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$(0,0)$ is a stationary point of inflexion A1","marks":5,"order":0,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"624dc372-5f76-4ffe-8be6-a7b7e2b83825","revisionId":"5e1e8f6f-317e-4c6c-830e-d136a1279b06","questionNumber":null,"questionType":"LA","specification":"","difficulty":6,"difficultyBand":"hard","tier":null,"totalMarks":12,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"dae5e12a-0102-493d-b970-86e30f6d24f0","label":"a","content":"Sketch the graph of a cubic function that has three distinct real roots and a positive coefficient of \$x^3\$.","markscheme":"- Correct 'up-down-up' cubic shape starting in the bottom-left quadrant and ending in the top-right quadrant. B1\n- Graph intersects the horizontal axis at exactly three distinct points. B1","marks":2,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"property_based","label":"Property-Based Sketching","steps":[{"type":"text","order":0,"title":"Analyze the question properties","content":"To sketch our cubic graph, we need to extract two key properties from the question:\n\n1. The coefficient of $x^3$ is positive (i.e., $a > 0$ for $y = ax^3 + bx^2 + cx + d$).\n2. The function has three distinct real roots.","highlights":[{"text":"three distinct real roots","location":"specification"},{"text":"positive coefficient of \$x^3\$","location":"specification"}]},{"type":"text","order":1,"title":"Determine the end behavior","content":"Let's look at the end behavior for a cubic function with a positive leading coefficient ($a > 0$):\n\n- As $x \\to \\infty$, the $ax^3$ term dominates, so $y \\to \\infty$.\n- As $x \\to -\\infty$, the $ax^3$ term dominates, so $y \\to -\\infty$.\n\nThis means our sketch must start in the **bottom-left quadrant** (third quadrant) and end in the **top-right quadrant** (first quadrant), giving it a characteristic **'up-down-up'** shape."},{"type":"text","order":2,"title":"Incorporate the roots","content":"The question specifies **three distinct real roots**.\n\nSince roots are the values of $x$ where $y = 0$, our curve must cross the horizontal $x$-axis at **exactly three separate points**."},{"type":"text","order":3,"title":"Sketch the final graph","content":"Now we combine these properties to sketch the curve:\n\n1. Draw a horizontal $x$-axis and a vertical $y$-axis.\n2. Mark three distinct crossing points on the $x$-axis.\n3. Draw a smooth, continuous 'up-down-up' curve that starts from the bottom-left, goes up through the first intercept, peaks, goes down through the second intercept, bottoms out, and then goes up through the third intercept towards the top-right.\n\n### Markscheme Breakdown\n- **1 Mark** (B1): Correct 'up-down-up' shape starting in the bottom-left and ending in the top-right.\n- **1 Mark** (B1): Graph intersects the horizontal axis at exactly three distinct points."},{"type":"text","order":4,"title":"Visualize with a calculator","content":"To build a visual intuition, you can plot a representative positive cubic equation with three distinct roots, such as $y = x^3 - 4x$, on your graphing calculator.","calculator":[{"gdc":{"expression":"x^3 - 4x","instructions":"1. Go to the Graph menu.\n2. Enter the function f(x) = x^3 - 4x.\n3. Adjust the view window to see the three intercepts and the characteristic cubic shape."},"ti84":{"expression":"x^3 - 4x","instructions":"1. Press [Y=] and enter Y1 = X^3 - 4*X.\n2. Press [WINDOW] and set Xmin = -4, Xmax = 4, Ymin = -5, Ymax = 5.\n3. Press [GRAPH] to observe the 'up-down-up' curve crossing the x-axis at -2, 0, and 2."},"desmos":{"expression":"y = x^3 - 4x","instructions":"Type y = x^3 - 4x into the input bar. Notice that the curve starts bottom-left, goes up, down, and up, crossing the x-axis at exactly three distinct points."},"description":"Visualize the graph of a representative positive cubic equation with three distinct roots: $y = x^3 - 4x$."