IBDP Maths AI: Topic: SL 2.3: The graph of a function: IB style Questions SL Paper2

Question

Given \(f (x) = x^2 − 3x^{−1}, x \in {\mathbb{R}}, – 5 \leqslant x \leqslant 5, x \ne 0\),

A football is kicked from a point A (a, 0), 0 < a < 10 on the ground towards a goal to the right of A.

The ball follows a path that can be modelled by part of the graph

\(y = − 0.021x^2 + 1.245x − 6.01, x \in {\mathbb{R}}, y \geqslant  0\).

x is the horizontal distance of the ball from the origin

y is the height above the ground

Both x and y are measured in metres.

Write down the equation of the vertical asymptote.[1]

i.a.

Find \(f ′(x)\).[2]

i.b.

Using your graphic display calculator or otherwise, write down the coordinates of any point where the graph of \(y = f (x)\) has zero gradient.[2]

i.c.

Write down all intervals in the given domain for which \(f (x)\) is increasing.[3]

i.d.

Using your graphic display calculator or otherwise, find the value of a.[1]

ii.a.

Find \(\frac{{dy}}{{dx}}\).[2]

ii.b.

(i) Use your answer to part (b) to calculate the horizontal distance the ball has travelled from A when its height is a maximum.

(ii) Find the maximum vertical height reached by the football.[4]

ii.c.

Draw a graph showing the path of the football from the point where it is kicked to the point where it hits the ground again. Use 1 cm to represent 5 m on the horizontal axis and 1 cm to represent 2 m on the vertical scale.[4]

ii.d.

The goal posts are 35 m from the point where the ball is kicked.

At what height does the ball pass over the goal posts?[2]

ii.e.
Answer/Explanation

Markscheme

equation of asymptote is x = 0     (A1)

(Must be an equation.)[1 mark]

i.a.

\(f ‘(x) = 2x + 3x^{-2}\)     (or equivalent)     (A1) for each term     (A1)(A1)[2 marks]

i.b.

stationary point (–1.14, 3.93)     (G1)(G1)(ft)

(-1,4) or similar error is awarded (G0)(G1)(ft). Here and also as follow through in part (d) accept exact values \( – {\left( {\frac{3}{2}} \right)^{\frac{1}{3}}}\)for the x coordinate and \(3{\left( {\frac{3}{2}} \right)^{\frac{2}{3}}}\) for the y coordinate.

OR \(2x + \frac{3}{{{x^2}}} = 0\) or equivalent

Correct coordinates as above     (M1)

Follow through from candidate’s \(f ′(x)\).     (A1)(ft)[2 marks]

i.c.

In all alternative answers for (d), follow through from candidate’s x coordinate in part (c).

Alternative answers include:

–1.14 ≤ x < 0,     0 < x < 5     (A1)(A1)(ft)(A1)

OR [–1.14,0), (0,5)

Accept alternative bracket notation for open interval ] [. (Union of these sets is not correct, award (A2) if all else is right in this case.)

OR \( – 1.14 \leqslant x < 5,x \ne 0\)

In all versions 0 must be excluded (A1). -1.14 must be the left bound . 5 must be the right bound (A1). For \(x \geqslant – 1.14\) or \(x > – 1.14\) alone, award (A1). For \( – 1.4 \leqslant x < 0\) together with \(x > 0\) award (A2).[3 marks]

i.d.

a = 5.30 (3sf)   (Allow (5.30, 0) but 5.3 receives an (AP).)     (A1)[1 mark]

ii.a.

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = – 0.042x + 1.245\)     (A1) for each term.     (A1)(A1)[2 marks]

ii.b.

Unit penalty (UP) is applicable where indicated in the left hand column.

(i) Maximum value when \(f ‘ (x) = 0\), \( – 0.042x + 1.245 = 0\),     (M1)

(M1) is for either of the above but at least one must be seen.

(x = 29.6.)

Football has travelled 29.6 – 5.30 = 24.3 m (3sf) horizontally.     (A1)(ft)

For answer of 24.3 m with no working or for correct subtraction of 5.3 from candidate’s x-coordinate at the maximum (if not 29.6), award (A1)(d).

