IB DP Mathematical Studies 7.3 Paper 2

Question

When Geraldine travels to work she can travel either by car (C), bus (B) or train (T). She travels by car on one day in five. She uses the bus 50 % of the time. The probabilities of her being late (L) when travelling by car, bus or train are 0.05, 0.12 and 0.08 respectively.

It is not necessary to use graph paper for this question.

Copy the tree diagram below and fill in all the probabilities, where NL represents not late, to represent this information.

[5]

i.a.

Find the probability that Geraldine travels by bus and is late.[1]

i.b.

Find the probability that Geraldine is late.[3]

i.c.

Find the probability that Geraldine travelled by train, given that she is late.[3]

i.d.

Sketch the curve of the function \(f (x) = x^3 − 2x^2 + x − 3\) for values of \(x\) from −2 to 4, giving the intercepts with both axes.[3]

ii.a.

On the same diagram, sketch the line \(y = 7 − 2x\) and find the coordinates of the point of intersection of the line with the curve.[3]

ii.b.

Find the value of the gradient of the curve where \(x = 1.7\) .[2]

ii.c.
Answer/Explanation

Markscheme

Award (A1) for 0.5 at B, (A1) for 0.3 at T, then (A1) for each correct pair. Accept fractions or percentages.     (A5)

[5 marks]

i.a.

0.06 (accept \(0.5 \times 0.12\) or 6%)     (A1)(ft)

[1 mark]

i.b.

for a relevant two-factor product, either \(C \times L\) or \(T \times L\)     (M1)

for summing three two-factor products     (M1)

\((0.2 \times 0.05 + 0.06 + 0.3 \times 0.08)\)

0.094     (A1)(ft)(G2)

[3 marks]

i.c.

\(\frac{{0.3 \times 0.08}}{{0.094}}\)     (M1)(A1)(ft)

award (M1) for substituted conditional probability formula seen, (A1)(ft) for correct substitution

= 0.255     (A1)(ft)(G2)

[3 marks]

i.d.

     (G3)

[3 marks]

ii.a.

line drawn with –ve gradient and +ve y-intercept     (G1)

(2.45, 2.11)     (G1)(G1)

[3 marks]

ii.b.

\(f ‘ (1.7) = 3(1.7)^2 – 4(1.7) + 1\)     (M1)

award (M1) for substituting in their \(f’ (x)\)

2.87     (A1)(G2)

[2 marks]

ii.c.

Question

The diagram below shows the graph of a line \(L\) passing through (1, 1) and (2 , 3) and the graph \(P\) of the function \(f (x) = x^2 − 3x − 4\)

Find the gradient of the line L.[2]

a.

Differentiate \(f (x)\) .[2]

b.

Find the coordinates of the point where the tangent to P is parallel to the line L.[3]

c.

Find the coordinates of the point where the tangent to P is perpendicular to the line L.[4]

d.

Find

(i) the gradient of the tangent to P at the point with coordinates (2, − 6).

(ii) the equation of the tangent to P at this point.[3]

e.

State the equation of the axis of symmetry of P.[1]

f.

Find the coordinates of the vertex of P and state the gradient of the curve at this point.[3]

g.
Answer/Explanation

Markscheme

for attempt at substituted \(\frac{{ydistance}}{{xdistance}}\)     (M1)

gradient = 2     (A1)(G2)

[2 marks]

a.

\(2x – 3\)     (A1)(A1)

(A1) for \(2x\) , (A1) for \(-3\)

[2 marks]

b.

for their \(2x – 3 =\) their gradient and attempt to solve     (M1)

\(x = 2.5\)     (A1)(ft)

\(y = -5.25\) ((ft) from their x value)     (A1)(ft)(G2)

[3 marks]

c.

for seeing \(\frac{{ – 1}}{{their(a)}}\)     (M1)

solving \(2x – 3 = – \frac{1}{2}\) (or their value)     (M1)

x = 1.25     (A1)(ft)(G1)

y = – 6.1875     (A1)(ft)(G1)

[4 marks]

d.

(i) \(2 \times 2 – 3 = 1\) ((ft) from (b))     (A1)(ft)(G1)

(ii) \(y = mx + c\) or equivalent method to find \(c \Rightarrow -6 = 2 + c\)     (M1)

\(y = x – 8\)     (A1)(ft)(G2) 

[3 marks]

e.

\(x = 1.5\)     (A1)

[1 mark]

f.

for substituting their answer to part (f) into the equation of the parabola (1.5, −6.25) accept x = 1.5, y = −6.25     (M1)(A1)(ft)(G2)

gradient is zero (accept \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\))     (A1)

[3 marks]

g.

Question

Consider the curve \(y = {x^3} + \frac{3}{2}{x^2} – 6x – 2\) .

(i)     Write down the value of \(y\) when \(x\) is \(2\).

(ii)    Write down the coordinates of the point where the curve intercepts the \(y\)-axis.[3]

a.

Sketch the curve for \( – 4 \leqslant x \leqslant 3\) and \( – 10 \leqslant y \leqslant 10\). Indicate clearly the information found in (a).[4]

b.

Find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) .[3]

c.

Let \({L_1}\) be the tangent to the curve at \(x = 2\).

Let \({L_2}\) be a tangent to the curve, parallel to \({L_1}\).

(i)     Show that the gradient of \({L_1}\) is \(12\).

(ii)    Find the \(x\)-coordinate of the point at which \({L_2}\) and the curve meet.

(iii)   Sketch and label \({L_1}\) and \({L_2}\) on the diagram drawn in (b).[8]

d.

It is known that \(\frac{{{\text{d}}y}}{{{\text{d}}x}} > 0\) for \(x < – 2\) and \(x > b\) where \(b\) is positive.

(i)     Using your graphic display calculator, or otherwise, find the value of \(b\).

(ii)    Describe the behaviour of the curve in the interval \( – 2 < x < b\) .

(iii)   Write down the equation of the tangent to the curve at \(x = – 2\).[5]

e.
Answer/Explanation

Markscheme

(i)     \(y = 0\)     (A1)

(ii)    \((0{\text{, }}{- 2})\)     (A1)(A1)

Note: Award (A1)(A0) if brackets missing.

OR

\(x = 0{\text{, }}y = – 2\)     (A1)(A1)

Note: If coordinates reversed award (A0)(A1)(ft). Two coordinates must be given.

[3 marks]

a.

     (A4)

Notes: (A1) for appropriate window. Some indication of scale on the \(x\)-axis must be present (for example ticks). Labels not required. (A1) for smooth curve and shape, (A1) for maximum and minimum in approximately correct position, (A1) for \(x\) and \(y\) intercepts found in (a) in approximately correct position.

[4 marks]

b.

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 3{x^2} + 3x – 6\)     (A1)(A1)(A1)

 Note: (A1) for each correct term. Award (A1)(A1)(A0) at most if any other term is present.

[3 marks]

c.

(i)     \(3 \times 4 + 3 \times 2 – 6 = 12\)     (M1)(A1)(AG)

Note: (M1) for using the derivative and substituting \(x = 2\) . (A1) for correct (and clear) substitution. The \(12\) must be seen.

(ii)    Gradient of \({L_2}\) is \(12\) (can be implied)     (A1)

\(3{x^2} + 3x – 6 = 12\)     (M1)

\(x = – 3\)     (A1)(G2)

Note: (M1) for equating the derivative to \(12\) or showing a sketch of the derivative together with a line at \(y = 12\) or a table of values showing the \(12\) in the derivative column.

(iii)   (A1) for \({L_1}\) correctly drawn at approx the correct point     (A1)

(A1) for \({L_2}\) correctly drawn at approx the correct point     (A1)

(A1) for 2 parallel lines     (A1)

Note: If lines are not labelled award at most (A1)(A1)(A0). Do not accept 2 horizontal or 2 vertical parallel lines.

[8 marks]

d.

(i)     \(b = 1\)     (G2)

(ii)    The curve is decreasing.     (A1)

Note: Accept any valid description.

(iii)   \(y = 8\)     (A1)(A1)(G2)

Note: (A1) for “\(y =\) a constant”, (A1) for \(8\).

[5 marks]

e.

Question

Consider the function \(f(x) = 3x + \frac{{12}}{{{x^2}}},{\text{ }}x \ne 0\).

Differentiate \(f (x)\) with respect to \(x\).[3]

a.

Calculate \(f ′(x)\) when \(x = 1\).[2]

b.

Use your answer to part (b) to decide whether the function, \(f\) , is increasing or decreasing at \(x = 1\). Justify your answer.[2]

c.

Solve the equation \(f ′(x) = 0\).[3]

d.

The graph of f has a local minimum at point P. Let T be the tangent to the graph of f at P.

Write down the coordinates of P.[2]

e, i.

The graph of f has a local minimum at point P. Let T be the tangent to the graph of f at P.

Write down the gradient of T.[1]

e, ii.

The graph of f has a local minimum at point P. Let T be the tangent to the graph of f at P.

Write down the equation of T.[2]

e, iii.

Sketch the graph of the function f, for −3 ≤ x ≤ 6 and −7 ≤ y ≤ 15. Indicate clearly the point P and any intercepts of the curve with the axes.[4]

f.

On your graph draw and label the tangent T.[2]

g, i.

T intersects the graph of f at a second point. Write down the x-coordinate of this point of intersection.[1]

g, ii.
Answer/Explanation

Markscheme

\(f’ (x) = 3 – \frac{24}{x^3}\)     (A1)(A1)(A1) 

Note: Award (A1) for 3, (A1) for –24, (A1) for x3 (or x−3). If extra terms present award at most (A1)(A1)(A0).

[3 marks]

a.

