IB DP Maths Topic 2.7 Solving equations, both graphically and analytically SL Paper 2

 

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Question

The following diagram shows the graphs of \(f(x) = \ln (3x – 2) + 1\) and \(g(x) = – 4\cos (0.5x) + 2\) , for \(1 \le x \le 10\) .


Let A be the area of the region enclosed by the curves of f and g.

(i)     Find an expression for A.

(ii)    Calculate the value of A.

[6]
a(i) and (ii).

(i)     Find \(f'(x)\) .

(ii)    Find \(g'(x)\) .

[4]
b(i) and (ii).

There are two values of x for which the gradient of f is equal to the gradient of g. Find both these values of x.

[4]
c.
Answer/Explanation

Markscheme

(i) intersection points \(x = 3.77\) , \(x = 8.30\) (may be seen as the limits)     (A1)(A1)

approach involving subtraction and integrals     (M1)

fully correct expression     A2

e.g. \(\int_{3.77}^{8.30} {(( – 4\cos (0.5x) + 2) – (\ln (3x – 2) + 1)){\rm{d}}x} \) , \(\int_{3.77}^{8.30} {g(x){\rm{d}}x – } \int_{3.77}^{8.30} {f(x){\rm{d}}x} \)

(ii) \(A = 6.46\)     A1     N1

[6 marks]

a(i) and (ii).

(i) \(f'(x) = \frac{3}{{3x – 2}}\)     A1A1     N2

Note: Award A1 for numerator (3), A1 for denominator (\({3x – 2}\)) , but penalize 1 mark for additional terms.

 

(ii) \(g'(x) = 2\sin (0.5x)\)     A1A1     N2

Note: Award A1 for 2, A1 for \(\sin (0.5x)\) , but penalize 1 mark for additional terms.

[4 marks]

b(i) and (ii).

evidence of using derivatives for gradients     (M1)

correct approach     (A1)

e.g. \(f'(x) = g'(x)\) , points of intersection

\(x = 1.43\) , \(x = 6.10\)     A1A1     N2N2

[4 marks]

c.

Question

A city is concerned about pollution, and decides to look at the number of people using taxis. At the end of the year 2000, there were 280 taxis in the city. After n years the number of taxis, T, in the city is given by\[T = 280 \times {1.12^n} .\]

(i)     Find the number of taxis in the city at the end of 2005.

(ii)    Find the year in which the number of taxis is double the number of taxis there were at the end of 2000.

[6]
a(i) and (ii).

At the end of 2000 there were \(25600\) people in the city who used taxis.

After n years the number of people, P, in the city who used taxis is given by\[P = \frac{{2560000}}{{10 + 90{{\rm{e}}^{ – 0.1n}}}} .\](i)     Find the value of P at the end of 2005, giving your answer to the nearest whole number.

(ii)    After seven complete years, will the value of P be double its value at the end of 2000? Justify your answer.

[6]
b(i) and (ii).

Let R be the ratio of the number of people using taxis in the city to the number of taxis. The city will reduce the number of taxis if \(R < 70\) .

(i)     Find the value of R at the end of 2000.

(ii)    After how many complete years will the city first reduce the number of taxis?

[5]
c(i) and (ii).
Answer/Explanation

Markscheme

(i) \(n = 5\)     (A1)

\(T = 280 \times {1.12^5}\)

\(T = 493\)     A1     N2

(ii) evidence of doubling     (A1)

e.g. 560

setting up equation     A1

e.g. \(280 \times {1.12^n} = 560\), \({1.12^n} = 2\)

\(n = 6.116 \ldots \)     (A1)

in the year 2007     A1     N3

[6 marks]

a(i) and (ii).

(i) \(P = \frac{{2560000}}{{10 + 90{{\rm{e}}^{ – 0.1(5)}}}}\)     (A1)

\(P = 39635.993 \ldots \)     (A1)

\(P = 39636\)     A1     N3

(ii) \(P = \frac{{2560000}}{{10 + 90{{\rm{e}}^{ – 0.1(7)}}}}\)

\(P = 46806.997 \ldots \)     A1

not doubled     A1     N0

valid reason for their answer     R1

e.g. \(P < 51200\)

[6 marks]

b(i) and (ii).

(i) correct value     A2     N2

e.g. \(\frac{{25600}}{{280}}\) , 91.4, \(640:7\)

(ii) setting up an inequality (accept an equation, or reversed inequality)     M1

e.g. \(\frac{P}{T} < 70\) , \(\frac{{2560000}}{{(10 + 90{{\rm{e}}^{ – 0.1n}})280 \times {{1.12}^n}}} < 70\)

finding the value \(9.31 \ldots \)     (A1)

after 10 years     A1     N2

[5 marks]

c(i) and (ii).

Question

In a geometric series, \({u_1} = \frac{1}{{81}}\) and \({u_4} = \frac{1}{3}\) .

Find the value of \(r\) .

[3]
a.

Find the smallest value of n for which \({S_n} > 40\) .

[4]
b.
Answer/Explanation

Markscheme

evidence of substituting into formula for \(n\)th term of GP     (M1)

e.g. \({u_4} = \frac{1}{{81}}{r^3}\)

setting up correct equation \(\frac{1}{{81}}{r^3} = \frac{1}{3}\)     A1

\(r = 3\)     A1     N2

[3 marks]

a.

METHOD 1

setting up an inequality (accept an equation)     M1

e.g. \(\frac{{\frac{1}{{81}}({3^n} – 1)}}{2} > 40\) , \(\frac{{\frac{1}{{81}}(1 – {3^n})}}{{ – 2}} > 40\) , \({3^n} > 6481\)

evidence of solving     M1

e.g. graph, taking logs

\(n > 7.9888 \ldots \)     (A1)

\(\therefore n = 8\)     A1     N2

METHOD 2

if \(n = 7\) , sum \( = 13.49 \ldots \) ; if \(n = 8\) , sum \( = 40.49 \ldots \)     A2

\(n = 8\) (is the smallest value)     A2     N2

[4 marks]

b.

