89:"UserProgressProvider" 8a:I[651451,["$1","$2","$3","$4","$5","$6","$7","$8","$9","$a","$b","$c","$d","$e","$f","$10","$11","$12","$13","$19","$1a","$1b","$1c","$1d","$1e","$1f","$20","$21","$22","$23","$24","$25","$26","$27","$28","$51","$52","$53","$54","$55","$56","$57","$58","$59","$5a","$5b","$5c","$5d","$5e","$5f","$60","$61","$62","$63","$64","$65","$66","$67","$68","$69","$6a","$6b","$6c","$6d","$6e"],"$89"] 8b:I[196793,["$1","$2","$3","$4","$5","$6","$7","$8","$9","$a","$b","$c","$d","$e","$f","$10","$11","$12","$13","$19","$1a","$1b","$1c","$1d","$1e","$1f","$20","$21","$22","$23","$24","$25","$26","$27","$28","$51","$52","$53","$54","$55","$56","$57","$58","$59","$5a","$5b","$5c","$5d","$5e","$5f","$60","$61","$62","$63","$64","$65","$66","$67","$68","$69","$6a","$6b","$6c","$6d","$6e"],"Gingerbread"] 8d:"TopicRecentlyViewed" 8e:I[716297,["$1","$2","$3","$4","$5","$6","$7","$8","$9","$a","$b","$c","$d","$e","$f","$10","$11","$12","$13","$19","$1a","$1b","$1c","$1d","$1e","$1f","$20","$21","$22","$23","$24","$25","$26","$27","$28","$51","$52","$53","$54","$55","$56","$57","$58","$59","$5a","$5b","$5c","$5d","$5e","$5f","$60","$61","$62","$63","$64","$65","$66","$67","$68","$69","$6a","$6b","$6c","$6d","$6e"],"$8d"] 8f:"RecordResourceResume" 90:I[647935,["$1","$2","$3","$4","$5","$6","$7","$8","$9","$a","$b","$c","$d","$e","$f","$10","$11","$12","$13","$19","$1a","$1b","$1c","$1d","$1e","$1f","$20","$21","$22","$23","$24","$25","$26","$27","$28","$51","$52","$53","$54","$55","$56","$57","$58","$59","$5a","$5b","$5c","$5d","$5e","$5f","$60","$61","$62","$63","$64","$65","$66","$67","$68","$69","$6a","$6b","$6c","$6d","$6e"],"$8f"] 91:"TopicQuestionBank" 92:I[426249,["$1","$2","$3","$4","$5","$6","$7","$8","$9","$a","$b","$c","$d","$e","$f","$10","$11","$12","$13","$19","$1a","$1b","$1c","$1d","$1e","$1f","$20","$21","$22","$23","$24","$25","$26","$27","$28","$51","$52","$53","$54","$55","$56","$57","$58","$59","$5a","$5b","$5c","$5d","$5e","$5f","$60","$61","$62","$63","$64","$65","$66","$67","$68","$69","$6a","$6b","$6c","$6d","$6e"],"$91"] 93:T554,We now simplify the expression to reach the required exact form. First, expand the fraction inside the bracket: $$\frac{dV}{dh} = \frac{\pi}{4} \cdot \frac{864^{2/3}}{\pi^{2/3}} = \frac{\pi^{1 - 2/3}}{4} \cdot 864^{2/3} = \frac{\pi^{1/3}}{4} \cdot 864^{2/3}$$ Next, let's simplify $864^{2/3}$ by writing $864$ in terms of its prime factors or perfect cubes: $$864 = 216 \times 4 = 6^3 \times 4$$ Thus: $$864^{2/3} = (6^3 \times 4)^{2/3} = (6^3)^{2/3} \times 4^{2/3} = 36 \times \sqrt[3]{16}$$ Substituting this back into our derivative expression: $$\frac{dV}{dh} = \frac{\pi^{1/3}}{4} \left(36 \sqrt[3]{16}\right) = 9 \pi^{1/3} \sqrt[3]{16}$$ Combining the cube roots: $$\frac{dV}{dh} = 9 \sqrt[3]{16\pi}$$ This final simplification yields the final accuracy mark (A1). ### Part 2: Find the rate of increase of height To find $\frac{dh}{dt}$ when $t = 6$, we use the chain rule: $$\frac{dh}{dt} = \frac{dh}{dV} \times \frac{dV}{dt}$$ We are given that wheat is dispensed at a constant rate, so $\frac{dV}{dt} = 12\text{ cm}^3\text{s}^{-1}$. Using our result for $\frac{dV}{dh}$ from Part 1: $$\frac{dh}{dt} = \frac{1}{\frac{dV}{dh}} \times 12 = \frac{12}{9 \sqrt[3]{16\pi}}$$ Using a calculator to evaluate this to 3 significant figures: $$\frac{dh}{dt} \approx 0.361\text{ cm s}^{-1}$$ This earns the final 3 marks of the question (M1, M1, A1).94:T484,Substitute $V = 72$ into our expression for $\frac{dV}{dh}$: $$\frac{dV}{dh} = \frac{\pi}{4} \left( \frac{12 \times 72}{\pi} \right)^{2/3}$$ Simplify the terms inside the parentheses first: $$12 \times 72 = 864 = 6^3 \times 4$$ Now, substitute this prime factorization back in: $$\begin{aligned} \frac{dV}{dh} &= \frac{\pi}{4} \left( \frac{864}{\pi} \right)^{2/3} \\ &= \frac{\pi^{1 - 2/3}}{4} \cdot (864)^{2/3} \\ &= \frac{\pi^{1/3}}{4} \left( 6^3 \times 4 \right)^{2/3} \\ &= \frac{\pi^{1/3}}{4} \left( (6^3)^{2/3} \times 4^{2/3} \right) \\ &= \frac{\pi^{1/3}}{4} \left( 36 \times 4^{2/3} \right) \\ &= 9 \pi^{1/3} 4^{2/3} \end{aligned}$$ Combine the base terms under the cube root: $$4^{2/3} \cdot \pi^{1/3} = (16)^{1/3} \cdot \pi^{1/3} = \sqrt[3]{16\pi}$$ This gives our final required result: $$\frac{dV}{dh} = 9 \sqrt[3]{16\pi}$$ ### Part 2: Find the rate of increase of height We apply the chain rule to find $\frac{dh}{dt}$ at $t = 6$: $$\frac{dh}{dt} = \frac{dV}{dt} \times \frac{dh}{dV}$$ Substituting $\frac{dV}{dt} = 12$ and the value of $\frac{dV}{dh}$: $$\frac{dh}{dt} = \frac{12}{9 \sqrt[3]{16\pi}} \approx 0.361\text{ cm s}^{-1}$$95:T554,We now simplify the expression to reach the required exact form. First, expand the fraction inside the bracket: $$\frac{dV}{dh} = \frac{\pi}{4} \cdot \frac{864^{2/3}}{\pi^{2/3}} = \frac{\pi^{1 - 2/3}}{4} \cdot 864^{2/3} = \frac{\pi^{1/3}}{4} \cdot 864^{2/3}$$ Next, let's simplify $864^{2/3}$ by writing $864$ in terms of its prime factors or perfect cubes: $$864 = 216 \times 4 = 6^3 \times 4$$ Thus: $$864^{2/3} = (6^3 \times 4)^{2/3} = (6^3)^{2/3} \times 4^{2/3} = 36 \times \sqrt[3]{16}$$ Substituting this back into our derivative expression: $$\frac{dV}{dh} = \frac{\pi^{1/3}}{4} \left(36 \sqrt[3]{16}\right) = 9 \pi^{1/3} \sqrt[3]{16}$$ Combining the cube roots: $$\frac{dV}{dh} = 9 \sqrt[3]{16\pi}$$ This final simplification yields the final accuracy mark (A1). ### Part 2: Find the rate of increase of height To find $\frac{dh}{dt}$ when $t = 6$, we use the chain rule: $$\frac{dh}{dt} = \frac{dh}{dV} \times \frac{dV}{dt}$$ We are given that wheat is dispensed at a constant rate, so $\frac{dV}{dt} = 12\text{ cm}^3\text{s}^{-1}$. Using our result for $\frac{dV}{dh}$ from Part 1: $$\frac{dh}{dt} = \frac{1}{\frac{dV}{dh}} \times 12 = \frac{12}{9 \sqrt[3]{16\pi}}$$ Using a calculator to evaluate this to 3 significant figures: $$\frac{dh}{dt} \approx 0.361\text{ cm s}^{-1}$$ This earns the final 3 marks of the question (M1, M1, A1).96:T484,Substitute $V = 72$ into our expression for $\frac{dV}{dh}$: $$\frac{dV}{dh} = \frac{\pi}{4} \left( \frac{12 \times 72}{\pi} \right)^{2/3}$$ Simplify the terms inside the parentheses first: $$12 \times 72 = 864 = 6^3 \times 4$$ Now, substitute this prime factorization back in: $$\begin{aligned} \frac{dV}{dh} &= \frac{\pi}{4} \left( \frac{864}{\pi} \right)^{2/3} \\ &= \frac{\pi^{1 - 2/3}}{4} \cdot (864)^{2/3} \\ &= \frac{\pi^{1/3}}{4} \left( 6^3 \times 4 \right)^{2/3} \\ &= \frac{\pi^{1/3}}{4} \left( (6^3)^{2/3} \times 4^{2/3} \right) \\ &= \frac{\pi^{1/3}}{4} \left( 36 \times 4^{2/3} \right) \\ &= 9 \pi^{1/3} 4^{2/3} \end{aligned}$$ Combine the base terms under the cube root: $$4^{2/3} \cdot \pi^{1/3} = (16)^{1/3} \cdot \pi^{1/3} = \sqrt[3]{16\pi}$$ This gives our final required result: $$\frac{dV}{dh} = 9 \sqrt[3]{16\pi}$$ ### Part 2: Find the rate of increase of height We apply the chain rule to find $\frac{dh}{dt}$ at $t = 6$: $$\frac{dh}{dt} = \frac{dV}{dt} \times \frac{dh}{dV}$$ Substituting $\frac{dV}{dt} = 12$ and the value of $\frac{dV}{dh}$: $$\frac{dh}{dt} = \frac{12}{9 \sqrt[3]{16\pi}} \approx 0.361\text{ cm s}^{-1}$$72:["$","$L8a",null,{"bundle":{"topicIds":["0c24674e-58ce-484a-b678-2a06e4e833be","d55f3cc0-127f-4dfa-af20-4acc04a4e08a","53b63989-f9dd-4345-9ca2-6c20ed806929","87b3f917-7320-4818-a085-bb3e7d5cf18a","3433c14a-bece-4e88-b54a-017b8ea98616","1572aa1e-6368-40c2-a01a-b308ded37abe","68360771-a573-4380-9a2b-d7954d73dde3","04b9dade-34aa-4c12-a50e-58b8bec05304","75039396-8c4a-422e-b8ee-8eb28fa3b932","2b685fd8-b028-404d-b217-e3285194898a","30c82f6a-e1f8-43f3-9904-ab0805f92da2","2aff6e78-66cf-43f8-bc40-5ba507e13645","6b211088-bfc8-4a27-8e9c-5d1b5b334c8e","a067c9a3-7710-4b9e-b2af-05ff94e429dc"],"paperIds":[],"stageIds":[],"stageTopicIds":[],"stageAssessmentIds":[],"topicNoteIds":[]},"children":[["$","$L8b",null,{"feature":"question-bank"}],"$L8c",["$","$L8e",null,{"subjectCode":"a_level_maths__aqa","subjectName":"Maths","topicSlug":"pure-mathematics","topicTitle":"1 Pure Mathematics","sectionTitle":null,"activity":"worksheet","resourceTitle":"1 Pure Mathematics Questions","resourceHref":"/a-level/maths/aqa/pure-mathematics/questions/q1","saveTrail":false}],["$","$L90",null,{"href":"/a-level/maths/aqa/pure-mathematics/questions/q1","label":"1 Pure Mathematics","subjectCode":"a_level_maths__aqa","surface":"questions"}],["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"LearningResource\",\"name\":\"Question 2319 – 1 Pure Mathematics - A Level Maths Questions\",\"description\":\"Practise 1 Pure Mathematics A Level Maths exam-style questions with worked solutions on MathsGenie.