}]}]},{"id":"algebraic_exemplar","label":"Using an Algebraic Exemplar","steps":[{"type":"text","order":0,"title":"Define an algebraic exemplar","content":"To understand the shape and features of a cubic graph with three distinct real roots and a positive $x^3$ coefficient, we can construct a specific algebraic exemplar in factored form:\n\n$$y = (x - 1)(x - 2)(x - 3)$$\n\nIf we expand this equation, we get:\n\n$$y = (x^2 - 3x + 2)(x - 3)$$\n\n$$y = x^3 - 6x^2 + 11x - 6$$\n\nWe see that the coefficient of $x^3$ is $+1$, which satisfies the condition of having a positive leading coefficient."},{"type":"text","order":1,"title":"Find the intercepts","content":"Next, let's find the points where our exemplar function crosses the coordinate axes.\n\n**1. $x$-intercepts (roots):**\nSet $y = 0$:\n\n$$(x - 1)(x - 2)(x - 3) = 0$$\n\nThis gives three distinct real solutions:\n$$x = 1, \\quad x = 2, \\quad x = 3$$\nSo, the graph intersects the $x$-axis at $(1, 0)$, $(2, 0)$, and $(3, 0)$.\n\n**2. $y$-intercept:**\nSet $x = 0$:\n\n$$y = (0 - 1)(0 - 2)(0 - 3) = -6$$\n\nSo, the graph crosses the $y$-axis below the origin at $(0, -6)$."},{"type":"text","order":2,"title":"Determine the end behaviour","content":"To determine where the curve starts and ends, we analyze the end behaviour as $x$ becomes very large in the positive and negative directions:\n\n- As $x \\to \\infty$, the term $x^3 \\to \\infty$. This means the graph rises to the top-right quadrant.\n- As $x \\to -\\infty$, the term $x^3 \\to -\\infty$. This means the graph falls to the bottom-left quadrant.\n\nSince the graph starts at the bottom-left, goes up through $(1, 0)$, turns around to pass down through $(2, 0)$, and turns again to go up through $(3, 0)$ to the top-right, it forms a classic 'up-down-up' shape."},{"type":"text","order":3,"title":"Visualize with a calculator","content":"You can verify this curve and its key features using a graphing calculator.","calculator":[{"gdc":{"expression":"(x-1)(x-2)(x-3)","instructions":"Go to the Graph menu, enter the function y = (x - 1)(x - 2)(x - 3), adjust your view window to show x from -1 to 5, and sketch."},"ti84":{"expression":"(X-1)(X-2)(X-3)","instructions":"Press [Y=], enter (X-1)(X-2)(X-3) in Y1. Press [WINDOW] and set Xmin = -1, Xmax = 5, Ymin = -8, Ymax = 4. Press [GRAPH] to view the curve."},"desmos":{"expression":"y = (x - 1)(x - 2)(x - 3)","instructions":"Type y = (x - 1)(x - 2)(x - 3) into the input panel to see the roots and y-intercept immediately."},"description":"Plot the cubic function to visualize its three distinct roots and standard positive cubic shape."}]},{"type":"text","order":4,"title":"Sketch the final graph","content":"Now, sketch the graph on a set of axes incorporating these features:\n\n1. Draw a smooth, continuous curve showing the **'up-down-up' shape** starting in the bottom-left and ending in the top-right.\n2. Ensure the curve clearly crosses the $x$-axis at **exactly three distinct points**."}]}],"generatedAt":"2026-06-09T05:58:15.849Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"4786c0e5-8567-471e-bbc6-65c6b6bc7888","label":"bi","content":"The vertical displacement, \$s\$, of a mechanical component is modelled by the function\n\\[ s(t) = k + 15at^2 - 2t^3 \\]\nwhere \$t \\ge 0\$ is time, and \$a\$ and \$k\$ are constants with \$a > 0\$.\n\nShow that the curve \$s(t)\$ has a stationary point at its \$s\$-intercept.","markscheme":"- Differentiates the displacement function to find $s'(t) = 30at - 6t^2$. M1\n- States that the $s$-intercept occurs at $t = 0$. B1\n- Evaluates $s'(0) = 30a(0) - 6(0)^2 = 0$, thus proving a stationary point exists at the intercept. A1","marks":3,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"direct_evaluation","label":"Direct Evaluation of Derivative","steps":[{"type":"text","order":0,"title":"Identify the s-intercept","content":"The $s$-intercept of a curve $s(t)$ occurs where the independent variable is equal to zero.