Question
The cubic curve \(y = 8{x^3} + b{x^2} + cx + d\) has two distinct points P and Q, where the gradient is zero.
(a) Show that \({b^2} > 24c\) .
(b) Given that the coordinates of P and Q are \(\left( {\frac{1}{2},{\text{ }} – 12} \right)\) and \(\left( { – \frac{3}{2},{\text{ }}20} \right)\) respectively, find the values of \(b\) , \(c\) and \(d\) .
Answer/Explanation
Markscheme
(a) \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 24{x^2} + 2bx + c\) (A1)
\(24{x^2} + 2bx + c = 0\) (M1)
\(\Delta = {\left( {2b} \right)^2} – 96\left( c \right)\) (A1)
\(4{b^2} – 96c > 0\) A1
\({b^2} > 24c\) AG
(b) \(1 + \frac{1}{4}b + \frac{1}{2}c + d = – 12\)
\(6 + b + c = 0\)
\( – 27 + \frac{9}{4}b – \frac{3}{2}c + d = 20\)
\(54 – 3b + c = 0\) A1A1A1
Note: Award A1 for each correct equation, up to \(3\), not necessarily simplified.
\(b = 12\), \(c = – 18\), \(d = – 7\) A1
[8 marks]
Examiners report
Many candidates throughout almost the whole mark range were able to score well on this question. It was pleasing that most candidates were aware of the discriminant condition for distinct real roots of a quadratic. Some who dropped marks on part (b) either didn’t write down a sufficient number of linear equations to determine the three unknowns or made arithmetic errors in their manual solution – few GDC solutions were seen.
Question
Consider the graphs \(y = {{\text{e}}^{ – x}}\) and \(y = {{\text{e}}^{ – x}}\sin 4x\) , for \(0 \leqslant x \leqslant \frac{{5\pi }}{4}\) .
(a) On the same set of axes draw, on graph paper, the graphs, for \(0 \leqslant x \leqslant \frac{{5\pi }}{4}\). Use a scale of \(1\) cm to \(\frac{\pi }{8}\) on your \(x\)-axis and \(5\) cm to \(1\) unit on your \(y\)-axis.
(b) Show that the \(x\)-intercepts of the graph \(y = {{\text{e}}^{ – x}}\) sin 4x are \(\frac{{n\pi }}{4}\) , \(n = 0\), \(1\), \(2\), \(3\), \(4\), \(5\).
(c) Find the \(x\)-coordinates of the points at which the graph of \(y = {{\text{e}}^{ – x}}\sin 4x\) meets the graph of \(y = {{\text{e}}^{ – x}}\) . Give your answers in terms of \( \pi\).
(d) (i) Show that when the graph of \(y = {{\text{e}}^{ – x}}\sin 4x\) meets the graph of \(y = {{\text{e}}^{ – x}}\) , their gradients are equal.
(ii) Hence explain why these three meeting points are not local maxima of the graph \(y = {{\text{e}}^{ – x}}\sin 4x\) .
(e) (i) Determine the \(y\)-coordinates, \({y_1}\) , \({y_2}\) and \({y_3}\), where \({y_1} > {y_2} > {y_3}\) , of the local maxima of \(y = {{\text{e}}^{ – x}}\sin 4x\) for \(0 \leqslant x \leqslant \frac{{5\pi }}{4}\) . You do not need to show that they are maximum values, but the values should be simplified.
(ii) Show that \({y_1}\) , \({y_2}\) and \({y_3}\) form a geometric sequence and determine the common ratio \(r\) .
Answer/Explanation
Markscheme
(a)
A3
Note: Award A1 for each correct shape,
A1 for correct relative position.
[3 marks]
(b) \({{\text{e}}^{ – x}}\sin \left( {4x} \right) = 0\) (M1)
\(\sin \left( {4x} \right) = 0\) A1
\(4x = 0\), \(\pi \), \(2\pi \), \(3\pi \), \(4\pi \), \(5\pi \) A1
\(x = 0\), \(\frac{\pi }{4}\), \(\frac{2\pi }{4}\), \(\frac{3\pi }{4}\), \(\frac{4\pi }{4}\), \(\frac{5\pi }{4}\) AG
[3 marks]
(c) \({{\text{e}}^{ – x}} = {{\text{e}}^{ – x}}\sin \left( {4x} \right) = 0\) or reference to graph
\(\sin 4x = 1\), M1
\(\sin 4x = 1\), \(frac{\pi }{2}\), \(frac{5\pi }{2}\), \(frac{9\pi }{2}\) A1
\(x = \frac{\pi }{8}\), \(\frac{5\pi }{8}\), \(\frac{9\pi }{8}\) A1 N3[3 marks](d) (i) \(y = {{\text{e}}^{ – x}}\sin 4x\)\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = – {{\text{e}}^{ – x}}\sin 4x + 4{{\text{e}}^{ – x}}\cos 4x\) M1A1\(y = {{\text{e}}^{ – x}}\)\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = – {{\text{e}}^{ – x}}\) A1verifying equality of gradients at one point R1verifying at the other two R1(ii) since \(\frac{{{\text{d}}y}}{{{\text{d}}x}} \ne 0\) at these points they cannot be local maxima R1[6 marks](e) (i) maximum when \(y’ = 4{{\text{e}}^{ – x}}\cos 4x – {{\text{e}}^{ – x}}\sin 4x = 0\) M1\(x = \frac{{\arctan \left( 4 \right)}}{4}\), \(\frac{{\arctan \left( 4 \right) + \pi }}{4}\), \(\frac{{\arctan \left( 4 \right) + 2\pi }}{4}\), …maxima occur at\(x = \frac{{\arctan \left( 4 \right)}}{4}\), \(\frac{{\arctan \left( 4 \right) + 2\pi }}{4}\), \(\frac{{\arctan \left( 4 \right) + 4\pi }}{4}\) A1so \({y_1} = {{\text{e}}^{ – \frac{1}{4}\left( {\arctan \left( 4 \right)} \right)}}\sin \left( {\arctan \left( 4 \right)} \right)\) (\( = 0.696\)) A1\({y_2} = {{\text{e}}^{ – \frac{1}{4}\left( {\arctan \left( 4 \right) + 2\pi } \right)}}\sin \left( {\arctan \left( 4 \right) + 2\pi } \right)\) A1\(\left( { = {{\text{e}}^{ – \frac{1}{4}\left( {\arctan \left( 4 \right) + 2\pi } \right)}}\sin \left( {\arctan \left( 4 \right)} \right) = 0.145} \right)\)\({y_3} = {{\text{e}}^{ – \frac{1}{4}\left( {\arctan \left( 4 \right) + 4\pi } \right)}}\sin \left( {\arctan \left( 4 \right) + 4\pi } \right)\) A1\(\left( { = {{\text{e}}^{ – \frac{1}{4}\left( {\arctan \left( 4 \right) + 4\pi } \right)}}\sin \left( {\arctan \left( 4 \right)} \right) = 0.0301} \right)\) N3(ii) for finding and comparing \(\frac{{{y_3}}}{{{y_2}}}\) and \(\frac{{{y_2}}}{{{y_1}}}\) M1\(r = {{\text{e}}^{ – \frac{\pi }{2}}}\) A1Note: Exact values must be used to gain the M1 and the A1.[7 marks]Total [22 marks]Examiners report
Although the final question on the paper it had parts accessible even to the weakest candidates. The vast majority of candidates earned marks on part (a), although some graphs were rather scruffy. Many candidates also tackled parts (b), (c) and (d). In part (b), however, as the answer was given, it should have been clear that some working was required rather than reference to a graph, which often had no scale indicated. In part d(i), although the functions were usually differentiated correctly, it was often the case that only one point was checked for the equality of the gradients. In part e(i) many candidates who got this far were able to determine the \(y\)-coordinates of the local maxima numerically using a GDC, and that was given credit. Only the exact values, however, could be used in part e(ii).
