Exam-Style Problems

A restricted inverse and a composite equation · Maths4U Original · 8 marks

16863

Notes & worked examples

Info

Maths4U Original · M4U-ORIG-REV-001

Let \(f(x)=2(x-1)^2+5\) for \(\displaystyle x\ge 1\), and let \(g(x)=3x-2\) for real x.

Find the range of f and its inverse, including the inverse domain. [3]
Solve \(f(g(x))=23\), retaining only valid inputs. [3]
Find all permitted x such that \(\displaystyle f(x)\le 13\). [2]

Solution

Hint

The outer function accepts inputs at least 1. Use that condition before accepting roots of the composite equation.

Worked solution

(a)

For \(\displaystyle x\ge 1, x-1\ge 0\) and f has minimum 5. Solving \(\displaystyle y=2(x-1)^{2}+5\) gives \(\displaystyle x- 1= \sqrt{\frac{y- 5}{2}}\), choosing the nonnegative branch. Hence the displayed inverse has input domain [5,∞) and output range [1,∞).

Answer: Range [5,∞); \(\displaystyle f^{- 1}(x)= 1+ \sqrt{\frac{x- 5}{2}}\), for \(\displaystyle x\ge 5\).

(b)

The composition requires \(\displaystyle 3x-2\ge 1\), so \(\displaystyle x\ge 1\). The equation becomes \(\displaystyle 2(3x-3)^{2}+5=23\), hence \(\displaystyle (3x-3)^{2}=9\). Its algebraic candidates are \(\displaystyle x=0\) and \(\displaystyle x=2\). Only \(\displaystyle x=2\) satisfies the composition domain; direct substitution gives \(\displaystyle f(g(2))=f(4)=23\).

Answer: \(x=2\).

(c)

The inequality gives \(\displaystyle (x-1)^{2}\le 4\). Together with \(\displaystyle x\ge 1\) this becomes \(\displaystyle 0\le x-1\le 2\), or \(\displaystyle 1\le x\le 3\). Equivalently the increasing branch runs from its vertex to \(\displaystyle f^{−1}(13)=3\).

Answer: \(1\le x\le3\).

Maths4U indicative marking guidance

Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.

(a) · 3 marks

1 mark: Give the range [5,∞).
1 mark: Give the correct positive-branch inverse.
1 mark: State its domain \(\displaystyle x\ge 5\), consistent with the original range.

Alternative methods: An equivalent exact square-root expression is accepted.

Common error: Keeping both square-root branches or confusing the two domains.

(b) · 3 marks

1 mark: Obtain composition domain \(\displaystyle x\ge 1\).
1 mark: Find both algebraic candidates 0 and 2.
1 mark: Retain only \(\displaystyle x=2\) with a valid domain check.

Alternative methods: Using the inverse \(\displaystyle f^{−1}(23)=4\) and then solving \(\displaystyle g(x)=4\) is accepted.

Common error: Accepting the discarded square-root branch as a composition solution.

(c) · 2 marks

1 mark: Solve the quadratic inequality with the stated branch restriction.
1 mark: Give the complete closed interval [1,3].

Alternative methods: Monotonicity and inverse-endpoint reasoning is accepted.

Common error: Giving the unrestricted interval [−1,3].

Total: 8 marks.

A line parameter controls intersection count · Maths4U Original · 8 marks

16864

Notes & worked examples

Info

Maths4U Original · M4U-ORIG-REV-002

Consider the parabola \(y=x^2-4x+7\) and the family of lines \(y=mx+3\), with m real.

Form the quadratic equation for the x-coordinates of intersections. [2]
Find all m for which there are two distinct real intersection points. [3]
Find both tangent lines in this family and their points of contact. [3]

Solution

Hint

Eliminate y first. A repeated intersection root identifies a tangent contact, whose coordinates still need to be found.

Worked solution

(a)

Equating the ordinates gives \(\displaystyle x^{2}\)−4x+7=mx+3. Collecting terms gives \(\displaystyle x^{2}-(m+4)x+4=0\).

Answer: \(x^2-(m+4)x+4=0\).

(b)

The discriminant is \(\displaystyle (m+4)^{2}-16\). Requiring it to be strictly positive gives |m+4|>4, hence m<−8 or m>0. At the excluded boundary values the intersection is repeated.