\n\nTherefore, the $s$-intercept occurs at:\n$$t = 0$$","highlights":[{"text":"s-intercept","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$s(t) = k + 15a\\htmlData{anchor=t2}{t^2} - 2\\htmlData{anchor=t3}{t^3}$"},{"latex":"$$s'(t) = 15a \\cdot \\htmlData{anchor=diff-t2}{\\color{@sky}{2t}} - 2 \\cdot \\htmlData{anchor=diff-t3}{\\color{@amber}{3t^2}}$"},{"latex":"$$s'(t) = 30at - 6t^2$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"diff-t2"},{"color":"amber","style":"text-color","anchor":"diff-t3"}],"connections":[{"to":"diff-t2","from":"t2","color":"sky","label":"power rule","style":"solid"},{"to":"diff-t3","from":"t3","color":"amber","label":"power rule","style":"solid"}]},"order":1,"title":"Differentiate the function","description":"To find the gradient of the curve, we differentiate the displacement function $s(t)$ with respect to $t$. Recall that $k$ and $a$ are constants, so the derivative of $k$ is $0$."},{"type":"text","order":2,"title":"Substitute t = 0","content":"A stationary point occurs where the gradient is zero, i.e., $s'(t) = 0$.\n\nTo see if a stationary point exists at the $s$-intercept, we substitute $t = 0$ into our derivative function $s'(t)$:\n\n$$s'(0) = 30a(0) - 6(0)^2 = 0$$\n\nSince $s'(0) = 0$, the gradient at the $s$-intercept is indeed zero, proving that there is a stationary point at this point."}]},{"id":"solve_derivative","label":"Solving for Stationary Points","steps":[{"type":"text","order":0,"title":"Differentiate the function","content":"To find the stationary points of the displacement function $s(t)$, we first need to find its derivative with respect to time $t$, denoted by $s'(t)$ or $\\frac{ds}{dt}$.\n\nThe function is:\n$$s(t) = k + 15at^2 - 2t^3$$\n\nUsing the power rule for differentiation:\n\n* The derivative of the constant $k$ is $0$.\n* The derivative of $15at^2$ is $2 \\times 15at = 30at$.\n* The derivative of $-2t^3$ is $-3 \\times 2t^2 = -6t^2$.\n\nCombining these gives:\n$$s'(t) = 30at - 6t^2$$"},{"type":"text","order":1,"title":"Find the stationary points","content":"A stationary point occurs where the gradient of the curve is zero, which means $s'(t) = 0$.\n\nSetting our derivative to zero:\n$$30at - 6t^2 = 0$$\n\nWe can factorise this equation by taking out a common factor of $6t$:\n$$6t(5a - t) = 0$$\n\nThis gives two solutions for the stationary points:\n$$t = 0 \\quad \\text{or} \\quad t = 5a$$"},{"type":"text","order":2,"title":"Identify the s-intercept","content":"The $s$-intercept of a function $s(t)$ is the point where the curve crosses the vertical axis, which occurs when the independent variable is zero, i.e., at:\n$$t = 0$$"},{"type":"text","order":3,"title":"Relate intercept to stationary point","content":"Comparing our findings:\n\n1. The $s$-intercept occurs at $t = 0$.\n2. One of the stationary points of the curve occurs at $t = 0$ (since $s'(0) = 30a(0) - 6(0)^2 = 0$).\n\nTherefore, we have shown that the curve $s(t)$ has a stationary point at its $s$-intercept."}]}],"generatedAt":"2026-06-09T05:59:15.897Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"12b55d23-5ca0-463d-9348-f9dee1a0a413","label":"bii","content":"Given that the polynomial equation \$s(t) = 0\$, when extended to all real values of \$t\$, has three distinct real roots, determine the range of possible values for \$k\$ in terms of \$a\$. You must use the second derivative to justify the nature of the stationary points used in your calculation.","markscheme":"- Sets $s'(t) = 0$ to find the other stationary point: $6t(5a - t) = 0$. M1\n- Identifies the stationary points are at $t = 0$ and $t = 5a$. A1\n- Finds the second derivative $s''(t) = 30a - 12t$. M1\n- Substitutes $t$ values into $s''(t)$ to determine