\(f ‘(1) = -21\)     (M1)(A1)(ft)(G2) 

Note: (ft) from their derivative only if working seen.

[2 marks]

b.

Derivative (gradient, slope) is negative. Decreasing.     (R1)(A1)(ft) 

Note: Do not award (R0)(A1).

[2 marks]

c.

\(3 – \frac{{24}}{{{x^3}}} = 0\)     (M1)

\(x^3 = 8\)     (A1)

\(x = 2\)     (A1)(ft)(G2)

[3 marks]

d.

(2, 9) (Accept x = 2, y = 9)     (A1)(A1)(G2)

Notes: (ft) from their answer in (d).

Award (A1)(A0) if brackets not included and not previously penalized.

[2 marks]

e, i.

0     (A1)

[1 mark]

e, ii.

y = 9     (A1)(A1)(ft)(G2)

Notes: Award (A1) for y = constant, (A1) for 9.

Award (A1)(ft) for their value of y in (e)(i).

[2 marks]

e, iii.

     (A4)

 

Notes: Award (A1) for labels and some indication of scale in the stated window.

Award (A1) for correct general shape (curve must be smooth and must not cross the y-axis).

Award (A1) for x-intercept seen in roughly the correct position.

Award (A1) for minimum (P).

[4 marks]

f.

Tangent drawn at P (line must be a tangent and horizontal).     (A1)

Tangent labeled T.     (A1)

Note: (ft) from their tangent equation only if tangent is drawn and answer is consistent with graph.

[2 marks]

g, i.

x = −1     (G1)(ft)

[1 mark]

g, ii.

Question

A function is defined by \(f(x) = \frac{5}{{{x^2}}} + 3x + c,{\text{ }}x \ne 0,{\text{ }}c \in \mathbb{Z}\).

Write down an expression for \(f ′(x)\).[4]

a.

Consider the graph of f. The graph of f passes through the point P(1, 4).

Find the value of c.[2]

b.

There is a local minimum at the point Q.

Find the coordinates of Q.[4]

c, i.

There is a local minimum at the point Q.

Find the set of values of x for which the function is decreasing.[3]

c, ii.

Let T be the tangent to the graph of f at P.

Show that the gradient of T is –7.[2]

d, i.

Let T be the tangent to the graph of f at P.

Find the equation of T.[2]

d, ii.

T intersects the graph again at R. Use your graphic display calculator to find the coordinates of R.[2]

e.
Answer/Explanation

Markscheme

\(f'(x) = \frac{{ – 10}}{{{x^3}}} + 3\)     (A1)(A1)(A1)(A1)

Note: Award (A1) for −10, (A1) for x3 (or x3), (A1) for 3, (A1) for no other constant term.

[4 marks]

a.

4 = 5 + 3 + c     (M1)

Note: Award (M1) for substitution in f (x).

c = −4     (A1)(G2)

[2 marks]

b.

\(f ‘(x) = 0\)     (M1)

\(0 = \frac{{ – 10}}{{{x^3}}} + 3\)     (A1)(ft)

(1.49, 2.72)   (accept x = 1.49  y = 2.72)     (A1)(ft)(A1)(ft)(G3)

Notes: If answer is given as (1.5, 2.7) award (A0)(AP)(A1).

Award at most (M1)(A1)(A1)(A0) if parentheses not included. (ft) from their (a).

If no working shown award (G2)(G0) if parentheses are not included.

OR

Award (M2) for sketch, (A1)(ft)(A1)(ft) for correct coordinates. (ft) from their (b).     (M2)(A1)(ft)(A1)(ft)

Note: Award at most (M2)(A1)(ft)(A0) if parentheses not included.

[4 marks]

c, i.

0 < x < 1.49  OR  0 < x ≤ 1.49     (A1)(A1)(ft)(A1)

Notes: Award (A1) for 0, (A1)(ft) for 1.49 and (A1) for correct inequality signs.

(ft) from their x value in (c) (i).

[3 marks]

c, ii.

For P(1, 4) \(f ‘(1) = – 10 + 3\)     (M1)(A1)

\(= -7\)     (AG)

Note: Award (M1) for substituting \(x = 1\) into their \(f ‘(x)\). (A1) for \(-10 + 3\).

\(-7\) must be seen for (A1) to be awarded.

[2 marks]

d, i.

\(4 = -7 \times 1 + c\)     \(11 = c\)     (A1)

\(y = -7 x + 11\)     (A1)

[2 marks]

d, ii.

Point of intersection is R(−0.5, 14.5)     (A1)(ft)(A1)(ft)(G2)(ft)

Notes: Award (A1) for the x coordinate, (A1) for the y coordinate.

Allow (ft) from candidate’s (d)(ii) equation and their (b) even with no working seen.

Award (A1)(ft)(A0) if brackets not included and not previously penalised.

[2 marks]

e.

Question

Sketch the graph of y = 2x for \( – 2 \leqslant x \leqslant 3\). Indicate clearly where the curve intersects the y-axis.[3]

A, a.

Write down the equation of the asymptote of the graph of y = 2x.[2]

A, b.

On the same axes sketch the graph of y = 3 + 2xx2. Indicate clearly where this curve intersects the x and y axes.[3]

A, c.

Using your graphic display calculator, solve the equation 3 + 2xx2 = 2x.[2]

A, d.

Write down the maximum value of the function f (x) = 3 + 2xx2.[1]

A, e.

Use Differential Calculus to verify that your answer to (e) is correct.[5]

A, f.

The curve y = px2 + qx − 4 passes through the point (2, –10).

Use the above information to write down an equation in p and q.[2]

B, a.

The gradient of the curve \(y = p{x^2} + qx – 4\) at the point (2, –10) is 1.

Find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\).[2]

B, b, i.

The gradient of the curve \(y = p{x^2} + qx – 4\) at the point (2, –10) is 1.

Hence, find a second equation in p and q.[1]

B, b, ii.

The gradient of the curve \(y = p{x^2} + qx – 4\) at the point (2, –10) is 1.

Solve the equations to find the value of p and of q.[3]

B, c.
Answer/Explanation

Markscheme

     (A1)(A1)(A1)

Note: Award (A1) for correct domain, (A1) for smooth curve, (A1) for y-intercept clearly indicated.

[3 marks]

A, a.

y = 0     (A1)(A1)

Note: Award (A1) for y = constant, (A1) for 0.

[2 marks]

A, b.

Note: Award (A1) for smooth parabola,

(A1) for vertex (maximum) in correct quadrant.

(A1) for all clearly indicated intercepts x = −1, x = 3 and y = 3.

The final mark is to be applied very strictly.     (A1)(A1)(A1)

[3 marks]

A, c.

x = −0.857   x = 1.77     (G1)(G1)

Note: Award a maximum of (G1) if x and y coordinates are both given.

[2 marks]

A, d.

4     (G1)

Note: Award (G0) for (1, 4).

[1 mark]

A, e.

\(f'(x) = 2 – 2x\)     (A1)(A1)

Note: Award (A1) for each correct term.

Award at most (A1)(A0) if any extra terms seen.

\(2 – 2x = 0\)     (M1)

Note: Award (M1) for equating their gradient function to zero.

\(x = 1\)     (A1)(ft)

\(f (1) = 3 + 2(1) – (1)^2 = 4\)     (A1)

Note: The final (A1) is for substitution of x = 1 into \(f (x)\) and subsequent correct answer. Working must be seen for final (A1) to be awarded.

[5 marks]

A, f.

22 × p + 2q − 4 = −10     (M1)

Note: Award (M1) for correct substitution in the equation.

4p + 2q = −6     or     2p + q = −3     (A1)

Note: Accept equivalent simplified forms.

[2 marks]

B, a.

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2px + q\)     (A1)(A1)

Note: Award (A1) for each correct term.

Award at most (A1)(A0) if any extra terms seen.

[2 marks]

B, b, i.

4p + q = 1     (A1)(ft)

[1 mark]

B, b, ii.

4p + 2q = −6

4p + q = 1     (M1)

Note: Award (M1) for sensible attempt to solve the equations.

p = 2, q = −7     (A1)(A1)(ft)(G3)

[3 marks]

B, c.

Question

The diagram shows a sketch of the function f (x) = 4x3 − 9x2 − 12x + 3.

Write down the values of x where the graph of f (x) intersects the x-axis.[3]

a.

Write down f ′(x).[3]

b.

Find the value of the local maximum of y = f (x).[4]

c.

Let P be the point where the graph of f (x) intersects the y axis.

Write down the coordinates of P.[1]

d.

Let P be the point where the graph of f (x) intersects the y axis.

Find the gradient of the curve at P.[2]

e.

The line, L, is the tangent to the graph of f (x) at P.

Find the equation of L in the form y = mx + c.[2]

f.

There is a second point, Q, on the curve at which the tangent to f (x) is parallel to L.

Write down the gradient of the tangent at Q.[1]

g.

There is a second point, Q, on the curve at which the tangent to f (x) is parallel to L.

Calculate the x-coordinate of Q.[3]

h.
Answer/Explanation

Markscheme

–1.10, 0.218, 3.13     (A1)(A1)(A1)

[3 marks]

a.

f ′(x) = 12x2 – 18x – 12     (A1)(A1)(A1)

Note: Award (A1) for each correct term and award maximum of (A1)(A1) if other terms seen.

[3 marks]

b.

′(x) = 0     (M1)
x = –0.5, 2
x = –0.5     (A1)

Note: If x = –0.5 not stated, can be inferred from working below.

y = 4(–0.5)3 – 9(–0.5)2 – 12(–0.5) + 3     (M1)
y = 6.25     (A1)(G3)

Note: Award (M1) for their value of x substituted into f (x).

