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Date May 2022 Marks available 3 Reference code 22M.3.AHL.TZ2.1
Level Additional Higher Level Paper Paper 3 Time zone Time zone 2
Command term Show that Question number 1 Adapted from N/A

Question

This question asks you to explore properties of a family of curves of the type y2=x3+ax+b for various values of a and b, where a, b.

On the same set of axes, sketch the following curves for -2x2 and -2y2, clearly indicating any points of intersection with the coordinate axes.

Now, consider curves of the form y2=x3+b, for x-b3, where b+.

Next, consider the curve y2=x3+x, x0.

The curve y2=x3+x has two points of inflexion. Due to the symmetry of the curve these points have the same x-coordinate.

P(x, y) is defined to be a rational point on a curve if x and y are rational numbers.

The tangent to the curve y2=x3+ax+b at a rational point P intersects the curve at another rational point Q.

Let C be the curve y2=x3+2, for x-23. The rational point P(-1, -1) lies on C.

y2=x3, x0

[2]
a.i.

y2=x3+1, x-1

[2]
a.ii.

Write down the coordinates of the two points of inflexion on the curve y2=x3+1.

[1]
b.i.

By considering each curve from part (a), identify two key features that would distinguish one curve from the other.

[1]
b.ii.

By varying the value of b, suggest two key features common to these curves.

[2]
c.

Show that dydx=±3x2+12x3+x, for x>0.

[3]
d.i.

Hence deduce that the curve y2=x3+x has no local minimum or maximum points.

[1]
d.ii.

Find the value of this x-coordinate, giving your answer in the form x=p3+qr, where p, q, r.

[7]
e.

Find the equation of the tangent to C at P.

[2]
f.i.

Hence, find the coordinates of the rational point Q where this tangent intersects C, expressing each coordinate as a fraction.

[2]
f.ii.

The point S(-1 , 1) also lies on C. The line [QS] intersects C at a further point. Determine the coordinates of this point.

[5]
g.

Markscheme

approximately symmetric about the x-axis graph of y2=x3         A1

including cusp/sharp point at (0, 0)         A1

 

[2 marks]

 

Note: Final A1 can be awarded if intersections are in approximate correct place with respect to the axes shown. Award A1A1A1A0 if graphs ‘merge’ or ‘cross’ or are discontinuous at x-axis but are otherwise correct. Award A1A0A0A0 if only one correct branch of both curves are seen.

Note: If they sketch graphs on separate axes, award a maximum of 2 marks for the ‘best’ response seen. This is likely to be A1A1A0A0.

a.i.

approximately symmetric about the x-axis graph of y2=x3+1 with approximately correct gradient at axes intercepts        A1
some indication of position of intersections at x=1, y=±1         A1

[2 marks]

 

Note: Final A1 can be awarded if intersections are in approximate correct place with respect to the axes shown. Award A1A1A1A0 if graphs ‘merge’ or ‘cross’ or are discontinuous at x-axis but are otherwise correct. Award A1A0A0A0 if only one correct branch of both curves are seen.

Note: If they sketch graphs on separate axes, award a maximum of 2 marks for the ‘best’ response seen. This is likely to be A1A1A0A0.

a.ii.

0, 1 and 0, -1       A1

 

[1 mark]

b.i.

Any two from:

y2=x3 has a cusp/sharp point, (the other does not)

graphs have different domains

y2=x3+1 has points of inflexion, (the other does not)

graphs have different x-axis intercepts (one goes through the origin, and the other does not)

graphs have different y-axis intercepts      A1

 

Note: Follow through from their sketch in part (a)(i). In accordance with marking rules, mark their first two responses and ignore any subsequent.

 

[1 mark]

b.ii.

Any two from:

as , x, y±

as x, y2=x3+b is approximated by y2=x3 (or similar)

they have x intercepts at x=-b3

they have y intercepts at y=±b

they all have the same range

y=0 (or x-axis) is a line of symmetry

they all have the same line of symmetry y=0

they have one x-axis intercept

they have two y-axis intercepts

they have two points of inflexion

at x-axis intercepts, curve is vertical/infinite gradient

there is no cusp/sharp point at x-axis intercepts     A1A1

 

Note: The last example is the only valid answer for things “not” present. Do not credit an answer of “they are all symmetrical” without some reference to the line of symmetry.

Note: Do not allow same/ similar shape or equivalent.

Note: In accordance with marking rules, mark their first two responses and ignore any subsequent.

 

[2 marks]

c.

METHOD 1

attempt to differentiate implicitly         M1

2ydydx=3x2+1         A1

dydx=3x2+12y  OR  ±2x3+xdydx=3x2+1         A1

dydx=±3x2+12x3+x         AG

 

METHOD 2

attempt to use chain rule y=±x3+x         M1

dydx=±12x3+x-123x2+1         A1A1

 

Note: Award A1 for ±12x3+x-12, A1 for 3x2+1

 

dydx=±3x2+12x3+x         AG

 

[3 marks]

d.i.