nature: $s''(0) = 30a > 0$ (Local Minimum) and $s''(5a) = 30a - 60a = -30a < 0$ (Local Maximum). A1\n- Calculates the vertical values at these points: $s(0) = k$ and $s(5a) = k + 15a(5a)^2 - 2(5a)^3 = k + 125a^3$. M1\n- States that for three distinct roots, the local minimum must be negative ($k < 0$) and the local maximum must be positive ($k + 125a^3 > 0$). M1\n- Concludes the range: $-125a^3 < k < 0$. A1","marks":7,"order":2,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"analytical_second_derivative","label":"Analytical Method with Second Derivative","steps":[{"type":"text","order":0,"title":"Find the stationary points","content":"To find the stationary points of the function $s(t)$, we set its derivative $s'(t)$ to zero:\n\n$$s'(t) = 6t(5a - t) = 0$$\n\nThis gives two factor equations to solve:\n\n$$6t = 0 \\implies t = 0$$\n\n$$5a - t = 0 \\implies t = 5a$$\n\nSince $a > 0$, we have two distinct stationary points at $t = 0$ and $t = 5a$."},{"type":"text","order":1,"title":"Find the second derivative","content":"To determine the nature of these stationary points, we need to find the second derivative, $s''(t)$.\n\nStarting from the expanded first derivative:\n\n$$s'(t) = 30at - 6t^2$$\n\nDifferentiating this with respect to $t$ gives:\n\n$$s''(t) = 30a - 12t$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\text{For } \\htmlData{anchor=t0}{t = 0}: \\quad s''(0) = 30a - 12(0) = \\htmlData{anchor=s2_0}{30a}"},{"latex":"\\text{Since } a > 0, \\quad \\htmlData{anchor=min-cond}{30a > 0} \\implies \\htmlData{anchor=min-nature}{\\text{Local Minimum}}"},{"latex":"\\text{For } \\htmlData{anchor=t5a}{t = 5a}: \\quad s''(5a) = 30a - 12(5a) = \\htmlData{anchor=s2_5a}{-30a}"},{"latex":"\\text{Since } a > 0, \\quad \\htmlData{anchor=max-cond}{-30a < 0} \\implies \\htmlData{anchor=max-nature}{\\text{Local Maximum}}"}],"highlights":[{"color":"green","style":"box","anchor":"min-nature"},{"color":"red","style":"box","anchor":"max-nature"}],"connections":[{"to":"s2_0","from":"t0","color":"sky","label":"substitute","style":"solid"},{"to":"s2_5a","from":"t5a","color":"amber","label":"substitute","style":"solid"}]},"order":2,"title":"Determine stationary point nature","description":"We substitute our critical values into the second derivative to classify them."},{"type":"text","order":3,"title":"Find stationary values","content":"Next, we calculate the function values at our local minimum and local maximum.\n\nFrom the equation $s(t) = -2t^3 + 15at^2 + k$:\n\n* **At the local minimum ($t = 0$):**\n\n$$s(0) = k$$\n\n* **At the local maximum ($t = 5a$):**\n\n$$s(5a) = -2(5a)^3 + 15a(5a)^2 + k$$\n\n$$s(5a) = -250a^3 + 375a^3 + k$$\n\n$$s(5a) = k + 125a^3$$"},{"type":"text","order":4,"title":"Apply three roots condition","content":"For a cubic equation to have **three distinct real roots**, the graph of the function must cross the horizontal axis three times.\n\nThis occurs if and only if the local minimum is below the axis and the local maximum is above the axis:\n\n$$\\text{Local Minimum} < 0 \\quad \\text{and} \\quad \\text{Local Maximum} > 0$$\n\nSubstituting our values into these inequalities:\n\n$$k < 0$$\n\n$$k + 125a^3 > 0 \\implies k > -125a^3$$"},{"type":"text","order":5,"title":"State the final range","content":"Combining both inequalities yields the range of possible values for $k$:\n\n$$-125a^3 < k < 0$$"}]}],"generatedAt":"2026-06-09T06:01:41.330Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"dae5e12a-0102-493d-b970-86e30f6d24f0:property_based","label":"(a) Property-Based Sketching","steps":[{"type":"text","order":0,"title":"Analyze the question properties","content":"To sketch our cubic graph, we need to extract two key properties from the question:\n\n1. The coefficient of $x^3$ is positive (i.e., $a > 0$ for $y = ax^3 + bx^2 + cx + d$).