Answer: \(m\lt -8\) or \(m\gt 0\).

(c)

Discriminant zero gives \(\displaystyle m=0\) or −8. The repeated root is \(\displaystyle x= \frac{m+ 4}{2}\). For \(\displaystyle m=0\) it is 2, with \(\displaystyle y=3\). For \(\displaystyle m=-8\) it is −2, with \(\displaystyle y=19\). The parabola derivative 2x−4 agrees with each stated line slope.

Answer: \(y=3\) at (2,3), and \(y=-8x+3\) at (−2,19).

Maths4U indicative marking guidance

Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.

(a) · 2 marks

1 mark: Equate the original curve and line.
1 mark: Obtain the complete intersection quadratic.

Alternative methods: An equivalent rearrangement is accepted.

Common error: Dropping the line constant or giving an equation only in m.

(b) · 3 marks

1 mark: Use the correct discriminant with a strict positive condition.
1 mark: Solve both outside branches of the inequality.
1 mark: Give m<−8 or m>0 with boundary values excluded.

Alternative methods: Factor m(m+8)>0 and use a sign chart.

Common error: Including tangency by using a non-strict inequality.

(c) · 3 marks

1 mark: Find both parameter values 0 and −8.
1 mark: Give tangent \(\displaystyle y=3\) and contact (2,3).
1 mark: Give tangent \(\displaystyle y=-8x+3\) and contact (−2,19).

Alternative methods: Solve the derivative slope condition together with the point lying on the corresponding line.

Common error: Giving the parameter values without line equations and contact points.

Total: 8 marks.

One tangent shared by two curves · Maths4U Original · 8 marks

16865

Notes & worked examples

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Maths4U Original · M4U-ORIG-REV-003

The point \(\displaystyle P=(2,2)\) lies on the parabola \(y=x^2-2x+2\). A circle has centre \(\displaystyle C=(4,1)\) and passes through P.

Find the tangent to the parabola at P. [3]
Find the equation of the circle. [2]
By substituting the line from (a) into the circle equation, prove that it is also tangent to the circle at P. [3]

Solution

Hint

Use the derivative for the parabola, distance for the circle, and substitution for a separate tangency check.

Worked solution

(a)

The derivative is 2x−2, so its value at \(\displaystyle x=2\) is 2. The tangent through (2,2) is \(\displaystyle y-2=2(x-2)\), hence \(\displaystyle y=2x-2\).

Answer: \(y=2x-2\).

(b)

The squared radius is \(\displaystyle CP^{2}=(2-4)^{2}+(2-1)^{2}=5\). With centre (4,1), the equation is \(\displaystyle (x-4)^{2}+(y-1)^{2}=5\).

Answer: \((x-4)^2+(y-1)^2=5\).

(c)

Substituting \(\displaystyle y=2x-2\) gives \(\displaystyle (x-4)^{2}+(2x-3)^{2}=5\). Expanding reduces it to \(\displaystyle 5x^{2}-20x+20=5(x-2)^{2}=0\). Thus \(\displaystyle x=2\) is the only, repeated root and the line gives \(\displaystyle y=2\). The unique circle contact is P, so the line is tangent there.

Answer: The equation reduces to \(5(x-2)^2=0\), giving only P.

Maths4U indicative marking guidance

Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.

(a) · 3 marks

1 mark: Find the derivative 2x−2.
1 mark: Use slope 2 and the point P.
1 mark: Give the correct tangent equation.

Alternative methods: A first-principles derivative is accepted.

Common error: Using the ordinate as a slope without differentiating the curve.

(b) · 2 marks

1 mark: Find the squared radius 5.
1 mark: Give the correct circle equation.

Alternative methods: An equivalent expanded circle equation is accepted.

Common error: Using 5 as the radius and squaring it again.

(c) · 3 marks

1 mark: Substitute the correct line into the circle.
1 mark: Obtain the repeated-root equation \(\displaystyle 5(x-2)^{2}=0\).
1 mark: Identify the unique point P and conclude tangency.

Alternative methods: An equivalent substitution followed by a zero discriminant is accepted.