EITHER

local minima/maxima occur whendydx=0

1+3x2=0 has no (real) solutions (or equivalent)         R1


OR

x20 3x2+1>0, so dydx0          R1


THEN

so, no local minima/maxima exist          AG

 

[1 mark]

d.ii.

EITHER

attempt to use quotient rule to find d2ydx2          M1

d2ydx2=±12xx+x3-1+3x2x+x3-121+3x24x+x3          A1A1


Note:
Award A1 for correct 12xx+x3 and correct denominator, A1 for correct -1+3x2x+x3-121+3x2.

Note: Future A marks may be awarded if the denominator is missing or incorrect.


stating or using d2ydx2=0 (may be seen anywhere)           (M1)

12xx+x3=1+3x2x+x3-121+3x2


OR

attempt to use product rule to find d2ydx2          M1

d2ydx2=123x2+1-123x2+1x3+x-32+3xx3+x-12          A1A1


Note:
Award A1 for correct first term, A1 for correct second term.


setting d2ydx2=0           (M1)


OR

attempts implicit differentiation on 2ydydx=3x2+1          M1

2dydx2+2yd2ydx2=6x          A1

recognizes that d2ydx2=0           (M1)

dydx=±3x

±3x2+12x3+x=±3x           (A1)


THEN

12xx+x3=1+3x22

12x2+12x4=9x4+6x2+1

3x4+6x2-1=0          A1

attempt to use quadratic formula or equivalent           (M1)

x2=-6±486

x>0x=23-33 p=2, q=-3, r=3          A1

 

Note: Accept any integer multiple of p, q and r (e.g. 4,-6 and 6).

 

[7 marks]

e.

attempt to find tangent line through -1, -1           (M1)

y+1=-32x+1  OR  y=-1.5x-2.5           A1

 

[2 marks]

f.i.

attempt to solve simultaneously with y2=x3+2           (M1)


Note: The M1 mark can be awarded for an unsupported correct answer in an incorrect format (e.g. (4.25, -8.875)).


obtain 174,-718           A1

 

[2 marks]

f.ii.

attempt to find equation of [QS]           (M1)

y-1x+1=-7942=-1.88095           (A1)

solve simultaneously with y2=x3+2           (M1)

x=0.28798=127441        A1

y=-1.4226=131759261        A1

0.228,-1.42

 

OR

attempt to find vector equation of [QS]           (M1)

xy=-11+λ214-798           (A1)

x=-1+214λ

y=1-798λ

attempt to solve 1-798λ2=-1+214λ3+2           (M1)

λ=0.2453

x=0.28798=127441        A1

y=-1.4226=131759261        A1

0.228,-1.42

 

[5 marks]

g.

Examiners report

This was a relatively straightforward start, though it was disappointing to see so many candidates sketch their graphs on two separate axes, despite the question stating they should be sketched on the same axes.

a.i.
[N/A]
a.ii.

Of those candidates producing clear sketches, the vast majority were able to recognise the points of inflexion and write down their coordinates. A small number embarked on a mostly fruitless algebraic approach rather than use their graphs as intended. The distinguishing features between curves tended to focus on points of intersection with the axes, which was accepted. Only a small number offered ideas such as y on both curves. A number of (incorrect) suggestions were seen, stating that both curves tended towards a linear asymptote.

b.i.
[N/A]
b.ii.

A majority of candidates' suggestions related to the number of intersection points with the coordinate axes, while the idea of the x-axis acting as a line of symmetry was also often seen.

c.

The required differentiation was straightforward for the majority of candidates.

d.i.
[N/A]
d.ii.

The majority employed the quotient rule here, often doing so successfully to find a correct expression for d2ydx2. Despite realising that d2ydx2=0, the resulting algebra to find the required solution proved a step too far for most. A number of slips were seen in candidates' working, though better candidates were able to answer the question confidently.

e.

Mistakes proved to be increasingly common by this stage of the paper. Various equations of lines were suggested, with the incorrect y=1.5x+2.5 appearing more than once. Only the better candidates were able to tackle the final part of the question with any success; it was pleasing to see a number of clear algebraic (only) approaches, though this was not necessary to obtain full marks.

f.i.
[N/A]
f.ii.

Significant work on this question part was rarely seen, and it may have been the case that many candidates chose to spend their remaining time on the second question, especially if they felt they were making little progress with part f. Having said that, correct final answers were seen from better candidates, though these were few and far between.

g.

Syllabus sections

Topic 5 —Calculus » SL 5.6—Differentiating polynomials n E Q. Chain, product and quotient rules
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