\n2. The function has three distinct real roots.","highlights":[{"text":"three distinct real roots","location":"specification"},{"text":"positive coefficient of \$x^3\$","location":"specification"}]},{"type":"text","order":1,"title":"Determine the end behavior","content":"Let's look at the end behavior for a cubic function with a positive leading coefficient ($a > 0$):\n\n- As $x \\to \\infty$, the $ax^3$ term dominates, so $y \\to \\infty$.\n- As $x \\to -\\infty$, the $ax^3$ term dominates, so $y \\to -\\infty$.\n\nThis means our sketch must start in the **bottom-left quadrant** (third quadrant) and end in the **top-right quadrant** (first quadrant), giving it a characteristic **'up-down-up'** shape."},{"type":"text","order":2,"title":"Incorporate the roots","content":"The question specifies **three distinct real roots**.\n\nSince roots are the values of $x$ where $y = 0$, our curve must cross the horizontal $x$-axis at **exactly three separate points**."},{"type":"text","order":3,"title":"Sketch the final graph","content":"Now we combine these properties to sketch the curve:\n\n1. Draw a horizontal $x$-axis and a vertical $y$-axis.\n2. Mark three distinct crossing points on the $x$-axis.\n3. Draw a smooth, continuous 'up-down-up' curve that starts from the bottom-left, goes up through the first intercept, peaks, goes down through the second intercept, bottoms out, and then goes up through the third intercept towards the top-right.\n\n### Markscheme Breakdown\n- **1 Mark** (B1): Correct 'up-down-up' shape starting in the bottom-left and ending in the top-right.\n- **1 Mark** (B1): Graph intersects the horizontal axis at exactly three distinct points."},{"type":"text","order":4,"title":"Visualize with a calculator","content":"To build a visual intuition, you can plot a representative positive cubic equation with three distinct roots, such as $y = x^3 - 4x$, on your graphing calculator.","calculator":[{"gdc":{"expression":"x^3 - 4x","instructions":"1. Go to the Graph menu.\n2. Enter the function f(x) = x^3 - 4x.\n3. Adjust the view window to see the three intercepts and the characteristic cubic shape."},"ti84":{"expression":"x^3 - 4x","instructions":"1. Press [Y=] and enter Y1 = X^3 - 4*X.\n2. Press [WINDOW] and set Xmin = -4, Xmax = 4, Ymin = -5, Ymax = 5.\n3. Press [GRAPH] to observe the 'up-down-up' curve crossing the x-axis at -2, 0, and 2."},"desmos":{"expression":"y = x^3 - 4x","instructions":"Type y = x^3 - 4x into the input bar. Notice that the curve starts bottom-left, goes up, down, and up, crossing the x-axis at exactly three distinct points."},"description":"Visualize the graph of a representative positive cubic equation with three distinct roots: $y = x^3 - 4x$."}]}]},{"id":"dae5e12a-0102-493d-b970-86e30f6d24f0:algebraic_exemplar","label":"(a) Using an Algebraic Exemplar","steps":[{"type":"text","order":0,"title":"Define an algebraic exemplar","content":"To understand the shape and features of a cubic graph with three distinct real roots and a positive $x^3$ coefficient, we can construct a specific algebraic exemplar in factored form:\n\n$$y = (x - 1)(x - 2)(x - 3)$$\n\nIf we expand this equation, we get:\n\n$$y = (x^2 - 3x + 2)(x - 3)$$\n\n$$y = x^3 - 6x^2 + 11x - 6$$\n\nWe see that the coefficient of $x^3$ is $+1$, which satisfies the condition of having a positive leading coefficient."},{"type":"text","order":1,"title":"Find the intercepts","content":"Next, let's find the points where our exemplar function crosses the coordinate axes.\n\n**1. $x$-intercepts (roots):**\nSet $y = 0$:\n\n$$(x - 1)(x - 2)(x - 3) = 0$$\n\nThis gives three distinct real solutions:\n$$x = 1, \\quad x = 2, \\quad x = 3$$\nSo, the graph intersects the $x$-axis at $(1, 0)$, $(2, 0)$, and $(3, 0)$.\n\n**2. $y$-intercept:**\nSet $x = 0$:\n\n$$y = (0 - 1)(0 - 2)(0 - 3) = -6$$\n\nSo, the graph crosses the $y$-axis below the origin at $(0, -6)$."