\",\"learningResourceType\":\"Practice questions\",\"educationalLevel\":\"A Level\",\"about\":{\"@type\":\"Thing\",\"name\":\"1 Pure Mathematics\"},\"url\":\"https://mathsgenie.co.uk/a-level/maths/aqa/pure-mathematics/questions/q2319\",\"inLanguage\":\"en-GB\",\"provider\":{\"@id\":\"https://mathsgenie.co.uk/#organization\"}}"}}],["$","div",null,{"className":"min-h-0 w-full flex-1 overflow-hidden bg-background","children":["$","$L92",null,{"backHref":"/a-level/maths/aqa/questions","basePath":"/a-level/maths/aqa/pure-mathematics/questions","formulaUrl":"https://filestore.aqa.org.uk/resources/mathematics/AQA-AS-A-MATHS-FORMULAE.PDF","initialIndex":2318,"initialQuestionWindow":[{"id":"dcb6f220-a63d-4134-b6a6-cdfd15c5382e","revisionId":"e0a78424-e1e6-431b-b9fe-dc91f75ccda0","questionNumber":null,"questionType":"LA","specification":"A specialized industrial centrifuge extracts a mineral from a solution such that the mass extracted each hour forms a geometric series. The theoretical maximum mass that can be extracted over an infinite time period is 150 grams. In the first hour, the centrifuge extracts \$a\$ grams, where \$a > 80\$. In the second hour, it extracts 36 grams.","difficultyBand":"hard","tier":null,"totalMarks":11,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"2526755c-b191-44f8-8920-501555a133a9","label":"a","content":"Determine the value of the first term, \$a\$, and the common ratio, \$r\$, of the series.","metadata":null,"markscheme":"- Uses $S_{\\infty} = \\frac{a}{1-r} = 150 \\implies a = 150(1-r)$ M1\n- Uses $u_2 = ar = 36$ B1\n- Substitutes to form a quadratic: $150r(1-r) = 36$ or $150r^2 - 150r + 36 = 0$ M1\n- Solves quadratic to find $r = 0.4$ or $r = 0.6$ (or equivalent fractions) M1\n- Since $a = \\frac{36}{r}$, if $r=0.6, a=60$; if $r=0.4, a=90$. Selects $a=90$ and $r=0.4$ as $a > 80$ A1","marks":5,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"substitution_for_r","label":"Substitution in terms of r","steps":[{"type":"text","order":0,"title":"Set up the equations","content":"We are given two pieces of information about the geometric series:\n\n1. The **theoretical maximum mass** (the sum to infinity, $S_{\\infty}$) is $150\\text{ g}$. The formula for the sum to infinity of a geometric series is:\n $$S_{\\infty} = \\frac{a}{1-r}$$\n Therefore, we can write:\n $$\\frac{a}{1-r} = 150$$\n\n2. The mass extracted in the **second hour** ($u_2$) is $36\\text{ g}$. The formula for the $n$-th term of a geometric series is $u_n = ar^{n-1}$, so:\n $$ar = 36$$","highlights":[{"text":"The theoretical maximum mass that can be extracted over an infinite time period is 150 grams.","location":"specification"},{"text":"In the second hour, it extracts 36 grams.","location":"specification"}]},{"type":"text","order":1,"title":"Express a in terms of r","content":"Using the sum to infinity equation, we can rearrange it to express the first term $a$ in terms of the common ratio $r$:\n\n$$a = 150(1-r)$$\n\nThis will allow us to substitute this expression into our second equation, reducing it to a single variable."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$a = \\htmlData{anchor=a-expr}{\\color{@sky}{150(1-r)}}$"},{"latex":"$$\\htmlData{anchor=a-var}{\\color{@sky}{a}}r = 36$"},{"latex":"$$\\htmlData{anchor=sub-eq}{\\color{@sky}{150(1-r)}}r = 36$"},{"latex":"$$150r - 150r^2 = 36 \\implies 150r^2 - 150r + 36 = 0$"}],"highlights":[{"color":"sky","style":"box","anchor":"a-expr"},{"color":"sky","style":"text-color","anchor":"a-var"}],"connections":[{"to":"a-var","from":"a-expr","color":"sky","label":"Substitute a","style":"solid"}]},"order":2,"title":"Substitute and form a quadratic","description":"We substitute our expression for $a$ into the second-term equation $ar = 36$ to obtain a quadratic equation in $r$."},{"type":"text","order":3,"title":"Solve the quadratic equation","content":"To solve the quadratic equation $150r^2 - 150r + 36 = 0$, first simplify it by dividing all terms by $6$:\n\n$$25r^2 - 25r + 6 = 0$$\n\nNow, we can factorise the quadratic equation:\n\n$$(5r - 2)(5r - 3) = 0$$\n\nThis gives two possible values for $r$:\n\n$$r = \\frac{2}{5} = 0.4 \\quad \\text{or} \\quad r = \\frac{3}{5} = 0.6$$","calculator":[{"gdc":{"expression":"150x^2 - 150x + 36 = 0","instructions":"1. Go to the Equation/Solver menu.\n2. Choose Polynomial -> Degree 2.\n3. Enter the coefficients: a = 150, b = -150, c = 36.\n4. Press Solve to get the roots: x = 0.4 and x = 0.6."},"ti84":{"expression":"150x^2 - 150x + 36 = 0","instructions":"1. Press [APPS] and select 'PlySmlt2'.\n2. Select '1: Polynomial Root Finder'.\n3. Set the order to 2 and press [NEXT] (Graph button).\n4. Enter the coefficients: a2 = 150, a1 = -150, a0 = 36.\n5. Press [SOLVE] (Graph button) to find the roots: x = 0.6 and x = 0.4."},"desmos":{"expression":"y = 150x^2 - 150x + 36","instructions":"1. Type the quadratic function into the input bar.\n2. Click on the x-intercepts of the plotted curve to view the roots."},"description":"You can solve this quadratic equation quickly using your calculator's equation solver or by finding the x-intercepts of the graph."}]},{"type":"text","order":4,"title":"Find the valid values","content":"Now we find the corresponding values of $a$ using the relation $a = \\frac{36}{r}$:\n\n* If $r = 0.6$:\n $$a = \\frac{36}{0.6} = 60$$\n\n* If $r = 0.4$:\n $$a = \\frac{36}{0.4} = 90$$\n\nSince the question specifies that $a > 80$, the value $a = 60$ is rejected.\n\nThus, the correct values are:\n$$\\mathbf{a = 90} \\quad \\text{and} \\quad \\mathbf{r = 0.4}$$","highlights":[{"text":"a > 80","location":"specification"}]}]},{"id":"substitution_for_a","label":"Substitution in terms of a","steps":[{"type":"text","order":0,"title":"Identify the given information","content":"From the problem statement, we are given:\n- The first term of the geometric series is $a$, with the constraint $a > 80$.\n- The second term is $u_2 = 36$ grams.\n- The sum to infinity is $S_{\\infty} = 150$ grams.\n\nWe need to find the values of $a$ and $r$.","highlights":[{"text":"theoretical maximum mass that can be extracted over an infinite time period is 150 grams","location":"specification"},{"text":"first hour, the centrifuge extracts \$a\$ grams, where \$a > 80\$","location":"specification"},{"text":"second hour, it extracts 36 grams","location":"specification"}]},{"type":"text","order":1,"title":"Set up the formulas","content":"Recall the standard formulas for a geometric series from your formula booklet:\n- The $n$-th term of a geometric series is:\n$$u_n = a r^{n-1}$$\n- The sum to infinity is:\n$$S_{\\infty} = \\frac{a}{1-r}$$\n\nApplying these to our given information, we get two equations:\n1) For the second term ($n=2$):\n$$ar = 36$$\n2) For the sum to infinity:\n$$\\frac{a}{1-r} = 150$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Using the second term equation: } \\htmlData{anchor=eq1}{\\color{@sky}{ar = 36}}$"},{"latex":"$$\\text{We can rearrange this to express } r \\text{ in terms of } a: \\htmlData{anchor=eq2}{\\color{@amber}{r = \\frac{36}{a}}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"eq1"},{"color":"amber","style":"box","anchor":"eq2"}],"connections":[{"to":"eq2","from":"eq1","color":"sky","label":"Rearrange for r","style":"solid"}]},"order":2,"title":"Express r in terms of a","description":"To follow our substitution method, we isolate the common ratio $r$ from our second term equation."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$S_{\\infty} = \\frac{a}{1-\\htmlData{anchor=r-sub}{\\color{@amber}{r}}} = 150$"},{"latex":"$$\\text{Substitute } \\htmlData{anchor=r-val}{\\color{@amber}{r = \\frac{36}{a}}} \\text{ into the formula:}$"},{"latex":"$$150 \\left(1 - \\htmlData{anchor=subbed}{\\color{@amber}{\\frac{36}{a}}}\\right) = a$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"r-sub"},{"color":"amber","style":"text-color","anchor":"r-val"},{"color":"amber","style":"box","anchor":"subbed"}],"connections":[{"to":"subbed","from":"r-val","color":"amber","label":"Substitute","style":"solid"}]},"order":3,"title":"Form a quadratic in a","description":"We substitute $r = \\frac{36}{a}$ into the rearranged sum to infinity equation, $a = 150(1-r)$, to form a single equation in terms of $a$."},{"type":"text","order":4,"title":"Solve the quadratic equation","content":"Let's expand and rearrange the equation to form a standard quadratic equation:\n\n$$\\begin{aligned} 150 \\left(1 - \\frac{36}{a}\\right) &= a \\\\[8pt] 150 - \\frac{5400}{a} &= a \\end{aligned}$$\n\nMultiply the entire equation by $a$ to remove the fraction:\n\n$$150a - 5400 = a^2$$\n\nRearranging into standard quadratic form $Ax^2 + Bx + C = 0$:\n\n$$a^2 - 150a + 5400 = 0$$"},{"type":"text","order":5,"title":"Find the values of a","content":"Now we solve the quadratic equation $a^2 - 150a + 5400 = 0$ by factoring:\n\nWe need two numbers that multiply to $5400$ and add up to $-150$. These numbers are $-90$ and $-60$:\n\n$$(a - 90)(a - 60) = 0$$\n\nThis gives two possible values for $a$:\n$$a = 90 \\quad \\text{or} \\quad a = 60$$"},{"type":"text","order":6,"title":"Determine the correct a","content":"The question specifies that the first term must satisfy the condition $a > 80$.\n\nComparing our two possible values:\n- For $a = 60$, this does not satisfy $a > 80$.\n- For $a = 90$, this satisfies $a > 80$.\n\nTherefore, we choose **$a = 90$**.","highlights":[{"text":"where \$a > 80\$","location":"specification"}]},{"type":"text","order":7,"title":"Calculate the common ratio r","content":"With the correct value of $a = 90$ identified, we substitute it back into our equation for $r$:\n\n$$r = \\frac{36}{a}$$\n\n$$r = \\frac{36}{90} = 0.4$$\n\nThus, the final solutions are **$a = 90$** and **$r = 0.4$**."