(UP) (ii) Maximum vertical height, f (29.6) = 12.4 m     (M1)(A1)(ft)(G2)

(M1) is for substitution into f of a value seen in part (c)(i). f(24.3) with or without evaluation is awarded (M1)(A0). For any other value without working, award (G0). If lines are seen on the graph in part (d) award (M1) and then (A1) for candidate’s value \( \pm 0.5\) (3sf not required.)[4 marks]

ii.c.

(not to scale)

     (A1)(A1)(A1)(ft)(A1)(ft)

Award (A1) for labels (units not required) and scale, (A1)(ft) for max(29.6,12.4), (A1)(ft) for x-intercepts at 5.30 and 53.9, (all coordinates can be within 0.5), (A1) for well-drawn parabola ending at the x-intercepts.[4 marks]

ii.d.

Unit penalty (UP) is applicable where indicated in the left hand column.

(UP) f (40.3) = 10.1 m (3sf).

Follow through from (a). If graph used, award (M1) for lines drawn and (A1) for candidate’s value \( \pm 0.5\). (3sf not required).     (M1)(A1)(ft)(G2)[2 marks]

ii.e.

Question

Consider the functions \(f(x) = \frac{{2x + 3}}{{x + 4}}\) and \(g(x) = x + 0.5\) .

Sketch the graph of the function \(f(x)\), for \( – 10 \leqslant x \leqslant 10\) . Indicating clearly the axis intercepts and any asymptotes.[6]

a.

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

b.

On the same diagram as part (a) sketch the graph of \(g(x) = x + 0.5\) .[2]

c.

Using your graphical display calculator write down the coordinates of one of the points of intersection on the graphs of \(f\) and \(g\), giving your answer correct to five decimal places.[3]

d.

Write down the gradient of the line \(g(x) = x + 0.5\) .[1]

e.

The line \(L\) passes through the point with coordinates \(( – 2{\text{, }} – 3)\) and is perpendicular to the line \(g(x)\) . Find the equation of \(L\).[3]

f.
Answer/Explanation

Markscheme

     (A6)

Notes: (A1) for labels and some idea of scale.
(A1) for \(x\)-intercept seen, (A1) for \(y\)-intercept seen in roughly the correct places (coordinates not required).
(A1) for vertical asymptote seen, (A1) for horizontal asymptote seen in roughly the correct places (equations of the lines not required).
(A1) for correct general shape.[6 marks]

a.

\(x = – 4\)     (A1)(A1)(ft)

Note: (A1) for \(x =\), (A1)(ft) for \( – 4\).[2 marks]

b.

     (A1)(A1)

Note: (A1) for correct axis intercepts, (A1) for straight line[2 marks]

c.

\(( – 2.85078{\text{, }} – 2.35078)\) OR \((0.35078{\text{, }}0.85078)\)     (G1)(G1)(A1)(ft)

Notes: (A1) for \(x\)-coordinate, (A1) for \(y\)-coordinate, (A1)(ft) for correct accuracy. Brackets required. If brackets not used award (G1)(G0)(A1)(ft).
Accept \(x = – 2.85078\), \(y = – 2.35078\) or \(x = 0.35078\), \(y = 0.85078\).[3 marks]

d.

\({\text{gradient}} = 1\)     (A1)[1 mark]

e.

\({\text{gradient of perpendicular}} = – 1\)     (A1)(ft)

(can be implied in the next step)

\(y = mx + c\)

\( – 3 = – 1 \times – 2 + c\)     (M1)

\(c = – 5\)

\(y = – x – 5\)     (A1)(ft)(G2)

OR

\(y + 3 = – (x + 2)\)     (M1)(A1)(ft)(G2)

Note: Award (G2) for correct answer with no working at all but (A1)(G1) if the gradient is mentioned as \( – 1\) then correct answer with no further working.[3 marks]

f.

Question

Consider the function \(f:x \mapsto \frac{{kx}}{{{2^x}}}\).

The cost per person, in euros, when \(x\) people are invited to a party can be determined by the function

\(C(x) = x + \frac{{100}}{x}\)

Given that \(f(1) = 2\), show that \(k = 4\).[2]

i.a.