Common error: Asserting that the parabola tangent is automatically a circle tangent without checking the circle.

Total: 8 marks.

A nearest point and its normal · Maths4U Original · 8 marks

16866

Notes & worked examples

Info

Maths4U Original · M4U-ORIG-REV-004

A point \(\displaystyle P=(x,x^{2})\), with \(\displaystyle x\ge 0\), moves on the parabola \(\displaystyle y=x^{2}\). The fixed point is \(\displaystyle A=(0,3)\).

Express the squared distance \(\displaystyle Q=AP^{2}\) as a polynomial in x. [2]
Find the point P for which AP is least and justify the global result. [3]
Find the least distance AP and show that AP is normal to the parabola there. [3]

Solution

Hint

Minimise \(\displaystyle AP^{2}\) before taking a square root. Explain why the chosen stationary value gives the global minimum.

Worked solution

(a)

\(\displaystyle Q=(x-0)^{2}+(x^{2}-3)^{2}=x^{2}+x^{4}-6x^{2}+9=x^{4}-5x^{2}+9\).

Answer: \(Q=x^4-5x^2+9\).

(b)

\(\displaystyle Q'=4x^{3}-10x=2x(2x^{2}-5)\). For positive x it is negative before \(\displaystyle \sqrt{\frac{5}{2}}\) and positive afterwards. Hence Q decreases from the boundary \(\displaystyle x=0\), then increases after the unique interior minimum. The nearest point has \(\displaystyle x= \sqrt{\frac{5}{2}}= \frac{\sqrt{10}}{2}\) and \(\displaystyle y= \frac{5}{2}\). Minimising Q minimises AP since the square root is increasing.

Answer: \(\displaystyle P= (\frac{\sqrt{10}}{2}, \frac{5}{2})\).

(c)

At the point, \(\displaystyle Q= \frac{5}{2}+ (- \frac{1}{2})^{2}= \frac{11}{4}\), so \(\displaystyle AP= \frac{\sqrt{11}}{2}\). The line AP has slope \(\displaystyle \frac{\frac{5}{2}- 3}{\frac{\sqrt{10}}{2}}= - \frac{1}{\sqrt{10}}\). The parabola tangent slope is \(\displaystyle 2x=\sqrt{10}\); their product is −1, proving AP is normal.

Answer: Least distance \(\displaystyle \frac{\sqrt{11}}{2}\); AP slope \(\displaystyle - \frac{1}{\sqrt{10}}\), tangent slope \(\sqrt{10}\).

Maths4U indicative marking guidance

Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.

(a) · 2 marks

1 mark: Use both coordinate differences in the distance-squared formula.
1 mark: Expand to \(\displaystyle x^{4}-5x^{2}+9\).

Alternative methods: An equivalent factored distance expression with the correct expansion is accepted.

Common error: Minimising only the vertical separation.

(b) · 3 marks

1 mark: Differentiate and obtain the feasible positive stationary input \(\displaystyle \sqrt{\frac{5}{2}}\).
1 mark: Give the corresponding point \(\displaystyle (\frac{\sqrt{10}}{2}, \frac{5}{2})\).
1 mark: Justify a global minimum on \(\displaystyle x\ge 0\) and the use of squared distance.

Alternative methods: Use \(\displaystyle Q= (x^{2}- \frac{5}{2})^{2}+ \frac{11}{4}\) as a global bound, with its equality value feasible.

Common error: Selecting the stationary boundary \(\displaystyle x=0\) without comparing behaviour or ignoring \(\displaystyle x\ge 0\).

(c) · 3 marks

1 mark: Give the distance \(\displaystyle \frac{\sqrt{11}}{2}\) rather than its square.
1 mark: Find the joining-line and tangent slopes correctly.
1 mark: Use their product −1 to conclude perpendicularity.

Alternative methods: A zero dot product of direction vectors is accepted.

Common error: Reporting \(\displaystyle \frac{11}{4}\) as the distance or checking perpendicularity at a different point.

Total: 8 marks.

Two angles with one triangle area · Maths4U Original · 8 marks

16867

Notes & worked examples

Info

Maths4U Original · M4U-ORIG-REV-005

Two radii of a circle of radius 6 cm enclose an angle θ with 0<θ<π radians. The triangle formed by the radii and their chord has area \(\displaystyle 9 \mathrm{cm}^{2}\).