}]},{"id":"graphical_intersection","label":"Graphical Approach","steps":[{"type":"text","order":0,"title":"Formulate the first equation","content":"The theoretical maximum mass extracted over an infinite time period is given by the sum to infinity of a geometric series, $S_{\\infty}$. The formula for $S_{\\infty}$ is:\n\n$$S_{\\infty} = \\frac{a}{1-r}$$\n\nwhere $a$ is the first term and $r$ is the common ratio. We are given $S_{\\infty} = 150$ grams:\n\n$$\\frac{a}{1-r} = 150$$\n\nRearranging this to make $a$ the subject gives our first equation as a function of $r$:\n\n$$a = 150(1-r)$$\n\nThis would earn the first mark M1.","highlights":[{"text":"The theoretical maximum mass that can be extracted over an infinite time period is 150 grams.","location":"specification"}]},{"type":"text","order":1,"title":"Formulate the second equation","content":"The mass extracted in the second hour represents the second term of the geometric series, $u_2$, which is defined as:\n\n$$u_2 = ar$$\n\nWe are given that $u_2 = 36$ grams:\n\n$$ar = 36$$\n\nRearranging this to isolate $a$ gives our second equation as a function of $r$:\n\n$$a = \\frac{36}{r}$$\n\nThis earns the second mark B1.","highlights":[{"text":"In the second hour, it extracts 36 grams.","location":"specification"}]},{"type":"text","order":2,"title":"Plot and find intersections","content":"To find the solution graphically, we can treat $r$ as the independent variable ($x$-axis) and $a$ as the dependent variable ($y$-axis). We define two functions:\n\n$$y_1 = 150(1-x)$$\n$$y_2 = \\frac{36}{x}$$\n\nUsing a graphing calculator or utility, we plot these functions and find their points of intersection. The intersection points representing values of $r$ and $a$ are:\n\n$$(0.4, 90) \\quad \\text{and} \\quad (0.6, 60)$$\n\nIdentifying these intersections earns M1 for forming the equivalence and M1 for solving to find $r = 0.4$ or $r = 0.6$.","calculator":[{"gdc":{"instructions":"1. Go to the Graph menu.\n2. Enter the functions \\\\f1(x) = 150(1-x)\\\\ and \\\\f2(x) = 36/x\\\\.\n3. Plot the graphs using an appropriate window scale (e.g., x-axis from 0 to 1, y-axis from 0 to 150).\n4. Use the G-Solv (Graph Solve) tool and select Intersect (INTSECT).\n5. The calculator display will show the two intersection points: (0.4, 90) and (0.6, 60)."},"ti84":{"instructions":"1. Press [Y=].\n2. Enter \\\\Y_1 = 150(1-X)\\\\.\n3. Enter \\\\Y_2 = 36/X\\\\.\n4. Press [WINDOW] and set Xmin = 0, Xmax = 1, Ymin = 0, Ymax = 150.\n5. Press [GRAPH].\n6. Press [2nd] [CALC] and select 5: intersect.\n7. Move the cursor near the first intersection and press [ENTER] three times to find the first point: (0.4, 90).\n8. Repeat the intersection step near the second point to find: (0.6, 60)."},"desmos":{"instructions":"1. Type \\\\y = 150(1-x)\\\\ in the first input bar.\n2. Type \\\\y = 36/x\\\\ in the second input bar.\n3. Click on the gray intersection dots on the plot to reveal their coordinates: (0.4, 90) and (0.6, 60)."},"description":"Find the intersection points of the two functions graphically."}]},{"type":"text","order":3,"title":"Apply constraint and state values","content":"We have two potential solutions from our graphical analysis:\n\n1. If $r = 0.6$, then $a = 60$\n2. If $r = 0.4$, then $a = 90$\n\nThe question specifies the constraint $a > 80$. Looking at our points:\n- For $(0.6, 60)$, $a = 60 \\ngtr 80$, so we reject this solution.\n- For $(0.4, 90)$, $a = 90 > 80$, which satisfies the condition.\n\nTherefore, the correct values are:\n\n$$a = 90 \\quad \\text{and} \\quad r = 0.4$$\n\nThis final selection earns the A1 mark.","highlights":[{"text":"In the first hour, the centrifuge extracts \$a\$ grams, where \$a > 80\$.","location":"specification"}]}]}],"generatedAt":"2026-06-09T10:06:58.418Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"40665fe7-7a74-4dff-bff1-c0fdd7f397cf","label":"bi","content":"Show that the mass extracted in the \$n\$th hour, \$M_n\$, can be written as\n\\[M_n = \\frac{2^n \\cdot 3^2}{5^{n-2}}\\]","metadata":null,"markscheme":"- State $M_n = 90(0.4)^{n-1}$ M1\n- Writes 90 and 0.4 as prime factors: $90 = 2 \\times 3^2 \\times 5$ and $0.4 = \\frac{2}{5}$ M1\n- Simplifies indices: $M_n = (2 \\cdot 3^2 \\cdot 5) \\cdot \\left(\\frac{2}{5}\\right)^{n-1} = 2^1 \\cdot 3^2 \\cdot 5^1 \\cdot 2^{n-1} \\cdot 5^{-(n-1)} = 2^{1+n-1} \\cdot 3^2 \\cdot 5^{1-n+1} = \\frac{2^n \\cdot 3^2}{5^{n-2}}$ A1","marks":3,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"algebraic","label":"Direct Algebraic Derivation","steps":[{"type":"text","order":0,"title":"Set up the equations","content":"From the question, we know:\n- The first term is $a$.\n- The second term is $ar = 36$.\n- The sum to infinity is $S_\\infty = \\frac{a}{1-r} = 150$.\n\nLet's express $r$ in terms of $a$ from the second term equation:\n$$r = \\frac{36}{a}$$","highlights":[{"text":"first hour, the centrifuge extracts \$a\$ grams","location":"specification"},{"text":"second hour, it extracts 36 grams","location":"specification"},{"text":"theoretical maximum mass that can be extracted over an infinite time period is 150 grams","location":"specification"}]},{"type":"text","order":1,"title":"Solve for first term","content":"Substitute $r = \\frac{36}{a}$ into the sum to infinity equation:\n$$\\frac{a}{1 - \\frac{36}{a}} = 150$$\n\nMultiply the numerator and denominator on the left by $a$:\n$$\\frac{a^2}{a - 36} = 150$$\n\nNow, multiply both sides by $(a - 36)$ to get a quadratic equation:\n$$a^2 = 150(a - 36)$$\n$$a^2 - 150a + 5400 = 0$$\n\nFactoring this quadratic equation:\n$$(a - 90)(a - 60) = 0$$\n\nThis gives two possible values: $a = 90$ or $a = 60$."},{"type":"text","order":2,"title":"Filter the first term","content":"The question specifies that $a > 80$. Therefore, we must choose:\n$$a = 90$$\n\nNow, we can find the common ratio $r$:\n$$r = \\frac{36}{a} = \\frac{36}{90} = 0.4$$","highlights":[{"text":"a > 80","location":"specification"}]},{"type":"text","order":3,"title":"Express general term","content":"The mass extracted in the $n$-th hour, $M_n$, is given by the formula for the $n$-th term of a geometric sequence:\n$$M_n = a r^{n-1}$$\n\nSubstituting our values for $a$ and $r$:\n$$M_n = 90(0.4)^{n-1}$$"},{"type":"text","order":4,"title":"Express as prime factors","content":"To show that $M_n$ can be written in the required form, we write the base numbers $90$ and $0.4$ as prime factors:\n- $90 = 2 \\times 3^2 \\times 5$\n- $0.4 = \\frac{2}{5} = 2 \\times 5^{-1}$\n\nNow substitute these prime factorizations into the equation for $M_n$:\n$$M_n = \\left(2 \\cdot 3^2 \\cdot 5\\right) \\cdot \\left(\\frac{2}{5}\\right)^{n-1}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$M_n = \\htmlData{anchor=factor-a}{\\color{@sky}{2^1}} \\cdot 3^2 \\cdot \\htmlData{anchor=factor-five}{\\color{@purple}{5^1}} \\cdot \\frac{\\htmlData{anchor=ratio-num}{\\color{@sky}{2^{n-1}}}}{\\htmlData{anchor=ratio-den}{\\color{@purple}{5^{n-1}}}}$"},{"latex":"$$M_n = \\htmlData{anchor=combined-two}{\\color{@sky}{2^{1 + n - 1}}} \\cdot 3^2 \\cdot \\htmlData{anchor=combined-five}{\\color{@purple}{5^{1 - (n - 1)}}}$"},{"latex":"$$M_n = 2^n \\cdot 3^2 \\cdot 5^{2-n} = \\frac{2^n \\cdot 3^2}{5^{n-2}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"factor-a"},{"color":"purple","style":"text-color","anchor":"factor-five"}],"connections":[{"to":"combined-two","from":"factor-a","color":"sky","label":"combine base 2","style":"solid"},{"to":"combined-five","from":"factor-five","color":"purple","label":"combine base 5","style":"solid"}]},"order":5,"title":"Apply index laws","description":"Combine the terms with like bases using index laws: $x^a \\cdot x^b = x^{a+b}$ and $\\frac{x^a}{x^b} = x^{a-b}$."}]}],"generatedAt":"2026-06-09T10:07:58.231Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"04893333-a5be-4b93-ba80-c4b8f5962917","label":"bii","content":"Hence show that\n\\[\\log_2 M_n = n(1 - \\log_2 5) + (2 \\log_2 3 + 2 \\log_2 5)\\]","metadata":null,"markscheme":"- Applies log laws to the result in (b)(i): $\\log_2 M_n = \\log_2 \\left( \\frac{2^n \\cdot 3^2}{5^{n-2}} \\right)$ M1\n- Uses $\\log(\\frac{xy}{z}) = \\log x + \\log y - \\log z$: $\\log_2(2^n) + \\log_2(3^2) - \\log_2(5^{n-2})$ M1\n- Uses $\\log a^k = k \\log a$ and $\\log_2 2 = 1$ to reach $n + 2 \\log_2 3 - (n-2) \\log_2 5$ and factorises to reach $n(1 - \\log_2 5) + (2 \\log_2 3 + 2 \\log_2 5)$ A1","marks":3,"order":2,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"forward_log_laws","label":"Direct Algebraic Expansion","steps":[{"type":"text","order":0,"title":"State the expression for $M_n$","content":"From the previous part of the question, we are given the expression for $M_n$:\n\n$$M_n = \\frac{2^n \\cdot 3^2}{5^{n-2}}$$\n\nTo find $\\log_2 M_n$, we start by taking the base-2 logarithm of both sides:\n\n$$\\log_2 M_n = \\log_2 \\left( \\frac{2^n \\cdot 3^2}{5^{n-2}} \\right)$$\n\nThis earns the first mark M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\log_2 M_n = \\log_2 \\left( \\frac{\\htmlData{anchor=num}{2^n \\cdot 3^2}}{\\htmlData{anchor=den}{5^{n-2}}} \\right)$$"},{"latex":"$$$\\phantom{\\log_2 M_n} = \\log_2(\\htmlData{anchor=num-sub}{2^n \\cdot 3^2}) - \\log_2(\\htmlData{anchor=den-sub}{5^{n-2}})$$"},{"latex":"$$$\\phantom{\\log_2 M_n} = \\log_2(2^n) + \\log_2(3^2) - \\log_2(5^{n-2})$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"num"},{"color":"pink","style":"text-color","anchor":"den"},{"color":"sky","style":"text-color","anchor":"num-sub"},{"color":"pink","style":"text-color","anchor":"den-sub"}],"connections":[{"to":"num-sub","from":"num","color":"sky","label":"Numerator","style":"solid"},{"to":"den-sub","from":"den","color":"pink","label":"Denominator","style":"solid"}]},"order":1,"title":"Apply logarithm laws","description":"We use the quotient law $\\log \\left(\\frac{x}{y}\\right) = \\log x - \\log y$, followed by the product law $\\log(ab) = \\log a + \\log b$ M1."},{"type":"text","order":2,"title":"Simplify using power laws","content":"Now, apply the power law of logarithms, $\\log(a^k) = k \\log a$:\n\n$$\\begin{aligned} \\log_2(2^n) &= n \\log_2 2 = n \\quad \\text{(since } \\log_2 2 = 1\\text{)} \\\\\\\\ \\log_2(3^2) &= 2 \\log_2 3 \\\\\\\\ \\log_2(5^{n-2}) &= (n-2) \\log_2 5 \\end{aligned}$$\n\nSubstituting these simplified terms back into our expression gives:\n\n$$\\log_2 M_n = n + 2 \\log_2 3 - (n - 2)\\log_2 5$$"},{"type":"text","order":3,"title":"Expand and factorise","content":"Expand the bracketed term containing $n$:\n\n$$-(n - 2)\\log_2 5 = -n\\log_2 5 + 2\\log_2 5$$\n\nSubstitute this back to group the terms:\n\n$$\\log_2 M_n = n + 2 \\log_2 3 - n\\log_2 5 + 2\\log_2 5$$\n\nNow, collect the terms containing $n$ together and the constant terms together:\n\n$$\\begin{aligned} \\log_2 M_n &= (n - n\\log_2 5) + (2 \\log_2 3 + 2\\log_2 5) \\\\\\\\ &= n(1 - \\log_2 5) + (2 \\log_2 3 + 2\\log_2 5) \\end{aligned}$$\n\nThis matches the required form A1."