},{"type":"text","order":2,"title":"Determine the end behaviour","content":"To determine where the curve starts and ends, we analyze the end behaviour as $x$ becomes very large in the positive and negative directions:\n\n- As $x \\to \\infty$, the term $x^3 \\to \\infty$. This means the graph rises to the top-right quadrant.\n- As $x \\to -\\infty$, the term $x^3 \\to -\\infty$. This means the graph falls to the bottom-left quadrant.\n\nSince the graph starts at the bottom-left, goes up through $(1, 0)$, turns around to pass down through $(2, 0)$, and turns again to go up through $(3, 0)$ to the top-right, it forms a classic 'up-down-up' shape."},{"type":"text","order":3,"title":"Visualize with a calculator","content":"You can verify this curve and its key features using a graphing calculator.","calculator":[{"gdc":{"expression":"(x-1)(x-2)(x-3)","instructions":"Go to the Graph menu, enter the function y = (x - 1)(x - 2)(x - 3), adjust your view window to show x from -1 to 5, and sketch."},"ti84":{"expression":"(X-1)(X-2)(X-3)","instructions":"Press [Y=], enter (X-1)(X-2)(X-3) in Y1. Press [WINDOW] and set Xmin = -1, Xmax = 5, Ymin = -8, Ymax = 4. Press [GRAPH] to view the curve."},"desmos":{"expression":"y = (x - 1)(x - 2)(x - 3)","instructions":"Type y = (x - 1)(x - 2)(x - 3) into the input panel to see the roots and y-intercept immediately."},"description":"Plot the cubic function to visualize its three distinct roots and standard positive cubic shape."}]},{"type":"text","order":4,"title":"Sketch the final graph","content":"Now, sketch the graph on a set of axes incorporating these features:\n\n1. Draw a smooth, continuous curve showing the **'up-down-up' shape** starting in the bottom-left and ending in the top-right.\n2. Ensure the curve clearly crosses the $x$-axis at **exactly three distinct points**."}]},{"id":"4786c0e5-8567-471e-bbc6-65c6b6bc7888:direct_evaluation","label":"(bi) Direct Evaluation of Derivative","steps":[{"type":"text","order":0,"title":"Identify the s-intercept","content":"The $s$-intercept of a curve $s(t)$ occurs where the independent variable is equal to zero.\n\nTherefore, the $s$-intercept occurs at:\n$$t = 0$$","highlights":[{"text":"s-intercept","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$s(t) = k + 15a\\htmlData{anchor=t2}{t^2} - 2\\htmlData{anchor=t3}{t^3}$"},{"latex":"$$s'(t) = 15a \\cdot \\htmlData{anchor=diff-t2}{\\color{@sky}{2t}} - 2 \\cdot \\htmlData{anchor=diff-t3}{\\color{@amber}{3t^2}}$"},{"latex":"$$s'(t) = 30at - 6t^2$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"diff-t2"},{"color":"amber","style":"text-color","anchor":"diff-t3"}],"connections":[{"to":"diff-t2","from":"t2","color":"sky","label":"power rule","style":"solid"},{"to":"diff-t3","from":"t3","color":"amber","label":"power rule","style":"solid"}]},"order":1,"title":"Differentiate the function","description":"To find the gradient of the curve, we differentiate the displacement function $s(t)$ with respect to $t$. Recall that $k$ and $a$ are constants, so the derivative of $k$ is $0$."},{"type":"text","order":2,"title":"Substitute t = 0","content":"A stationary point occurs where the gradient is zero, i.e., $s'(t) = 0$.\n\nTo see if a stationary point exists at the $s$-intercept, we substitute $t = 0$ into our derivative function $s'(t)$:\n\n$$s'(0) = 30a(0) - 6(0)^2 = 0$$\n\nSince $s'(0) = 0$, the gradient at the $s$-intercept is indeed zero, proving that there is a stationary point at this point."}]},{"id":"4786c0e5-8567-471e-bbc6-65c6b6bc7888:solve_derivative","label":"(bi) Solving for Stationary Points","steps":[{"type":"text","order":0,"title":"Differentiate the function","content":"To find the stationary points of the displacement function $s(t)$, we first need to find its derivative with respect to time $t$, denoted by $s'(t)$ or $\\frac{ds}{dt}$.