{"latex":"$$f(2) = \\htmlData{anchor=final}{\\color{@green}{5}}$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"input-val"},{"color":"purple","style":"text-color","anchor":"x-sub"},{"color":"orange","style":"text-color","anchor":"mult"},{"color":"green","style":"box","anchor":"final"}],"connections":[{"to":"x-sub","from":"input-val","color":"purple","label":"substitute","style":"solid"},{"to":"mult","from":"x-sub","color":"orange","label":"multiply","style":"solid"},{"to":"final","from":"mult","color":"green","label":"subtract","style":"solid"}]},"order":2,"title":"Evaluate the outer function","description":"Now, we substitute the result from the inner function, $2$, as the input into the outer function $f(x) = 4x - 3$."},{"type":"text","order":3,"title":"State the final answer","content":"By evaluating the inside function first and then the outside function, we find:\n\n$$fg(-2) = 5$$"}]},{"id":"algebraic_composition","label":"Algebraic Composition First","steps":[{"type":"text","order":0,"title":"Understand composite functions","content":"To find the composite function $fg(x)$, we substitute the entire function $g(x)$ into the function $f(x)$. \n\nWe are given:\n- $$f(x) = 4x - 3$$\n- $$g(x) = \\frac{6}{x+5}$$","highlights":[{"text":"f: x \\rightarrow 4x - 3","location":"specification"},{"text":"g: x \\rightarrow \\frac{6}{x+5}","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$g(x) = \\htmlData{anchor=g-expr}{\\color{@sky}{\\frac{6}{x+5}}}$"},{"latex":"$$fg(x) = f\\left( \\htmlData{anchor=g-sub}{\\color{@sky}{\\frac{6}{x+5}}} \\right)$"},{"latex":"$$fg(x) = 4\\left( \\color{@sky}{\\frac{6}{x+5}} \\right) - 3$"},{"latex":"$$\\phantom{fg(x)} = \\htmlData{anchor=fg-expr}{\\color{@purple}{\\frac{24}{x+5} - 3}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"g-expr"},{"color":"purple","style":"box","anchor":"fg-expr"}],"connections":[{"to":"g-sub","from":"g-expr","color":"sky","label":"substitute g(x)","style":"solid"}]},"order":1,"title":"Find the composite expression","description":"We substitute the expression for $g(x)$ into $f(x)$ wherever $x$ occurs."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$fg(x) = \\frac{24}{\\htmlData{anchor=x-var}{\\color{@amber}{x}} + 5} - 3$"},{"latex":"$$fg(-2) = \\frac{24}{\\htmlData{anchor=x-sub}{\\color{@amber}{-2}} + 5} - 3$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"x-var"},{"color":"amber","style":"text-color","anchor":"x-sub"}],"connections":[{"to":"x-sub","from":"x-var","color":"amber","label":"x = -2","style":"solid"}]},"order":2,"title":"Substitute the value","description":"Now, we substitute $x = -2$ into our simplified algebraic expression for $fg(x)$."},{"type":"text","order":3,"title":"Evaluate the final result","content":"Simplify the fraction to get the final answer:\n\n$$\\begin{aligned} fg(-2) &= \\frac{24}{3} - 3 \\\\ &= 8 - 3 \\\\ &= 5 \\end{aligned}$$\n\n**Note on Markscheme:** Even though the markscheme shows the working by calculating $g(-2)$ first (which gives $2$) and then finding $f(2)$, our algebraic composition method is equally valid and arrives at the same final correct answer of $5$!"}]}],"generatedAt":"2026-06-11T03:54:50.000Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"5ec45b13-b55f-4131-a0da-2b4f032bdfc0","label":"b","content":"Solve the equation $gf(x) = 3$","markscheme":"- $gf(x) = \\frac{6}{(4x-3)+5} = \\frac{6}{4x+2}$ M1\n- $\\frac{6}{4x+2} = 3 \\Rightarrow 6 = 3(4x+2)$ M1\n- $6 = 12x + 6 \\Rightarrow 12x = 0$ M1\n- $x = 0$ A1","marks":4,"order":1,"markingRubric":{"totalMarks":4,"points":[{"id":"p1","marks":1,"description":"Finding algebraic expression for $gf(x)$.","code":"M"},{"id":"p2","marks":1,"description":"Equating composition to 3 and removing denominator.","code":"M"},{"id":"p3","marks":1,"description":"Expansion and gathering $x$ terms.","code":"M"},{"id":"p4","marks":1,"description":"Final