Award (M1)(G2) if sketch shown as method. If coordinate pair given then award (M1)(A1)(M1)(A0). If coordinate pair given with no working award (G2).

[4 marks]

c.

(0, 3)     (A1)

Note: Accept x = 0, y = 3.

[1 mark]

d.

′(0) = –12     (M1)(A1)(ft)(G2)

Note: Award (M1) for substituting x = 0 into their derivative.

[2 marks]

e.

Tangent: y = –12x + 3     (A1)(ft)(A1)(G2)

Note: Award (A1)(ft) for their gradient, (A1) for intercept = 3.

Award (A1)(A0) if y = not seen.

[2 marks]

f.

–12     (A1)(ft)

Note: Follow through from their part (e).

[1 mark]

g.

12x2 – 18x – 12 = –12     (M1)

12x2 – 18x = 0     (M1)

x = 1.5, 0

At Q x = 1.5     (A1)(ft)(G2)

Note: Award (M1)(G2) for 12x2 – 18x – 12 = –12 followed by x = 1.5.

Follow through from their part (g).

[3 marks]

h.

Question

Consider the function f (x) = x3 3x– 24x + 30.

Write down f (0).[1]

a.

Find \(f'(x)\).[3]

b.

Find the gradient of the graph of f (x) at the point where x = 1.[2]

c.

(i) Use f ‘(x) to find the x-coordinate of M and of N.

(ii) Hence or otherwise write down the coordinates of M and of N.[5]

d.

Sketch the graph of f (x) for \( – 5 \leqslant x \leqslant 7\) and \( – 60 \leqslant y \leqslant 60\). Mark clearly M and N on your graph.[4]

e.

Lines L1 and L2 are parallel, and they are tangents to the graph of f (x) at points A and B respectively. L1 has equation y = 21x + 111.

(i) Find the x-coordinate of A and of B.

(ii) Find the y-coordinate of B.[6]

f.
Answer/Explanation

Markscheme

30     (A1)

[1 mark]

a.

f (x) = 3x2 – 6x – 24     (A1)(A1)(A1)

Note: Award (A1) for each term. Award at most (A1)(A1) if extra terms present.

[3 marks]

b.

f ‘(1) = –27     (M1)(A1)(ft)(G2)

Note: Award (M1) for substituting x = 1 into their derivative.

[2 marks]

c.

(i) f ‘(x) = 0

3x2 – 6x – 24 = 0     (M1)

x = 4; x = –2     (A1)(ft)(A1)(ft)

Notes: Award (M1) for either f ‘(x) = 0 or 3x2 – 6x – 24 = 0 seen. Follow through from their derivative. Do not award the two answer marks if derivative not used.

(ii) M(–2, 58) accept x = –2, y = 58     (A1)(ft)

N(4, – 50) accept x = 4, y = –50     (A1)(ft)

Note: Follow through from their answer to part (d) (i).

[5 marks]

d.

(A1) for window

(A1) for a smooth curve with the correct shape

(A1) for axes intercepts in approximately the correct positions

(A1) for M and N marked on diagram and in approximately
correct position     (A4)

Note: If window is not indicated award at most (A0)(A1)(A0)(A1)(ft).

[4 marks]

e.

(i) 3x2 – 6x – 24 = 21     (M1)

3x2 – 6x – 45 = 0     (M1)

x = 5; x = –3     (A1)(ft)(A1)(ft)(G3)

Note: Follow through from their derivative.

OR

Award (A1) for L1 drawn tangent to the graph of f on their sketch in approximately the correct position (x = –3), (A1) for a second tangent parallel to their L1, (A1) for x = –3, (A1) for x = 5 .     (A1)(ft)(A1)(ft)(A1)(A1)

Note: If only x = –3 is shown without working award (G2). If both answers are shown irrespective of workingaward (G3).

(ii) f (5) = –40     (M1)(A1)(ft)(G2)

Notes: Award (M1) for attempting to find the image of their x = 5. Award (A1) only for (5, –40). Follow through from their x-coordinate of B only if it has been clearly identified in (f) (i).

[6 marks]

f.

Question

Consider the function \(f(x) = {x^3} + \frac{{48}}{x}{\text{, }}x \ne 0\).

Calculate \(f(2)\) .[2]

a.

Sketch the graph of the function \(y = f(x)\) for \( – 5 \leqslant x \leqslant 5\) and \( – 200 \leqslant y \leqslant 200\) .[4]

b.

Find \(f'(x)\) .[3]

c.

Find \(f'(2)\) .[2]

d.

Write down the coordinates of the local maximum point on the graph of \(f\) .[2]

e.
Find the range of \(f\) .[3]
f.

Find the gradient of the tangent to the graph of \(f\) at \(x = 1\).[2]

g.

There is a second point on the graph of \(f\) at which the tangent is parallel to the tangent at \(x = 1\).

Find the \(x\)-coordinate of this point.[2]

h.
Answer/Explanation

Markscheme

\(f(2) = {2^3} + \frac{{48}}{2}\)     (M1)

\(= 32\)     (A1)(G2)

[2 marks]

a.

(A1) for labels and some indication of scale in an appropriate window
(A1) for correct shape of the two unconnected and smooth branches
(A1) for maximum and minimum in approximately correct positions
(A1) for asymptotic behaviour at \(y\)-axis     (A4)

Notes: Please be rigorous.
The axes need not be drawn with a ruler.
The branches must be smooth: a single continuous line that does not deviate from its proper direction.
The position of the maximum and minimum points must be symmetrical about the origin.
The \(y\)-axis must be an asymptote for both branches. Neither branch should touch the axis nor must the curve approach the
asymptote then deviate away later.

[4 marks]

b.

\(f'(x) = 3{x^2} – \frac{{48}}{{{x^2}}}\)     (A1)(A1)(A1)

Notes: Award (A1) for \(3{x^2}\) , (A1) for \( – 48\) , (A1) for \({x^{ – 2}}\) . Award a maximum of (A1)(A1)(A0) if extra terms seen.

[3 marks]

c.

\(f'(2) = 3{(2)^2} – \frac{{48}}{{{{(2)}^2}}}\)     (M1)

Note: Award (M1) for substitution of \(x = 2\) into their derivative.

\(= 0\)     (A1)(ft)(G1)

[2 marks]

d.

\(( – 2{\text{, }} – 32)\) or \(x = – 2\), \(y = – 32\)     (G1)(G1)

Notes: Award (G0)(G0) for \(x = – 32\), \(y = – 2\) . Award at most (G0)(G1) if parentheses are omitted.

[2 marks]

e.

\(\{ y \geqslant 32\}  \cup \{ y \leqslant – 32\} \)     (A1)(A1)(ft)(A1)(ft)

Notes: Award (A1)(ft) \(y \geqslant 32\) or \(y > 32\) seen, (A1)(ft) for \(y \leqslant – 32\) or \(y < – 32\) , (A1) for weak (non-strict) inequalities used in both of the above.
Accept use of \(f\) in place of \(y\). Accept alternative interval notation.
Follow through from their (a) and (e).
If domain is given award (A0)(A0)(A0).
Award (A0)(A1)(ft)(A1)(ft) for \([ – 200{\text{, }} – 32]\) , \([32{\text{, }}200]\).
Award (A0)(A1)(ft)(A1)(ft) for \(] – 200{\text{, }} – 32]\) , \([32{\text{, }}200[\).

[3 marks]

f.

\(f'(1) = – 45\)     (M1)(A1)(ft)(G2)

Notes: Award (M1) for \(f'(1)\) seen or substitution of \(x = 1\) into their derivative. Follow through from their derivative if working is seen.

[2 marks]

g.

\(x = – 1\)     (M1)(A1)(ft)(G2)

Notes: Award (M1) for equating their derivative to their \( – 45\) or for seeing parallel lines on their graph in the approximately correct position.

[2 marks]

h.

Question

The function \(f(x)\) is defined by \(f(x) = 1.5x + 4 + \frac{6}{x}{\text{, }}x \ne 0\) .

Write down the equation of the vertical asymptote.[2]

a.

Find \(f'(x)\) .[3]

b.

Find the gradient of the graph of the function at \(x = – 1\).[2]

c.

Using your answer to part (c), decide whether the function \(f(x)\) is increasing or decreasing at \(x = – 1\). Justify your answer.[2]

d.

Sketch the graph of \(f(x)\) for \( – 10 \leqslant x \leqslant 10\) and \( – 20 \leqslant y \leqslant 20\) .[4]

e.

\({{\text{P}}_1}\) is the local maximum point and \({{\text{P}}_2}\) is the local minimum point on the graph of \(f(x)\) .

Using your graphic display calculator, write down the coordinates of

(i)     \({{\text{P}}_1}\) ;

(ii)    \({{\text{P}}_2}\) .[4]

f.

Using your sketch from (e), determine the range of the function \(f(x)\) for \( – 10 \leqslant x \leqslant 10\) .[3]

g.
Answer/Explanation

Markscheme

\(x = 0\)     (A1)(A1)

Note: Award (A1) for \(x = {\text{constant}}\), (A1) for \(0\).

[2 marks]

a.

\(f'(x) = 1.5 – \frac{6}{{{x^2}}}\)     (A1)(A1)(A1)

Notes: Award (A1) for \(1.5\), (A1) for \( – 6\), (A1) for \({x^{ – 2}}\) . Award (A1)(A1)(A0) at most if any other term present.

[3 marks]

b.

\(1.5 – \frac{6}{{( – 1)}}\)     (M1)

\( =  – 4.5\)     (A1)(ft)(G2)

Note: Follow through from their derivative function.

[2 marks]

c.

Decreasing, the derivative (gradient or slope) is negative (at \(x = – 1\))     (A1)(R1)(ft)

Notes: Do not award (A1)(R0). Follow through from their answer to part (c).