Question

Solve the equation \({{\rm{e}}^x} = 4\sin x\) , for \(0 \le x \le 2\pi \) .

Answer/Explanation

Markscheme

evidence of appropriate approach     M1

e.g. a sketch, writing \({{\rm{e}}^x} – 4\sin x = 0\) 

\(x = 0.371\) , \(x = 1.36\)     A2A2     N2N2

[5 marks]

Question

Let \(f(x) = 3\sin x + 4\cos x\) , for \( – 2\pi  \le x \le 2\pi \) .

Sketch the graph of f .

[3]
a.

Write down

(i)     the amplitude;

(ii)    the period;

(iii)   the x-intercept that lies between \( – \frac{\pi }{2}\) and 0.

[3]
b.

Hence write \(f(x)\) in the form \(p\sin (qx + r)\) .

[3]
c.

Write down one value of x such that \(f'(x) = 0\) .

[2]
d.

Write down the two values of k for which the equation \(f(x) = k\) has exactly two solutions.

[2]
e.

Let \(g(x) = \ln (x + 1)\) , for \(0 \le x \le \pi \) . There is a value of x, between \(0\) and \(1\), for which the gradient of f is equal to the gradient of g. Find this value of x.

[5]
f.
Answer/Explanation

Markscheme

 


     A1A1A1     N3

 

Note: Award A1 for approximately sinusoidal shape, A1 for end points approximately correct \(( – 2\pi {\text{, }}4)\) \((2\pi {\text{, }}4)\), A1 for approximately correct position of graph, (y-intercept \((0{\text{, }}4)\), maximum to right of y-axis).

[3 marks]

a.

(i) 5     A1     N1

(ii) \(2\pi \)  (6.28)     A1     N1

(iii) \( – 0.927\)     A1     N1

[3 marks]

b.

\(f(x) = 5\sin (x + 0.927)\) (accept \(p = 5\) , \(q = 1\) , \(r = 0.927\) )     A1A1A1     N3

[3 marks]

c.

evidence of correct approach     (M1)

e.g. max/min, sketch of \(f'(x)\) indicating roots


one 3 s.f. value which rounds to one of \( – 5.6\), \( – 2.5\), \(0.64\), \(3.8\)     A1     N2

 

[2 marks]

 

 

d.

\(k = – 5\) , \(k = 5\)     A1A1     N2

[2 marks]

e.

METHOD 1

graphical approach (but must involve derivative functions)     M1

e.g.


each curve     A1A1

\(x = 0.511\)     A2     N2

METHOD 2

\(g'(x) = \frac{1}{{x + 1}}\)     A1

\(f'(x) = 3\cos x – 4\sin x\)     \((5\cos (x + 0.927))\)     A1

evidence of attempt to solve \(g'(x) = f'(x)\)     M1

\(x = 0.511\)     A2     N2

[5 marks]

f.

Question

Let \(f(x) = \cos ({x^2})\) and \(g(x) = {{\rm{e}}^x}\) , for \( – 1.5 \le x \le 0.5\) .

Find the area of the region enclosed by the graphs of f and g .

Answer/Explanation

Markscheme

evidence of finding intersection points     (M1)

e.g. \(f(x) = g(x)\) , \(\cos {x^2} = {{\rm{e}}^x}\) , sketch showing intersection

\(x = – 1.11\) , \(x = 0\) (may be seen as limits in the integral)     A1A1

evidence of approach involving integration and subtraction (in any order)     (M1)

e.g. \(\int_{ – 1.11}^0 {\cos {x^2} – {{\rm{e}}^x}} \) , \(\int {(\cos {x^2} – {{\rm{e}}^x}){\rm{d}}x} \) , \(\int {g – f} \)

\({\text{area}} = 0.282\)     A2     N3

[6 marks]

Question

Let \(f(x) = 2{x^2} – 8x – 9\) .

(i)     Write down the coordinates of the vertex.

(ii)    Hence or otherwise, express the function in the form \(f(x) = 2{(x – h)^2} + k\) .

[4]
a(i) and (ii).

Solve the equation \(f(x) = 0\) .

[3]
b.
Answer/Explanation

Markscheme

(i) \((2{\text{, }} – 17)\) or \(x = 2\) , \(y = – 17\)     A1A1     N2

(ii) evidence of valid approach     (M1)

e.g. graph, completing the square, equating coefficients

\(f(x) = 2{(x – 2)^2} – 17\)     A1     N2

[4 marks]

a(i) and (ii).

evidence of valid approach     (M1)

e.g. graph, quadratic formula

\( – 0.9154759 \ldots \) , \(4.915475 \ldots \)

\(x = – 0.915\) , \(4.92\)     A1A1     N3

[3 marks]

b.

Question

Let \(f(x) = \frac{{100}}{{(1 + 50{{\rm{e}}^{ – 0.2x}})}}\) . Part of the graph of \(f\) is shown below.

Write down \(f(0)\) .

[1]
a.

Solve \(f(x) = 95\) .

[2]
b.

Find the range of \(f\) .

[3]
c.

Show that \(f'(x) = \frac{{1000{{\rm{e}}^{ – 0.2x}}}}{{{{(1 + 50{{\rm{e}}^{ – 0.2x}})}^2}}}\) .

[5]
d.

Find the maximum rate of change of \(f\) .

[4]
e.
Answer/Explanation

Markscheme

\(f(0) = \frac{{100}}{{51}}\) (exact), \(1.96\)     A1     N1

[1 mark]

a.

setting up equation     (M1)

eg   \(95 = \frac{{100}}{{1 + 50{{\rm{e}}^{ – 0.2x}}}}\) , sketch of graph with horizontal line at \(y = 95\)

\(x = 34.3\)     A1     N2

[2 marks]

b.

upper bound of \(y\) is \(100\)     (A1)

lower bound of \(y\) is \(0\)     (A1)

range is \(0 < y < 100\)     A1     N3

[3 marks]

c.