solution $x=0$.","code":"A"}],"finalAnswers":["0"],"generatedAt":"2026-05-06T14:04:47.436Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"composite_first","label":"Forming the Composite Function First","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f(x) = \blank{\\htmlData{anchor=f-expr}{\\color{@sky}{4x - 3}}}$"},{"latex":"$$g(x) = \\frac{6}{\\htmlData{anchor=g-x}{\\color{@sky}{x}} + 5}$"},{"latex":"$$gf(x) = \\frac{6}{(\\htmlData{anchor=gf-sub}{\\color{@sky}{4x - 3}}) + 5}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"f-expr"},{"color":"sky","style":"text-color","anchor":"gf-sub"}],"connections":[{"to":"gf-sub","from":"f-expr","color":"sky","label":null,"style":"solid"}]},"order":0,"title":"Form the composite function","description":"To find the composite function $gf(x)$, we substitute the entire expression for $f(x)$ into $g(x)$ everywhere there is an $x$. This gives us the expression shown above."},{"type":"text","order":1,"title":"Simplify the denominator","content":"Now, we simplify the denominator of our composite function $gf(x)$:\n\n$$\\begin{aligned} gf(x) &= \\frac{6}{4x - 3 + 5} \\\\[1em] &= \\frac{6}{4x + 2} \\end{aligned}$$\n\nThis simplified expression earns you the first mark (**M1**)."},{"type":"text","order":2,"title":"Set the equation equal to 3","content":"Next, we set our composite function equal to $3$ as required by the equation $gf(x) = 3$:\n\n$$\\frac{6}{4x + 2} = 3$$\n\nTo solve this, multiply both sides by the denominator $(4x + 2)$ to clear the fraction:\n\n$$6 = 3(4x + 2)$$\n\nWriting this equation earns the second mark (**M1**)."},{"type":"text","order":3,"title":"Solve for x","content":"Now we expand the right-hand side and solve the linear equation for $x$:\n\n$$\\begin{aligned} 6 &= 12x + 6 \\\\[0.5em] 12x &= 0 \\\\[0.5em] x &= 0 \\end{aligned}$$\n\nShowing the expanded linear equation earns the third mark (**M1**), and finding the final correct value of $x = 0$ earns the final accuracy mark (**A1**)."}]},{"id":"working_backwards","label":"Working Backwards","steps":[{"type":"text","order":0,"title":"Understand the method","content":"To solve the composite equation $gf(x) = 3$ using the **Working Backwards** method, we break the problem down into two simpler steps:\n\n1. Replace the inner function $f(x)$ with an intermediate variable $y$, turning the equation into $g(y) = 3$.\n2. Solve $g(y) = 3$ to find the value of $y$.\n3. Set $f(x) = y$ and solve for $x$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Let } \\htmlData{anchor=y-var}{\\color{@sky}{y}} = f(x)$"},{"text":"Substitute this into $g(x) = 3$ to solve for $y$:"},{"latex":"$$g(\\htmlData{anchor=y-in}{\\color{@sky}{y}}) = \\frac{6}{\\htmlData{anchor=y-denom}{\\color{@sky}{y}} + 5} = 3$"},{"latex":"$$6 = 3(\\htmlData{anchor=y-mult}{\\color{@sky}{y}} + 5)$"},{"latex":"$$6 = 3\\htmlData{anchor=y-term}{\\color{@sky}{y}} + 15 \\implies 3\\color{@sky}{y} = -9 \\implies \\htmlData{anchor=y-ans}{\\color{@sky}{y}} = -3$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y-var"},{"color":"sky","style":"text-color","anchor":"y-in"},{"color":"sky","style":"text-color","anchor":"y-ans"}],"connections":[{"to":"y-in","from":"y-var","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Solve for the intermediate variable","description":"We define an intermediate variable $y$ representing the output of $f(x)$, so we can find what input $y$ into the function $g$ yields an output of $3$."},{"type":"text","order":2,"title":"Solve for x","content":"Now that we know the intermediate value must be $y = -3$, we set the inner function $f(x)$ equal to $-3$ and solve for $x$:\n\n$$\\begin{aligned} f(x) &= -3 \\\\ 4x - 3 &= -3 \\\\ 4x &= 0 \\\\ x &= 0 \\end{aligned}$$\n\nThis gives us our final solution."