[2 marks]

d.

     (A4)
Notes: Award (A1) for labels and some indication of scales and an appropriate window.

Award (A1) for correct shape of the two unconnected, and smooth branches.

Award (A1) for the maximum and minimum points in the approximately correct positions.

Award (A1) for correct asymptotic behaviour at \(x = 0\) .

Notes: Please be rigorous.

The axes need not be drawn with a ruler.

The branches must be smooth and single continuous lines that do not deviate from their proper direction.

The max and min points must be symmetrical about point \((0{\text{, }}4)\) .

The \(y\)-axis must be an asymptote for both branches.

[4 marks]

e.

(i)     \(( – 2{\text{, }} – 2)\) or \(x = – 2\), \(y = – 2\)     (G1)(G1)

(ii)    \((2{\text{, }}10)\) or \(x = 2\), \(y = 10\)     (G1)(G1)

[4 marks]

f.

\(\{  – 2 \geqslant y\} \) or \(\{ y \geqslant 10\} \)     (A1)(A1)(ft)(A1)

Notes: (A1)(ft) for \(y > 10\) or \(y \geqslant 10\) . (A1)(ft) for \(y < – 2\) or \(y \leqslant – 2\) . (A1) for weak (non-strict) inequalities used in both of the above. Follow through from their (e) and (f).

[3 marks]

g.

Question

The diagram shows part of the graph of \(f(x) = {x^2} – 2x + \frac{9}{x}\) , where \(x \ne 0\) .

Write down

(i)     the equation of the vertical asymptote to the graph of \(y = f (x)\) ;

(ii)    the solution to the equation \(f (x) = 0\) ;

(iii)   the coordinates of the local minimum point.[5]

a.

Find  \(f'(x)\) .[4]

b.

Show that \(f'(x)\) can be written as \(f'(x) = \frac{{2{x^3} – 2{x^2} – 9}}{{{x^2}}}\) .[2]

c.

Find the gradient of the tangent to \(y = f (x)\) at the point \({\text{A}}(1{\text{, }}8)\) .[2]

d.

The line, \(L\), passes through the point A and is perpendicular to the tangent at A.

Write down the gradient of \(L\) .[1]

e.

The line, \(L\) , passes through the point A and is perpendicular to the tangent at A.

Find the equation of \(L\) . Give your answer in the form \(y = mx + c\) .[3]

f.

The line, \(L\) , passes through the point A and is perpendicular to the tangent at A.

\(L\) also intersects the graph of \(y = f (x)\) at points B and C . Write down the x-coordinate of B and of C .[2]

g.
Answer/Explanation

Markscheme

(i)     \(x = 0\)     (A1)(A1)

Note: Award (A1) for \(x = \) a constant, (A1) for the constant in their equation being \(0\).

(ii)    \( – 1.58\) (\( – 1.58454 \ldots \))     (G1)

Note: Accept \( – 1.6\), do not accept \( – 2\) or \( – 1.59\).

(iii)   \((2.06{\text{, }}4.49)\) \((2.06020 \ldots {\text{, }}4.49253 \ldots )\)     (G1)(G1)

Note: Award at most (G1)(G0) if brackets not used. Award (G0)(G1)(ft) if coordinates are reversed.

Note: Accept \(x = 2.06\), \(y = 4.49\) .

Note: Accept \(2.1\), do not accept \(2.0\) or \(2\). Accept \(4.5\), do not accept \(5\) or \(4.50\).

[5 marks]

a.

\(f'(x) = 2x – 2 – \frac{9}{{{x^2}}}\)     (A1)(A1)(A1)(A1)

Notes: Award (A1) for \(2x\), (A1) for \( – 2\), (A1) for \( – 9\), (A1) for \({x^{ – 2}}\) . Award a maximum of (A1)(A1)(A1)(A0) if there are extra terms present.

[4 marks]

b.

\(f'(x) = \frac{{{x^2}(2x – 2)}}{{{x^2}}} – \frac{9}{{{x^2}}}\)     (M1)

Note: Award (M1) for taking the correct common denominator.

\( = \frac{{(2{x^3} – 2{x^2})}}{{{x^2}}} – \frac{9}{{{x^2}}}\)     (M1)

Note: Award (M1) for multiplying brackets or equivalent.

\( = \frac{{2{x^3} – 2{x^2} – 9}}{{{x^2}}}\)     (AG)

Note: The final (M1) is not awarded if the given answer is not seen.

[2 marks]

c.

\(f'(1) = \frac{{2{{(1)}^3} – 2(1) – 9}}{{{{(1)}^2}}}\)     (M1)

\( = – 9\)     (A1)(G2)

Note: Award (M1) for substitution into given (or their correct) \(f'(x)\) . There is no follow through for use of their incorrect derivative.

[2 marks]

d.

\(\frac{1}{9}\)     (A1)(ft)

Note: Follow through from part (d).

[1 mark]

e.

\(y – 8 = \frac{1}{9}(x – 1)\)     (M1)(M1)

Notes: Award (M1) for substitution of their gradient from (e), (M1) for substitution of given point. Accept all forms of straight line.

\(y = \frac{1}{9}x + \frac{{71}}{9}\) (\(y = 0.111111 \ldots x + 7.88888 \ldots \))     (A1)(ft)(G3)

Note: Award the final (A1)(ft) for a correctly rearranged formula of their straight line in (f). Accept \(0.11x\), do not accept \(0.1x\). Accept \(7.9\), do not accept \(7.88\), do not accept \(7.8\).

[3 marks]

f.

\( – 2.50\), \(3.61\) (\( – 2.49545 \ldots \), \(3.60656 \ldots \))     (A1)(ft)(A1)(ft)

Notes: Follow through from their line \(L\) from part (f) even if no working shown. Award at most (A0)(A1)(ft) if their correct coordinate pairs given.

Note: Accept \( – 2.5\), do not accept \( – 2.49\). Accept \(3.6\), do not accept \(3.60\).

[2 marks]

g.

Question

Consider the function \(f(x) =  – \frac{1}{3}{x^3} + \frac{5}{3}{x^2} – x – 3\).

Sketch the graph of y = f (x) for −3 ≤ x ≤ 6 and −10 ≤ y ≤ 10 showing clearly the axes intercepts and local maximum and minimum points. Use a scale of 2 cm to represent 1 unit on the x-axis, and a scale of 1 cm to represent 1 unit on the y-axis.[4]

a.

Find the value of f (−1).[2]

b.

Write down the coordinates of the y-intercept of the graph of f (x).[1]

c.

Find f ‘(x).[3]

d.

Show that \(f'( – 1) =  – \frac{{16}}{3}\).[1]

e.

Explain what f (−1) represents.[2]

f.

Find the equation of the tangent to the graph of f (x) at the point where x is –1.[2]

g.

Sketch the tangent to the graph of f (x) at x = −1 on your diagram for (a).[2]

h.

P and Q are points on the curve such that the tangents to the curve at these points are horizontal. The x-coordinate of P is a, and the x-coordinate of Q is b, b > a.

Write down the value of

(i) a ;

(ii) b .[2]

i.

P and Q are points on the curve such that the tangents to the curve at these points are horizontal. The x-coordinate of P is a, and the x-coordinate of Q is b, b > a.

Describe the behaviour of f (x) for a < x < b.[1]

j.
Answer/Explanation

Markscheme

(A1) for indication of window and labels. (A1) for smooth curve that does not enter the first quadrant, the curve must consist of one line only.

(A1) for x and y intercepts in approximately correct positions (allow ±0.5).

(A1) for local maximum and minimum in approximately correct position. (minimum should be 0 ≤ x ≤ 1 and –2 ≤ y ≤ –4 ), the y-coordinate of the maximum should be 0 ± 0.5.     (A4)

[4 marks]

a.

\(-\frac{1}{3}(-1)^3 + \frac{5}{3}(-1)^2 – (-1) – 3 \)     (M1)

Note: Award (M1) for substitution of –1 into f (x)

= 0     (A1)(G2)

[2 marks]

b.

(0, –3)     (A1)

OR

x = 0, y = –3     (A1)

Note: Award (A0) if brackets are omitted.

[1 mark]

c.

\(f'(x) =  – {x^2} + \frac{{10}}{3}x – 1\)     (A1)(A1)(A1)

Note: Award (A1) for each correct term. Award (A1)(A1)(A0) at most if there are extra terms.

[3 marks]

d.

\(f'( – 1) =  – {( – 1)^2} + \frac{{10}}{3}( – 1) – 1\)     (M1)

\(= -\frac{16}{3}\)     (AG)

Note: Award (M1) for substitution of x = –1 into correct derivative only. The final answer must be seen.

[1 mark]

e.

f (–1) gives the gradient of the tangent to the curve at the point with x = –1.     (A1)(A1)

Note: Award (A1) for “gradient (of curve)”, (A1) for “at the point with x = –1”. Accept “the instantaneous rate of change of y” or “the (first) derivative”.

[2 marks]

f.

\(y = – \frac{16}{3} x + c\)     (M1)

Note: Award (M1) for \(-\frac{16}{3}\) substituted in equation.

\(0 = – \frac{16}{3} \times (-1) + c \)

\(c = – \frac{16}{3}\)

\(y =  – \frac{{16}}{3}x – \frac{{16}}{3}\)     (A1)(G2)

Note: Accept y = –5.33x – 5.33.

OR

\((y – 0) = \frac{{-16}}{3}(x + 1)\)     (M1)(A1)(G2)

Note: Award (M1) for \( – \frac{{16}}{3}\) substituted in equation, (A1) for correct equation. Follow through from their answer to part (b). Accept y = –5.33 (x +1). Accept equivalent equations.