\n\nThe function is:\n$$s(t) = k + 15at^2 - 2t^3$$\n\nUsing the power rule for differentiation:\n\n* The derivative of the constant $k$ is $0$.\n* The derivative of $15at^2$ is $2 \\times 15at = 30at$.\n* The derivative of $-2t^3$ is $-3 \\times 2t^2 = -6t^2$.\n\nCombining these gives:\n$$s'(t) = 30at - 6t^2$$"},{"type":"text","order":1,"title":"Find the stationary points","content":"A stationary point occurs where the gradient of the curve is zero, which means $s'(t) = 0$.\n\nSetting our derivative to zero:\n$$30at - 6t^2 = 0$$\n\nWe can factorise this equation by taking out a common factor of $6t$:\n$$6t(5a - t) = 0$$\n\nThis gives two solutions for the stationary points:\n$$t = 0 \\quad \\text{or} \\quad t = 5a$$"},{"type":"text","order":2,"title":"Identify the s-intercept","content":"The $s$-intercept of a function $s(t)$ is the point where the curve crosses the vertical axis, which occurs when the independent variable is zero, i.e., at:\n$$t = 0$$"},{"type":"text","order":3,"title":"Relate intercept to stationary point","content":"Comparing our findings:\n\n1. The $s$-intercept occurs at $t = 0$.\n2. One of the stationary points of the curve occurs at $t = 0$ (since $s'(0) = 30a(0) - 6(0)^2 = 0$).\n\nTherefore, we have shown that the curve $s(t)$ has a stationary point at its $s$-intercept."}]},{"id":"12b55d23-5ca0-463d-9348-f9dee1a0a413:analytical_second_derivative","label":"(bii) Analytical Method with Second Derivative","steps":[{"type":"text","order":0,"title":"Find the stationary points","content":"To find the stationary points of the function $s(t)$, we set its derivative $s'(t)$ to zero:\n\n$$s'(t) = 6t(5a - t) = 0$$\n\nThis gives two factor equations to solve:\n\n$$6t = 0 \\implies t = 0$$\n\n$$5a - t = 0 \\implies t = 5a$$\n\nSince $a > 0$, we have two distinct stationary points at $t = 0$ and $t = 5a$."},{"type":"text","order":1,"title":"Find the second derivative","content":"To determine the nature of these stationary points, we need to find the second derivative, $s''(t)$.\n\nStarting from the expanded first derivative:\n\n$$s'(t) = 30at - 6t^2$$\n\nDifferentiating this with respect to $t$ gives:\n\n$$s''(t) = 30a - 12t$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\text{For } \\htmlData{anchor=t0}{t = 0}: \\quad s''(0) = 30a - 12(0) = \\htmlData{anchor=s2_0}{30a}"},{"latex":"\\text{Since } a > 0, \\quad \\htmlData{anchor=min-cond}{30a > 0} \\implies \\htmlData{anchor=min-nature}{\\text{Local Minimum}}"},{"latex":"\\text{For } \\htmlData{anchor=t5a}{t = 5a}: \\quad s''(5a) = 30a - 12(5a) = \\htmlData{anchor=s2_5a}{-30a}"},{"latex":"\\text{Since } a > 0, \\quad \\htmlData{anchor=max-cond}{-30a < 0} \\implies \\htmlData{anchor=max-nature}{\\text{Local Maximum}}"}],"highlights":[{"color":"green","style":"box","anchor":"min-nature"},{"color":"red","style":"box","anchor":"max-nature"}],"connections":[{"to":"s2_0","from":"t0","color":"sky","label":"substitute","style":"solid"},{"to":"s2_5a","from":"t5a","color":"amber","label":"substitute","style":"solid"}]},"order":2,"title":"Determine stationary point nature","description":"We substitute our critical values into the second derivative to classify them."},{"type":"text","order":3,"title":"Find stationary values","content":"Next, we calculate the function values at our local minimum and local maximum.\n\nFrom the equation $s(t) = -2t^3 + 15at^2 + k$:\n\n* **At the local minimum ($t = 0$):**\n\n$$s(0) = k$$\n\n* **At the local maximum ($t = 5a$):**\n\n$$s(5a) = -2(5a)^3 + 15a(5a)^2 + k$$\n\n$$s(5a) = -250a^3 + 375a^3 + k$$\n\n$$s(5a) = k + 125a^3$$"},{"type":"text","order":4,"title":"Apply three roots condition","content":"For a cubic equation to have **three distinct real roots**, the graph of the function must cross the horizontal axis three times.