},{"type":"text","order":3,"title":"Mark scheme breakdown","content":"Here is how the marks are distributed for this question:\n\n* **M1**: Correctly setting up the equation, either by finding the full composite function $gf(x) = \\frac{6}{4x+2}$ or using our working backwards approach with $g(y) = 3$.\n* **M1**: Rearranging the equation to eliminate the fraction (e.g., multiplying both sides by the denominator).\n* **M1**: Simplifying the equation down to a basic linear format (e.g., $12x = 0$).\n* **A1**: Reaching the correct final answer of $x = 0$."}]}],"generatedAt":"2026-06-11T03:55:18.594Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"42324282-2795-4f5e-8b28-76df8f60c575:inside_out_evaluation","label":"(a) Inside-Out Evaluation","steps":[{"type":"text","order":0,"title":"Understand the composite function","content":"To evaluate the composite function $fg(-2)$ using the **Inside-Out Evaluation** approach, we start from the \"inside\" function and work our way \"out\".\n\nIn the expression $fg(-2)$, which can also be written as $f(g(-2))$:\n* The **inner function** is $g(x)$.\n* The **outer function** is $f(x)$.\n\nFirst, we will calculate the numerical value of the inner function, $g(-2)$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"g(\\color{@sky}{-2}) = \\frac{6}{\\htmlData{anchor=x-val}{\\color{@sky}{-2}} + 5}"},{"latex":"$$g(-2) = \\frac{6}{\\htmlData{anchor=denom}{\\color{@amber}{3}}}$"},{"latex":"$$g(-2) = \\htmlData{anchor=inner-res}{\\color{@purple}{2}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"x-val"},{"color":"amber","style":"text-color","anchor":"denom"},{"color":"purple","style":"text-color","anchor":"inner-res"}],"connections":[{"to":"denom","from":"x-val","color":"amber","label":"simplify","style":"solid"},{"to":"inner-res","from":"denom","color":"purple","label":"divide","style":"solid"}]},"order":1,"title":"Evaluate the inner function","description":"We substitute $x = -2$ into the definition of $g(x) = \\frac{6}{x+5}$ and simplify to get our intermediate result."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f(\\htmlData{anchor=input-val}{\\color{@purple}{2}}) = 4(\\htmlData{anchor=x-sub}{\\color{@purple}{2}}) - 3$"},{"latex":"$$f(2) = \\htmlData{anchor=mult}{\\color{@orange}{8}} - 3$"},{"latex":"$$f(2) = \\htmlData{anchor=final}{\\color{@green}{5}}$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"input-val"},{"color":"purple","style":"text-color","anchor":"x-sub"},{"color":"orange","style":"text-color","anchor":"mult"},{"color":"green","style":"box","anchor":"final"}],"connections":[{"to":"x-sub","from":"input-val","color":"purple","label":"substitute","style":"solid"},{"to":"mult","from":"x-sub","color":"orange","label":"multiply","style":"solid"},{"to":"final","from":"mult","color":"green","label":"subtract","style":"solid"}]},"order":2,"title":"Evaluate the outer function","description":"Now, we substitute the result from the inner function, $2$, as the input into the outer function $f(x) = 4x - 3$."},{"type":"text","order":3,"title":"State the final answer","content":"By evaluating the inside function first and then the outside function, we find:\n\n$$fg(-2) = 5$$"}]},{"id":"42324282-2795-4f5e-8b28-76df8f60c575:algebraic_composition","label":"(a) Algebraic Composition First","steps":[{"type":"text","order":0,"title":"Understand composite functions","content":"To find the composite function $fg(x)$, we substitute the entire function $g(x)$ into the function $f(x)$. \n\nWe are given:\n- $$f(x) = 4x - 3$$\n- $$g(x) = \\frac{6}{x+5}$$","highlights":[{"text":"f: x \\rightarrow 4x - 3","location":"specification"},{"text":"g: x \\rightarrow \\frac{6}{x+5}","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$g(x) = \\htmlData{anchor=g-expr}{\\color{@sky}{\\frac{6}{x+5}}}$"},{"latex":"$$fg(x) = f\\left( \\htmlData{anchor=g-sub}{\\color{@sky}{\\frac{6}{x+5}}} \\right)$"},{"latex":"$$fg(x) = 4\\left( \\color{@sky}{\\frac{6}{x+5}} \\right) - 3$"},{"latex":"$$\\phantom{fg(x)} = \\htmlData{anchor=fg-expr}{\\color{@purple}{\\frac{24}{x+5} - 3}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"g-expr"},{"color":"purple","style":"box","anchor":"fg-expr"}],"connections":[{"to":"g-sub","from":"g-expr","color":"sky","label":"substitute g(x)","style":"solid"}]},"order":1,"title":"Find the composite expression","description":"We substitute the expression for $g(x)$ into $f(x)$ wherever $x$ occurs."