[2 marks]

g.

(A1)(ft) for a tangent to their curve drawn.

(A1)(ft) for their tangent drawn at the point x = –1.     (A1)(ft)(A1)(ft)

Note: Follow through from their graph. The tangent must be a straight line otherwise award at most (A0)(A1).

[2 marks]

h.

(i) \(a = \frac{1}{3}\)     (G1)

(ii) \(b = 3\)     (G1)

Note: If a and b are reversed award (A0)(A1).

[2 marks]

i.

f (x) is increasing     (A1)

[1 mark]

j.

Question

Consider the function \(g(x) = bx – 3 + \frac{1}{{{x^2}}},{\text{ }}x \ne 0\).

Write down the equation of the vertical asymptote of the graph of y = g(x) .[2]

a.

Write down g′(x) .[3]

b.

The line T is the tangent to the graph of y = g(x) at the point where x = 1. The gradient of T is 3.

Show that b = 5.[2]

c.

The line T is the tangent to the graph of y = g(x) at the point where x = 1. The gradient of T is 3.

Find the equation of T.[3]

d.

Using your graphic display calculator find the coordinates of the point where the graph of y = g(x) intersects the x-axis.[2]

e.

(i) Sketch the graph of y = g(x) for −2 ≤ x ≤ 5 and −15 ≤ y ≤ 25, indicating clearly your answer to part (e).

(ii) Draw the line T on your sketch.[6]

f.

Using your graphic display calculator find the coordinates of the local minimum point of y = g(x) .[2]

g.

Write down the interval for which g(x) is increasing in the domain 0 < x < 5 .[2]

h.
Answer/Explanation

Markscheme

x = 0     (A1)(A1)

Notes: Award (A1) for x=constant, (A1) for 0. Award (A0)(A0) if answer is not an equation.

[2 marks]

a.

\(b – \frac{2}{{{x^3}}}\)     (A1)(A1)(A1)

Note: Award (A1) for b, (A1) for −2, (A1) for \(\frac{1}{{{x^3}}}\) (or x−3). Award at most (A1)(A1)(A0) if extra terms seen.

[3 marks]

b.

\(3 = b – \frac{2}{{{{(1)}^3}}}\)     (M1)(M1)

Note: Award (M1) for substituting 1 into their gradient function, (M1) for equating their gradient function to 3.

b = 5     (AG)

Note: Award at most (M1)(A0) if final line is not seen or b does not equal 5.

[2 marks]

c.

g(1) = 3 or (1, 3) (seen or implied from the line below)     (A1)

3 = 3 × 1 + c     (M1)

Note: Award (M1) for correct substitution of their point (1, 3) and gradient 3 into equation y = mx + c. Follow through from their point of tangency.

y = 3x     (A1)(ft)(G2)

OR

y − 3 = 3(x − 1)     (M1)(A1)(ft)(G2) 

Note: Award (M1) for substitution of gradient 3 and their point (1, 3) into y y1 = m(x x1), (A1)(ft) for correct substitutions. Follow through from their point of tangency. Award at most (A1)(M1)(A0)(ft) if further incorrect working seen.

[3 marks]

d.

(−0.439, 0)   ((−0.438785…, 0))     (G1)(G1)

Notes: If no parentheses award at most (G1)(G0). Accept x = 0.439, y = 0.

[2 marks]

e.

(i)

Award (A1) for labels and some indication of scale in the stated window.

Award (A1) for correct general shape (curve must be smooth and must not cross the y-axis)

Award (A1)(ft) for x-intercept consistent with their part (e).

Award (A1) for local minimum in the first quadrant.     (A1)(A1)(A1)(ft)(A1)

(ii) Tangent to curve drawn at approximately x = 1     (A1)(A1) 

Note: Award (A1) for a line tangent to curve approximately at x = 1. Must be a straight line for the mark to be awarded. Award (A1)(ft) for line passing through the origin. Follow through from their answer to part (d).

[6 marks]

f.

(0.737, 2.53) ((0.736806…, 2.52604…))     (G1)(G1)

Notes: Do not penalize for lack of parentheses if already penalized in (e). Accept x = 0.737, y = 2.53.

[2 marks]

g.

0.737 < x < 5  OR  (0.737;5)     (A1)(A1)(ft)

Notes: Award (A1) for correct strict or weak inequalities with x seen if the interval is given as inequalities, (A1)(ft) for 0.737 and 5 or their value from part (g).

[2 marks]

h.

Question

The graph of the function \(f(x) = \frac{{14}}{x} + x – 6\), for 1 ≤ x ≤ 7 is given below.

Calculate \(f (1)\).[2]

a.

Find \(f ′(x)\).[3]

b.

Use your answer to part (b) to show that the x-coordinate of the local minimum point of the graph of \(f\) is 3.7 correct to 2 significant figures.[3]

c.

Find the range of \(f\).[3]

d.

Points A and B lie on the graph of \(f\). The x-coordinates of A and B are 1 and 7 respectively.

Write down the y-coordinate of B.[1]

e.

Points A and B lie on the graph of f . The x-coordinates of A and B are 1 and 7 respectively.

Find the gradient of the straight line passing through A and B.[2]

f.

M is the midpoint of the line segment AB.

Write down the coordinates of M.[2]

g.

L is the tangent to the graph of the function \(y = f (x)\), at the point on the graph with the same x-coordinate as M.

Find the gradient of L.[2]

h.

Find the equation of L. Give your answer in the form \(y = mx + c\).[3]

i.
Answer/Explanation

Markscheme

\(\frac{{14}}{{(1)}} + (1) – 6\)     (M1)

Note: Award (M1) for substituting \(x = 1\) into \(f\).

\(= 9\)     (A1)(G2)

a.

\( – \frac{{14}}{{{x^2}}} + 1\)     (A3)

Note: Award (A1) for \(-14\), (A1) for \(\frac{{14}}{{{x^2}}}\) or for \(x^{-2}\), (A1) for \(1\).

   Award at most (A2) if any extra terms are present.

b.

\( – \frac{{14}}{{{x^2}}} + 1 = 0\) or \(f ‘ (x) = 0 \)     (M1)

Note: Award (M1) for equating their derivative in part (b) to 0.

\(\frac{{14}}{{{x^2}}} = 1\) or \({x^2} = 14\) or equivalent     (M1)

Note: Award (M1) for correct rearrangement of their equation.

\(x = 3.74165…(\sqrt {14})\)     (A1)

\(x = 3.7\)     (AG)

Notes: Both the unrounded and rounded answers must be seen to award the (A1). This is a “show that” question; appeals to their GDC are not accepted –award a maximum of (M1)(M0)(A0).

Specifically, \( – \frac{{14}}{{{x^2}}} + 1 = 0\) followed by \(x = 3.74165…, x = 3.7\) is awarded (M1)(M0)(A0).

c.

\(1.48 \leqslant y \leqslant 9\)     (A1)(A1)(ft)(A1)
Note: Accept alternative notations, for example [1.48,9]. (\(x = \sqrt{14}\) leads to answer 1.48331…)

Note: Award (A1) for 1.48331…seen, accept 1.48378… from using the given answer \(x = 3.7\), (A1)(ft) for their 9 from part (a) seen, (A1) for the correct notation for their interval (accept \( \leqslant y \leqslant \) or \( \leqslant f \leqslant \) ).

d.

3     (A1)

Note: Do not accept a coordinate pair.

e.

\(\frac{{3 – 9}}{{7 – 1}}\)     (M1)

Note: Award (M1) for their correct substitution into the gradient formula.

\(= -1\)     (A1)(ft)(G2)

Note: Follow through from their answers to parts (a) and (e).

f.

(4, 6)     (A1)(ft)(A1)

Note: Accept \(x = 4\), \(y = 6\). Award at most (A1)(A0) if parentheses not seen.

If coordinates reversed award (A0)(A1)(ft).

Follow through from their answers to parts (a) and (e).

g.

\( – \frac{{14}}{{{4^2}}} + 1\)     (M1)

Note: Award (M1) for substitution into their gradient function.

Follow through from their answers to parts (b) and (g).

\( = \frac{1}{8}(0.125)\)     (A1)(ft)(G2)

h.

\(y – 1.5 = \frac{1}{8}(x – 4)\)     (M1)(ft)(M1)

Note: Award (M1) for substituting their (4, 1.5) in any straight line formula,

(M1) for substituting their gradient in any straight line formula.

\(y = \frac{x}{8} + 4\)     (A1)(ft)(G2)

Note: The form of the line has been specified in the question.

i.

Question

Consider the function \(f(x) = \frac{3}{4}{x^4} – {x^3} – 9{x^2} + 20\).

Find \(f( – 2)\).[2]

a.

Find \(f'(x)\).[3]

b.

The graph of the function \(f(x)\) has a local minimum at the point where \(x =  – 2\).

Using your answer to part (b), show that there is a second local minimum at \(x = 3\).[5]

c.

The graph of the function \(f(x)\) has a local minimum at the point where \(x =  – 2\).

Sketch the graph of the function \(f(x)\) for \( – 5 \leqslant x \leqslant 5\) and \( – 40 \leqslant y \leqslant 50\). Indicate on your

sketch the coordinates of the \(y\)-intercept.[4]

d.

The graph of the function \(f(x)\) has a local minimum at the point where \(x =  – 2\).

Write down the coordinates of the local maximum.[2]

e.

Let \(T\) be the tangent to the graph of the function \(f(x)\) at the point \((2, –12)\).

Find the gradient of \(T\).[2]

f.

The line \(L\) passes through the point \((2, −12)\) and is perpendicular to \(T\).