METHOD 1

setting function ready to apply the chain rule     (M1)

eg   \(100{(1 + 50{{\rm{e}}^{ – 0.2x}})^{ – 1}}\) 

evidence of correct differentiation (must be substituted into chain rule)     (A1)(A1)

eg   \(u’ = – 100{(1 + 50{{\rm{e}}^{ – 0.2x}})^{ – 2}}\) , \(v’ = (50{{\rm{e}}^{ – 0.2x}})( – 0.2)\) 

correct chain rule derivative     A1

eg   \(f'(x) = – 100{(1 + 50{{\rm{e}}^{ – 0.2x}})^{ – 2}}(50{{\rm{e}}^{ – 0.2x}})( – 0.2)\) 

correct working clearly leading to the required answer     A1

eg   \(f'(x) = 1000{{\rm{e}}^{ – 0.2x}}{(1 + 50{{\rm{e}}^{ – 0.2x}})^{ – 2}}\) 

\(f'(x) = \frac{{1000{{\rm{e}}^{ – 0.2x}}}}{{{{(1 + 50{{\rm{e}}^{ – 0.2x}})}^2}}}\)     AG     N0

METHOD 2

attempt to apply the quotient rule (accept reversed numerator terms)     (M1)

eg   \(\frac{{vu’ – uv’}}{{{v^2}}}\) , \(\frac{{uv’ – vu’}}{{{v^2}}}\)

evidence of correct differentiation inside the quotient rule     (A1)(A1)

eg   \(f'(x) = \frac{{(1 + 50{{\rm{e}}^{ – 0.2x}})(0) – 100(50{{\rm{e}}^{ – 0.2x}} \times  – 0.2)}}{{{{(1 + 50{{\rm{e}}^{ – 0.2x}})}^2}}}\) , \(\frac{{100( – 10){{\rm{e}}^{ – 0.2x}} – 0}}{{{{(1 + 50{{\rm{e}}^{ – 0.2x}})}^2}}}\)

any correct expression for derivative (\(0\) may not be explicitly seen)     (A1)

eg   \(\frac{{ – 100(50{{\rm{e}}^{ – 0.2x}} \times  – 0.2)}}{{{{(1 + 50{{\rm{e}}^{ – 0.2x}})}^2}}}\)

correct working clearly leading to the required answer     A1

eg   \(f'(x) = \frac{{0 – 100( – 10){{\rm{e}}^{ – 0.2x}}}}{{{{(1 + 50{{\rm{e}}^{ – 0.2x}})}^2}}}\) , \(\frac{{ – 100( – 10){{\rm{e}}^{ – 0.2x}}}}{{{{(1 + 50{{\rm{e}}^{ – 0.2x}})}^2}}}\)

\(f'(x) = \frac{{{\rm{1000}}{{\rm{e}}^{ – 0.2x}}}}{{{{(1 + 50{{\rm{e}}^{ – 0.2x}})}^2}}}\)     AG     N0

[5 marks]

d.

METHOD 1

sketch of \(f'(x)\)     (A1)

eg

recognizing maximum on \(f'(x)\)     (M1)

eg dot on max of sketch

finding maximum on graph of \(f'(x)\)     A1

eg   (\(19.6\), \(5\)) , \(x = 19.560 \ldots \)

maximum rate of increase is \(5\)     A1 N2

METHOD 2

recognizing \(f”(x) = 0\)     (M1)

finding any correct expression for  \(f”(x) = 0\)     (A1)

eg   \(\frac{{{{(1 + 50{{\rm{e}}^{ – 0.2x}})}^2}( – 200{{\rm{e}}^{ – 0.2x}}) – (1000{{\rm{e}}^{ – 0.2x}})(2(1 + 50{{\rm{e}}^{ – 0.2x}})( – 10{{\rm{e}}^{ – 0.2x}}))}}{{{{(1 + 50{{\rm{e}}^{ – 0.2x}})}^4}}}\)

finding \(x = 19.560 \ldots \)     A1

maximum rate of increase is \(5\)     A1     N2

[4 marks]

e.

Question

The following diagram shows a circle with centre O and radius \(r\) cm.


Points A and B are on the circumference of the circle and \({\rm{A}}\hat {\rm{O}}{\rm{B}} = 1.4\) radians .

The point C is on [OA] such that \({\rm{B}}\hat {\rm{C}}{\rm{O}} = \frac{\pi }{2}\) radians .

Show that \({\rm{OC}} = r\cos 1.4\) .

[1]
a.

The area of the shaded region is \(25\) cm2 . Find the value of \(r\) .

[7]
b.
Answer/Explanation

Markscheme

use right triangle trigonometry     A1

eg   \(\cos 1.4 = \frac{{{\rm{OC}}}}{r}\)

\({\rm{OC}} = r\cos 1.4\)     AG     N0

[1 mark]

a.

correct value for BC

eg   \({\rm{BC}} = r\sin 1.4\) , \(\sqrt {{r^2} – {{(r\cos 1.4)}^2}} \)     (A1)

area of \(\Delta {\rm{OBC}} = \frac{1}{2}r\sin 1.4 \times r\cos 1.4\) \(\left( { = \frac{1}{2}{r^2}\sin 1.4 \times \cos 1.4} \right)\)     A1

area of sector \({{\rm{OAB}} = \frac{1}{2}{r^2} \times 1.4}\)     A1

attempt to subtract in any order     (M1)

eg sector – triangle, \({\frac{1}{2}{r^2}\sin 1.4 \times \cos 1.4 – 0.7{r^2}}\)

correct equation     A1

eg   \({0.7{r^2} – \frac{1}{2}r\sin 1.4 \times r\cos 1.4 = 25}\)

attempt to solve their equation     (M1)

eg sketch, writing as quadratic, \(\frac{{25}}{{0.616 \ldots }}\)

\(r = 6.37\)     A1     N4

[7 marks]

Note: Exception to FT rule. Award A1FT for a correct FT answer from a quadratic equation involving two trigonometric functions.

b.

Question

Let \(f(x) = {{\rm{e}}^{\frac{x}{4}}}\) and \(g(x) = mx\) , where \(m \ge 0\) , and \( – 5 \le x \le 5\) . Let \(R\) be the region enclosed by the \(y\)-axis, the graph of \(f\) , and the graph of \(g\) .