Find all possible θ exactly. [3]
Find the corresponding sector area for each angle. [2]
For the larger angle, find the minor-segment area exactly. [3]

Solution

Hint

The triangle area uses sin θ, whereas the sector area uses θ itself. Keep the two angle branches separate.

Worked solution

(a)

The triangle area is \(\displaystyle (\frac{1}{2})(6^{2})\sin \theta = 18\sin \theta\). Thus \(\displaystyle \sin \theta = \frac{1}{2}\). On the stated interval both \(\displaystyle \frac{\pi }{6}\) and \(\displaystyle \frac{5\pi }{6}\) are permitted.

Answer: \(\displaystyle \theta = \frac{\pi }{6}\) or \(\displaystyle \frac{5\pi }{6}\).

(b)

The sector area is 18θ. It gives \(\displaystyle 18(\frac{\pi }{6})= 3\pi \mathrm{cm}^{2}\) and \(\displaystyle 18(\frac{5\pi }{6})= 15\pi \mathrm{cm}^{2}\). The equal triangle areas do not make the sectors equal.

Answer: \(3\pi\) \(\displaystyle \mathrm{cm}^{2}\) for \(\displaystyle \frac{\pi }{6}\); \(15\pi\) \(\displaystyle \mathrm{cm}^{2}\) for \(\displaystyle \frac{5\pi }{6}\).

(c)

The larger angle \(\displaystyle \frac{5\pi }{6}\) is still less than π, so this is a minor segment. Subtract its triangle from its sector: \(\displaystyle S=18(\theta -\sin \theta )=15\pi -9 \mathrm{cm}^{2}\).

Answer: \(15\pi-9\) \(\displaystyle \mathrm{cm}^{2}\).

Maths4U indicative marking guidance

Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.

(a) · 3 marks

1 mark: Form \(\displaystyle 18\sin \theta =9\).
1 mark: Obtain \(\displaystyle \sin \theta = \frac{1}{2}\).
1 mark: Give both and only permitted exact radian angles.

Alternative methods: A unit-circle or reference-angle argument is accepted.

Common error: Keeping only the inverse-calculator principal angle.

(b) · 2 marks

1 mark: Use the radian sector formula and obtain \(\displaystyle 3\pi \mathrm{cm}^{2}\) for the smaller angle.
1 mark: Obtain \(\displaystyle 15\pi \mathrm{cm}^{2}\) for the larger angle with correct matching.

Alternative methods: Equivalent exact expressions are accepted.

Common error: Putting sin θ into the sector formula.

(c) · 3 marks

1 mark: Identify sector minus the enclosed triangle as the segment area.
1 mark: Use the larger permitted angle and its matching triangle area.
1 mark: Give \(\displaystyle 15\pi -9 \mathrm{cm}^{2}\).

Alternative methods: Subtracting the known triangle area directly from 15π is accepted.

Common error: Subtracting the small-angle sector or adding the triangle area.

Total: 8 marks.

A sector constraint becomes a quadratic · Maths4U Original · 8 marks

16868

Notes & worked examples

Info

Maths4U Original · M4U-ORIG-REV-006

A proper circular sector has radius r>0 cm and central angle 0<θ<2π radians. Its boundary consists of two distinct radii and the connecting arc, with total length 20 cm.

Express θ and the sector area A in terms of r, and state the feasible interval for r. [3]
Use differentiation to find the greatest sector area, stating the corresponding r and θ and justifying global optimality. [3]
When \(\displaystyle \theta = \frac{\pi }{2}\) instead, find r and A exactly. [2]

Solution

Hint

Eliminate θ and use r as the optimisation variable. The proper-sector angle range imposes an extra lower bound on r.

Worked solution

(a)

The perimeter equation is \(\displaystyle r\theta +2r=20\), so \(\displaystyle \theta = \frac{20}{r}- 2\). Then \(\displaystyle A= \frac{r^{2}\theta }{2}= 10r- r^{2}\). The inequalities \(\displaystyle 0< \frac{20}{r}- 2< 2\pi\), with r>0, give \(\displaystyle \frac{10}{1+ \pi }< r< 10\). Both boundaries are excluded.