}]},{"id":"backward_log_laws","label":"Working Backward","steps":[{"type":"text","order":0,"title":"Identify the target","content":"To prove the statement using the **Working Backward** approach, we will start with the right-hand side (RHS) of the equation and use the laws of logarithms to combine it into a single logarithm:\n\n$$\\text{RHS} = n(1 - \\log_2 5) + (2 \\log_2 3 + 2 \\log_2 5)$$\n\nOur goal is to show that this simplifies to the expression for $\\log_2 M_n$, which we know from previous parts of the question is:\n\n$$\\log_2 M_n = \\log_2 \\left( \\frac{2^n \\cdot 3^2}{5^{n-2}} \\right)$$\n\nLet's expand and rearrange the RHS to prepare for combining the terms."},{"type":"text","order":1,"title":"Expand and group","content":"First, let's expand the terms in the RHS expression:\n\n$$n(1 - \\log_2 5) + 2 \\log_2 3 + 2 \\log_2 5 = n - n \\log_2 5 + 2 \\log_2 3 + 2 \\log_2 5$$\n\nNow, group the $\\log_2 5$ terms together to make it easier to apply the laws of logarithms:\n\n$$= n + 2 \\log_2 3 - (n - 2) \\log_2 5$$"},{"type":"text","order":2,"title":"Convert constant to log","content":"To combine all terms into a single base-2 logarithm, we need to express the linear term $n$ as a logarithm with base 2.\n\nSince $\\log_2 2 = 1$, we can write:\n\n$$n = n \\log_2 2$$\n\nUsing the power law of logarithms ($k \\log_b a = \\log_b a^k$):\n\n$$n \\log_2 2 = \\log_2(2^n)$$\n\nSubstituting this back into our expression gives:\n\n$$\\log_2(2^n) + 2 \\log_2 3 - (n - 2) \\log_2 5$$"},{"type":"text","order":3,"title":"Apply power laws","content":"Next, let's use the power law of logarithms on the remaining terms to move the coefficients inside the logarithms:\n\n* For the second term:\n $$2 \\log_2 3 = \\log_2(3^2)$$\n\n* For the third term:\n $$(n - 2) \\log_2 5 = \\log_2(5^{n-2})$$\n\nSubstituting these back into our expression yields:\n\n$$\\log_2(2^n) + \\log_2(3^2) - \\log_2(5^{n-2})$$"},{"type":"text","order":4,"title":"Combine into one logarithm","content":"Now, we can combine these terms into a single logarithm using the product rule, $\\log_b x + \\log_b y = \\log_b(xy)$, and the quotient rule, $\\log_b x - \\log_b y = \\log_b\\left(\\frac{x}{y}\\right)$:\n\n$$\\begin{aligned} \\log_2(2^n) + \\log_2(3^2) - \\log_2(5^{n-2}) &= \\log_2(2^n \\cdot 3^2) - \\log_2(5^{n-2}) \\\\\\\\ &= \\log_2 \\left( \\frac{2^n \\cdot 3^2}{5^{n-2}} \\right) \\end{aligned}$$\n\nSince $M_n = \\frac{2^n \\cdot 3^2}{5^{n-2}}$, this is exactly:\n\n$$\\log_2 M_n$$\n\nThus, we have successfully shown that:\n\n$$\\log_2 M_n = n(1 - \\log_2 5) + (2 \\log_2 3 + 2 \\log_2 5)$$"}]}],"generatedAt":"2026-06-09T10:09:22.149Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"2526755c-b191-44f8-8920-501555a133a9:substitution_for_r","label":"(a) Substitution in terms of r","steps":[{"type":"text","order":0,"title":"Set up the equations","content":"We are given two pieces of information about the geometric series:\n\n1. The **theoretical maximum mass** (the sum to infinity, $S_{\\infty}$) is $150\\text{ g}$. The formula for the sum to infinity of a geometric series is:\n $$S_{\\infty} = \\frac{a}{1-r}$$\n Therefore, we can write:\n $$\\frac{a}{1-r} = 150$$\n\n2. The mass extracted in the **second hour** ($u_2$) is $36\\text{ g}$. The formula for the $n$-th term of a geometric series is $u_n = ar^{n-1}$, so:\n $$ar = 36$$","highlights":[{"text":"The theoretical maximum mass that can be extracted over an infinite time period is 150 grams.","location":"specification"},{"text":"In the second hour, it extracts 36 grams.","location":"specification"}]},{"type":"text","order":1,"title":"Express a in terms of r","content":"Using the sum to infinity equation, we can rearrange it to express the first term $a$ in terms of the common ratio $r$:\n\n$$a = 150(1-r)$$\n\nThis will allow us to substitute this expression into our second equation, reducing it to a single variable."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$a = \\htmlData{anchor=a-expr}{\\color{@sky}{150(1-r)}}$"},{"latex":"$$\\htmlData{anchor=a-var}{\\color{@sky}{a}}r = 36$"},{"latex":"$$\\htmlData{anchor=sub-eq}{\\color{@sky}{150(1-r)}}r = 36$"},{"latex":"$$150r - 150r^2 = 36 \\implies 150r^2 - 150r + 36 = 0$"}],"highlights":[{"color":"sky","style":"box","anchor":"a-expr"},{"color":"sky","style":"text-color","anchor":"a-var"}],"connections":[{"to":"a-var","from":"a-expr","color":"sky","label":"Substitute a","style":"solid"}]},"order":2,"title":"Substitute and form a quadratic","description":"We substitute our expression for $a$ into the second-term equation $ar = 36$ to obtain a quadratic equation in $r$."},{"type":"text","order":3,"title":"Solve the quadratic equation","content":"To solve the quadratic equation $150r^2 - 150r + 36 = 0$, first simplify it by dividing all terms by $6$:\n\n$$25r^2 - 25r + 6 = 0$$\n\nNow, we can factorise the quadratic equation:\n\n$$(5r - 2)(5r - 3) = 0$$\n\nThis gives two possible values for $r$:\n\n$$r = \\frac{2}{5} = 0.4 \\quad \\text{or} \\quad r = \\frac{3}{5} = 0.6$$","calculator":[{"gdc":{"expression":"150x^2 - 150x + 36 = 0","instructions":"1. Go to the Equation/Solver menu.\n2. Choose Polynomial -> Degree 2.\n3. Enter the coefficients: a = 150, b = -150, c = 36.\n4. Press Solve to get the roots: x = 0.4 and x = 0.6."},"ti84":{"expression":"150x^2 - 150x + 36 = 0","instructions":"1. Press [APPS] and select 'PlySmlt2'.\n2. Select '1: Polynomial Root Finder'.\n3. Set the order to 2 and press [NEXT] (Graph button).\n4. Enter the coefficients: a2 = 150, a1 = -150, a0 = 36.\n5. Press [SOLVE] (Graph button) to find the roots: x = 0.6 and x = 0.4."},"desmos":{"expression":"y = 150x^2 - 150x + 36","instructions":"1. Type the quadratic function into the input bar.\n2. Click on the x-intercepts of the plotted curve to view the roots."},"description":"You can solve this quadratic equation quickly using your calculator's equation solver or by finding the x-intercepts of the graph."}]},{"type":"text","order":4,"title":"Find the valid values","content":"Now we find the corresponding values of $a$ using the relation $a = \\frac{36}{r}$:\n\n* If $r = 0.6$:\n $$a = \\frac{36}{0.6} = 60$$\n\n* If $r = 0.4$:\n $$a = \\frac{36}{0.4} = 90$$\n\nSince the question specifies that $a > 80$, the value $a = 60$ is rejected.\n\nThus, the correct values are:\n$$\\mathbf{a = 90} \\quad \\text{and} \\quad \\mathbf{r = 0.4}$$","highlights":[{"text":"a > 80","location":"specification"}]}]},{"id":"2526755c-b191-44f8-8920-501555a133a9:substitution_for_a","label":"(a) Substitution in terms of a","steps":[{"type":"text","order":0,"title":"Identify the given information","content":"From the problem statement, we are given:\n- The first term of the geometric series is $a$, with the constraint $a > 80$.\n- The second term is $u_2 = 36$ grams.\n- The sum to infinity is $S_{\\infty} = 150$ grams.\n\nWe need to find the values of $a$ and $r$.","highlights":[{"text":"theoretical maximum mass that can be extracted over an infinite time period is 150 grams","location":"specification"},{"text":"first hour, the centrifuge extracts \$a\$ grams, where \$a > 80\$","location":"specification"},{"text":"second hour, it extracts 36 grams","location":"specification"}]},{"type":"text","order":1,"title":"Set up the formulas","content":"Recall the standard formulas for a geometric series from your formula booklet:\n- The $n$-th term of a geometric series is:\n$$u_n = a r^{n-1}$$\n- The sum to infinity is:\n$$S_{\\infty} = \\frac{a}{1-r}$$\n\nApplying these to our given information, we get two equations:\n1) For the second term ($n=2$):\n$$ar = 36$$\n2) For the sum to infinity:\n$$\\frac{a}{1-r} = 150$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Using the second term equation: } \\htmlData{anchor=eq1}{\\color{@sky}{ar = 36}}$"},{"latex":"$$\\text{We can rearrange this to express } r \\text{ in terms of } a: \\htmlData{anchor=eq2}{\\color{@amber}{r = \\frac{36}{a}}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"eq1"},{"color":"amber","style":"box","anchor":"eq2"}],"connections":[{"to":"eq2","from":"eq1","color":"sky","label":"Rearrange for r","style":"solid"}]},"order":2,"title":"Express r in terms of a","description":"To follow our substitution method, we isolate the common ratio $r$ from our second term equation."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$S_{\\infty} = \\frac{a}{1-\\htmlData{anchor=r-sub}{\\color{@amber}{r}}} = 150$"},{"latex":"$$\\text{Substitute } \\htmlData{anchor=r-val}{\\color{@amber}{r = \\frac{36}{a}}} \\text{ into the formula:}$"},{"latex":"$$150 \\left(1 - \\htmlData{anchor=subbed}{\\color{@amber}{\\frac{36}{a}}}\\right) = a$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"r-sub"},{"color":"amber","style":"text-color","anchor":"r-val"},{"color":"amber","style":"box","anchor":"subbed"}],"connections":[{"to":"subbed","from":"r-val","color":"amber","label":"Substitute","style":"solid"}]},"order":3,"title":"Form a quadratic in a","description":"We substitute $r = \\frac{36}{a}$ into the rearranged sum to infinity equation, $a = 150(1-r)$, to form a single equation in terms of $a$."