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$fg(x) = \\frac{24}{\\htmlData{anchor=x-var}{\\color{@amber}{x}} + 5} - 3$"},{"latex":"$$fg(-2) = \\frac{24}{\\htmlData{anchor=x-sub}{\\color{@amber}{-2}} + 5} - 3$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"x-var"},{"color":"amber","style":"text-color","anchor":"x-sub"}],"connections":[{"to":"x-sub","from":"x-var","color":"amber","label":"x = -2","style":"solid"}]},"order":2,"title":"Substitute the value","description":"Now, we substitute $x = -2$ into our simplified algebraic expression for $fg(x)$."},{"type":"text","order":3,"title":"Evaluate the final result","content":"Simplify the fraction to get the final answer:\n\n$$\\begin{aligned} fg(-2) &= \\frac{24}{3} - 3 \\\\ &= 8 - 3 \\\\ &= 5 \\end{aligned}$$\n\n**Note on Markscheme:** Even though the markscheme shows the working by calculating $g(-2)$ first (which gives $2$) and then finding $f(2)$, our algebraic composition method is equally valid and arrives at the same final correct answer of $5$!"}]},{"id":"5ec45b13-b55f-4131-a0da-2b4f032bdfc0:composite_first","label":"(b) Forming the Composite Function First","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"f(x) = \\blank{\\htmlData{anchor=f-expr}{\\color{@sky}{4x - 3}}}"},{"latex":"$$g(x) = \\frac{6}{\\htmlData{anchor=g-x}{\\color{@sky}{x}} + 5}$"},{"latex":"$$gf(x) = \\frac{6}{(\\htmlData{anchor=gf-sub}{\\color{@sky}{4x - 3}}) + 5}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"f-expr"},{"color":"sky","style":"text-color","anchor":"gf-sub"}],"connections":[{"to":"gf-sub","from":"f-expr","color":"sky","label":null,"style":"solid"}]},"order":0,"title":"Form the composite function","description":"To find the composite function $gf(x)$, we substitute the entire expression for $f(x)$ into $g(x)$ everywhere there is an $x$. This gives us the expression shown above."},{"type":"text","order":1,"title":"Simplify the denominator","content":"Now, we simplify the denominator of our composite function $gf(x)$:\n\n$$\\begin{aligned} gf(x) &= \\frac{6}{4x - 3 + 5} \\\\[1em] &= \\frac{6}{4x + 2} \\end{aligned}$$\n\nThis simplified expression earns you the first mark (**M1**)."},{"type":"text","order":2,"title":"Set the equation equal to 3","content":"Next, we set our composite function equal to $3$ as required by the equation $gf(x) = 3$:\n\n$$\\frac{6}{4x + 2} = 3$$\n\nTo solve this, multiply both sides by the denominator $(4x + 2)$ to clear the fraction:\n\n$$6 = 3(4x + 2)$$\n\nWriting this equation earns the second mark (**M1**)."},{"type":"text","order":3,"title":"Solve for x","content":"Now we expand the right-hand side and solve the linear equation for $x$:\n\n$$\\begin{aligned} 6 &= 12x + 6 \\\\[0.5em] 12x &= 0 \\\\[0.5em] x &= 0 \\end{aligned}$$\n\nShowing the expanded linear equation earns the third mark (**M1**), and finding the final correct value of $x = 0$ earns the final accuracy mark (**A1**)."}]},{"id":"5ec45b13-b55f-4131-a0da-2b4f032bdfc0:working_backwards","label":"(b) Working Backwards","steps":[{"type":"text","order":0,"title":"Understand the method","content":"To solve the composite equation $gf(x) = 3$ using the **Working Backwards** method, we break the problem down into two simpler steps:\n\n1. Replace the inner function $f(x)$ with an intermediate variable $y$, turning the equation into $g(y) = 3$.\n2. Solve $g(y) = 3$ to find the value of $y$.