Answer: \(\displaystyle \theta = \frac{20}{r}- 2\), \(A=10r-r^2\), and \(\displaystyle \frac{10}{1+ \pi }\lt r\lt 10\).

(b)

\(\displaystyle A'=10-2r\) vanishes at \(\displaystyle r=5\). It gives \(\displaystyle \theta = \frac{20}{5}- 2= 2\) and \(\displaystyle A=25\). The candidate is feasible because 0<2<2π. Also \(\displaystyle A=25-(r-5)^{2}\le 25\) on the entire feasible domain, with equality at \(\displaystyle r=5\), so this is the global maximum.

Answer: \(r=5\) cm, \(\theta=2\) rad, greatest area \(\displaystyle 25 \mathrm{cm}^{2}\).

(c)

The perimeter gives \(\displaystyle r= \frac{20}{2+ \frac{\pi }{2}}= \frac{40}{4+ \pi }\). The area is \(\displaystyle \frac{r^{2}\pi }{4}= \frac{400\pi }{(4+ \pi )^{2}}\mathrm{cm}^{2}\). This angle is feasible but differs from the optimal angle 2 rad.

Answer: \(\displaystyle r= \frac{40}{4+ \pi }\) cm; \(\displaystyle A= \frac{400\pi }{(4+ \pi )^{2}}\) \(\displaystyle \mathrm{cm}^{2}\).

Maths4U indicative marking guidance

Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.

(a) · 3 marks

1 mark: Obtain \(\displaystyle \theta = \frac{20}{r}- 2\).
1 mark: Obtain \(\displaystyle A=10r-r^{2}\).
1 mark: Give the complete strict radius interval.

Alternative methods: Use \(\displaystyle r= \frac{20}{\theta + 2}\) to derive the equivalent constraints before expressing the requested model in r.

Common error: Using only r<10 or including a full-turn boundary.

(b) · 3 marks

1 mark: Differentiate and find the feasible candidate \(\displaystyle r=5\) with \(\displaystyle \theta =2\).
1 mark: Give the greatest area \(\displaystyle 25 \mathrm{cm}^{2}\).
1 mark: Justify the global result and admissibility on the stated domain.

Alternative methods: A complete derivative-sign or concavity argument on the feasible interval is accepted.

Common error: Keeping an algebraic stationary point without checking the sector angle.

(c) · 2 marks

1 mark: Give the exact radius \(\displaystyle \frac{40}{4+ \pi }\).
1 mark: Give the exact sector area \(\displaystyle \frac{400\pi }{(4+ \pi )^{2}}\mathrm{cm}^{2}\).

Alternative methods: Substitution into \(\displaystyle A=10r-r^{2}\) is accepted if simplified to an equivalent exact result.

Common error: Using \(\displaystyle \frac{\pi }{2}\) degrees or changing the perimeter model when the angle changes.

Total: 8 marks.

Binomial coefficients in an arithmetic progression · Maths4U Original · 8 marks

16869

Notes & worked examples

Info

Maths4U Original · M4U-ORIG-REV-007

In the expansion of \((1+kx)^6\), with k real, the coefficients of \(\displaystyle x^{0}, x^{1}\) and \(\displaystyle x^{2}\), in that order, form three consecutive terms of an arithmetic progression.

Show that k satisfies \(15k^2-12k+1=0\). [3]
Find both possible k exactly. [2]
It is now given that the progression is increasing. Select k and find the common difference exactly. [3]

Solution

Hint

The sequence uses coefficients, not the polynomial terms. An increasing progression requires a positive common difference.

Worked solution

(a)

The first three coefficients are 1, 6k and \(\displaystyle 15k^{2}\). Twice the middle AP term equals the sum of the outer terms, so \(\displaystyle 12k=1+15k^{2}\), giving the stated quadratic.

Answer: \(2(6k)=1+15k^2\), hence \(15k^2-12k+1=0\).

(b)

The quadratic formula gives \(\displaystyle k= \frac{12\pm \sqrt{144- 60}}{30}= \frac{12\pm 2\sqrt{21}}{30}= \frac{6\pm \sqrt{21}}{15}\). Both are real and both are positive, so positivity alone cannot select a branch.