Let \(m = 1\).

(i)     Sketch the graphs of \(f\) and \(g\) on the same axes.

(ii)     Find the area of \(R\) .

[7]
a.

Find the area of \(R\) .

[5]
a.ii.

Consider all values of \(m\) such that the graphs of \(f\) and \(g\) intersect. Find the value of \(m\) that gives the greatest value for the area of \(R\) .

[8]
b.
Answer/Explanation

Markscheme

 (i)

   A1A1     N2

Notes: Award A1 for the graph of \(f\) positive, increasing and concave up.

    Award A1 for graph of \(g\) increasing and linear with \(y\)-intercept of \(0\).

    Penalize one mark if domain is not [\( – 5\), \(5\)] and/or if \(f\) and \(g\) do not intersect in the first quadrant.

[2 marks]

(ii)
attempt to find intersection of the graphs of \(f\) and \(g\)     (M1)
eg   \({{\rm{e}}^{\frac{x}{4}}} = x\)
\(x = 1.42961 \ldots \)     A1
valid attempt to find area of \(R\)     (M1)
eg   \(\int {(x – {{\rm{e}}^{\frac{x}{4}}}} ){\rm{d}}x\) ,  \(\int_0^1 {(g – f)} \) , \(\int {(f – g)} \)
area \( = 0.697\)     A2     N3

[5 marks]

a.

attempt to find intersection of the graphs of \(f\) and \(g\)     (M1)

eg   \({{\rm{e}}^{\frac{x}{4}}} = x\)

\(x = 1.42961 \ldots \)     A1

valid attempt to find area of \(R\)     (M1)

eg   \(\int {(x – {{\rm{e}}^{\frac{x}{4}}}} ){\rm{d}}x\) ,  \(\int_0^1 {(g – f)} \) , \(\int {(f – g)} \)

area \( = 0.697\)     A2     N3

[5 marks]

a.ii.

recognize that area of \(R\) is a maximum at point of tangency     (R1)

eg   \(m = f'(x)\)

equating functions     (M1)

eg   \(f(x) = g(x)\) , \({{\rm{e}}^{\frac{x}{4}}} = mx\)

\(f'(x) = \frac{1}{4}{{\rm{e}}^{\frac{x}{4}}}\)     (A1)

equating gradients     (A1)

eg   \(f'(x) = g'(x)\) , \(\frac{1}{4}{{\rm{e}}^{\frac{x}{4}}} = m\)

attempt to solve system of two equations for \(x\)     (M1)

eg   \(\frac{1}{4}{{\rm{e}}^{\frac{x}{4}}} \times x = {{\rm{e}}^{\frac{x}{4}}}\)

\(x = 4\)     (A1)

attempt to find \(m\)     (M1)

eg   \(f'(4)\) , \(\frac{1}{4}{{\rm{e}}^{\frac{4}{4}}}\)

\(m = \frac{1}{4}e\) (exact), \(0.680\)     A1     N3

[8 marks]

b.

Question

The number of bacteria in two colonies, \(\rm{A}\) and \(\rm{B}\), starts increasing at the same time.

The number of bacteria in colony \(\rm{A}\) after \(t\) hours is modelled by the function \(\rm{A}(t) = 12{{\text{e}}^{0.4t}}\).

Find the number of bacteria in colony \({\text{A}}\) after four hours.

[2]
a.

Find the number of bacteria in colony \({\text{A}}\) after four hours.

[3]
b.

How long does it take for the number of bacteria in colony \({\text{A}}\) to reach \(400\)?

[3]
c.

The number of bacteria in colony \({\text{B}}\) after \(t\) hours is modelled by the function \(B(t) = 24{{\text{e}}^{kt}}\).

After four hours, there are \(60\) bacteria in colony \({\text{B}}\). Find the value of \(k\).

[3]
d.

The number of bacteria in colony \({\text{B}}\) after \(t\) hours is modelled by the function \(B(t) = 24{{\text{e}}^{kt}}\).

The number of bacteria in colony \({\text{A}}\) first exceeds the number of bacteria in colony \({\text{B}}\) after \(n\) hours, where \(n \in \mathbb{Z}\). Find the value of \(n\).

[4]
e.
Answer/Explanation

Markscheme

correct substitution into formula     (A1)

eg     \(12{{\text{e}}^{0.4(0)}}\)

\(12\) bacteria in the dish     A1     N2

[2 marks]

a.

correct substitution into formula     (A1)

eg     \(12{{\text{e}}^{0.4(4)}}\)

\(59.4363\)     (A1)

\(59\) bacteria in the dish (integer answer only)     A1     N3

[3 marks]

b.

correct equation     (A1)

eg     \(A(t) = 400,{\text{ }}12{{\text{e}}^{0.4t}} = 400\)

valid attempt to solve     (M1)

eg     graph, use of logs

\(8.76639\)

\(8.77\) (hours)     A1     N3

[3 marks]

c.

valid attempt to solve     (M1)

eg     \(n(4) = 60,{\text{ }}60 = 24{{\text{e}}^{4k}}\), use of logs

correct working     (A1)

eg     sketch of intersection, \(4k = \ln 2.5\)

\(k = 0.229072\)

\(k = \frac{{\ln 2.5}}{4}\) (exact), \(k = 0.229\)     A1     N3

[3 marks]

d.

METHOD 1

setting up an equation or inequality (accept any variable for \(n\))     (M1)

eg     \(A(t) > B(t),{\text{ }}12{{\text{e}}^{0.4n}} = 24{{\text{e}}^{0.229n}},{\text{ }}{{\text{e}}^{0.4n}} = 2{{\text{e}}^{0.229n}}\)

correct working     (A1)

eg     sketch of intersection, \({{\text{e}}^{0.171n}} = 2\)

\(4.05521\)   (accept \(4.05349\))     (A1)

\(n = 5\)   (integer answer only)     A1     N3

METHOD 2

\(A(4) = 59,{\text{ }}B(4) = 60\)   (from earlier work)

\(A(5) = 88.668,{\text{ }}B(5) = 75.446\)     A1A1

valid reasoning     (R1)

eg     \(A(4) < B(4)\) and \(A(5) > B(5)\)

\(n = 5\)   (integer answer only)     A1     N3

[4 marks]

e.