Write down the values of \(q\) and \(r\) for the following table.

[2]

i.b.

As \(x\) increases from \( – 1\), the graph of \(y = f(x)\) reaches a maximum value and then decreases, behaving asymptotically.

Draw the graph of \(y = f(x)\) for \( – 1 \leqslant x \leqslant 8\). Use a scale of \({\text{1 cm}}\) to represent 1 unit on both axes. The position of the maximum, \({\text{M}}\), the \(y\)-intercept and the asymptotic behaviour should be clearly shown.[4]

i.c.

Using your graphic display calculator, find the coordinates of \({\text{M}}\), the maximum point on the graph of \(y = f(x)\).[2]

i.d.

Write down the equation of the horizontal asymptote to the graph of \(y = f(x)\).[2]

i.e.

(i) Draw and label the line \( y = 1\) on your graph.

(ii) The equation \(f(x) = 1\) has two solutions. One of the solutions is \(x = 4\). Use your graph to find the other solution.[4]

i.f.

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

ii.a.

Show that the cost per person is a minimum when \(10\) people are invited to the party.[2]

ii.b.

Calculate the minimum cost per person.[2]

ii.c.
Answer/Explanation

Markscheme

\(f(1) = \frac{k}{{{2^1}}}\)     (M1)


Note: (M1)
for substituting \(x = 1\) into the formula.

\(\frac{k}{2} = 2\)     (M1)


Note: (M1)
for equating to 2.

\(k = 4\)     (AG)[2 marks]

i.a.

\(q = 2\), \(r = 0.125\)     (A1)(A1)[2 marks]

i.b.

     (A4)

Notes: (A1) for scales and labels.
(A1) for accurate smooth curve passing through \((0, 0)\) drawn at least in the given domain.
(A1) for asymptotic behaviour (curve must not go up or cross the \(x\)-axis).
(A1) for indicating the position of the maximum point.[4 marks]

i.c.

\({\text{M}}\) (\(1.44\), \(2.12\))     (G1)(G1)

Note: Brackets required, if missing award (G1)(G0). Accept \(x = 1.44\) and \(y = 2.12\).[2 marks]

i.d.

\(y = 0\)     (A1)(A1)

Note: (A1) for ‘\(y = \)’ provided the right hand side is a constant. (A1) for 0.[2 marks]

i.e.

(i) See graph     (A1)(A1)


Note: (A1)
for correct line, (A1) for label.

(ii) \(x = 0.3\) (ft) from candidate’s graph.     (A2)(ft)

Notes: Accept \( \pm 0.1\) from their x. For \(0.310\) award (G1)(G0). For other answers taken from the GDC and not given correct to 3 significant figures award (G0)(AP)(G0) or (G1)(G0) if (AP) already applied.[4 marks]

i.f.

\(C'(x) = 1 – \frac{{100}}{{{x^2}}}\)     (A1)(A1)(A1)

Note: (A1) for 1, (A1) for \( – 100\) , (A1) for \({x^2}\) as denominator or \({{x^{ – 2}}}\) as numerator. Award a maximum of (A2) if an extra term is seen.[3 marks]

ii.a.

For studying signs of the derivative at either side of \(x = 10\)     (M1)

For saying there is a change of sign of the derivative     (M1)(AG)

OR

For putting \(x = 10\) into \(C’\) and getting zero     (M1)

For clear sketch of the function or for mentioning that the function changes from decreasing to increasing at \(x = 10\)     (M1)(AG)

OR

For solving \(C'(x) = 0\) and getting \(10\)     (M1)

For clear sketch of the function or for mentioning that the function changes from decreasing to increasing at \(x = 10\)     (M1)(AG)

Note: For a sketch with a clear indication of the minimum or for a table with values of \(x\) at either side of \(x = 10\) award (M1)(M0).[2 marks]

ii.b.

\(C(10) = 10 + \frac{{100}}{{10}}\)     (M1)

\(C(10) = 20\)     (A1)(G2)[2 marks]

ii.c.

Question

The temperature in \(^ \circ {\text{C}}\) of a pot of water removed from the cooker is given by \(T(m) = 20 + 70 \times {2.72^{ – 0.4m}}\), where \(m\) is the number of minutes after the pot is removed from the cooker.