Answer: \(\displaystyle k= \frac{6\pm \sqrt{21}}{15}\).

(c)

The difference is \(\displaystyle d= 6k- 1= \frac{7\pm 2\sqrt{21}}{5}\). Since \(\displaystyle 2\sqrt{21}>7\), the minus branch has negative difference and the plus branch positive difference. Increasing order therefore selects the plus value of k and \(\displaystyle d= \frac{7+ 2\sqrt{21}}{5}\).

Answer: \(\displaystyle k= \frac{6+ \sqrt{21}}{15}\), \(\displaystyle d= \frac{7+ 2\sqrt{21}}{5}\).

Maths4U indicative marking guidance

Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.

(a) · 3 marks

1 mark: Extract the three correct coefficient values.
1 mark: Use the AP relation between them.
1 mark: Obtain \(\displaystyle 15k^{2}-12k+1=0\).

Alternative methods: Equating successive differences is accepted.

Common error: Including x and \(\displaystyle x^{2}\) in the progression data or using a GP product condition.

(b) · 2 marks

1 mark: Use the quadratic formula or complete the square correctly.
1 mark: Give both exact roots \(\displaystyle \frac{6\pm \sqrt{21}}{15}\).

Alternative methods: Equivalent exact surd expressions are accepted.

Common error: Rejecting a branch without using the remaining condition.

(c) · 3 marks

1 mark: Express the common difference using the coefficient values.
1 mark: Select the plus parameter branch using the sign of the difference.
1 mark: Give the exact difference \(\displaystyle \frac{7+ 2\sqrt{21}}{5}\) with the rejected branch correctly accounted for.

Alternative methods: Compare both pairs of consecutive coefficients directly.

Common error: Selecting the plus branch solely because k must be positive; both roots are positive.

Total: 8 marks.

A coefficient determines a convergent ratio · Maths4U Original · 8 marks

16870

Notes & worked examples

Info

Maths4U Original · M4U-ORIG-REV-008

A GP has first term a>0, ratio 0<r<1 and infinite sum 12. The coefficient of \(\displaystyle x^{2}\) in \((1+rx)^5\) is \(\displaystyle \frac{5}{2}\).

Find r and a. [3]
Find the first-four-term sum and the omitted tail after those four terms. [2]
Find the least positive integer n for which the omitted tail is less than 0.02. [3]

Solution

Hint

The ratio restriction selects a root of the coefficient equation. For the final part, compare tails at consecutive integer indices.

Worked solution

(a)

The coefficient is \(\displaystyle 10r^{2}= \frac{5}{2}\), so \(\displaystyle r^{2}= \frac{1}{4}\). Since \(\displaystyle 0< r< 1, r= \frac{1}{2}\). From \(\displaystyle \frac{a}{1- r}= 12\) we get \(\displaystyle a=6\).

Answer: \(\displaystyle r= \frac{1}{2}\), \(a=6\).

(b)

\(\displaystyle S_{4}= \frac{6(1- (\frac{1}{2})^{4})}{1- \frac{1}{2}}= 12(\frac{15}{16})= \frac{45}{4}\). Subtracting from the infinite sum gives \(\displaystyle 12- \frac{45}{4}= \frac{3}{4}\).

Answer: \(\displaystyle S_{4}= \frac{45}{4}\); tail \(\displaystyle \frac{3}{4}\).

(c)

The tail is \(\displaystyle R_{n}=12\cdot 2^{−n}\). Requiring \(\displaystyle R_{n}< \frac{1}{50}\) gives \(\displaystyle 2^{n}>600\). At \(\displaystyle n=9\) the tail is \(\displaystyle \frac{3}{128}\approx 0.0234375\), while at \(\displaystyle n=10\) it is \(\displaystyle \frac{3}{256}\approx 0.01171875\). The tails decrease strictly, so the least successful n is 10.

Answer: \(n=10\).

Maths4U indicative marking guidance

Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.

(a) · 3 marks

1 mark: Use the correct binomial coefficient equation.
1 mark: Select \(\displaystyle r= \frac{1}{2}\) using the given ratio domain.
1 mark: Use the infinite sum to obtain \(\displaystyle a=6\).