Question

Let \(f(x) =  – {x^4} + 2{x^3} – 1\), for \(0 \le x \le 2\).

Sketch the graph of \(f\) on the following grid.

[3]
a.

Solve \(f(x) = 0\).

[2]
b.

The region enclosed by the graph of \(f\) and the \(x\)-axis is rotated \(360°\) about the \(x\)-axis.

Find the volume of the solid formed.

[3]
c.
Answer/Explanation

Markscheme

   A1A1A1     N3

Note:     Award A1 for both endpoints in circles,

A1 for approximately correct shape (concave up to concave down).

Only if this A1 for shape is awarded, award A1 for maximum point in circle.

a.

\(x = 1\;\;\;x = 1.83928\)

\(x = 1{\text{ (exact)}}\;\;\;x = 1.84{\text{ }}[1.83,{\text{ }}1.84]\)     A1A1     N2

[2 marks]

b.

attempt to substitute either (FT ) limits or function into formula with \({f^2}\)     (M1)

eg\(\;\;\;\)\(V = \pi \int_1^{1.84} {{f^2},{\text{ }}\int {{{( – {x^4} + 2{x^3} – 1)}^2}{\text{d}}x} } \)

\(0.636581\)

\(V = 0.637{\text{ }}[0.636,{\text{ }}0.637]\)     A2     N3

[3 marks]

Total [8 marks]

c.

Question

The first two terms of a geometric sequence \({u_n}\) are \({u_1} = 4\) and \({u_2} = 4.2\).

(i)     Find the common ratio.

(ii)     Hence or otherwise, find \({u_5}\).

[5]
a.

Another sequence \({v_n}\) is defined by \({v_n} = a{n^k}\), where \(a,{\text{ }}k \in \mathbb{R}\), and \(n \in {\mathbb{Z}^ + }\), such that \({v_1} = 0.05\) and \({v_2} = 0.25\).

(i)     Find the value of \(a\).

(ii)     Find the value of \(k\).

[5]
b.

Find the smallest value of \(n\) for which \({v_n} > {u_n}\).

[5]
c.
Answer/Explanation

Markscheme

(i)     valid approach     (M1)

eg\(\;\;\;\)\(r = \frac{{{u_2}}}{{{u_1}}},{\text{ }}\frac{4}{{4.2}}\)

\(r = 1.05\;\;\;{\text{(exact)}}\)     A1     N2

(ii)     attempt to substitute into formula, with their \(r\)     (M1)

eg\(\;\;\;\)\(4 \times {1.05^n},{\text{ }}4 \times 1.05 \times 1.05 \ldots \)

correct substitution     (A1)

eg\(\;\;\;\)\(4 \times {1.05^4},{\text{ }}4 \times 1.05 \times 1.05 \times 1.05 \times 1.05\)

\({u_5} = 4.862025{\text{ (exact), }}4.86{\text{ }}[4.86,{\text{ }}4.87]{\text{ }}\)     A1     N2

[5 marks]

a.

(i)     attempt to substitute \(n = 1\)     (M1)

eg\(\;\;\;\)\(0.05 = a \times {1^k}\)

\(a = 0.05\)     A1     N2

(ii)     correct substitution of \(n = 2\) into \({v_2}\)     A1

eg\(\;\;\;\)\(0.25 = a \times {2^k}\)

correct work     (A1)

eg\(\;\;\;\)finding intersection point, \(k = {\log _2}\left( {\frac{{0.25}}{{0.05}}} \right),{\text{ }}\frac{{\log 5}}{{\log 2}}\)

\(2.32192\)

\(k = {\log _2}5\;\;\;{\text{(exact), }}2.32{\text{ }}[2.32,{\text{ }}2.33]\)     A1     N2

[5 marks]

b.

correct expression for \({u_n}\)     (A1)

eg\(\;\;\;\)\(4 \times {1.05^{n – 1}}\)

EITHER

correct substitution into inequality (accept equation)     (A1)

eg\(\;\;\;\)\(0.05 \times {n^k} > 4 \times {1.05^{n – 1}}\)

valid approach to solve inequality (accept equation)     (M1)

eg\(\;\;\;\)finding point of intersection, \(n = 7.57994{\text{ }}(7.59508{\text{ from 2.32)}}\)

\(n = 8\;\;\;\)(must be an integer)     A1     N2

OR

table of values

when \(n = 7,{\text{ }}{u_7} = 5.3604,{\text{ }}{v_7} = 4.5836\)     A1

when \(n = 8,{\text{ }}{u_8} = 5.6284,{\text{ }}{v_8} = 6.2496\)     A1

\(n = 8\;\;\;\)(must be an integer)     A1     N2

[4 marks]

Total [14 marks]

c.

Question

The following diagram shows part of the graph of \(f(x) =  – 2{x^3} + 5.1{x^2} + 3.6x – 0.4\).

Find the coordinates of the local minimum point.

[2]
a.

The graph of \(f\) is translated to the graph of \(g\) by the vector \(\left( {\begin{array}{*{20}{c}} 0 \\ k \end{array}} \right)\). Find all values of \(k\) so that \(g(x) = 0\) has exactly one solution.

[5]
b.
Answer/Explanation

Markscheme

\(( – 0.3,{\text{ }} – 0.967)\)

\(x =  – 0.3\) (exact), \(y =  – 0.967\) (exact)     A1A1     N2

[2 marks]

a.

\(y\)-coordinate of local maximum is \(y = 11.2\)     (A1)

negating the \(y\)-coordinate of one of the max/min     (M1)

eg\(\;\;\;y = 0.967,{\text{ }}y =  – 11.2\)

recognizing that the solution set has two intervals     R1

eg\(\;\;\;\)two answers,

\(k <  – 11.2,{\text{ }}k > 0.967\)     A1A1     N3N2

[5 marks]

Notes:     If working shown, do not award the final mark if strict inequalities are not used.