},{"type":"text","order":4,"title":"Solve the quadratic equation","content":"Let's expand and rearrange the equation to form a standard quadratic equation:\n\n$$\\begin{aligned} 150 \\left(1 - \\frac{36}{a}\\right) &= a \\\\[8pt] 150 - \\frac{5400}{a} &= a \\end{aligned}$$\n\nMultiply the entire equation by $a$ to remove the fraction:\n\n$$150a - 5400 = a^2$$\n\nRearranging into standard quadratic form $Ax^2 + Bx + C = 0$:\n\n$$a^2 - 150a + 5400 = 0$$"},{"type":"text","order":5,"title":"Find the values of a","content":"Now we solve the quadratic equation $a^2 - 150a + 5400 = 0$ by factoring:\n\nWe need two numbers that multiply to $5400$ and add up to $-150$. These numbers are $-90$ and $-60$:\n\n$$(a - 90)(a - 60) = 0$$\n\nThis gives two possible values for $a$:\n$$a = 90 \\quad \\text{or} \\quad a = 60$$"},{"type":"text","order":6,"title":"Determine the correct a","content":"The question specifies that the first term must satisfy the condition $a > 80$.\n\nComparing our two possible values:\n- For $a = 60$, this does not satisfy $a > 80$.\n- For $a = 90$, this satisfies $a > 80$.\n\nTherefore, we choose **$a = 90$**.","highlights":[{"text":"where \$a > 80\$","location":"specification"}]},{"type":"text","order":7,"title":"Calculate the common ratio r","content":"With the correct value of $a = 90$ identified, we substitute it back into our equation for $r$:\n\n$$r = \\frac{36}{a}$$\n\n$$r = \\frac{36}{90} = 0.4$$\n\nThus, the final solutions are **$a = 90$** and **$r = 0.4$**."}]},{"id":"2526755c-b191-44f8-8920-501555a133a9:graphical_intersection","label":"(a) Graphical Approach","steps":[{"type":"text","order":0,"title":"Formulate the first equation","content":"The theoretical maximum mass extracted over an infinite time period is given by the sum to infinity of a geometric series, $S_{\\infty}$. The formula for $S_{\\infty}$ is:\n\n$$S_{\\infty} = \\frac{a}{1-r}$$\n\nwhere $a$ is the first term and $r$ is the common ratio. We are given $S_{\\infty} = 150$ grams:\n\n$$\\frac{a}{1-r} = 150$$\n\nRearranging this to make $a$ the subject gives our first equation as a function of $r$:\n\n$$a = 150(1-r)$$\n\nThis would earn the first mark M1.","highlights":[{"text":"The theoretical maximum mass that can be extracted over an infinite time period is 150 grams.","location":"specification"}]},{"type":"text","order":1,"title":"Formulate the second equation","content":"The mass extracted in the second hour represents the second term of the geometric series, $u_2$, which is defined as:\n\n$$u_2 = ar$$\n\nWe are given that $u_2 = 36$ grams:\n\n$$ar = 36$$\n\nRearranging this to isolate $a$ gives our second equation as a function of $r$:\n\n$$a = \\frac{36}{r}$$\n\nThis earns the second mark B1.","highlights":[{"text":"In the second hour, it extracts 36 grams.","location":"specification"}]},{"type":"text","order":2,"title":"Plot and find intersections","content":"To find the solution graphically, we can treat $r$ as the independent variable ($x$-axis) and $a$ as the dependent variable ($y$-axis). We define two functions:\n\n$$y_1 = 150(1-x)$$\n$$y_2 = \\frac{36}{x}$$\n\nUsing a graphing calculator or utility, we plot these functions and find their points of intersection. The intersection points representing values of $r$ and $a$ are:\n\n$$(0.4, 90) \\quad \\text{and} \\quad (0.6, 60)$$\n\nIdentifying these intersections earns M1 for forming the equivalence and M1 for solving to find $r = 0.4$ or $r = 0.6$.","calculator":[{"gdc":{"instructions":"1. Go to the Graph menu.\n2. Enter the functions \\\\f1(x) = 150(1-x)\\\\ and \\\\f2(x) = 36/x\\\\.\n3. Plot the graphs using an appropriate window scale (e.g., x-axis from 0 to 1, y-axis from 0 to 150).\n4. Use the G-Solv (Graph Solve) tool and select Intersect (INTSECT).\n5. The calculator display will show the two intersection points: (0.4, 90) and (0.6, 60)."},"ti84":{"instructions":"1. Press [Y=].\n2. Enter \\\\Y_1 = 150(1-X)\\\\.\n3. Enter \\\\Y_2 = 36/X\\\\.\n4. Press [WINDOW] and set Xmin = 0, Xmax = 1, Ymin = 0, Ymax = 150.\n5. Press [GRAPH].\n6. Press [2nd] [CALC] and select 5: intersect.\n7. Move the cursor near the first intersection and press [ENTER] three times to find the first point: (0.4, 90).\n8. Repeat the intersection step near the second point to find: (0.6, 60)."},"desmos":{"instructions":"1. Type \\\\y = 150(1-x)\\\\ in the first input bar.\n2. Type \\\\y = 36/x\\\\ in the second input bar.\n3. Click on the gray intersection dots on the plot to reveal their coordinates: (0.4, 90) and (0.6, 60)."},"description":"Find the intersection points of the two functions graphically."}]},{"type":"text","order":3,"title":"Apply constraint and state values","content":"We have two potential solutions from our graphical analysis:\n\n1. If $r = 0.6$, then $a = 60$\n2. If $r = 0.4$, then $a = 90$\n\nThe question specifies the constraint $a > 80$. Looking at our points:\n- For $(0.6, 60)$, $a = 60 \\ngtr 80$, so we reject this solution.\n- For $(0.4, 90)$, $a = 90 > 80$, which satisfies the condition.\n\nTherefore, the correct values are:\n\n$$a = 90 \\quad \\text{and} \\quad r = 0.4$$\n\nThis final selection earns the A1 mark.","highlights":[{"text":"In the first hour, the centrifuge extracts \$a\$ grams, where \$a > 80\$.","location":"specification"}]}]},{"id":"40665fe7-7a74-4dff-bff1-c0fdd7f397cf:algebraic","label":"(bi) Direct Algebraic Derivation","steps":[{"type":"text","order":0,"title":"Set up the equations","content":"From the question, we know:\n- The first term is $a$.\n- The second term is $ar = 36$.\n- The sum to infinity is $S_\\infty = \\frac{a}{1-r} = 150$.\n\nLet's express $r$ in terms of $a$ from the second term equation:\n$$r = \\frac{36}{a}$$","highlights":[{"text":"first hour, the centrifuge extracts \$a\$ grams","location":"specification"},{"text":"second hour, it extracts 36 grams","location":"specification"},{"text":"theoretical maximum mass that can be extracted over an infinite time period is 150 grams","location":"specification"}]},{"type":"text","order":1,"title":"Solve for first term","content":"Substitute $r = \\frac{36}{a}$ into the sum to infinity equation:\n$$\\frac{a}{1 - \\frac{36}{a}} = 150$$\n\nMultiply the numerator and denominator on the left by $a$:\n$$\\frac{a^2}{a - 36} = 150$$\n\nNow, multiply both sides by $(a - 36)$ to get a quadratic equation:\n$$a^2 = 150(a - 36)$$\n$$a^2 - 150a + 5400 = 0$$\n\nFactoring this quadratic equation:\n$$(a - 90)(a - 60) = 0$$\n\nThis gives two possible values: $a = 90$ or $a = 60$."},{"type":"text","order":2,"title":"Filter the first term","content":"The question specifies that $a > 80$. Therefore, we must choose:\n$$a = 90$$\n\nNow, we can find the common ratio $r$:\n$$r = \\frac{36}{a} = \\frac{36}{90} = 0.4$$","highlights":[{"text":"a > 80","location":"specification"}]},{"type":"text","order":3,"title":"Express general term","content":"The mass extracted in the $n$-th hour, $M_n$, is given by the formula for the $n$-th term of a geometric sequence:\n$$M_n = a r^{n-1}$$\n\nSubstituting our values for $a$ and $r$:\n$$M_n = 90(0.4)^{n-1}$$"},{"type":"text","order":4,"title":"Express as prime factors","content":"To show that $M_n$ can be written in the required form, we write the base numbers $90$ and $0.4$ as prime factors:\n- $90 = 2 \\times 3^2 \\times 5$\n- $0.4 = \\frac{2}{5} = 2 \\times 5^{-1}$\n\nNow substitute these prime factorizations into the equation for $M_n$:\n$$M_n = \\left(2 \\cdot 3^2 \\cdot 5\\right) \\cdot \\left(\\frac{2}{5}\\right)^{n-1}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$M_n = \\htmlData{anchor=factor-a}{\\color{@sky}{2^1}} \\cdot 3^2 \\cdot \\htmlData{anchor=factor-five}{\\color{@purple}{5^1}} \\cdot \\frac{\\htmlData{anchor=ratio-num}{\\color{@sky}{2^{n-1}}}}{\\htmlData{anchor=ratio-den}{\\color{@purple}{5^{n-1}}}}$"},{"latex":"$$M_n = \\htmlData{anchor=combined-two}{\\color{@sky}{2^{1 + n - 1}}} \\cdot 3^2 \\cdot \\htmlData{anchor=combined-five}{\\color{@purple}{5^{1 - (n - 1)}}}$"},{"latex":"$$M_n = 2^n \\cdot 3^2 \\cdot 5^{2-n} = \\frac{2^n \\cdot 3^2}{5^{n-2}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"factor-a"},{"color":"purple","style":"text-color","anchor":"factor-five"}],"connections":[{"to":"combined-two","from":"factor-a","color":"sky","label":"combine base 2","style":"solid"},{"to":"combined-five","from":"factor-five","color":"purple","label":"combine base 5","style":"solid"}]},"order":5,"title":"Apply index laws","description":"Combine the terms with like bases using index laws: $x^a \\cdot x^b = x^{a+b}$ and $\\frac{x^a}{x^b} = x^{a-b}$."}]},{"id":"04893333-a5be-4b93-ba80-c4b8f5962917:forward_log_laws","label":"(bii) Direct Algebraic Expansion","steps":[{"type":"text","order":0,"title":"State the expression for $M_n$","content":"From the previous part of the question, we are given the expression for $M_n$:\n\n$$M_n = \\frac{2^n \\cdot 3^2}{5^{n-2}}$$\n\nTo find $\\log_2 M_n$, we start by taking the base-2 logarithm of both sides:\n\n$$\\log_2 M_n = \\log_2 \\left( \\frac{2^n \\cdot 3^2}{5^{n-2}} \\right)$$\n\nThis earns the first mark M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\log_2 M_n = \\log_2 \\left( \\frac{\\htmlData{anchor=num}{2^n \\cdot 3^2}}{\\htmlData{anchor=den}{5^{n-2}}} \\right)$$"},{"latex":"$$$\\phantom{\\log_2 M_n} = \\log_2(\\htmlData{anchor=num-sub}{2^n \\cdot 3^2}) - \\log_2(\\htmlData{anchor=den-sub}{5^{n-2}})$$"},{"latex":"$$$\\phantom{\\log_2 M_n} = \\log_2(2^n) + \\log_2(3^2) - \\log_2(5^{n-2})$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"num"},{"color":"pink","style":"text-color","anchor":"den"},{"color":"sky","style":"text-color","anchor":"num-sub"},{"color":"pink","style":"text-color","anchor":"den-sub"}],"connections":[{"to":"num-sub","from":"num","color":"sky","label":"Numerator","style":"solid"},{"to":"den-sub","from":"den","color":"pink","label":"Denominator","style":"solid"}]},"order":1,"title":"Apply logarithm laws","description":"We use the quotient law $\\log \\left(\\frac{x}{y}\\right) = \\log x - \\log y$, followed by the product law $\\log(ab) = \\log a + \\log b$ M1."