\(L\) has equation \(x + by + c = 0\), where \(b\) and \(c \in \mathbb{Z}\).

Find

(i)     the gradient of \(L\);

(ii)     the value of \(b\) and the value of \(c\).[5]

g.
Answer/Explanation

Markscheme

\(\frac{3}{4}{( – 2)^4} – {( – 2)^3} – 9{( – 2)^2} + 20\)     (M1)

Note: Award (M1) for substituting \(x =  – 2\) in the function.

\(= 4\)     (A1)(G2)

Note: If the coordinates \(( – 2,{\text{ }}4)\) are given as the answer award, at most, (M1)(A0). If no working shown award (G1).

     If \(x =  – 2,{\text{ }}y = 4\) seen then award full marks.

[2 marks]

a.

\(3{x^3} – 3{x^2} – 18x\)     (A1)(A1)(A1)

Note: Award (A1) for each correct term, award at most (A1)(A1)(A0) if extra terms seen.

[3 marks]

b.

\(f'(3) = 3 \times {(3)^3} – 3 \times {(3)^2} – 18 \times 3\)     (M1)

Note: Award (M1) for substitution in their \(f'(x)\) of \(x = 3\).

\( = 0\)     (A1)

OR

\(3{x^3} – 3{x^2} – 18x = 0\)     (M1)

Note: Award (M1) for equating their \(f'(x)\) to zero.

\(x = 3\)     (A1)

\(f'({x_1}) = 3 \times {({x_1})^3} – 3 \times {({x_1})^2} – 18 \times {x_1} < 0\) where \(0 < {x_1} < 3\)     (M1)

Note: Award (M1) for substituting a value of \({x_1}\) in the range \(0 < {x_1} < 3\) into their \(f’\) and showing it is negative (decreasing).

\(f'({x_2}) = 3 \times {({x_2})^3} – 3 \times {({x_2})^2} – 18 \times {x_2} > 0\) where \({x_2} > 3\)     (M1)

Note: Award (M1) for substituting a value of \({x_2}\) in the range \({x_2} > 3\) into their \(f’\) and showing it is positive (increasing).

OR

With or without a sketch:

Showing \(f({x_1}) > f(3)\) where \({x_1} < 3\) and \({x_1}\) is close to 3.     (M1)

Showing \(f({x_2}) > f(3)\) where \({x_2} > 3\) and \({x_2}\) is close to 3.     (M1)

Note: If a sketch of \(f(x)\) is drawn in this part of the question and \(x = 3\) is identified as a stationary point on the curve, then

     (i) award, at most, (M1)(A1)(M1)(M0) if the stationary point has been found;

     (ii) award, at most, (M0)(A0)(M1)(M0) if the stationary point has not been previously found.

Since the gradients go from negative (decreasing) through zero to positive (increasing) it is a local minimum     (R1)(AG)

Note: Only award (R1) if the first two marks have been awarded ie \(f'(3)\) has been shown to be equal to \(0\).

[5 marks]

c.

     (A1)(A1)(A1)(A1)

Notes: Award (A1) for labelled axes and indication of scale on both axes.

     Award (A1) for smooth curve with correct shape.

     Award (A1) for local minima in \({2^{{\text{nd}}}}\) and \({4^{{\text{th}}}}\) quadrants.

     Award (A1) for y intercept \((0, 20)\) seen and labelled. Accept \(20\) on \(y\)axis.

     Do not award the third (A1) mark if there is a turning point on the \(x\)-axis.

     If the derivative function is sketched then award, at most, (A1)(A0)(A0)(A0).

     For a smooth curve (with correct shape) there should be ONE continuous thin line, no part of which is straight and no (one to many) mappings of \(x\).

[4 marks]

d.

\((0, 20)\)     (G1)(G1)

Note: If parentheses are omitted award (G0)(G1).

OR

\(x = 0,{\text{ }}y = 20\)     (G1)(G1)

Note: If the derivative function is sketched in part (d), award (G1)(ft)(G1)(ft) for \((–1.12, 12.2)\).

[2 marks]

e.

\(f'(2) = 3{(2)^3} – 3{(2)^2} – 18(2)\)     (M1)

Notes: Award (M1) for substituting \(x = 2\) into their \(f'(x)\).

\( =  – 24\)     (A1)(ft)(G2)

[2 marks]

f.

(i)     Gradient of perpendicular \( = \frac{1}{{24}}\)   \((0.0417, 0.041666…)\)     (A1)(ft)(G1)

Note: Follow through from part (f).

(ii)     \(y + 12 = \frac{1}{{24}}(x – 2)\)     (M1)(M1)

Note: Award (M1) for correct substitution of \((2, –12)\), (M1) for correct substitution of their perpendicular gradient into equation of line.

OR

\( – 12 = \frac{1}{{24}} \times 2 + d\)     (M1)

\(d =  – \frac{{145}}{{12}}\)

\(y = \frac{1}{{24}}x – \frac{{145}}{{12}}\)     (M1)

Note: Award (M1) for correct substitution of \((2, –12)\) and gradient into equation of a straight line, (M1) for correct substitution of the perpendicular gradient and correct substitution of \(d\)into equation of line.

\(b =  – 24,{\text{ }}c =  – 290\)     (A1)(ft)(A1)(ft)(G3)

Note: Follow through from parts (f) and g(i).

     To award (ft) marks, \(b\) and \(c\) must be integers.

     Where candidate has used \(0.042\) from g(i), award (A1)(ft) for \(–288\).

[5 marks]

g.

Question

Consider the function \(f(x) = \frac{{96}}{{{x^2}}} + kx\), where \(k\) is a constant and \(x \ne 0\).

Write down \(f'(x)\).[3]

a.

The graph of \(y = f(x)\) has a local minimum point at \(x = 4\).

Show that \(k = 3\).[2]

b.

The graph of \(y = f(x)\) has a local minimum point at \(x = 4\).

Find \(f(2)\).[2]

c.

The graph of \(y = f(x)\) has a local minimum point at \(x = 4\).

Find \(f'(2)\)[2]

d.

The graph of \(y = f(x)\) has a local minimum point at \(x = 4\).

Find the equation of the normal to the graph of \(y = f(x)\) at the point where \(x = 2\).

Give your answer in the form \(ax + by + d = 0\) where \(a,{\text{ }}b,{\text{ }}d \in \mathbb{Z}\).[3]

e.

The graph of \(y = f(x)\) has a local minimum point at \(x = 4\).

Sketch the graph of \(y = f(x)\), for \( – 5 \leqslant x \leqslant 10\) and \( – 10 \leqslant y \leqslant 100\).[4]

f.

The graph of \(y = f(x)\) has a local minimum point at \(x = 4\).

Write down the coordinates of the point where the graph of \(y = f(x)\) intersects the \(x\)-axis.[2]

g.

The graph of \(y = f(x)\) has a local minimum point at \(x = 4\).

State the values of \(x\) for which \(f(x)\) is decreasing.[2]

h.
Answer/Explanation

Markscheme

\(\frac{{ – 192}}{{{x^3}}} + k\)     (A1)(A1)(A1)

Note: Award (A1) for \(-192\), (A1) for \({x^{ – 3}}\), (A1) for \(k\) (only).

a.

at local minimum \(f'(x) = 0\)     (M1)

Note: Award (M1) for seeing \(f'(x) = 0\) (may be implicit in their working).

\(\frac{{ – 192}}{{{4^3}}} + k = 0\)     (A1)

\(k = 3\)     (AG)

Note: Award (A1) for substituting \(x = 4\) in their \(f'(x) = 0\), provided it leads to \(k = 3\). The conclusion \(k = 3\) must be seen for the (A1) to be awarded.

b.

\(\frac{{96}}{{{2^2}}} + 3(2)\)     (M1)

Note: Award (M1) for substituting \(x = 2\) and \(k = 3\) in \(f(x)\).

\( = 30\)     (A1)(G2)

c.

\(\frac{{ – 192}}{{{2^3}}} + 3\)     (M1)

Note: Award (M1) for substituting \(x = 2\) and \(k = 3\) in their \(f'(x)\).

\( =  – 21\)     (A1)(ft)(G2)

Note: Follow through from part (a).

d.

\(y – 30 = \frac{1}{{21}}(x – 2)\)     (A1)(ft)(M1)

Notes: Award (A1)(ft) for their \(\frac{1}{{21}}\) seen, (M1) for the correct substitution of their point and their normal gradient in equation of a line.

Follow through from part (c) and part (d).

OR

gradient of normal \( = \frac{1}{{21}}\)     (A1)(ft)

\(30 = \frac{1}{{21}} \times 2 + c\)     (M1)

\(c = 29\frac{{19}}{{21}}\)

\(y = \frac{1}{{21}}x + 29\frac{{19}}{{21}}\;\;\;(y = 0.0476x + 29.904)\)

\(x – 21y + 628 = 0\)     (A1)(ft)(G2)

Notes:  Accept equivalent answers.

e.

     (A1)(A1)(A1)(A1)

Notes: Award (A1) for correct window (at least one value, other than zero, labelled on each axis), the axes must also be labelled; (A1) for a smooth curve with the correct shape (graph should not touch \(y\)-axis and should not curve away from the \(y\)-axis), on the given domain; (A1) for axis intercept in approximately the correct position (nearer \(-5\) than zero); (A1) for local minimum in approximately the correct position (first quadrant, nearer the \(y\)-axis than \(x = 10\)).

If there is no scale, award a maximum of (A0)(A1)(A0)(A1) – the final (A1) being awarded for the zero and local minimum in approximately correct positions relative to each other.

f.

\(( – 3.17,{\text{ }}0)\;\;\;\left( {( – 3.17480 \ldots ,{\text{ 0)}}} \right)\)     (G1)(G1)

Notes: If parentheses are omitted award (G0)(G1)(ft).