If no working shown, award N2 for \(k \le  – 11.2\) or N1 for \(k \ge 0.967\)

Total [7 marks]

b.

Question

Let \(f(x) = k{x^2} + kx\) and \(g(x) = x – 0.8\). The graphs of \(f\) and \(g\) intersect at two distinct points.

Find the possible values of \(k\).

Answer/Explanation

Markscheme

attempt to set up equation     (M1)

eg\(\;\;\;f = g,{\text{ }}k{x^2} + kx = x – 0.8\)

rearranging their equation to equal zero     M1

eg\(\;\;\;k{x^2} + kx – x + 0.8 = 0,{\text{ }}k{x^2} + x(k – 1) + 0.8 = 0\)

evidence of discriminant (if seen explicitly, not just in quadratic formula)     (M1)

eg\(\;\;\;{b^2} – 4ac,{\text{ }}\Delta  = {(k – 1)^2} – 4k \times 0.8,{\text{ }}D = 0\)

correct discriminant     (A1)

eg\(\;\;\;{(k – 1)^2} – 4k \times 0.8,{\text{ }}{k^2} – 5.2k + 1\)

evidence of correct discriminant greater than zero     R1

eg\(\;\;\;{k^2} – 5.2k + 1 > 0,{\text{ }}{(k – 1)^2} – 4k \times 0.8 > 0\), correct answer

both correct values     (A1)

eg\(\;\;\;0.2,{\text{ }}5\)

correct answer     A2     N3

eg\(\;\;\;k < 0.2,{\text{ }}k \ne 0,{\text{ }}k > 5\)

[8 marks]

Question

The first three terms of a geometric sequence are \({u_1} = 0.64,{\text{ }}{u_2} = 1.6\), and \({u_3} = 4\).

Find the value of \(r\).

[2]
a.

Find the value of \({S_6}\).

[2]
b.

Find the least value of \(n\) such that \({S_n} > 75\,000\).

[3]
c.
Answer/Explanation

Markscheme

valid approach     (M1)

eg\(\;\;\;\frac{{{u_1}}}{{{u_2}}},{\text{ }}\frac{4}{{1.6}},{\text{ }}1.6 = r(0.64)\)

\(r = 2.5\;\;\;\left( { = \frac{5}{2}} \right)\)     A1     N2

[2 marks]

a.

correct substitution into \({S_6}\)     (A1)

eg\(\;\;\;\frac{{0.64({{2.5}^6} – 1)}}{{2.5 – 1}}\)

\({S_6} = 103.74\) (exact), \(104\)     A1     N2

[2 marks]

b.

METHOD 1 (analytic)

valid approach     (M1)

eg\(\;\;\;\frac{{0.64({{2.5}^n} – 1)}}{{2.5 – 1}} > 75\,000,{\text{ }}\frac{{0.64({{2.5}^n} – 1)}}{{2.5 – 1}} = 75\,000\)

correct inequality (accept equation)     (A1)

eg\(\;\;\;n > 13.1803,{\text{ }}n = 13.2\)

\(n = 14\)     A1     N1

METHOD 2 (table of values)

both crossover values     A2

eg\(\;\;\;{S_{13}} = 63577.8,{\text{ }}{S_{14}} = 158945\)

\(n = 14\)     A1     N1

[3 marks]

Total [7 marks]

c.

Question

An environmental group records the numbers of coyotes and foxes in a wildlife reserve after \(t\) years, starting on 1 January 1995.

Let \(c\) be the number of coyotes in the reserve after \(t\) years. The following table shows the number of coyotes after \(t\) years.

The relationship between the variables can be modelled by the regression equation \(c = at + b\).

Find the value of \(a\) and of \(b\).

[3]
a.

Use the regression equation to estimate the number of coyotes in the reserve when \(t = 7\).

[3]
b.

Let \(f\) be the number of foxes in the reserve after \(t\) years. The number of foxes can be modelled by the equation \(f = \frac{{2000}}{{1 + 99{{\text{e}}^{ – kt}}}}\), where \(k\) is a constant.

Find the number of foxes in the reserve on 1 January 1995.

[3]
c.

After five years, there were 64 foxes in the reserve. Find \(k\).

[3]
d.

During which year were the number of coyotes the same as the number of foxes?

[4]
e.
Answer/Explanation

Markscheme

evidence of setup     (M1)

eg\(\;\;\;\)correct value for \(a\) or \(b\)

\(13.3823\), \(137.482\)

\(a{\rm{ }} = {\rm{ }}13.4\), \(b{\rm{ }} = {\rm{ }}137\)     A1A1     N3

[3 marks]

a.

correct substitution into their regression equation

eg\(\;\;\;13.3823 \times 7 + 137.482\)     (A1)

correct calculation

\(231.158\)     (A1)

\(231\) (coyotes) (must be an integer)     A1     N2

[3 marks]

b.

recognizing \(t = 0\)     (M1)

eg\(\;\;\;f(0)\)

correct substitution into the model

eg\(\;\;\;\frac{{2000}}{{1 + 99{{\text{e}}^{ – k(0)}}}},{\text{ }}\frac{{2000}}{{100}}\)     (A1)

\(20\) (foxes)     A1     N2

[3 marks]

c.

recognizing \((5,{\text{ }}64)\) satisfies the equation     (M1)

eg\(\;\;\;f(5) = 64\)

correct substitution into the model

eg\(\;\;\;64 = \frac{{2000}}{{1 + 99{{\text{e}}^{ – k(5)}}}},{\text{ }}64(1 + 99\(e\)^{ – 5k}}) = 2000\)     (A1)

\(0.237124\)

\(k =  – \frac{1}{5}\ln \left( {\frac{{11}}{{36}}} \right){\text{ (exact), }}0.237{\text{ }}[0.237,{\text{ }}0.238]\)     A1     N2

[3 marks]

d.

valid approach     (M1)

eg\(\;\;\;c = f\), sketch of graphs

correct working     (A1)

eg\(\;\;\;\frac{{2000}}{{1 + 99{{\text{e}}^{ – 0.237124t}}}} = 13.382t + 137.482\), sketch of graphs, table of values

\(t = 12.0403\)     (A1)

\(2007\)     A1     N2

Note:     Exception to the FT rule. Award A1FT on their value of \(t\).