},{"type":"text","order":2,"title":"Simplify using power laws","content":"Now, apply the power law of logarithms, $\\log(a^k) = k \\log a$:\n\n$$\\begin{aligned} \\log_2(2^n) &= n \\log_2 2 = n \\quad \\text{(since } \\log_2 2 = 1\\text{)} \\\\\\\\ \\log_2(3^2) &= 2 \\log_2 3 \\\\\\\\ \\log_2(5^{n-2}) &= (n-2) \\log_2 5 \\end{aligned}$$\n\nSubstituting these simplified terms back into our expression gives:\n\n$$\\log_2 M_n = n + 2 \\log_2 3 - (n - 2)\\log_2 5$$"},{"type":"text","order":3,"title":"Expand and factorise","content":"Expand the bracketed term containing $n$:\n\n$$-(n - 2)\\log_2 5 = -n\\log_2 5 + 2\\log_2 5$$\n\nSubstitute this back to group the terms:\n\n$$\\log_2 M_n = n + 2 \\log_2 3 - n\\log_2 5 + 2\\log_2 5$$\n\nNow, collect the terms containing $n$ together and the constant terms together:\n\n$$\\begin{aligned} \\log_2 M_n &= (n - n\\log_2 5) + (2 \\log_2 3 + 2\\log_2 5) \\\\\\\\ &= n(1 - \\log_2 5) + (2 \\log_2 3 + 2\\log_2 5) \\end{aligned}$$\n\nThis matches the required form A1."}]},{"id":"04893333-a5be-4b93-ba80-c4b8f5962917:backward_log_laws","label":"(bii) Working Backward","steps":[{"type":"text","order":0,"title":"Identify the target","content":"To prove the statement using the **Working Backward** approach, we will start with the right-hand side (RHS) of the equation and use the laws of logarithms to combine it into a single logarithm:\n\n$$\\text{RHS} = n(1 - \\log_2 5) + (2 \\log_2 3 + 2 \\log_2 5)$$\n\nOur goal is to show that this simplifies to the expression for $\\log_2 M_n$, which we know from previous parts of the question is:\n\n$$\\log_2 M_n = \\log_2 \\left( \\frac{2^n \\cdot 3^2}{5^{n-2}} \\right)$$\n\nLet's expand and rearrange the RHS to prepare for combining the terms."},{"type":"text","order":1,"title":"Expand and group","content":"First, let's expand the terms in the RHS expression:\n\n$$n(1 - \\log_2 5) + 2 \\log_2 3 + 2 \\log_2 5 = n - n \\log_2 5 + 2 \\log_2 3 + 2 \\log_2 5$$\n\nNow, group the $\\log_2 5$ terms together to make it easier to apply the laws of logarithms:\n\n$$= n + 2 \\log_2 3 - (n - 2) \\log_2 5$$"},{"type":"text","order":2,"title":"Convert constant to log","content":"To combine all terms into a single base-2 logarithm, we need to express the linear term $n$ as a logarithm with base 2.\n\nSince $\\log_2 2 = 1$, we can write:\n\n$$n = n \\log_2 2$$\n\nUsing the power law of logarithms ($k \\log_b a = \\log_b a^k$):\n\n$$n \\log_2 2 = \\log_2(2^n)$$\n\nSubstituting this back into our expression gives:\n\n$$\\log_2(2^n) + 2 \\log_2 3 - (n - 2) \\log_2 5$$"},{"type":"text","order":3,"title":"Apply power laws","content":"Next, let's use the power law of logarithms on the remaining terms to move the coefficients inside the logarithms:\n\n* For the second term:\n $$2 \\log_2 3 = \\log_2(3^2)$$\n\n* For the third term:\n $$(n - 2) \\log_2 5 = \\log_2(5^{n-2})$$\n\nSubstituting these back into our expression yields:\n\n$$\\log_2(2^n) + \\log_2(3^2) - \\log_2(5^{n-2})$$"},{"type":"text","order":4,"title":"Combine into one logarithm","content":"Now, we can combine these terms into a single logarithm using the product rule, $\\log_b x + \\log_b y = \\log_b(xy)$, and the quotient rule, $\\log_b x - \\log_b y = \\log_b\\left(\\frac{x}{y}\\right)$:\n\n$$\\begin{aligned} \\log_2(2^n) + \\log_2(3^2) - \\log_2(5^{n-2}) &= \\log_2(2^n \\cdot 3^2) - \\log_2(5^{n-2}) \\\\\\\\ &= \\log_2 \\left( \\frac{2^n \\cdot 3^2}{5^{n-2}} \\right) \\end{aligned}$$\n\nSince $M_n = \\frac{2^n \\cdot 3^2}{5^{n-2}}$, this is exactly:\n\n$$\\log_2 M_n$$\n\nThus, we have successfully shown that:\n\n$$\\log_2 M_n = n(1 - \\log_2 5) + (2 \\log_2 3 + 2 \\log_2 5)$$"}]}],"generatedAt":"2026-06-09T10:09:40.872Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"dcc073a0-0326-43a6-9582-71f2b1da9911","revisionId":"fab70473-c063-4763-a833-de077f1eef82","questionNumber":null,"questionType":"LA","specification":"The curve $C$ has equation\n\n$y = x^5 + 3x + \\sin x$\n\nwhere $x$ is measured in radians.","difficultyBand":"hard","tier":null,"totalMarks":5,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"5df6a541-8df5-44f4-97b3-9ba75d68548b","label":null,"content":"Use proof by contradiction to prove that $C$ has no stationary points.","metadata":null,"markscheme":"- $\\frac{dy}{dx} = 5x^4 + 3 + \\cos x$ B1\n- Assume $C$ has a stationary point, so $5x^4 + 3 + \\cos x = 0$ for some real $x$ M1\n- Rearranges to $5x^4 = -3 - \\cos x$ M1\n- Since $-1 \\leq \\cos x \\leq 1$, the right-hand side satisfies $-4 \\leq -3 - \\cos x \\leq -2$, so $5x^4 \\leq -2$ A1\n- But $5x^4 \\geq 0$ for all real $x$, a contradiction, so $C$ has no stationary points A1\n\nAccept the equivalent bounding argument: $5x^4 \\geq 0$ and $3 + \\cos x \\geq 2$, so $\\frac{dy}{dx} \\geq 2 > 0$.","marks":5,"order":0,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"dcfda464-e53d-480a-aeb3-72da124a9328","revisionId":"d9c9b958-b0b9-4d29-91a9-7b47456cf7ec","questionNumber":null,"questionType":"LA","specification":"A grain hopper dispenses wheat at a constant rate of $12\\text{ cm}^3\\text{s}^{-1}$ onto a level floor, where it forms a conical pile. After $t$ seconds, the pile has a height of $h\\text{ cm}$ and a volume of $V\\text{ cm}^3$. The volume is modeled by the equation:\n\n\\[V = \\frac{\\pi h^3}{12}\\]","difficultyBand":"hard","tier":null,"totalMarks":8,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"09a9226b-a40b-4b43-adb8-0a800fc78d91","label":"a","content":"Show that when $t = 6$,\n\n\\[\\frac{dV}{dh} = 9 \\sqrt[3]{16\\pi}\\]","metadata":null,"markscheme":"- Differentiates $V$ with respect to $h$ to obtain $\\frac{dV}{dh} = \\frac{3\\pi h^2}{12} = \\frac{\\pi h^2}{4}$ B1\n- Calculates the volume of the pile at $t = 6$ using $V = 12 \\times 6 = 72$ M1\n- Equates the volume formula to the calculated volume: $\\frac{\\pi h^3}{12} = 72 \\Rightarrow h^3 = \\frac{864}{\\pi}$ M1\n- Substitutes $h^2 = \\left(\\frac{864}{\\pi}\\right)^{2/3}$ into the expression for $\\frac{dV}{dh}$ to get $\\frac{\\pi}{4} \\left(\\frac{864}{\\pi}\\right)^{2/3}$ M1\n- Simplifies $\\frac{\\pi^{1/3}}{4} (864)^{2/3} = \\frac{\\pi^{1/3}}{4} (6^3 \\times 4)^{2/3} = \\frac{\\pi^{1/3}}{4} (36 \\times 4^{2/3})$ to reach the required form $9 \\sqrt[3]{16\\pi}$ A1","marks":5,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"direct_evaluation_of_h","label":"Finding height first","steps":[{"type":"text","order":0,"title":"Differentiate the volume function","content":"First, we need to find the rate of change of the volume $V$ with respect to the height $h$, which is $\\frac{dV}{dh}$. We are given the formula:\n\n$$V = \\frac{\\pi h^3}{12}$$\n\nDifferentiating $V$ with respect to $h$ using the power rule gives:\n\n$$\\frac{dV}{dh} = \\frac{3\\pi h^2}{12} = \\frac{\\pi h^2}{4}$$\n\nThis step earns the first mark (B1) in the markscheme."},{"type":"text","order":1,"title":"Calculate the volume at t = 6","content":"The hopper dispenses wheat at a constant rate of $12\\text{ cm}^3\\text{s}^{-1}$. Since the pile starts from nothing, the volume $V$ after $t$ seconds is given by:\n\n$$V = 12t$$\n\nSubstituting $t = 6$ to find the volume of the pile at that specific instant:\n\n$$V = 12 \\times 6 = 72\\text{ cm}^3$$\n\nThis calculation earns the second mark (M1)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=vol-eq}{\\frac{\\pi h^3}{12}} = \\htmlData{anchor=vol-val}{72}$"},{"latex":"$$\\htmlData{anchor=h3-eq}{h^3} = \\frac{12 \\times 72}{\\pi} = \\htmlData{anchor=h3-val}{\\frac{864}{\\pi}}$"},{"latex":"$$\\htmlData{anchor=h-eq}{h} = \\htmlData{anchor=h-val}{\\left(\\frac{864}{\\pi}\\right)^{1/3}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"vol-eq"},{"color":"purple","style":"underline","anchor":"h3-eq"}],"connections":[{"to":"h3-val","from":"vol-val","color":"sky","label":"multiply by 12","style":"solid"},{"to":"h-val","from":"h3-val","color":"purple","label":"cube root","style":"solid"}]},"order":2,"title":"Solve for the height h","description":"Equating our two expressions for the volume at $t=6$ lets us solve for $h$ to find the height of the cone at this instant. This earns the third mark (M1)."},{"type":"text","order":3,"title":"Substitute h into the derivative","content":"Now substitute $h = \\left(\\frac{864}{\\pi}\\right)^{1/3}$ into our expression for $\\frac{dV}{dh}$:\n\n$$\\frac{dV}{dh} = \\frac{\\pi}{4} h^2$$\n\n$$\\frac{dV}{dh} = \\frac{\\pi}{4} \\left[\\left(\\frac{864}{\\pi}\\right)^{1/3}\\right]^2 = \\frac{\\pi}{4} \\left(\\frac{864}{\\pi}\\right)^{2/3}$$\n\nThis substitution earns the fourth mark (M1)."},{"type":"text","order":4,"title":"Simplify to show the required form","content":"$93","calculator":[{"gdc":{"expression":"12 / (9 * (16 * π)^(1/3))","instructions":"Use the fraction template. Input 12 in the numerator and 9 * (16 * π)^(1/3) in the denominator."},"ti84":{"expression":"12 / (9 * nroot(16 * π, 3))","instructions":"Press [alpha] [y=] and select 1:n/d to insert a fraction. Enter 12 in the numerator. In the denominator, enter 9 * cubeRoot(16 * π). To get the cube root, press [math] and select 4:³√(. Press [second] [^] to insert π. Press [enter] to evaluate."