Question
The diagram below shows two concentric circles with centre O and radii 2 cm and 4 cm.
The points P and Q lie on the larger circle and \({\rm{P}}\hat {\text{O}}{\text{Q}} = x\) , where \(0 < x < \frac{\pi }{2}\) .
(a) Show that the area of the shaded region is \(8\sin x – 2x\) .
(b) Find the maximum area of the shaded region.
Answer/Explanation
Markscheme
(a) shaded area area of triangle area of sector, i.e. (M1)
\(\left( {\frac{1}{2} \times {4^2}\sin x} \right) – \left( {\frac{1}{2}{2^2}x} \right) = 8\sin x – 2x\) A1A1AG
(b) EITHER
any method from GDC gaining \(x \approx 1.32\) (M1)(A1)
maximum value for given domain is \(5.11\) A2
OR
\(\frac{{{\text{d}}A}}{{{\text{d}}x}} = 8\cos x – 2\) A1
set \(\frac{{{\text{d}}A}}{{{\text{d}}x}} = 0\), hence \(8\cos x – 2 = 0\) M1
\(\cos x = \frac{1}{4} \Rightarrow x \approx 1.32\) A1
hence \({A_{\max }} = 5.11\) A1
[7 marks]
Examiners report
Generally a well answered question.
Question
The function f is defined by
\[f(x) = {({x^3} + 6{x^2} + 3x – 10)^{\frac{1}{2}}},{\text{ for }}x \in D,\]
where \(D \subseteq \mathbb{R}\) is the greatest possible domain of f.
(a) Find the roots of \(f(x) = 0\).
(b) Hence specify the set D.
(c) Find the coordinates of the local maximum on the graph \(y = f(x)\).
(d) Solve the equation \(f(x) = 3\).
(e) Sketch the graph of \(\left| y \right| = f(x),{\text{ for }}x \in D\).
(f) Find the area of the region completely enclosed by the graph of \(\left| y \right| = f(x)\)
Answer/Explanation
Markscheme
(a) solving to obtain one root: 1, – 2 or – 5 A1
obtain other roots A1
[2 marks]
(b) \(D = x \in [ – 5,{\text{ }} – 2] \cup [1,{\text{ }}\infty {\text{)}}\) (or equivalent) M1A1
Note: M1 is for 1 finite and 1 infinite interval.
[2 marks]
(c) coordinates of local maximum \( – 3.73 – 2 – \sqrt 3 ,{\text{ }}3.22\sqrt {6\sqrt 3 } \) A1A1
[2 marks]
(d) use GDC to obtain one root: 1.41, – 3.18 or – 4.23 A1
obtain other roots A1
[2 marks]
(e)
A1A1A1
Note: Award A1 for shape, A1 for max and for min clearly in correct places, A1 for all intercepts.
Award A1A0A0 if only the complete top half is shown.
[3 marks]
(f) required area is twice that of \(y = f(x)\) between – 5 and – 2 M1A1
answer 14.9 A1 N3
Note: Award M1A0A0 for \(\int_{ – 5}^{ – 2} {f(x){\text{d}}x = 7.47 \ldots } \) or N1 for 7.47.
[3 marks]
Total [14 marks]
Examiners report
This was a multi-part question that was well answered by many candidates. The main difficulty was sketching the graph and this meant that the last part was not well answered.
Question
Let \(f(x) = \frac{{a + b{{\text{e}}^x}}}{{a{{\text{e}}^x} + b}}\), where \(0 < b < a\).
(a) Show that \(f'(x) = \frac{{({b^2} – {a^2}){{\text{e}}^x}}}{{{{(a{{\text{e}}^x} + b)}^2}}}\).
(b) Hence justify that the graph of f has no local maxima or minima.
(c) Given that the graph of f has a point of inflexion, find its coordinates.
(d) Show that the graph of f has exactly two asymptotes.
(e) Let a = 4 and b =1. Consider the region R enclosed by the graph of \(y = f(x)\), the y-axis and the line with equation \(y = \frac{1}{2}\).
Find the volume V of the solid obtained when R is rotated through \(2\pi \) about the x-axis.
Answer/Explanation
Markscheme
(a) \(f'(x) = \frac{{b{{\text{e}}^x}(a{{\text{e}}^x} + b) – a{{\text{e}}^x}(a + b{{\text{e}}^x})}}{{{{(a{{\text{e}}^x} + b)}^2}}}\) M1A1
\( = \frac{{ab{{\text{e}}^{2x}} + {b^2}{{\text{e}}^x} – {a^2}{{\text{e}}^x} – ab{{\text{e}}^{2x}}}}{{{{(a{{\text{e}}^x} + b)}^2}}}\) A1
\( = \frac{{({b^2} – {a^2}){{\text{e}}^x}}}{{{{(a{{\text{e}}^x} + b)}^2}}}\) AG
[3 marks]
(b) EITHER
\(f'(x) = 0 \Rightarrow ({b^2} – {a^2}){{\text{e}}^x} = 0 \Rightarrow b = \pm a{\text{ or }}{{\text{e}}^x} = 0\) A1
which is impossible as \(0 < b < a\) and \({{\text{e}}^x} > 0\) for all \(x \in \mathbb{R}\) R1
OR
\(f'(x) < 0\) for all \(x \in \mathbb{R}\) since \(0 < b < a\) and \({{\text{e}}^x} > 0\) for all \(x \in \mathbb{R}\) A1R1
OR
\(f'(x)\) cannot be equal to zero because \({{\text{e}}^x}\) is never equal to zero A1R1
[2 marks]
(c) EITHER
\(f”(x) = \frac{{({b^2} – {a^2}){{\text{e}}^x}{{(a{{\text{e}}^x} + b)}^2} – 2a{{\text{e}}^x}(a{{\text{e}}^x} + b)({b^2} – {a^2}){{\text{e}}^x}}}{{{{(a{{\text{e}}^x} + b)}^4}}}\) M1A1A1
Note: Award A1 for each term in the numerator.