[4 marks]

Total [16 marks]

e.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.

[N/A]

d.

[N/A]

e.

Question

Let \(f(x) = {x^2}\) and \(g(x) = 3\ln (x + 1)\), for \(x >  – 1\).

Solve \(f(x) = g(x)\).

[3]
a.

Find the area of the region enclosed by the graphs of \(f\) and \(g\).

[3]
b.
Answer/Explanation

Markscheme

valid approach     (M1)

eg sketch

0, 1.73843

\(x = 0,{\text{ }}x = 1.74{\text{ }}\left( {{\text{accept }}(0,{\text{ }}0){\text{ and }}(1.74,{\text{ }}3.02)} \right)\)     A1A1     N3

[3 marks]

a.

integrating and subtracting functions (in any order)     (M1)

eg\(\,\,\,\,\,\)\(\int {g – f} \)

correct substitution of their limits or function (accept missing \({\text{d}}x\))

(A1)

eg\(\,\,\,\,\,\)\(\int_0^{1.74} {g – f,{\text{ }}\int {3\ln (x + 1) – {x^2}} } \)

Note:     Do not award A1 if there is an error in the substitution.

1.30940

1.31     A1     N3

[3 marks]

b.

Question

A population of rare birds, \({P_t}\), can be modelled by the equation \({P_t} = {P_0}{{\text{e}}^{kt}}\), where \({P_0}\) is the initial population, and \(t\) is measured in decades. After one decade, it is estimated that \(\frac{{{P_1}}}{{{P_0}}} = 0.9\).

(i)     Find the value of \(k\).

(ii)     Interpret the meaning of the value of \(k\).

[3]
a.

Find the least number of whole years for which \(\frac{{{P_t}}}{{{P_0}}} < 0.75\).

[5]
b.
Answer/Explanation

Markscheme

(i)     valid approach     (M1)

eg\(\,\,\,\,\,\)\(0.9 = {{\text{e}}^{k(1)}}\)

\(k =  – 0.105360\)

\(k = \ln 0.9{\text{ (exact), }} – 0.105\)    A1     N2

(ii)     correct interpretation     R1     N1

eg\(\,\,\,\,\,\)population is decreasing, growth rate is negative

[3 marks]

a.

METHOD 1

valid approach (accept an equality, but do not accept 0.74)     (M1)

eg\(\,\,\,\,\,\)\(P < 0.75{P_0},{\text{ }}{P_0}{{\text{e}}^{kt}} < 0.75{P_0},{\text{ }}0.75 = {{\text{e}}^{t\ln 0.9}}\)

valid approach to solve their inequality     (M1)

eg\(\,\,\,\,\,\)logs, graph

\(t > 2.73045{\text{ }}({\text{accept }}t = 2.73045){\text{ }}(2.73982{\text{ from }} – 0.105)\)    A1

28 years     A2     N2

METHOD 2

valid approach which gives both crossover values accurate to at least 2 sf     A2

eg\(\,\,\,\,\,\)\(\frac{{{P_{2.7}}}}{{{P_0}}} = 0.75241 \ldots ,{\text{ }}\frac{{{P_{2.8}}}}{{{P_0}}} = 0.74452 \ldots \)

\(t = 2.8\)    (A1)

28 years     A2     N2

[5 marks]

b.

Question

A particle P moves along a straight line so that its velocity, \(v\,{\text{m}}{{\text{s}}^{ – 1}}\), after \(t\) seconds, is given by \(v = \cos 3t – 2\sin t – 0.5\), for \(0 \leqslant t \leqslant 5\). The initial displacement of P from a fixed point O is 4 metres.

The following sketch shows the graph of \(v\).

M16/5/MATME/SP2/ENG/TZ1/09.b+c+d+e

Find the displacement of P from O after 5 seconds.

[5]
a.

Find when P is first at rest.

[2]
b.

Write down the number of times P changes direction.

[2]
c.

Find the acceleration of P after 3 seconds.

[2]
d.

Find the maximum speed of P.

[3]
e.
Answer/Explanation

Markscheme

METHOD 1

recognizing \(s = \int v \)     (M1)

recognizing displacement of P in first 5 seconds (seen anywhere)     A1

(accept missing \({\text{d}}t\))

eg\(\,\,\,\,\,\)\(\int_0^5 {v{\text{d}}t,{\text{ }} – 3.71591} \)

valid approach to find total displacement     (M1)

eg\(\,\,\,\,\,\)\(4 + ( – 3.7159),{\text{ }}s = 4 + \int_0^5 v \)

0.284086

0.284 (m)     A2     N3

METHOD 2

recognizing \(s = \int v \)     (M1)

correct integration     A1

eg\(\,\,\,\,\,\)\(\frac{1}{3}\sin 3t + 2\cos t – \frac{t}{2} + c\) (do not penalize missing “\(c\)”)

attempt to find \(c\)     (M1)

eg\(\,\,\,\,\,\)\(4 = \frac{1}{3}\sin (0) + 2\cos (0)–\frac{0}{2} + c,{\text{ }}4 = \frac{1}{3}\sin 3t + 2\cos t – \frac{t}{2} + c,{\text{ }}2 + c = 4\)

attempt to substitute \(t = 5\) into their expression with \(c\)     (M1)

eg\(\,\,\,\,\,\)\(s(5),{\text{ }}\frac{1}{3}\sin (15) + 2\cos (5)5–\frac{5}{2} + 2\)

0.284086

0.284 (m)     A1     N3

[5 marks]

a.

recognizing that at rest, \(v = 0\)     (M1)

\(t = 0.179900\)

\(t = 0.180{\text{ (secs)}}\)     A1     N2

[2 marks]

b.