\n\nThis occurs if and only if the local minimum is below the axis and the local maximum is above the axis:\n\n$$\\text{Local Minimum} < 0 \\quad \\text{and} \\quad \\text{Local Maximum} > 0$$\n\nSubstituting our values into these inequalities:\n\n$$k < 0$$\n\n$$k + 125a^3 > 0 \\implies k > -125a^3$$"},{"type":"text","order":5,"title":"State the final range","content":"Combining both inequalities yields the range of possible values for $k$:\n\n$$-125a^3 < k < 0$$"}]}],"generatedAt":"2026-06-09T06:01:50.033Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths__ccea","summaries":[{"id":"271629d4-4d79-40bc-9d73-e2a69e4036eb","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"2f04e762-4586-414e-bebe-f889e9acfb44","difficultyBand":"easy","calculatorUse":null,"totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"43ee0cc3-2875-40e6-a2d3-05db031f0926","difficultyBand":"hard","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"7e520f0b-0f26-441d-bcf8-d8772c82816b","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"ab53f349-394c-4b37-bad6-d4335063076a","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"a4b21e08-0489-45ba-ab5d-ed8649a1545e","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"6f9c79e5-881d-407e-9885-7700f9a2feb6","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"f043ba0f-4bdf-4210-92b5-44c6e30454ba","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"f21883f5-ff65-4677-a258-451f367f8f12","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"c55851ef-0fe0-4d7e-a59a-71ac0e1071fb","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"24e55a38-714e-4bfd-90a4-40809c096bdb","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"3f25dfef-0089-4d94-b516-812bde5b0b8f","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"df6d7373-4429-4681-9f98-9849a3e2830a","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"540b49d2-f1b8-44f1-91cc-1e869ba99e81","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"d9772c72-c014-451b-8b39-0c80b4c7894f","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"1988e260-3dd6-4d40-a3a0-29f80bf07e36","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"36a3d7b5-ff9e-4cb5-8a36-77923b67b622","difficultyBand":"hard","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"bad6ae49-1fc1-4a13-ad6b-1e95d1d8e4b7","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"d4eb891b-aa87-4878-8b69-97659a0e2e91","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"9fb410a6-9095-4bc7-b817-dd6f3b20aea8","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"0ce72054-4cfb-498c-9c59-f0032919f2c9","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"037eaafa-64de-4214-aef0-0f0cfb011866","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"01425972-1ed3-4db5-950c-3ff2baeadef5","difficultyBand":"hard","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"02494252-4d0f-4a78-96a8-f60aab966de8","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"04d7eb5e-4403-48b9-b1a2-08ee7e153fca","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"11ff4a7b-71fe-4f70-8a94-35a4aac7a4b8","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"154d1b17-037a-4a68-96cb-06b6987ed59a","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":false},{"id":"168b68c1-b891-4e24-b580-e25fab75e10d","difficultyBand":"easy","calculatorUse":null,"totalMarks":1,"questionType":"LA","isCalculation":false},{"id":"179f5472-e005-4723-a7e2-6d59af55df0f","difficultyBand":"hard","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"17acfb93-eb1d-4d72-9558-4d69d5288c45","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"1874b7f1-1d88-45a5-be42-94eaeae824f5","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"1931b42d-e9fb-4354-91a3-da4818aaa52a","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":false},{"id":"21c0ba59-a5a9-49ac-b2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1.6 Differentiation

Question 10

1.6 Differentiation Questions

1.6 Differentiation

Question 10

1.6 Differentiation Questions