\( = \frac{{({b^2} – {a^2}){{\text{e}}^x}(a{{\text{e}}^x} + b – 2a{{\text{e}}^x})}}{{{{(a{{\text{e}}^x} + b)}^3}}}\)
\( = \frac{{({b^2} – {a^2})(b – a{{\text{e}}^x}){{\text{e}}^x}}}{{{{(a{{\text{e}}^x} + b)}^3}}}\)
OR
\(f'(x) = ({b^2} – {a^2}){{\text{e}}^x}{(a{{\text{e}}^x} + b)^{ – 2}}\)
\(f”(x) = ({b^2} – {a^2}){{\text{e}}^x}{(a{{\text{e}}^x} + b)^{ – 2}} + ({b^2} – {a^2}){{\text{e}}^x}( – 2a{{\text{e}}^x}){(a{{\text{e}}^x} + b)^{ – 3}}\) M1A1A1
Note: Award A1 for each term.
\( = ({b^2} – {a^2}){{\text{e}}^x}{(a{{\text{e}}^x} + b)^{ – 3}}\left( {(a{{\text{e}}^x} + b) – 2a{{\text{e}}^x}} \right)\)
\( = ({b^2} – {a^2}){{\text{e}}^x}{(a{{\text{e}}^x} + b)^{ – 3}}(b – a{{\text{e}}^x})\)
THEN
\(f”(x) = 0 \Rightarrow b – a{{\text{e}}^x} = 0 \Rightarrow x = \ln \frac{b}{a}\) M1A1
\(f\left( {\ln \frac{b}{a}} \right) = \frac{{{a^2} + {b^2}}}{{2ab}}\) A1
coordinates are \(\left( {\ln \frac{b}{a},\frac{{{a^2} + {b^2}}}{{2ab}}} \right)\)
[6 marks]
(d) \(\mathop {\lim }\limits_{x – \infty } f(x) = \frac{a}{b} \Rightarrow y = \frac{a}{b}\) horizontal asymptote A1
\(\mathop {\lim }\limits_{x \to + \infty } f(x) = \frac{b}{a} \Rightarrow y = \frac{b}{a}\) horizontal asymptote A1
\(0 < b < a \Rightarrow a{{\text{e}}^x} + b > 0\) for all \(x \in \mathbb{R}\) (accept \(a{{\text{e}}^x} + b \ne 0\))
so no vertical asymptotes R1
Note: Statement on vertical asymptote must be seen for R1.
[3 marks]
(e) \(y = \frac{{4 + {{\text{e}}^x}}}{{4{{\text{e}}^x} + 1}}\)
\(y = \frac{1}{2} \Leftrightarrow x = \ln \frac{7}{2}\) (or 1.25 to 3 sf) (M1)(A1)
\(V = \pi \int_0^{\ln \frac{7}{2}} {\left( {{{\left( {\frac{{4 + {{\text{e}}^x}}}{{4{{\text{e}}^x} + 1}}} \right)}^2} – \frac{1}{4}} \right){\text{d}}x} \) (M1)A1
\( = 1.09\) (3 sf) A1 N4
[5 marks]
Total [19 marks]
Examiners report
This question was well attempted by many candidates. In some cases, candidates who skipped other questions still answered, with some success, parts of this question. Part (a) was in general well done but in (b) candidates found difficulty in justifying that f’(x) was non-zero. Performance in part (c) was mixed: it was pleasing to see good levels of algebraic ability of good candidates who successfully answered this question; weaker candidates found the simplification required difficult. There were very few good answers to part (d) which showed the weaknesses of most candidates in dealing with the concept of asymptotes. In part (e) there were a large number of good attempts, with many candidates evaluating correctly the limits of the integral and a smaller number scoring full marks in this part.
Question
Consider the function \(f(x) = {x^3} – 3{x^2} – 9x + 10\) , \(x \in \mathbb{R}\).
Find the equation of the straight line passing through the maximum and minimum points of the graph \(y = f (x)\) .
Show that the point of inflexion of the graph \(y = f (x)\) lies on this straight line.
Answer/Explanation
Markscheme
\(f'(x) = 3{x^2} – 6x – 9\) (\(= 0\)) (M1)
\(\left( {x + 1} \right)\left( {x – 3} \right) = 0\)
\(x = – 1\); \(x = 3\)
(max) (−1, 15); (min) (3, −17) A1A1
Note: The coordinates need not be explicitly stated but the values need to be seen.
\(y = – 8x + 7\) A1 N2
[4 marks]
\(f”(x) = 6x – 6 = 0 \Rightarrow \) inflexion (1, −1) A1
which lies on \(y = – 8x + 7\) R1AG
[2 marks]
Examiners report
There were a significant number of completely correct answers to this question. Many candidates demonstrated a good understanding of basic differential calculus in the context of coordinate geometry whilst other used technology to find the turning points.
There were a significant number of completely correct answers to this question. Many candidates demonstrated a good understanding of basic differential calculus in the context of coordinate geometry whilst other used technology to find the turning points. There were many correct demonstrations of the “show that” in (b).
Question
A triangle is formed by the three lines \(y = 10 – 2x,{\text{ }}y = mx\) and \(y = -\frac{1}{m}x\), where \(m > \frac{1}{2}\).
Find the value of m for which the area of the triangle is a minimum.
Answer/Explanation
Markscheme
attempt to find intersections M1
intersections are \(\left( {\frac{{10}}{{m + 2}},\frac{{10m}}{{m + 2}}} \right){\text{ and }}\left( {\frac{{10m}}{{2m – 1}}, – \frac{{10}}{{2m – 1}}} \right)\) A1A1
area of triangle \( = \frac{1}{2} \times \frac{{\sqrt {100 + 100{m^2}} }}{{(m + 2)}} \times \frac{{\sqrt {100 + 100{m^2}} }}{{(2m – 1)}}\) M1A1A1
\( = \frac{{50(1 + {m^2})}}{{(m + 2)(2m – 1)}}\)
minimum when \(m = 3\) (M1)A1
[8 marks]
Examiners report
Most candidates had difficulties with this question and did not go beyond the determination of the intersection points of the lines; in a few cases candidates set up the expression of the area, in some cases using unsimplified expressions of the coordinates.