},"desmos":{"expression":"12 / (9 * (16 * \\pi)^{1/3})","instructions":"Type '12 / (9 * nthroot(16 * pi, 3))' into the input bar."},"description":"Calculate the decimal value of the rate of change of height."}]}]},{"id":"derivative_in_terms_of_v","label":"Expressing derivative in terms of volume","steps":[{"type":"text","order":0,"title":"Differentiate V with respect to h","content":"First, we find the rate of change of volume with respect to height, which is represented by the derivative $\\frac{dV}{dh}$. \n\nWe are given the volume formula:\n$$V = \\frac{\\pi h^3}{12}$$\n\nDifferentiating this with respect to $h$ using the power rule:\n$$\\frac{dV}{dh} = \\frac{3\\pi h^2}{12} = \\frac{\\pi h^2}{4}$$"},{"type":"text","order":1,"title":"Express h in terms of V","content":"To express the derivative solely as a function of volume $V$, we rearrange the volume equation to solve for $h$:\n\n$$\\begin{aligned} V &= \\frac{\\pi h^3}{12} \\\\ 12V &= \\pi h^3 \\\\ h^3 &= \\frac{12V}{\\pi} \\\\ h &= \\left(\\frac{12V}{\\pi}\\right)^{1/3} \\end{aligned}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$h = \\htmlData{anchor=h-expr}{\\color{@sky}{\\left(\\frac{12V}{\\pi}\\right)^{1/3}}}$"},{"latex":"$$\\frac{dV}{dh} = \\frac{\\pi}{4} \\htmlData{anchor=h-sq}{\\color{@sky}{h^2}}$"},{"latex":"$$\\frac{dV}{dh} = \\frac{\\pi}{4} \\htmlData{anchor=dv-dh-v}{\\color{@sky}{\\left(\\frac{12V}{\\pi}\\right)^{2/3}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"h-expr"},{"color":"sky","style":"text-color","anchor":"dv-dh-v"}],"connections":[{"to":"h-sq","from":"h-expr","color":"sky","label":"Substitute h","style":"solid"}]},"order":2,"title":"Substitute h into the derivative","description":"Now, we substitute our expression for $h$ back into the derivative $\\frac{dV}{dh} = \\frac{\\pi h^2}{4}$ to obtain $\\frac{dV}{dh}$ purely in terms of $V$."},{"type":"text","order":3,"title":"Calculate volume at t = 6","content":"The grain is dispensed at a constant rate of $12\\text{ cm}^3\\text{s}^{-1}$. Therefore, after $t = 6$ seconds, the total volume of the cone $V$ is:\n\n$$V = 12 \\times 6 = 72\\text{ cm}^3$$"},{"type":"text","order":4,"title":"Evaluate and simplify the derivative","content":"$94","calculator":[{"gdc":{"expression":"12 / (9 * (16 * π)^(1/3))","instructions":"Use the fraction template. Input 12 in the numerator and 9 * (16 * π)^(1/3) in the denominator."},"ti84":{"expression":"12 / (9 * nroot(16 * π, 3))","instructions":"Press [alpha] [y=] and select 1:n/d to insert a fraction. Enter 12 in the numerator. In the denominator, enter 9 * cubeRoot(16 * π). To get the cube root, press [math] and select 4:³√(. Press [second] [^] to insert π. Press [enter] to evaluate."},"desmos":{"expression":"12 / (9 * (16 * \\pi)^{1/3})","instructions":"Type '12 / (9 * nthroot(16 * pi, 3))' into the input bar."},"description":"Calculate the decimal value of the rate of change of height."}]}]}],"generatedAt":"2026-06-09T09:33:59.938Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"df984b01-7552-429f-8c35-0b9d6b85b33d","label":"b","content":"Hence, find the rate at which the height of the pile is increasing when $t = 6$. Give your answer in $\\text{cm s}^{-1}$ to three significant figures.","metadata":null,"markscheme":"- Applies the chain rule $\\frac{dh}{dt} = \\frac{dV}{dt} \\times \\frac{dh}{dV}$ or equivalent M1\n- Substitutes $\\frac{dV}{dt} = 12$ and the value of $\\frac{dV}{dh}$ from part (a) to get $\\frac{dh}{dt} = \\frac{12}{9 \\sqrt[3]{16\\pi}}$ M1\n- Obtains $0.361$ (accept $0.361$ only) A1","marks":3,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"related_rates","label":"Related Rates using Chain Rule","steps":[{"type":"text","order":0,"title":"Differentiate volume formula","content":"To show the required expression for $\\frac{dV}{dh}$, we first differentiate the given volume equation with respect to $h$:\n\n$$V = \\frac{\\pi h^3}{12}$$\n\nApplying the power rule, we get:\n\n$$\\frac{dV}{dh} = \\frac{3\\pi h^2}{12} = \\frac{\\pi h^2}{4}$$"},{"type":"text","order":1,"title":"Find volume when t = 6","content":"Since wheat is dispensed at a constant rate of $12\\text{ cm}^3\\text{s}^{-1}$, the volume $V$ at any time $t$ is:\n\n$$V = 12t$$\n\nSubstituting $t = 6$, we find the volume of the pile at this instant:\n\n$$V = 12 \\times 6 = 72\\text{ cm}^3$$","highlights":[{"text":"constant rate of 12","location":"specification"},{"text":"t = 6","location":"specification"}]},{"type":"text","order":2,"title":"Express height in terms of pi","content":"We equate the volume formula to our calculated volume at $t = 6$ to find $h$:\n\n$$\\frac{\\pi h^3}{12} = 72$$\n\nMultiply both sides by $12$:\n\n$$\\pi h^3 = 864 \\implies h^3 = \\frac{864}{\\pi}$$\n\nTaking the cube root gives:\n\n$$h = \\left(\\frac{864}{\\pi}\\right)^{1/3}$$"},{"type":"text","order":3,"title":"Substitute and simplify dV/dh","content":"Now we substitute $h = \\left(\\frac{864}{\\pi}\\right)^{1/3}$ into our derivative $\\frac{dV}{dh} = \\frac{\\pi h^2}{4}$:\n\n$$\\frac{dV}{dh} = \\frac{\\pi}{4} \\left(\\frac{864}{\\pi}\\right)^{2/3} = \\frac{\\pi^{1/3}}{4} \\left(864\\right)^{2/3}$$\n\nWe can express $864$ as $6^3 \\times 4$:\n\n$$\\begin{aligned} \\frac{dV}{dh} &= \\frac{\\pi^{1/3}}{4} \\left(6^3 \\times 4\\right)^{2/3} \\\\ &= \\frac{\\pi^{1/3}}{4} \\left(36 \\times 4^{2/3}\\right) \\\\ &= 9 \\pi^{1/3} \\times 16^{1/3} \\\\ &= 9 \\sqrt[3]{16\\pi} \\end{aligned}$$"},{"type":"text","order":4,"title":"Calculate rate of change","content":"To find the rate at which the height of the pile is increasing, we use the chain rule:\n\n$$\\frac{dh}{dt} = \\frac{dh}{dV} \\times \\frac{dV}{dt} = \\frac{1}{\\frac{dV}{dh}} \\times \\frac{dV}{dt}$$\n\nSubstitute $\\frac{dV}{dt} = 12$ and our value for $\\frac{dV}{dh}$ at $t = 6$:\n\n$$\\frac{dh}{dt} = \\frac{12}{9 \\sqrt[3]{16\\pi}} = \\frac{4}{3 \\sqrt[3]{16\\pi}}$$\n\nEvaluating this numerically:\n\n$$\\frac{dh}{dt} \\approx 0.361\\text{ cm s}^{-1}\\text{ (to 3 s.f.)}$$","highlights":[{"text":"three significant figures","location":"specification"}]}]},{"id":"direct_differentiation","label":"Direct Differentiation of $h(t)$","steps":[{"type":"text","order":0,"title":"Express volume as a function","content":"The grain is dispensed at a constant rate of $12\\text{ cm}^3\\text{s}^{-1}$. Assuming the pile starts with zero volume at $t = 0$, the volume $V$ at any time $t$ seconds is given by:\n\n$$V = 12t$$","highlights":[{"text":"constant rate of 12","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=eq1}{\\frac{\\pi h^3}{12}} = \\htmlData{anchor=eq2}{12t}$"},{"latex":"$$\\htmlData{anchor=eq3}{h^3} = \\htmlData{anchor=eq4}{\\frac{144t}{\\pi}}$"},{"latex":"$$\\htmlData{anchor=eq5}{h(t)} = \\htmlData{anchor=eq6}{\\left(\\frac{144}{\\pi}\\right)^{\\frac{1}{3}} t^{\\frac{1}{3}}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"eq1"},{"color":"sky","style":"box","anchor":"eq3"},{"color":"amber","style":"box","anchor":"eq5"}],"connections":[{"to":"eq3","from":"eq1","color":"sky","label":"multiply by 12/pi","style":"solid"},{"to":"eq5","from":"eq3","color":"amber","label":"take cube root","style":"solid"}]},"order":1,"title":"Express height as a function","description":"We substitute $V = 12t$ into the given volume formula $V = \\frac{\\pi h^3}{12}$ to find an explicit equation for $h$ in terms of $t$."},{"type":"text","order":2,"title":"Differentiate height with respect to time","content":"Now, we differentiate $h(t)$ directly with respect to $t$ using the power rule:\n\n$$\\frac{dh}{dt} = \\frac{1}{3} \\left(\\frac{144}{\\pi}\\right)^{\\frac{1}{3}} t^{-\\frac{2}{3}}$$\n\nWe can simplify this expression to make the evaluation easier:\n\n$$\\frac{dh}{dt} = \\frac{1}{3} \\left(\\frac{144}{\\pi t^2}\\right)^{\\frac{1}{3}}$$"},{"type":"text","order":3,"title":"Evaluate at t = 6","content":"Substitute $t = 6$ into the derivative expression to find the rate of change of the height at that exact instant:\n\n$$\\begin{aligned} \\frac{dh}{dt} &= \\frac{1}{3} \\left(\\frac{144}{\\pi (6)^2}\\right)^{\\frac{1}{3}} \\\\ &= \\frac{1}{3} \\left(\\frac{144}{36\\pi}\\right)^{\\frac{1}{3}} \\\\ &= \\frac{1}{3} \\left(\\frac{4}{\\pi}\\right)^{\\frac{1}{3}} \\end{aligned}$$\n\nUsing a calculator to compute this value:\n\n$$\\frac{dh}{dt} \\approx 0.361\\text{ cm s}^{-1}$$","highlights":[{"text":"t = 6","location":"specification"},{"text":"three significant figures","location":"specification"}]}]}],"generatedAt":"2026-06-09T09:35:45.479Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"09a9226b-a40b-4b43-adb8-0a800fc78d91:direct_evaluation_of_h","label":"(a) Finding height first","steps":[{"type":"text","order":0,"title":"Differentiate the volume function","content":"First, we need to find the rate of change of the volume $V$ with respect to the height $h$, which is $\\frac{dV}{dh}$. We are given the formula:\n\n$$V = \\frac{\\pi h^3}{12}$$\n\nDifferentiating $V$ with respect to $h$ using the power rule gives:\n\n$$\\frac{dV}{dh} = \\frac{3\\pi h^2}{12} = \\frac{\\pi h^2}{4}$$\n\nThis step earns the first mark (B1) in the markscheme."