Show that the temperature of the water when it is removed from the cooker is \({90^ \circ }{\text{C}}\).[2]

a.

The following table shows values for \(m\) and \(T(m)\).

(i)     Write down the value of \(s\).

(ii)    Draw the graph of \(T(m)\) for \(0 \leqslant m \leqslant 10\) . Use a scale of \(1{\text{ cm}}\) to represent \(1\) minute on the horizontal axis and a scale of \(1{\text{ cm}}\) to represent \({10^ \circ }{\text{C}}\) on the vertical axis.

(iii)   Use your graph to find how long it takes for the temperature to reach \({56^ \circ }{\text{C}}\). Show your method clearly.

(iv)   Write down the temperature approached by the water after a long time. Justify your answer.[9]

b.

Consider the function \(S(m) = 20m – 40\) for \(2 \leqslant m \leqslant 6\) .

The function \(S(m)\) represents the temperature of soup in a pot placed on the cooker two minutes after the water has been removed. The soup is then heated.

Draw the graph of \(S(m)\) on the same set of axes used for part (b).[2]

c.

Consider the function \(S(m) = 20m – 40\) for \(2 \leqslant m \leqslant 6\) .

The function \(S(m)\) represents the temperature of soup in a pot placed on the cooker two minutes after the water has been removed. The soup is then heated.

Comment on the meaning of the constant \(20\) in the formula for \(S(m)\) in relation to the temperature of the soup.[1]

d.

Consider the function \(S(m) = 20m – 40\) for \(2 \leqslant m \leqslant 6\) .

The function \(S(m)\) represents the temperature of soup in a pot placed on the cooker two minutes after the water has been removed. The soup is then heated.

(i)     Use your graph to solve the equation \(S(m) = T(m)\) . Show your method clearly.

(ii)    Hence describe by using inequalities the set of values of \(m\) for which \(S(m) > T(m)\).[4]

e.
Answer/Explanation

Markscheme

\(T(0) = 20 + 70 \times {2.72^{ – 0.4 \times 0}} = 90\)     (M1)(A1)(AG)

Note: (M1) for taking \(m = 0\) , (A1) for substituting \(0\) into the formula. For the A mark to be awarded \(90\) must be justified by correct method.[2 marks]

a.

(i)    21.3     (A1)

(ii)  

     (A4)(ft)

Note: Scales and labels (A1). Smooth curve (A1). All points correct including the \(y\)-intercept (A2), 1 point incorrect (A1), otherwise (A0). Follow through from their value of \(s\).

(iii)    \(m = 1.7{\text{ minutes}}\) (Accept \( \pm 0.2\) )     (A2)(ft)


Note:
Follow through from candidate’s graph. Accept answers in minutes and seconds if consistent with graph. If answer incorrect and correct line(s) seen on graph award (M1)(A0).

(iv)    \({20^ \circ }{\text{C}}\)     (A1)(ft)

The curve behaves asymptotically to the line \(y = 20\) or similar.     (A1)

OR

The room temperature is 20 or similar

OR

When \({\text{m}}\) is a very large number the term \(70 \times {2.72^{ – 0.4{\text{m}}}}\) tends to zero or similar.

Note: Follow through from their graph if appropriate.[9 marks]

b.

     (A1)(A1)


Notes: (A1) for correct line, (A1) for domain. If line not drawn on same set of axes award at most (A1)(A0).
[2 marks]

c.

It indicates by how much the temperature increases per minute.     (A1)[1 mark]

d.

(i)     \(m = 3.8\) (Accept \( \pm 0.1\) )     (A2)(ft)

Note: Follow through from candidate’s graph. Accept answers in minutes and seconds if consistent with graph. If answer incorrect and correct line(s) seen on graph award (M1)(A0).

(ii)    \(3.8 < m \leqslant 6\)     (A1)(A1)(ft)

Note: (A1) for \(m > 3.8\) and (A1) for \(m \leqslant 6\). Follow through from candidate’s answer to part (e)(i). If candidate was already penalized in (c) for domain and does not state \(m \leqslant 6\) then award (A2)(ft).

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.
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