\n3. Set $f(x) = y$ and solve for $x$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Let } \\htmlData{anchor=y-var}{\\color{@sky}{y}} = f(x)$"},{"text":"Substitute this into $g(x) = 3$ to solve for $y$:"},{"latex":"$$g(\\htmlData{anchor=y-in}{\\color{@sky}{y}}) = \\frac{6}{\\htmlData{anchor=y-denom}{\\color{@sky}{y}} + 5} = 3$"},{"latex":"$$6 = 3(\\htmlData{anchor=y-mult}{\\color{@sky}{y}} + 5)$"},{"latex":"$$6 = 3\\htmlData{anchor=y-term}{\\color{@sky}{y}} + 15 \\implies 3\\color{@sky}{y} = -9 \\implies \\htmlData{anchor=y-ans}{\\color{@sky}{y}} = -3$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"y-var"},{"color":"sky","style":"text-color","anchor":"y-in"},{"color":"sky","style":"text-color","anchor":"y-ans"}],"connections":[{"to":"y-in","from":"y-var","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Solve for the intermediate variable","description":"We define an intermediate variable $y$ representing the output of $f(x)$, so we can find what input $y$ into the function $g$ yields an output of $3$."},{"type":"text","order":2,"title":"Solve for x","content":"Now that we know the intermediate value must be $y = -3$, we set the inner function $f(x)$ equal to $-3$ and solve for $x$:\n\n$$\\begin{aligned} f(x) &= -3 \\\\ 4x - 3 &= -3 \\\\ 4x &= 0 \\\\ x &= 0 \\end{aligned}$$\n\nThis gives us our final solution."},{"type":"text","order":3,"title":"Mark scheme breakdown","content":"Here is how the marks are distributed for this question:\n\n* **M1**: Correctly setting up the equation, either by finding the full composite function $gf(x) = \\frac{6}{4x+2}$ or using our working backwards approach with $g(y) = 3$.\n* **M1**: Rearranging the equation to eliminate the fraction (e.g., multiplying both sides by the denominator).\n* **M1**: Simplifying the equation down to a basic linear format (e.g., $12x = 0$).\n* **A1**: Reaching the correct final answer of $x = 0$."}]}],"generatedAt":"2026-06-11T03:55:18.641Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths__ocr_mei","summaries":[{"id":"58009e5f-cb73-4f77-aacc-62f15f67f802","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"a848e687-ea93-416b-917d-2021e2674d27","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"1debb36a-2bd6-4c5f-8b6f-ab7ddb284ef3","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"84cd14c2-de99-4b55-bbfa-cbebf9d3c204","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"a8d550b4-2216-4e88-b4bf-8092a0ef3a9b","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"2e72a13d-0133-49dd-bb77-e5bdec80bf1b","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"98874a9e-d554-40b4-9ecf-6a130e82838e","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"fac15d6f-33ea-49de-8341-30b527130c30","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"63b534da-10a4-4c73-870b-70428325dca9","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"63ff0a99-985c-45db-9356-486541697e4c","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"6dc8afb1-3d30-4a58-83d2-93219d674f5c","difficultyBand":"medium","calculatorUse":null,"totalMarks":15,"questionType":"LA","isCalculation":true},{"id":"87832fe6-e976-455e-856e-e25b02be24c0","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"c5b3c735-5cc8-4332-a912-38afc92cd365","difficultyBand":"medium","calculatorUse":null,"totalMarks":11,"questionType":"LA","isCalculation":true},{"id":"d2c3d172-e6d3-41a7-b9cc-13ee76c0b3f3","difficultyBand":"medium","calculatorUse":null,"totalMarks":11,"questionType":"LA","isCalculation":true},{"id":"e110d95e-eddc-47c7-9418-bf756eea7cf9","difficultyBand":"medium","calculatorUse":null,"totalMarks":11,"questionType":"LA","isCalculation":true}],"title":"1.3.4 Composite functions (A-level only)","topicId":"2a53c63b-714a-4b92-83d4-cacc5e0717cb","topicSidebar":"$L3c","topicSlug":"composite-functions","topicTitle":"1.3.4 Composite functions (A-level only)"}]}],"$L3d","$L3e"]}]