Alternative methods: Equivalent elimination using the convergent GP formula is accepted.

Common error: Keeping the negative ratio root despite the stated positive domain.

(b) · 2 marks

1 mark: Give the exact finite sum \(\displaystyle \frac{45}{4}\).
1 mark: Give the distinct omitted tail \(\displaystyle \frac{3}{4}\).

Alternative methods: Adding \(\displaystyle 6+ 3+ \frac{3}{2}+ \frac{3}{4}\) directly is accepted.

Common error: Treating the fourth term as the whole omitted tail.

(c) · 3 marks

1 mark: Form the correct strict tail inequality.
1 mark: Obtain \(\displaystyle n=10\).
1 mark: Check the consecutive bounds and justify minimality.

Alternative methods: A logarithm calculation followed by a valid strict integer check is accepted.

Common error: Using an individual term threshold or rounding a real index down.

Total: 8 marks.

One recovered curve, two geometric quantities · Maths4U Original · 9 marks

16871

Notes & worked examples

Info

Maths4U Original · M4U-ORIG-REV-009

A curve has \(\displaystyle \frac{dy}{dx}= 6x- 6\) for real x and passes through (0,8). Coordinates are measured in cm.

Find the curve equation and its stationary point, stating the nature of the point. [3]
Find the area enclosed by the curve and the line \(\displaystyle y=8\). [3]
The enclosed region is rotated about the x-axis. Find its volume exactly. [3]

Solution

Hint

Recover y first. For the enclosed region, the horizontal line is above a positive curve, so the area and rotated volume use different integrands.

Worked solution

(a)

Integration gives \(\displaystyle y=3x^{2}-6x+C\), with \(\displaystyle C=8\) from the point. The derivative vanishes at \(\displaystyle x=1\), where \(\displaystyle y=5\). The second derivative is 6>0, so this is a minimum.

Answer: \(y=3x^2-6x+8\); minimum at (1,5).

(b)

Intersections satisfy \(\displaystyle 3x^{2}-6x+8=8\), giving \(\displaystyle x=0\) and 2. On (0,2), the line is above the curve. Thus \(\displaystyle A=\int _{0}^{2}(6x-3x^{2})dx=[3x^{2}-x^{3}]_{0}^{2}=4 \mathrm{cm}^{2}\).

Answer: \(\displaystyle 4 \mathrm{cm}^{2}\).

(c)

The outer radius is 8 and inner radius \(\displaystyle y=3x^{2}-6x+8\), which is at least 5. The washer integral is \(\displaystyle \pi \int _{0}^{2}{64-(3x^{2}-6x+8)^{2}}dx\). Its integrand expands to \(\displaystyle -9x^{4}+36x^{3}-84x^{2}+96x\). The integral is \(\displaystyle \pi {- \frac{288}{5}+ 144- 224+ 192}= \frac{272\pi }{5}\mathrm{cm}^{3}\).

Answer: \(\displaystyle \frac{272\pi }{5}\) \(\displaystyle \mathrm{cm}^{3}\).

Maths4U indicative marking guidance

Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.

(a) · 3 marks

1 mark: Recover the complete curve using the point condition.
1 mark: Find the stationary point (1,5).
1 mark: Classify a minimum with a valid derivative or completed-square argument.

Alternative methods: Write \(\displaystyle y=3(x-1)^{2}+5\) to confirm the classification.

Common error: Using the given gradient itself as the curve ordinate.

(b) · 3 marks

1 mark: Find both intersections 0 and 2 and the correct upper curve.
1 mark: Set up the area as upper minus lower.
1 mark: Evaluate \(\displaystyle 4 \mathrm{cm}^{2}\).

Alternative methods: A complete symmetric area calculation is accepted.

Common error: Integrating the curve alone instead of the vertical separation.

(c) · 3 marks

1 mark: Use the correct squared outer and inner radii over [0,2].
1 mark: Expand and integrate the washer expression correctly.
1 mark: Give \(\displaystyle \frac{272\pi }{5}\mathrm{cm}^{3}\) exactly.

Alternative methods: Subtract the inner solid volume from the enclosing cylinder volume.