Accept \(x =  – 3.17,{\text{ }}y = 0\). Award (G1) for \(-3.17\) seen.

g.

\(0 < x \leqslant 4{\text{ or }}0 < x < 4\)     (A1)(A1)

Notes: Award (A1) for correct end points of interval, (A1) for correct notation (note: lower inequality must be strict).

Award a maximum of (A1)(A0) if \(y\) or \(f(x)\) used in place of \(x\).

h.

Question

Consider the function \(f(x) = 0.5{x^2} – \frac{8}{x},{\text{ }}x \ne 0\).

Find \(f( – 2)\).[2]

a.

Find \(f'(x)\).[3]

b.

Find the gradient of the graph of \(f\) at \(x =  – 2\).[2]

c.

Let \(T\) be the tangent to the graph of \(f\) at \(x =  – 2\).

Write down the equation of \(T\).[2]

d.

Let \(T\) be the tangent to the graph of \(f\) at \(x =  – 2\).

Sketch the graph of \(f\) for \( – 5 \leqslant x \leqslant 5\) and \( – 20 \leqslant y \leqslant 20\).

[4]
e.

Let \(T\) be the tangent to the graph of \(f\) at \(x =  – 2\).

Draw \(T\) on your sketch.[2]

f.

The tangent, \(T\), intersects the graph of \(f\) at a second point, P.

Use your graphic display calculator to find the coordinates of P.[2]

g.
Answer/Explanation

Markscheme

\(0.5 \times {( – 2)^2} – \frac{8}{{ – 2}}\)     (M1)

Note: Award (M1) for substitution of \(x =  – 2\) into the formula of the function.

\(6\)     (A1)(G2)

a.

\(f'(x) = x + 8{x^{ – 2}}\)     (A1)(A1)(A1)

Notes: Award (A1) for \(x\), (A1) for \(8\), (A1) for \({x^{ – 2}}\) or \(\frac{1}{{{x^2}}}\) (each term must have correct sign). Award at most (A1)(A1)(A0) if there are additional terms present or further incorrect simplifications are seen.

b.

\(f'( – 2) =  – 2 + 8{( – 2)^{ – 2}}\)     (M1)

Note: Award (M1) for \(x =  – 2\) substituted into their \(f'(x)\) from part (b).

\( = 0\)     (A1)(ft)(G2)

Note: Follow through from their derivative function.

c.

\(y = 6\;\;\;\)OR\(\;\;\;y = 0x + 6\;\;\;\)OR\(\;\;\;y – 6 = 0(x + 2)\)     (A1)(ft)(A1)(ft)(G2)

Notes: Award (A1)(ft) for their gradient from part (c), (A1)(ft) for their answer from part (a). Answer must be an equation.

Award (A0)(A0) for \(x = 6\).

d.

     (A1)(A1)(A1)(A1)

Notes: Award (A1) for labels and some indication of scales in the stated window. The point \((-2,{\text{ }}6)\) correctly labelled, or an \(x\)-value and a \(y\)-value on their axes in approximately the correct position, are acceptable indication of scales.

Award (A1) for correct general shape (curve must be smooth and must not cross the \(y\)-axis).

Award (A1) for \(x\)-intercept in approximately the correct position.

Award (A1) for local minimum in the second quadrant.

e.

Tangent to graph drawn approximately at \(x =  – 2\)     (A1)(ft)(A1)(ft)

Notes: Award (A1)(ft) for straight line tangent to curve at approximately \(x =  – 2\), with approximately correct gradient. Tangent must be straight for the (A1)(ft) to be awarded.

Award (A1)(ft) for (extended) line passing through approximately their \(y\)-intercept from (d). Follow through from their gradient in part (c) and their equation in part (d).

f.

\((4,{\text{ }}6)\;\;\;\)OR\(\;\;\;x = 4,{\text{ }}y = 6\)     (G1)(ft)(G1)(ft)

Notes: Follow through from their tangent from part (d). If brackets are missing then award (G0)(G1)(ft).

If line intersects their graph at more than one point (apart from \(( – 2,{\text{ }}6)\)), follow through from the first point of intersection (to the right of \( – 2\)).

Award (G0)(G0) for \(( – 2,{\text{ }}6)\).

g.

Question

A function, \(f\) , is given by

\[f(x) = 4 \times {2^{ – x}} + 1.5x – 5.\]

Calculate \(f(0)\)[2]

a.

Use your graphic display calculator to solve \(f(x) = 0.\)[2]

b.

Sketch the graph of \(y = f(x)\) for \( – 2 \leqslant x \leqslant 6\) and \( – 4 \leqslant y \leqslant 10\) , showing the \(x\) and \(y\) intercepts. Use a scale of \(2\,{\text{cm}}\) to represent \(2\) units on both the horizontal axis, \(x\) , and the vertical axis, \(y\) .[4]

c.

The function \(f\) is the derivative of a function \(g\) . It is known that \(g(1) = 3.\)

i)     Calculate \(g'(1).\)

ii)    Find the equation of the tangent to the graph of \(y = g(x)\) at \(x = 1.\) Give your answer in the form \(y = mx + c.\)[4]

d.
Answer/Explanation

Markscheme

\(4 \times {2^{ – 0}} + 1.5 \times 0 – 5\)       (M1)

Note: Award (M1) for substitution of \(0\) into the expression for \(f(x)\) .

\( =  – 1\)        (A1)(G2)

a.

\( – 0.538\,\,\,( – 0.537670…)\) and \(3\)        (A1)(A1)

Note: Award at most (A0)(A1)(ft) if answer is given as pairs of coordinates.

b.

(A1)(A1)(A1)(ft)(A1)(ft)

Note: Award (A1) for labels and some indication of scale in the correct given window.

Award (A1) for smooth curve with correct general shape with \(f( – 2) > f(6)\) and minimum to the right of the \(y\)-axis.

Award (A1)(ft) for correct \(y\)-intercept (consistent with their part (a)).

Award (A1)(ft) for approximately correct \(x\)-intercepts (consistent with their part (b), one zero between \( – 1\) and \(0\), the other between \(2.5\) and \(3.5\)).

c.

i)     \(g'(1) = f(1) = 4 \times {2^{ – 1}} + 1.5 – 5\)       (M1)

Note: Award (M1) for substitution of \(1\) into \(f(x)\).

\( =  – 1.5\)        (A1)(G2)

ii)    \(3 =  – 1.5 \times 1 + c\) OR \((y – 3) =  – 1.5\,(x – 1)\)       (M1)

Note: Award (M1) for correct substitution of gradient and the point \((1,\,\,3)\) into the equation of a line. Follow through from (d)(i).

\(y =  – 1.5x + 4.5\)        (A1)(ft)(G2)

d.

Question

Consider the function \(g(x) = {x^3} + k{x^2} – 15x + 5\).

The tangent to the graph of \(y = g(x)\) at \(x = 2\) is parallel to the line \(y = 21x + 7\).

Find \(g'(x)\).[3]

a.

Show that \(k = 6\).[2]

b.i.

Find the equation of the tangent to the graph of \(y = g(x)\) at \(x = 2\). Give your answer in the form \(y = mx + c\).[3]

b.ii.

Use your answer to part (a) and the value of \(k\), to find the \(x\)-coordinates of the stationary points of the graph of \(y = g(x)\).[3]

c.

Find \(g’( – 1)\).[2]

d.i.

Hence justify that \(g\) is decreasing at \(x =  – 1\).[1]

d.ii.

Find the \(y\)-coordinate of the local minimum.[2]

e.
Answer/Explanation

Markscheme

\(3{x^2} + 2kx – 15\)     (A1)(A1)(A1)

Note:     Award (A1) for \(3{x^2}\), (A1) for \(2kx\) and (A1) for \( – 15\). Award at most (A1)(A1)(A0) if additional terms are seen.

[3 marks]

a.

\(21 = 3{(2)^2} + 2k(2) – 15\)     (M1)(M1)

Note:     Award (M1) for equating their derivative to 21. Award (M1) for substituting 2 into their derivative. The second (M1) should only be awarded if correct working leads to the final answer of \(k = 6\).

Substituting in the known value, \(k = 6\), invalidates the process; award (M0)(M0).

\(k = 6\)     (AG)

[2 marks]

b.i.

\(g(2) = {(2)^3} + (6){(2)^2} – 15(2) + 5{\text{ }}( = 7)\)     (M1)

Note:     Award (M1) for substituting 2 into \(g\).

\(7 = 21(2) + c\)     (M1)

Note:     Award (M1) for correct substitution of 21, 2 and their 7 into gradient intercept form.

OR

\(y – 7 = 21(x – 2)\)     (M1)

Note:     Award (M1) for correct substitution of 21, 2 and their 7 into gradient point form.

\(y = 21x – 35\)     (A1)     (G2)

[3 marks]

b.ii.

\(3{x^2} + 12x – 15 = 0\) (or equivalent)     (M1)

Note:     Award (M1) for equating their part (a) (with \(k = 6\) substituted) to zero.

\(x =  – 5,{\text{ }}x = 1\)     (A1)(ft)(A1)(ft)

Note:     Follow through from part (a).

[3 marks]

c.

\(3{( – 1)^2} + 12( – 1) – 15\)     (M1)

Note:     Award (M1) for substituting \( – 1\) into their derivative, with \(k = 6\) substituted. Follow through from part (a).

\( =  – 24\)     (A1)(ft)     (G2)

[2 marks]

d.i.

\(g’( – 1) < 0\) (therefore \(g\) is decreasing when \(x =  – 1\))     (R1)

[1 marks]

d.ii.