},{"type":"text","order":1,"title":"Calculate the volume at t = 6","content":"The hopper dispenses wheat at a constant rate of $12\\text{ cm}^3\\text{s}^{-1}$. Since the pile starts from nothing, the volume $V$ after $t$ seconds is given by:\n\n$$V = 12t$$\n\nSubstituting $t = 6$ to find the volume of the pile at that specific instant:\n\n$$V = 12 \\times 6 = 72\\text{ cm}^3$$\n\nThis calculation earns the second mark (M1)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=vol-eq}{\\frac{\\pi h^3}{12}} = \\htmlData{anchor=vol-val}{72}$"},{"latex":"$$\\htmlData{anchor=h3-eq}{h^3} = \\frac{12 \\times 72}{\\pi} = \\htmlData{anchor=h3-val}{\\frac{864}{\\pi}}$"},{"latex":"$$\\htmlData{anchor=h-eq}{h} = \\htmlData{anchor=h-val}{\\left(\\frac{864}{\\pi}\\right)^{1/3}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"vol-eq"},{"color":"purple","style":"underline","anchor":"h3-eq"}],"connections":[{"to":"h3-val","from":"vol-val","color":"sky","label":"multiply by 12","style":"solid"},{"to":"h-val","from":"h3-val","color":"purple","label":"cube root","style":"solid"}]},"order":2,"title":"Solve for the height h","description":"Equating our two expressions for the volume at $t=6$ lets us solve for $h$ to find the height of the cone at this instant. This earns the third mark (M1)."},{"type":"text","order":3,"title":"Substitute h into the derivative","content":"Now substitute $h = \\left(\\frac{864}{\\pi}\\right)^{1/3}$ into our expression for $\\frac{dV}{dh}$:\n\n$$\\frac{dV}{dh} = \\frac{\\pi}{4} h^2$$\n\n$$\\frac{dV}{dh} = \\frac{\\pi}{4} \\left[\\left(\\frac{864}{\\pi}\\right)^{1/3}\\right]^2 = \\frac{\\pi}{4} \\left(\\frac{864}{\\pi}\\right)^{2/3}$$\n\nThis substitution earns the fourth mark (M1)."},{"type":"text","order":4,"title":"Simplify to show the required form","content":"$95","calculator":[{"gdc":{"expression":"12 / (9 * (16 * π)^(1/3))","instructions":"Use the fraction template. Input 12 in the numerator and 9 * (16 * π)^(1/3) in the denominator."},"ti84":{"expression":"12 / (9 * nroot(16 * π, 3))","instructions":"Press [alpha] [y=] and select 1:n/d to insert a fraction. Enter 12 in the numerator. In the denominator, enter 9 * cubeRoot(16 * π). To get the cube root, press [math] and select 4:³√(. Press [second] [^] to insert π. Press [enter] to evaluate."},"desmos":{"expression":"12 / (9 * (16 * \\pi)^{1/3})","instructions":"Type '12 / (9 * nthroot(16 * pi, 3))' into the input bar."},"description":"Calculate the decimal value of the rate of change of height."}]}]},{"id":"09a9226b-a40b-4b43-adb8-0a800fc78d91:derivative_in_terms_of_v","label":"(a) Expressing derivative in terms of volume","steps":[{"type":"text","order":0,"title":"Differentiate V with respect to h","content":"First, we find the rate of change of volume with respect to height, which is represented by the derivative $\\frac{dV}{dh}$. \n\nWe are given the volume formula:\n$$V = \\frac{\\pi h^3}{12}$$\n\nDifferentiating this with respect to $h$ using the power rule:\n$$\\frac{dV}{dh} = \\frac{3\\pi h^2}{12} = \\frac{\\pi h^2}{4}$$"},{"type":"text","order":1,"title":"Express h in terms of V","content":"To express the derivative solely as a function of volume $V$, we rearrange the volume equation to solve for $h$:\n\n$$\\begin{aligned} V &= \\frac{\\pi h^3}{12} \\\\ 12V &= \\pi h^3 \\\\ h^3 &= \\frac{12V}{\\pi} \\\\ h &= \\left(\\frac{12V}{\\pi}\\right)^{1/3} \\end{aligned}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$h = \\htmlData{anchor=h-expr}{\\color{@sky}{\\left(\\frac{12V}{\\pi}\\right)^{1/3}}}$"},{"latex":"$$\\frac{dV}{dh} = \\frac{\\pi}{4} \\htmlData{anchor=h-sq}{\\color{@sky}{h^2}}$"},{"latex":"$$\\frac{dV}{dh} = \\frac{\\pi}{4} \\htmlData{anchor=dv-dh-v}{\\color{@sky}{\\left(\\frac{12V}{\\pi}\\right)^{2/3}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"h-expr"},{"color":"sky","style":"text-color","anchor":"dv-dh-v"}],"connections":[{"to":"h-sq","from":"h-expr","color":"sky","label":"Substitute h","style":"solid"}]},"order":2,"title":"Substitute h into the derivative","description":"Now, we substitute our expression for $h$ back into the derivative $\\frac{dV}{dh} = \\frac{\\pi h^2}{4}$ to obtain $\\frac{dV}{dh}$ purely in terms of $V$."},{"type":"text","order":3,"title":"Calculate volume at t = 6","content":"The grain is dispensed at a constant rate of $12\\text{ cm}^3\\text{s}^{-1}$. Therefore, after $t = 6$ seconds, the total volume of the cone $V$ is:\n\n$$V = 12 \\times 6 = 72\\text{ cm}^3$$"},{"type":"text","order":4,"title":"Evaluate and simplify the derivative","content":"$96","calculator":[{"gdc":{"expression":"12 / (9 * (16 * π)^(1/3))","instructions":"Use the fraction template. Input 12 in the numerator and 9 * (16 * π)^(1/3) in the denominator."},"ti84":{"expression":"12 / (9 * nroot(16 * π, 3))","instructions":"Press [alpha] [y=] and select 1:n/d to insert a fraction. Enter 12 in the numerator. In the denominator, enter 9 * cubeRoot(16 * π). To get the cube root, press [math] and select 4:³√(. Press [second] [^] to insert π. Press [enter] to evaluate."},"desmos":{"expression":"12 / (9 * (16 * \\pi)^{1/3})","instructions":"Type '12 / (9 * nthroot(16 * pi, 3))' into the input bar."},"description":"Calculate the decimal value of the rate of change of height."}]}]},{"id":"df984b01-7552-429f-8c35-0b9d6b85b33d:related_rates","label":"(b) Related Rates using Chain Rule","steps":[{"type":"text","order":0,"title":"Differentiate volume formula","content":"To show the required expression for $\\frac{dV}{dh}$, we first differentiate the given volume equation with respect to $h$:\n\n$$V = \\frac{\\pi h^3}{12}$$\n\nApplying the power rule, we get:\n\n$$\\frac{dV}{dh} = \\frac{3\\pi h^2}{12} = \\frac{\\pi h^2}{4}$$"},{"type":"text","order":1,"title":"Find volume when t = 6","content":"Since wheat is dispensed at a constant rate of $12\\text{ cm}^3\\text{s}^{-1}$, the volume $V$ at any time $t$ is:\n\n$$V = 12t$$\n\nSubstituting $t = 6$, we find the volume of the pile at this instant:\n\n$$V = 12 \\times 6 = 72\\text{ cm}^3$$","highlights":[{"text":"constant rate of 12","location":"specification"},{"text":"t = 6","location":"specification"}]},{"type":"text","order":2,"title":"Express height in terms of pi","content":"We equate the volume formula to our calculated volume at $t = 6$ to find $h$:\n\n$$\\frac{\\pi h^3}{12} = 72$$\n\nMultiply both sides by $12$:\n\n$$\\pi h^3 = 864 \\implies h^3 = \\frac{864}{\\pi}$$\n\nTaking the cube root gives:\n\n$$h = \\left(\\frac{864}{\\pi}\\right)^{1/3}$$"},{"type":"text","order":3,"title":"Substitute and simplify dV/dh","content":"Now we substitute $h = \\left(\\frac{864}{\\pi}\\right)^{1/3}$ into our derivative $\\frac{dV}{dh} = \\frac{\\pi h^2}{4}$:\n\n$$\\frac{dV}{dh} = \\frac{\\pi}{4} \\left(\\frac{864}{\\pi}\\right)^{2/3} = \\frac{\\pi^{1/3}}{4} \\left(864\\right)^{2/3}$$\n\nWe can express $864$ as $6^3 \\times 4$:\n\n$$\\begin{aligned} \\frac{dV}{dh} &= \\frac{\\pi^{1/3}}{4} \\left(6^3 \\times 4\\right)^{2/3} \\\\ &= \\frac{\\pi^{1/3}}{4} \\left(36 \\times 4^{2/3}\\right) \\\\ &= 9 \\pi^{1/3} \\times 16^{1/3} \\\\ &= 9 \\sqrt[3]{16\\pi} \\end{aligned}$$"},{"type":"text","order":4,"title":"Calculate rate of change","content":"To find the rate at which the height of the pile is increasing, we use the chain rule:\n\n$$\\frac{dh}{dt} = \\frac{dh}{dV} \\times \\frac{dV}{dt} = \\frac{1}{\\frac{dV}{dh}} \\times \\frac{dV}{dt}$$\n\nSubstitute $\\frac{dV}{dt} = 12$ and our value for $\\frac{dV}{dh}$ at $t = 6$:\n\n$$\\frac{dh}{dt} = \\frac{12}{9 \\sqrt[3]{16\\pi}} = \\frac{4}{3 \\sqrt[3]{16\\pi}}$$\n\nEvaluating this numerically:\n\n$$\\frac{dh}{dt} \\approx 0.361\\text{ cm s}^{-1}\\text{ (to 3 s.f.)}$$","highlights":[{"text":"three significant figures","location":"specification"}]}]},{"id":"df984b01-7552-429f-8c35-0b9d6b85b33d:direct_differentiation","label":"(b) Direct Differentiation of $h(t)$","steps":[{"type":"text","order":0,"title":"Express volume as a function","content":"The grain is dispensed at a constant rate of $12\\text{ cm}^3\\text{s}^{-1}$. Assuming the pile starts with zero volume at $t = 0$, the volume $V$ at any time $t$ seconds is given by:\n\n$$V = 12t$$","highlights":[{"text":"constant rate of 12","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=eq1}{\\frac{\\pi h^3}{12}} = \\htmlData{anchor=eq2}{12t}$"},{"latex":"$$\\htmlData{anchor=eq3}{h^3} = \\htmlData{anchor=eq4}{\\frac{144t}{\\pi}}$"},{"latex":"$$\\htmlData{anchor=eq5}{h(t)} = \\htmlData{anchor=eq6}{\\left(\\frac{144}{\\pi}\\right)^{\\frac{1}{3}} t^{\\frac{1}{3}}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"eq1"},{"color":"sky","style":"box","anchor":"eq3"},{"color":"amber","style":"box","anchor":"eq5"}],"connections":[{"to":"eq3","from":"eq1","color":"sky","label":"multiply by 12/pi","style":"solid"},{"to":"eq5","from":"eq3","color":"amber","label":"take cube root","style":"solid"}]},"order":1,"title":"Express height as a function","description":"We substitute $V = 12t$ into the given volume formula $V = \\frac{\\pi h^3}{12}$ to find an explicit equation for $h$ in terms of $t$."},{"type":"text","order":2,"title":"Differentiate height with respect to time","content":"Now, we differentiate $h(t)$ directly with respect to $t$ using the power rule:\n\n$$\\frac{dh}{dt} = \\frac{1}{3} \\left(\\frac{144}{\\pi}\\right)^{\\frac{1}{3}} t^{-\\frac{2}{3}}$$\n\nWe can simplify this expression to make the evaluation easier:\n\n$$\\frac{dh}{dt} = \\frac{1}{3} \\left(\\frac{144}{\\pi t^2}\\right)^{\\frac{1}{3}}$$"},{"type":"text","order":3,"title":"Evaluate at t = 6","content":"Substitute $t = 6$ into the derivative expression to find the rate of change of the height at that exact instant:\n\n$$\\begin{aligned} \\frac{dh}{dt} &= \\frac{1}{3} \\left(\\frac{144}{\\pi (6)^2}\\right)^{\\frac{1}{3}} \\\\ &= \\frac{1}{3} \\left(\\frac{144}{36\\pi}\\right)^{\\frac{1}{3}} \\\\ &= \\frac{1}{3} \\left(\\frac{4}{\\pi}\\right)^{\\frac{1}{3}} \\end{aligned}$$\n\nUsing a calculator to compute this 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1 Pure Mathematics

Question 432

1 Pure Mathematics Questions

1 Pure Mathematics

Question 432

1 Pure Mathematics Questions