Question
The diagram shows the plan of an art gallery a metres wide. [AB] represents a doorway, leading to an exit corridor b metres wide. In order to remove a painting from the art gallery, CD (denoted by L ) is measured for various values of \(\alpha \) , as represented in the diagram.
If \(\alpha \) is the angle between [CD] and the wall, show that \(L = \frac{a }{{\sin \alpha }} + \frac{b}{{\cos \alpha }}{\text{, }}0 < \alpha < \frac{\pi }{2}\).
If a = 5 and b = 1, find the maximum length of a painting that can be removed through this doorway.
Let a = 3k and b = k .
Find \(\frac{{{\text{d}}L}}{{{\text{d}}\alpha }}\).
Let a = 3k and b = k .
Find, in terms of k , the maximum length of a painting that can be removed from the gallery through this doorway.
Let a = 3k and b = k .
Find the minimum value of k if a painting 8 metres long is to be removed through this doorway.
Answer/Explanation
Markscheme
\(L = {\text{CA}} + {\text{AD}}\) M1
\({\text{sin}}\alpha {\text{ = }}\frac{a}{{{\text{CA}}}} \Rightarrow {\text{CA}} = \frac{a}{{\sin \alpha }}\) A1
\(\cos \alpha = \frac{b}{{{\text{AD}}}} \Rightarrow {\text{AD}} = \frac{b}{{\cos \alpha }}\) A1
\(L = \frac{a}{{\sin \alpha }} + \frac{b}{{\cos \alpha }}\) AG
[2 marks]
\(a = 5{\text{ and }}b = 1 \Rightarrow L = \frac{5}{{\sin \alpha }} + \frac{1}{{\cos \alpha }}\)
METHOD 1
(M1)
minimum from graph \( \Rightarrow L = 7.77\) (M1)A1
minimum of L gives the max length of the painting R1
[4 marks]
METHOD 2
\(\frac{{{\text{d}}L}}{{{\text{d}}\alpha }} = \frac{{ – 5\cos \alpha }}{{{{\sin }^2}\alpha }} + \frac{{\sin \alpha }}{{{{\cos }^2}\alpha }}\) (M1)
\(\frac{{{\text{d}}L}}{{{\text{d}}\alpha }} = 0 \Rightarrow \frac{{{{\sin }^3}\alpha }}{{{{\cos }^3}\alpha }} = 5 \Rightarrow \tan \alpha = \sqrt[{3{\text{ }}}]{5}{\text{ }}(\alpha = 1.0416…)\) (M1)
minimum of L gives the max length of the painting R1
maximum length = 7.77 A1
[4 marks]
\(\frac{{{\text{d}}L}}{{{\text{d}}\alpha }} = \frac{{ – 3k\cos \alpha }}{{{{\sin }^2}\alpha }} + \frac{{k\sin \alpha }}{{{{\cos }^2}\alpha }}\,\,\,\,\,{\text{(or equivalent)}}\) M1A1A1
[3 marks]
\(\frac{{{\text{d}}L}}{{{\text{d}}\alpha }} = \frac{{ – 3k{{\cos }^3}\alpha + k{{\sin }^3}\alpha }}{{{{\sin }^2}\alpha {{\cos }^2}\alpha }}\) (A1)
\(\frac{{{\text{d}}L}}{{{\text{d}}\alpha }} = 0 \Rightarrow \frac{{{{\sin }^3}\alpha }}{{{{\cos }^3}\alpha }} = \frac{{3k}}{k} \Rightarrow \tan \alpha = \sqrt[3]{3}\,\,\,\,\,(\alpha = 0.96454…)\) M1A1
\(\tan \alpha = \sqrt[3]{3} \Rightarrow \frac{1}{{\cos \alpha }} = \sqrt {1 + \sqrt[3]{9}} \,\,\,\,\,(1.755…)\) (A1)
\({\text{and }}\frac{1}{{\sin \alpha }} = \frac{{\sqrt {1 + \sqrt[3]{9}} }}{{\sqrt[3]{3}}}\,\,\,\,\,(1.216…)\) (A1)
\(L = 3k\left( {\frac{{\sqrt {1 + \sqrt[3]{9}} }}{{\sqrt[3]{3}}}} \right) + k\sqrt {1 + \sqrt[3]{9}} \,\,\,\,\,(L = 5.405598…k)\) A1 N4
[6 marks]
\(L \leqslant 8 \Rightarrow k \geqslant 1.48\) M1A1
the minimum value is 1.48
[2 marks]
Examiners report
Part (a) was very well done by most candidates. Parts (b), (c) and (d) required a subtle balance between abstraction, differentiation skills and use of GDC.
In part (b), although candidates were asked to justify their reasoning, very few candidates offered an explanation for the maximum. Therefore most candidates did not earn the R1 mark in part (b). Also not as many candidates as anticipated used a graphical approach, preferring to use the calculus with varying degrees of success. In part (c), some candidates calculated the derivatives of inverse trigonometric functions. Some candidates had difficulty with parts (d) and (e). In part (d), some candidates erroneously used their alpha value from part (b). In part (d) many candidates used GDC to calculate decimal values for \(\alpha \) and L. The premature rounding of decimals led sometimes to inaccurate results. Nevertheless many candidates got excellent results in this question.
Part (a) was very well done by most candidates. Parts (b), (c) and (d) required a subtle balance between abstraction, differentiation skills and use of GDC.
In part (b), although candidates were asked to justify their reasoning, very few candidates offered an explanation for the maximum. Therefore most candidates did not earn the R1 mark in part (b). Also not as many candidates as anticipated used a graphical approach, preferring to use the calculus with varying degrees of success. In part (c), some candidates calculated the derivatives of inverse trigonometric functions. Some candidates had difficulty with parts (d) and (e). In part (d), some candidates erroneously used their alpha value from part (b). In part (d) many candidates used GDC to calculate decimal values for \(\alpha \) and L. The premature rounding of decimals led sometimes to inaccurate results. Nevertheless many candidates got excellent results in this question.
Part (a) was very well done by most candidates. Parts (b), (c) and (d) required a subtle balance between abstraction, differentiation skills and use of GDC.
In part (b), although candidates were asked to justify their reasoning, very few candidates offered an explanation for the maximum. Therefore most candidates did not earn the R1 mark in part (b). Also not as many candidates as anticipated used a graphical approach, preferring to use the calculus with varying degrees of success. In part (c), some candidates calculated the derivatives of inverse trigonometric functions. Some candidates had difficulty with parts (d) and (e). In part (d), some candidates erroneously used their alpha value from part (b). In part (d) many candidates used GDC to calculate decimal values for \(\alpha \) and L. The premature rounding of decimals led sometimes to inaccurate results. Nevertheless many candidates got excellent results in this question.