\(g(1) = {(1)^3} + (6){(1)^2} – 15(1) + 5\)     (M1)

Note:     Award (M1) for correctly substituting 6 and their 1 into \(g\).

\( =  – 3\)     (A1)(ft)     (G2)

Note:     Award, at most, (M1)(A0) or (G1) if answer is given as a coordinate pair. Follow through from part (c).

[2 marks]

e.

Question

Consider the function \(f(x) =  – {x^4} + a{x^2} + 5\), where \(a\) is a constant. Part of the graph of \(y = f(x)\) is shown below.

M17/5/MATSD/SP2/ENG/TZ2/06

It is known that at the point where \(x = 2\) the tangent to the graph of \(y = f(x)\) is horizontal.

There are two other points on the graph of \(y = f(x)\) at which the tangent is horizontal.

Write down the \(y\)-intercept of the graph.[1]

a.

Find \(f'(x)\).[2]

b.

Show that \(a = 8\).[2]

c.i.

Find \(f(2)\).[2]

c.ii.

Write down the \(x\)-coordinates of these two points;[2]

d.i.

Write down the intervals where the gradient of the graph of \(y = f(x)\) is positive.[2]

d.ii.

Write down the range of \(f(x)\).[2]

e.

Write down the number of possible solutions to the equation \(f(x) = 5\).[1]

f.

The equation \(f(x) = m\), where \(m \in \mathbb{R}\), has four solutions. Find the possible values of \(m\).[2]

g.
Answer/Explanation

Markscheme

5     (A1)

Note:     Accept an answer of \((0,{\text{ }}5)\).

[1 mark]

a.

\(\left( {f'(x) = } \right) – 4{x^3} + 2ax\)     (A1)(A1)

Note:     Award (A1) for \( – 4{x^3}\) and (A1) for \( + 2ax\). Award at most (A1)(A0) if extra terms are seen.

[2 marks]

b.

\( – 4 \times {2^3} + 2a \times 2 = 0\)     (M1)(M1)

Note:     Award (M1) for substitution of \(x = 2\) into their derivative, (M1) for equating their derivative, written in terms of \(a\), to 0 leading to a correct answer (note, the 8 does not need to be seen).

\(a = 8\)     (AG)

[2 marks]

c.i.

\(\left( {f(2) = } \right) – {2^4} + 8 \times {2^2} + 5\)     (M1)

Note:     Award (M1) for correct substitution of \(x = 2\) and  \(a = 8\) into the formula of the function.

21     (A1)(G2)

[2 marks]

c.ii.

\((x = ){\text{ }} – 2,{\text{ }}(x = ){\text{ 0}}\)     (A1)(A1)

Note:     Award (A1) for each correct solution. Award at most (A0)(A1)(ft) if answers are given as \(( – 2{\text{ }},21)\) and \((0,{\text{ }}5)\) or \(( – 2,{\text{ }}0)\) and \((0,{\text{ }}0)\).

[2 marks]

d.i.

\(x <  – 2,{\text{ }}0 < x < 2\)     (A1)(ft)(A1)(ft)

Note:     Award (A1)(ft) for \(x <  – 2\), follow through from part (d)(i) provided their value is negative.

Award (A1)(ft) for \(0 < x < 2\), follow through only from their 0 from part (d)(i); 2 must be the upper limit.

Accept interval notation.

[2 marks]

d.ii.

\(y \leqslant 21\)     (A1)(ft)(A1)

Notes:     Award (A1)(ft) for 21 seen in an interval or an inequality, (A1) for “\(y \leqslant \)”.

Accept interval notation.

Accept \( – \infty  < y \leqslant 21\) or \(f(x) \leqslant 21\).

Follow through from their answer to part (c)(ii). Award at most (A1)(ft)(A0) if \(x\) is seen instead of \(y\). Do not award the second (A1) if a (finite) lower limit is seen.

[2 marks]

e.

3 (solutions)     (A1)

[1 mark]

f.

\(5 < m < 21\) or equivalent     (A1)(ft)(A1)

Note:     Award (A1)(ft) for 5 and 21 seen in an interval or an inequality, (A1) for correct strict inequalities. Follow through from their answers to parts (a) and (c)(ii).

Accept interval notation.

[2 marks]

g.

Question

Consider the function \(f\left( x \right) = \frac{{48}}{x} + k{x^2} – 58\), where x > 0 and k is a constant.

The graph of the function passes through the point with coordinates (4 , 2).

P is the minimum point of the graph of f (x).

Find the value of k.[2]

a.

Using your value of k , find f ′(x).[3]

b.

Use your answer to part (b) to show that the minimum value of f(x) is −22 .[3]

c.

Write down the two values of x which satisfy f (x) = 0.[2]

d.

Sketch the graph of y = f (x) for 0 < x ≤ 6 and −30 ≤ y ≤ 60.
Clearly indicate the minimum point P and the x-intercepts on your graph.[4]

e.
Answer/Explanation

Markscheme

\(\frac{{48}}{4} + k \times {4^2} – 58 = 2\)    (M1)
Note: Award (M1) for correct substitution of x = 4 and y = 2 into the function.

k = 3     (A1) (G2)

[2 marks]

a.

\(\frac{{ – 48}}{{{x^2}}} + 6x\)     (A1)(A1)(A1)(ft) (G3)

Note: Award (A1) for −48 , (A1) for x−2, (A1)(ft) for their 6x. Follow through from part (a). Award at most (A1)(A1)(A0) if additional terms are seen.

[3 marks]

b.

\(\frac{{ – 48}}{{{x^2}}} + 6x = 0\)     (M1)

Note: Award (M1) for equating their part (b) to zero.

x = 2     (A1)(ft)

Note: Follow through from part (b). Award (M1)(A1) for \(\frac{{ – 48}}{{{{\left( 2 \right)}^2}}} + 6\left( 2 \right) = 0\) seen.

Award (M0)(A0) for x = 2 seen either from a graphical method or without working.

\(\frac{{48}}{2} + 3 \times {2^2} – 58\,\,\,\left( { =  – 22} \right)\)   (M1)

Note: Award (M1) for substituting their 2 into their function, but only if the final answer is −22. Substitution of the known result invalidates the process; award (M0)(A0)(M0).

−22     (AG)

[3 marks]

c.

0.861  (0.860548…), 3.90  (3.90307…)     (A1)(ft)(A1)(ft) (G2)

Note: Follow through from part (a) but only if the answer is positive. Award at most (A1)(ft)(A0) if answers are given as coordinate pairs or if extra values are seen. The function f (x) only has two x-intercepts within the domain. Do not accept a negative x-intercept.

[2 marks]

d.

(A1)(A1)(ft)(A1)(ft)(A1)(ft)

Note: Award (A1) for correct window. Axes must be labelled.
(A1)(ft) for a smooth curve with correct shape and zeros in approximately correct positions relative to each other.
(A1)(ft) for point P indicated in approximately the correct position. Follow through from their x-coordinate in part (c). (A1)(ft) for two x-intercepts identified on the graph and curve reflecting asymptotic properties.

[4 marks]

e.

Question

Consider the curve y = 2x3 − 9x2 + 12x + 2, for −1 < x < 3

Sketch the curve for −1 < x < 3 and −2 < y < 12.[4]

a.

A teacher asks her students to make some observations about the curve.

Three students responded.
Nadia said “The x-intercept of the curve is between −1 and zero”.
Rick said “The curve is decreasing when x < 1 ”.
Paula said “The gradient of the curve is less than zero between x = 1 and x = 2 ”.

State the name of the student who made an incorrect observation.[1]

b.

Find the value of y when x = 1 .[2]

c.

Find \(\frac{{{\text{dy}}}}{{{\text{dx}}}}\).[3]

d.

Show that the stationary points of the curve are at x = 1 and x = 2.[2]

e.

Given that y = 2x3 − 9x2 + 12x + 2 = k has three solutions, find the possible values of k.[3]

f.
Answer/Explanation

Markscheme

(A1)(A1)(A1)(A1)

Note: Award (A1) for correct window (condone a window which is slightly off) and axes labels. An indication of window is necessary. −1 to 3 on the x-axis and −2 to 12 on the y-axis and a graph in that window.
(A1) for correct shape (curve having cubic shape and must be smooth).
(A1) for both stationary points in the 1st quadrant with approximate correct position,
(A1) for intercepts (negative x-intercept and positive y intercept) with approximate correct position.

[4 marks]

a.

Rick     (A1)

Note: Award (A0) if extra names stated.

[1 mark]

b.

2(1)3 − 9(1)2 + 12(1) + 2     (M1)

Note: Award (M1) for correct substitution into equation.

= 7     (A1)(G2)

[2 marks]

c.

6x2 − 18x + 12     (A1)(A1)(A1)

Note: Award (A1) for each correct term. Award at most (A1)(A1)(A0) if extra terms seen.

[3 marks]

d.

6x2 − 18x + 12 = 0     (M1)

Note: Award (M1) for equating their derivative to 0. If the derivative is not explicitly equated to 0, but a subsequent solving of their correct equation is seen, award (M1).

6( − 1)(x − 2) = 0  (or equivalent)      (M1)

Note: Award (M1) for correct factorization. The final (M1) is awarded only if answers are clearly stated.

Award (M0)(M0) for substitution of 1 and of 2 in their derivative.

x = 1, x = 2 (AG)

[2 marks]

e.

6 < k < 7     (A1)(A1)(ft)(A1)

Note: Award (A1) for an inequality with 6, award (A1)(ft) for an inequality with 7 from their part (c) provided it is greater than 6, (A1) for their correct strict inequalities. Accept ]6, 7[ or (6, 7).

[3 marks]

f.
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