Common error: Squaring the area integrand (8−y) instead of using \(\displaystyle 64-y^{2}\).

Total: 9 marks.

The largest volume and its fastest growth · Maths4U Original · 9 marks

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Notes & worked examples

Info

Maths4U Original · M4U-ORIG-REV-010

For \(\displaystyle 0\le k\le 4\), the region under \(y=x(4-x)\) between \(\displaystyle x=0\) and \(\displaystyle x=k\) is rotated about the x-axis. Coordinates are in cm and the resulting volume is V(k).

Find an explicit expression for V(k). [3]
Find the volume when \(\displaystyle k=4\) exactly. [2]
Find where V grows most rapidly with respect to k, including the greatest growth value. Explain why this differs from where V is greatest. [4]

Solution

Hint

Square the radius before integrating. The derivative with respect to the cut-off coordinate k measures cross-sectional area, not elapsed time.

Worked solution

(a)

The radius is nonnegative on [0,4]. Its square is \(\displaystyle x^{2}(4-x)^{2}=16x^{2}-8x^{3}+x^{4}\). Hence \(\displaystyle V(k)= \pi \int_{0}^{k} (16x^{2}- 8x^{3}+ x^{4})dx= \pi (\frac{16k^{3}}{3}- 2k^{4}+ \frac{k^{5}}{5})\). The lower limit fixes \(\displaystyle V(0)=0\).

Answer: \(\displaystyle V(k)= \pi (\frac{16k^{3}}{3}- 2k^{4}+ \frac{k^{5}}{5})\).

(b)

\(\displaystyle V(4)= \pi (\frac{1024}{3}- 512+ \frac{1024}{5})= \frac{512\pi }{15}\mathrm{cm}^{3}\).

Answer: \(\displaystyle \frac{512\pi }{15}\) \(\displaystyle \mathrm{cm}^{3}\).

(c)

\(\displaystyle V'(k)=\pi k^{2}(4-k)^{2}\) is the disk area. Its derivative is 4πk(k−2)(k−4), positive for 0<k<2 and negative for 2<k<4. With endpoint values zero, the global maximum disk area is \(\displaystyle V'(2)=16\pi \mathrm{cm}^{2}\). However V′ is positive throughout (0,4), so V is strictly increasing and its greatest value occurs at \(\displaystyle k=4\). This is growth per unit length, not a time rate.

Answer: Greatest \(V^{\prime}=16\pi\) \(\displaystyle \mathrm{cm}^{2}\) at \(\displaystyle k=2 \mathrm{cm}\); V itself is greatest at \(\displaystyle k=4 \mathrm{cm}\).

Maths4U indicative marking guidance

Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.

(a) · 3 marks

1 mark: Expand the squared radius as \(\displaystyle 16x^{2}-8x^{3}+x^{4}\).
1 mark: Integrate every term correctly.
1 mark: Evaluate the variable limits and retain π without a free constant.

Alternative methods: An equivalent exact antiderivative expression evaluated from 0 to k is accepted.

Common error: Using the unsquared profile or introducing an arbitrary volume constant.

(b) · 2 marks

1 mark: Substitute the upper cut-off 4 into the complete expression.
1 mark: Give \(\displaystyle \frac{512\pi }{15}\mathrm{cm}^{3}\) exactly.

Alternative methods: A separate definite integral of the squared radius over [0,4] is accepted.

Common error: Rounding intermediate fractions or omitting π.

(c) · 4 marks

1 mark: Identify V′(k) as \(\displaystyle \pi k^{2}(4-k)^{2}\).
1 mark: Find the maximum-growth candidate \(\displaystyle k=2\) with valid differentiation or an equivalent bound.
1 mark: Give the global greatest growth value \(\displaystyle 16\pi \mathrm{cm}^{2}\) with endpoint/sign reasoning.
1 mark: Explain that V remains increasing and therefore is greatest at \(\displaystyle k=4\).

Alternative methods: Bound the nonnegative radius by 4 using \(\displaystyle x(4-x)=4-(x-2)^{2}\), then distinguish the accumulated volume.

Common error: Claiming V is largest at \(\displaystyle k=2\) because its derivative is largest there.

Total: 9 marks.