Part (a) was very well done by most candidates. Parts (b), (c) and (d) required a subtle balance between abstraction, differentiation skills and use of GDC.
In part (b), although candidates were asked to justify their reasoning, very few candidates offered an explanation for the maximum. Therefore most candidates did not earn the R1 mark in part (b). Also not as many candidates as anticipated used a graphical approach, preferring to use the calculus with varying degrees of success. In part (c), some candidates calculated the derivatives of inverse trigonometric functions. Some candidates had difficulty with parts (d) and (e). In part (d), some candidates erroneously used their alpha value from part (b). In part (d) many candidates used GDC to calculate decimal values for \(\alpha \) and L. The premature rounding of decimals led sometimes to inaccurate results. Nevertheless many candidates got excellent results in this question.
Part (a) was very well done by most candidates. Parts (b), (c) and (d) required a subtle balance between abstraction, differentiation skills and use of GDC.
In part (b), although candidates were asked to justify their reasoning, very few candidates offered an explanation for the maximum. Therefore most candidates did not earn the R1 mark in part (b). Also not as many candidates as anticipated used a graphical approach, preferring to use the calculus with varying degrees of success. In part (c), some candidates calculated the derivatives of inverse trigonometric functions. Some candidates had difficulty with parts (d) and (e). In part (d), some candidates erroneously used their alpha value from part (b). In part (d) many candidates used GDC to calculate decimal values for \(\alpha \) and L. The premature rounding of decimals led sometimes to inaccurate results. Nevertheless many candidates got excellent results in this question.
Question
An electricity station is on the edge of a straight coastline. A lighthouse is located in the sea 200 m from the electricity station. The angle between the coastline and the line joining the lighthouse with the electricity station is 60°. A cable needs to be laid connecting the lighthouse to the electricity station. It is decided to lay the cable in a straight line to the coast and then along the coast to the electricity station. The length of cable laid along the coastline is x metres. This information is illustrated in the diagram below.
The cost of laying the cable along the sea bed is US$80 per metre, and the cost of laying it on land is US$20 per metre.
Find, in terms of x, an expression for the cost of laying the cable.
Find the value of x, to the nearest metre, such that this cost is minimized.
Answer/Explanation
Markscheme
let the distance the cable is laid along the seabed be y
\({y^2} = {x^2} + {200^2} – 2 \times x \times 200\cos 60^\circ \) (M1)
(or equivalent method)
\({y^2} = {x^2} – 200x + 40000\) (A1)
cost = C = 80y + 20x (M1)
\(C = 80{({x^2} – 200x + 40000)^{\frac{1}{2}}} + 20x\) A1
[4 marks]
\(x = 55.2786 \ldots = 55\) (m to the nearest metre) (A1)A1
\(\left( {x = 100 – \sqrt {2000} } \right)\)
[2 marks]
Examiners report
Some surprising misconceptions were evident here, using right angled trigonometry in non right angled triangles etc. Those that used the cosine rule, usually managed to obtain the correct answer to part (a).
Some surprising misconceptions were evident here, using right angled trigonometry in non right angled triangles etc. Many students attempted to find the value of the minimum algebraically instead of the simple calculator solution.
Question
A straight street of width 20 metres is bounded on its parallel sides by two vertical walls, one of height 13 metres, the other of height 8 metres. The intensity of light at point P at ground level on the street is proportional to the angle \(\theta \) where \(\theta = {\rm{A\hat PB}}\), as shown in the diagram.
Find an expression for \(\theta \) in terms of x, where x is the distance of P from the base of the wall of height 8 m.
(i) Calculate the value of \(\theta \) when x = 0.
(ii) Calculate the value of \(\theta \) when x = 20.
Sketch the graph of \(\theta \), for \(0 \leqslant x \leqslant 20\).
Show that \(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = \frac{{5(744 – 64x – {x^2})}}{{({x^2} + 64)({x^2} – 40x + 569)}}\).
Using the result in part (d), or otherwise, determine the value of x corresponding to the maximum light intensity at P. Give your answer to four significant figures.
The point P moves across the street with speed \(0.5{\text{ m}}{{\text{s}}^{ – 1}}\). Determine the rate of change of \(\theta \) with respect to time when P is at the midpoint of the street.
Answer/Explanation
Markscheme
EITHER
\(\theta = \pi – \arctan \left( {\frac{8}{x}} \right) – \arctan \left( {\frac{{13}}{{20 – x}}} \right)\) (or equivalent) M1A1
Note: Accept \(\theta = 180^\circ – \arctan \left( {\frac{8}{x}} \right) – \arctan \left( {\frac{{13}}{{20 – x}}} \right)\) (or equivalent).
OR
\(\theta = \arctan \left( {\frac{x}{8}} \right) + \arctan \left( {\frac{{20 – x}}{{13}}} \right)\) (or equivalent) M1A1
[2 marks]
(i) \(\theta = 0.994{\text{ }}\left( { = \arctan \frac{{20}}{{13}}} \right)\) A1
(ii) \(\theta = 1.19{\text{ }}\left( { = \arctan \frac{5}{2}} \right)\) A1
[2 marks]
correct shape. A1
correct domain indicated. A1
[2 marks]
attempting to differentiate one \(\arctan \left( {f(x)} \right)\) term M1
EITHER
\(\theta = \pi – \arctan \left( {\frac{8}{x}} \right) – \arctan \left( {\frac{{13}}{{20 – x}}} \right)\)
\(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = \frac{8}{{{x^2}}} \times \frac{1}{{1 + {{\left( {\frac{8}{x}} \right)}^2}}} – \frac{{13}}{{{{(20 – x)}^2}}} \times \frac{1}{{1 + {{\left( {\frac{{13}}{{20 – x}}} \right)}^2}}}\) A1A1
OR
\(\theta = \arctan \left( {\frac{x}{8}} \right) + \arctan \left( {\frac{{20 – x}}{{13}}} \right)\)
\(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = \frac{{\frac{1}{8}}}{{1 + {{\left( {\frac{x}{8}} \right)}^2}}} + \frac{{ – \frac{1}{{13}}}}{{1 + {{\left( {\frac{{20 – x}}{{13}}} \right)}^2}}}\) A1A1
THEN
\( = \frac{8}{{{x^2} + 64}} – \frac{{13}}{{569 – 40x + {x^2}}}\) A1
\( = \frac{{8(569 – 40x + {x^2}) – 13({x^{2}} + 64)}}{{({x^2} + 64)({x^2} – 40x + 569)}}\) M1A1
\( = \frac{{5(744 – 64x – {x^2})}}{{({x^2} + 64)({x^2} – 40x + 569)}}\) AG
[6 marks]
Maximum light intensity at P occurs when \(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = 0\). (M1)
either attempting to solve \(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = 0\) for x or using the graph of either \(\theta \) or \(\frac{{{\text{d}}\theta }}{{{\text{d}}x}}\) (M1)
x = 10.05 (m) A1
[3 marks]
\(\frac{{{\text{d}}x}}{{{\text{d}}t}} = 0.5\) (A1)
At x = 10, \(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = 0.000453{\text{ }}\left( { = \frac{5}{{11029}}} \right)\). (A1)
use of \(\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{{\text{d}}\theta }}{{{\text{d}}x}} \times \frac{{{\text{d}}x}}{{{\text{d}}t}}\) M1
\(\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = 0.000227{\text{ }}\left( { = \frac{5}{{22058}}} \right){\text{ (rad }}{{\text{s}}^{ – 1}})\) A1
Note: Award (A1) for \(\frac{{{\text{d}}x}}{{{\text{d}}t}} = – 0.5\) and A1 for \(\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = – 0.000227{\text{ }}\left( { = – \frac{5}{{22058}}} \right){\text{ }}\).
Note: Implicit differentiation can be used to find \(\frac{{{\text{d}}\theta }}{{{\text{d}}t}}\). Award as above.
[4 marks]
Examiners report
Part (a) was reasonably well done. While many candidates exhibited sound trigonometric knowledge to correctly express θ in terms of x, many other candidates were not able to use elementary trigonometry to formulate the required expression for θ.
In part (b), a large number of candidates did not realize that θ could only be acute and gave obtuse angle values for θ. Many candidates also demonstrated a lack of insight when substituting endpoint x-values into θ.
In part (c), many candidates sketched either inaccurate or implausible graphs.
In part (d), a large number of candidates started their differentiation incorrectly by failing to use the chain rule correctly.
For a question part situated at the end of the paper, part (e) was reasonably well done. A large number of candidates demonstrated a sound knowledge of finding where the maximum value of θ occurred and rejected solutions that were not physically feasible.
In part (f), many candidates were able to link the required rates, however only a few candidates were able to successfully apply the chain rule in a related rates context.
Question
Consider the triangle \({\text{PQR}}\) where \({\rm{Q\hat PR = 30^\circ }}\), \({\text{PQ}} = (x + 2){\text{ cm}}\) and \({\text{PR}} = {(5 – x)^2}{\text{ cm}}\), where \( – 2 < x < 5\).
Show that the area, \(A\;{\text{c}}{{\text{m}}^2}\), of the triangle is given by \(A = \frac{1}{4}({x^3} – 8{x^2} + 5x + 50)\).
(i) State \(\frac{{{\text{d}}A}}{{{\text{d}}x}}\).
(ii) Verify that \(\frac{{{\text{d}}A}}{{{\text{d}}x}} = 0\) when \(x = \frac{1}{3}\).
(i) Find \(\frac{{{{\text{d}}^2}A}}{{{\text{d}}{x^2}}}\) and hence justify that \(x = \frac{1}{3}\) gives the maximum area of triangle \(PQR\).
(ii) State the maximum area of triangle \(PQR\).
(iii) Find \(QR\) when the area of triangle \(PQR\) is a maximum.
Answer/Explanation
Markscheme
use of \(A = \frac{1}{2}qr\sin \theta \) to obtain \(A = \frac{1}{2}(x + 2){(5 – x)^2}\sin 30^\circ \) M1
\( = \frac{1}{4}(x + 2)(25 – 10x + {x^2})\) A1
\(A = \frac{1}{4}({x^3} – 8{x^2} + 5x + 50)\) AG
[2 marks]
(i) \(\frac{{{\text{d}}A}}{{{\text{d}}x}} = \frac{1}{4}(3{x^2} – 16x + 5) = \frac{1}{4}(3x – 1)(x – 5)\) A1
(ii) METHOD 1
EITHER
\(\frac{{{\text{d}}A}}{{{\text{d}}x}} = \frac{1}{4}\left( {3{{\left( {\frac{1}{3}} \right)}^2} – 16\left( {\frac{1}{3}} \right) + 5} \right) = 0\) M1A1
OR
\(\frac{{{\text{d}}A}}{{{\text{d}}x}} = \frac{1}{4}\left( {3\left( {\frac{1}{3}} \right) – 1} \right)\left( {\left( {\frac{1}{3}} \right) – 5} \right) = 0\) M1A1
THEN
so \(\frac{{{\text{d}}A}}{{{\text{d}}x}} = 0\) when \(x = \frac{1}{3}\) AG
METHOD 2
solving \(\frac{{{\text{d}}A}}{{{\text{d}}x}} = 0\) for \(x\) M1
\( – 2 < x < 5 \Rightarrow x = \frac{1}{3}\) A1
so \(\frac{{{\text{d}}A}}{{{\text{d}}x}} = 0\) when \(x = \frac{1}{3}\) AG
METHOD 3
a correct graph of \(\frac{{{\text{d}}A}}{{{\text{d}}x}}\) versus \(x\) M1
the graph clearly showing that \(\frac{{{\text{d}}A}}{{{\text{d}}x}} = 0\) when \(x = \frac{1}{3}\) A1
so \(\frac{{{\text{d}}A}}{{{\text{d}}x}} = 0\) when \(x = \frac{1}{3}\) AG
[3 marks]
(i) \(\frac{{{{\text{d}}^2}A}}{{{\text{d}}{x^2}}} = \frac{1}{2}(3x – 8)\) A1
for \(x = \frac{1}{3},{\text{ }}\frac{{{{\text{d}}^2}A}}{{{\text{d}}{x^2}}} = – 3.5{\text{ }}( < 0)\) R1
so \(x = \frac{1}{3}\) gives the maximum area of triangle \(PQR\) AG
(ii) \({A_{\max }} = \frac{{343}}{{27}}{\text{ }}( = 12.7){\text{ (c}}{{\text{m}}^2}{\text{)}}\) A1
(iii) \({\text{PQ}} = \frac{7}{3}{\text{ (cm)}}\) and \({\text{PR}} = {\left( {\frac{{14}}{3}} \right)^2}{\text{ (cm)}}\) (A1)
\({\text{Q}}{{\text{R}}^2} = {\left( {\frac{7}{3}} \right)^2} + {\left( {\frac{{14}}{3}} \right)^4} – 2\left( {\frac{7}{3}} \right){\left( {\frac{{14}}{3}} \right)^2}\cos 30^\circ \) (M1)(A1)
\( = 391.702 \ldots \)
\({\text{QR = 19.8 (cm)}}\) A1
[7 marks]
Total [12 marks]
Examiners report
This question was generally well done. Parts (a) and (b) were straightforward and well answered.
This question was generally well done. Parts (a) and (b) were straightforward and well answered.
This question was generally well done. Parts (c) (i) and (ii) were also well answered with most candidates correctly applying the second derivative test and displaying sound reasoning skills.
Part (c) (iii) required the use of the cosine rule and was reasonably well done. The most common error committed by candidates in attempting to find the value of \(QR\) was to use \({\text{PR}} = \frac{{14}}{3}{\text{ (cm)}}\) rather than \({\text{PR}} = {\left( {\frac{{14}}{3}} \right)^2}{\text{ (cm)}}\). The occasional candidate used \(\cos 30^\circ = \frac{1}{2}\).
Question
The following diagram shows a vertical cross section of a building. The cross section of the roof of the building can be modelled by the curve \(f(x) = 30{{\text{e}}^{ – \frac{{{x^2}}}{{400}}}}\), where \( – 20 \le x \le 20\).
Ground level is represented by the \(x\)-axis.
Find \(f”(x)\).
Show that the gradient of the roof function is greatest when \(x = – \sqrt {200} \).
The cross section of the living space under the roof can be modelled by a rectangle \(CDEF\) with points \({\text{C}}( – a,{\text{ }}0)\) and \({\text{D}}(a,{\text{ }}0)\), where \(0 < a \le 20\).
Show that the maximum area \(A\) of the rectangle \(CDEF\) is \(600\sqrt 2 {{\text{e}}^{ – \frac{1}{2}}}\).
A function \(I\) is known as the Insulation Factor of \(CDEF\). The function is defined as \(I(a) = \frac{{P(a)}}{{A(a)}}\) where \({\text{P}} = {\text{Perimeter}}\) and \({\text{A}} = {\text{Area of the rectangle}}\).
(i) Find an expression for \(P\) in terms of \(a\).
(ii) Find the value of \(a\) which minimizes \(I\).
(iii) Using the value of \(a\) found in part (ii) calculate the percentage of the cross sectional area under the whole roof that is not included in the cross section of the living space.
Answer/Explanation
Markscheme
\(f'(x) = 30{{\text{e}}^{ – \frac{{{x^2}}}{{400}}}} \bullet – \frac{{2x}}{{400}}\;\;\;\left( { = – \frac{{3x}}{{20}}{{\text{e}}^{ – \frac{{{x^2}}}{{400}}}}} \right)\) M1A1
Note: Award M1 for attempting to use the chain rule.
\(f”(x) = – \frac{3}{{20}}{{\text{e}}^{ – \frac{{{x^2}}}{{400}}}} + \frac{{3{x^2}}}{{4000}}{{\text{e}}^{ – \frac{{{x^2}}}{{400}}}}\;\;\;\left( { = \frac{3}{{20}}{{\text{e}}^{ – \frac{{{x^2}}}{{400}}}}\left( {\frac{{{x^2}}}{{200}} – 1} \right)} \right)\) M1A1
Note: Award M1 for attempting to use the product rule.
[4 marks]
the roof function has maximum gradient when \(f”(x) = 0\) (M1)
Note: Award (M1) for attempting to find \(f”\left( { – \sqrt {200} } \right)\).
EITHER
\( = 0\) A1
OR
\(f”(x) = 0 \Rightarrow x = \pm \sqrt {200} \) A1
THEN
valid argument for maximum such as reference to an appropriate graph or change in the sign of \(f”(x)\) eg \(f”( – 15) = 0.010 \ldots ( > 0)\) and \(f”( – 14) = – 0.001 \ldots ( < 0)\) R1
\( \Rightarrow x = – \sqrt {200} \) AG
[3 marks]
\(A = 2a \bullet 30{{\text{e}}^{ – \frac{{{a^2}}}{{400}}}}\;\;\;\left( { = 60a{{\text{e}}^{ – \frac{{{a^2}}}{{400}}}} = – 400g'(a)} \right)\) (M1)(A1)
EITHER
\(\frac{{{\text{d}}A}}{{{\text{d}}a}} = 60a{{\text{e}}^{ – \frac{{{a^2}}}{{400}}}} \bullet – \frac{a}{{200}} + 60{{\text{e}}^{ – \frac{{{a^2}}}{{400}}}} = 0 \Rightarrow a = \sqrt {200} {\text{ }}\left( { – 400f”(a) = 0 \Rightarrow a = \sqrt {200} } \right)\) M1A1
OR
by symmetry eg \(a = – \sqrt {200} \) found in (b) or \({A_{{\text{max}}}}\) coincides with \(f”(a) = 0\) R1
\( \Rightarrow a = \sqrt {200} \) A1
Note: Award A0(M1)(A1)M0M1 for candidates who start with \(a = \sqrt {200} \) and do not provide any justification for the maximum area. Condone use of \(x\).
THEN
\({A_{{\text{max}}}} = 60 \bullet \sqrt {200} {{\text{e}}^{ – \frac{{200}}{{400}}}}\) M1
\( = 600\sqrt 2 {{\text{e}}^{ – \frac{1}{2}}}\) AG
[5 marks]
(i) perimeter \( = 4a + 60{{\text{e}}^{ – \frac{{{a^2}}}{{400}}}}\) A1A1
Note: Condone use of \(x\).
(ii) \(I(a) = \frac{{4a + 60{{\text{e}}^{ – \frac{{{a^2}}}{{400}}}}}}{{60a{{\text{e}}^{ – \frac{{{a^2}}}{{400}}}}}}\) (A1)
graphing \(I(a)\) or other valid method to find the minimum (M1)
\(a = 12.6\) A1
(iii) area under roof \( = \int_{ – 20}^{20} {30{{\text{e}}^{ – \frac{{{x^2}}}{{400}}}}} {\text{d}}x\) M1
\( = 896.18 \ldots \) (A1)
area of living space \( = 60 \cdot (12.6…) \cdot e – {\frac{{(12.6…)}}{{400}}^2} = 508.56…\)
percentage of empty space \( = 43.3\% \) A1
[9 marks]
Total [21 marks]
Examiners report
[N/A]
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[N/A]
Question
Let \(f(x) = {x^4} + 0.2{x^3} – 5.8{x^2} – x + 4,{\text{ }}x \in \mathbb{R}\).
The domain of \(f\) is now restricted to \([0,{\text{ }}a]\).
Let \(g(x) = 2\sin (x – 1) – 3,{\text{ }} – \frac{\pi }{2} + 1 \leqslant x \leqslant \frac{\pi }{2} + 1\).
Find the solutions of \(f(x) > 0\).
For the curve \(y = f(x)\).
(i) Find the coordinates of both local minimum points.
(ii) Find the \(x\)-coordinates of the points of inflexion.
Write down the largest value of \(a\) for which \(f\) has an inverse. Give your answer correct to 3 significant figures.
For this value of a sketch the graphs of \(y = f(x)\) and \(y = {f^{ – 1}}(x)\) on the same set of axes, showing clearly the coordinates of the end points of each curve.
Solve \({f^{ – 1}}(x) = 1\).
Find an expression for \({g^{ – 1}}(x)\), stating the domain.
Solve \(({f^{ – 1}} \circ g)(x) < 1\).
Answer/Explanation
Markscheme
valid method eg, sketch of curve or critical values found (M1)
\(x < – 2.24,{\text{ }}x > 2.24,\) A1
\( – 1 < x < 0.8\) A1
Note: Award M1A1A0 for correct intervals but with inclusive inequalities.
[3 marks]
(i) \((1.67,{\text{ }} – 5.14),{\text{ }}( – 1.74,{\text{ }} – 3.71)\) A1A1
Note: Award A1A0 for any two correct terms.
(ii) \(f'(x) = 4{x^3} + 0.6{x^2} – 11.6x – 1\)
\(f”(x) = 12{x^2} + 1.2x – 11.6 = 0\) (M1)
\( – 1.03,{\text{ }}0.934\) A1A1
Note: M1 should be awarded if graphical method to find zeros of \(f”(x)\) or turning points of \(f'(x)\) is shown.
[5 marks]
1.67 A1
[2 marks]
M1A1A1
Note: Award M1 for reflection of their \(y = f(x)\) in the line \(y = x\) provided their \(f\) is one-one.
A1 for \((0,{\text{ }}4)\), \((4,{\text{ }}0)\) (Accept axis intercept values) A1 for the other two sets of coordinates of other end points
[2 marks]
\(x = f(1)\) M1
\( = – 1.6\) A1
[2 marks]
\(y = 2\sin (x – 1) – 3\)
\(x = 2\sin (y – 1) – 3\) (M1)
\(\left( {{g^{ – 1}}(x) = } \right){\text{ }}\arcsin \left( {\frac{{x + 3}}{2}} \right) + 1\) A1
\( – 5 \leqslant x \leqslant – 1\) A1A1
Note: Award A1 for −5 and −1, and A1 for correct inequalities if numbers are reasonable.
[8 marks]
\({f^{ – 1}}\left( {g(x)} \right) < 1\)
\(g(x) > – 1.6\) (M1)
\(x > {g^{ – 1}}( – 1.6) = 1.78\) (A1)
Note: Accept = in the above.
\(1.78 < x \leqslant \frac{\pi }{2} + 1\) A1A1
Note: A1 for \(x > 1.78\) (allow ≥) and A1 for \(x \leqslant \frac{\pi }{2} + 1\).
[4 marks]
Examiners report
Parts (a) and (b) were well answered, with considerably less success in part (c). Surprisingly few students were able to reflect the curve in \(y = x\) satisfactorily, and many were not making their sketch using the correct domain.
Parts (a) and (b) were well answered, with considerably less success in part (c). Surprisingly few students were able to reflect the curve in \(y = x\) satisfactorily, and many were not making their sketch using the correct domain.
Parts (a) and (b) were well answered, with considerably less success in part (c). Surprisingly few students were able to reflect the curve in \(y = x\) satisfactorily, and many were not making their sketch using the correct domain.
Parts (a) and (b) were well answered, with considerably less success in part (c). Surprisingly few students were able to reflect the curve in \(y = x\) satisfactorily, and many were not making their sketch using the correct domain.
Parts (a) and (b) were well answered, with considerably less success in part (c). Surprisingly few students were able to reflect the curve in \(y = x\) satisfactorily, and many were not making their sketch using the correct domain.
Part d(i) was generally well done, but there were few correct answers for d(ii).
Part d(i) was generally well done, but there were few correct answers for d(ii).
Question
The following graph shows the two parts of the curve defined by the equation \({x^2}y = 5 – {y^4}\), and the normal to the curve at the point P(2 , 1).
Show that there are exactly two points on the curve where the gradient is zero.
Find the equation of the normal to the curve at the point P.
The normal at P cuts the curve again at the point Q. Find the \(x\)-coordinate of Q.
The shaded region is rotated by 2\(\pi \) about the \(y\)-axis. Find the volume of the solid formed.
Answer/Explanation
Markscheme
differentiating implicitly: M1
\(2xy + {x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} = – 4{y^3}\frac{{{\text{d}}y}}{{{\text{d}}x}}\) A1A1
Note: Award A1 for each side.
if \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) then either \(x = 0\) or \(y = 0\) M1A1
\(x = 0 \Rightarrow \) two solutions for \(y\left( {y = \pm \sqrt[4]{5}} \right)\) R1
\(y = 0\) not possible (as 0 ≠ 5) R1
hence exactly two points AG
Note: For a solution that only refers to the graph giving two solutions at \(x = 0\) and no solutions for \(y = 0\) award R1 only.
[7 marks]
at (2, 1) \(4 + 4\frac{{{\text{d}}y}}{{{\text{d}}x}} = – 4\frac{{{\text{d}}y}}{{{\text{d}}x}}\) M1
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = – \frac{1}{2}\) (A1)
gradient of normal is 2 M1
1 = 4 + c (M1)
equation of normal is \(y = 2x – 3\) A1
[5 marks]
substituting (M1)
\({x^2}\left( {2x – 3} \right) = 5 – {\left( {2x – 3} \right)^4}\) or \({\left( {\frac{{y + 3}}{2}} \right)^2}\,y = 5 – {y^4}\) (A1)
\(x = 0.724\) A1
[3 marks]
recognition of two volumes (M1)
volume \(1 = \pi \int_1^{\sqrt[4]{5}} {\frac{{5 – {y^4}}}{y}} {\text{d}}y\left( { = 101\pi = 3.178 \ldots } \right)\) M1A1A1
Note: Award M1 for attempt to use \(\pi \int {{x^2}} {\text{d}}y\), A1 for limits, A1 for \({\frac{{5 – {y^4}}}{y}}\) Condone omission of \(\pi \) at this stage.
volume 2
EITHER
\( = \frac{1}{3}\pi \times {2^2} \times 4\left( { = 16.75 \ldots } \right)\) (M1)(A1)
OR
\( = \pi \int_{ – 3}^1 {{{\left( {\frac{{y + 3}}{2}} \right)}^2}} {\text{d}}y\left( { = \frac{{16\pi }}{3} = 16.75 \ldots } \right)\) (M1)(A1)
THEN
total volume = 19.9 A1
[7 marks]
Examiners report
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