recognizing when change of direction occurs     (M1)

eg\(\,\,\,\,\,\)\(v\) crosses \(t\) axis

2 (times)     A1     N2

[2 marks]

c.

acceleration is \({v’}\) (seen anywhere)     (M1)

eg\(\,\,\,\,\,\)\(v'(3)\)

0.743631

\(0.744{\text{ }}({\text{m}}{{\text{s}}^{ – 2}})\)     A1     N2

[2 marks]

d.

valid approach involving max or min of \(v\)     (M1)

eg\(\,\,\,\,\,\)\(v\prime  = 0,{\text{ }}a = 0\), graph

one correct co-ordinate for min     (A1)

eg\(\,\,\,\,\,\)\(1.14102,{\text{ }}-3.27876\)

\(3.28{\text{ }}({\text{m}}{{\text{s}}^{ – 1}})\)     A1     N2

[3 marks]

e.

Question

The points A and B lie on a line \(L\), and have position vectors \(\left( {\begin{array}{*{20}{c}} { – 3} \\ { – 2} \\ 2 \end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { – 1} \end{array}} \right)\) respectively. Let O be the origin. This is shown on the following diagram.

M16/5/MATME/SP2/ENG/TZ1/10

The point C also lies on \(L\), such that \(\overrightarrow {{\text{AC}}}  = 2\overrightarrow {{\text{CB}}} \).

Let \(\theta \) be the angle between \(\overrightarrow {{\text{AB}}} \) and \(\overrightarrow {{\text{OC}}} \).

Let D be a point such that \(\overrightarrow {{\text{OD}}}  = k\overrightarrow {{\text{OC}}} \), where \(k > 1\). Let E be a point on \(L\) such that \({\rm{C\hat ED}}\) is a right angle. This is shown on the following diagram.

M16/5/MATME/SP2/ENG/TZ1/10.d

Find \(\overrightarrow {{\text{AB}}} \).

[2]
a.

Show that \(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ 0 \end{array}} \right)\).

[[N/A]]
b.

Find \(\theta \).

[5]
c.

(i)     Show that \(\left| {\overrightarrow {{\text{DE}}} } \right| = (k – 1)\left| {\overrightarrow {{\text{OC}}} } \right|\sin \theta \).

(ii)     The distance from D to line \(L\) is less than 3 units. Find the possible values of \(k\).

[6]
d.
Answer/Explanation

Markscheme

valid approach (addition or subtraction)     (M1)

eg\(\,\,\,\,\,\)\({\text{AO}} + {\text{OB}},{\text{ B}} – {\text{A}}\)

\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 9 \\ 6 \\ { – 3} \end{array}} \right)\)    A1     N2

[2 marks]

a.

METHOD 1

valid approach using \(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right)\)     (M1)

eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} {x + 3} \\ {y + 2} \\ {z – 2} \end{array}} \right),{\text{ }}\overrightarrow {{\text{CB}}} = \left( {\begin{array}{*{20}{c}} {6 – x} \\ {4 – y} \\ { – 1 – z} \end{array}} \right)\)

correct working     A1

eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} {x + 3} \\ {y + 2} \\ {z – 2} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {12 – 2x} \\ {8 – 2y} \\ { – 2 – 2z} \end{array}} \right)\)

all three equations     A1

eg\(\,\,\,\,\,\)\(x + 3 = 12 – 2x,{\text{ }}y + 2 = 8 – 2y,{\text{ }}z – 2 =  – 2 – 2z\),

\(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ 0 \end{array}} \right)\)     AG     N0

METHOD 2

valid approach     (M1)

eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{OC}}}  – \overrightarrow {{\text{OA}}}  = 2\left( {\overrightarrow {{\text{OB}}}  – \overrightarrow {{\text{OC}}} } \right)\)

correct working     A1

eg\(\,\,\,\,\,\)\(3\overrightarrow {{\text{OC}}}  = 2\overrightarrow {{\text{OB}}}  + \overrightarrow {{\text{OA}}} \)

correct substitution of \(\overrightarrow {{\text{OB}}} \) and \(\overrightarrow {{\text{OA}}} \)     A1

eg\(\,\,\,\,\,\)\(3\overrightarrow {{\text{OC}}} = 2\left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { – 1} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} { – 3} \\ { – 2} \\ 2 \end{array}} \right),{\text{ }}3\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 9 \\ 6 \\ 0 \end{array}} \right)\)

\(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ 0 \end{array}} \right)\)    AG     N0

METHOD 3

valid approach     (M1)

eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{AC}}}  = \frac{2}{3}\overrightarrow {{\text{AB}}} \), diagram, \(\overrightarrow {{\text{CB}}}  = \frac{1}{3}\overrightarrow {{\text{AB}}} \)

M16/5/MATME/SP2/ENG/TZ1/10.b/M

correct working     A1

eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { – 2} \end{array}} \right),{\text{ }}\overrightarrow {{\text{CB}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ { – 1} \end{array}} \right)\)

correct working involving \(\overrightarrow {{\text{OC}}} \)     A1

eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} { – 3} \\ { – 2} \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { – 2} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { – 1} \end{array}} \right) – \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ { – 1} \end{array}} \right)\)

\(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ 0 \end{array}} \right)\)    AG     N0

[3 marks]

b.

finding scalar product and magnitudes     (A1)(A1)(A1)

scalar product \( = (9 \times 3) + (6 \times 2) + ( – 3 \times 0){\text{ }}( = 39)\)

magnitudes \(\sqrt {81 + 36 + 9} {\text{ }}( = 11.22),{\text{ }}\sqrt {9 + 4} {\text{ }}( = 3.605)\)

substitution into formula     M1

eg\(\,\,\,\,\,\)\(\cos \theta  = \frac{{(9 \times 3) + 12}}{{\sqrt {126}  \times \sqrt {13} }}\)

\(\theta  = 0.270549{\text{ }}({\text{accept }}15.50135^\circ )\)

\(\theta  = 0.271{\text{ }}({\text{accept }}15.5^\circ )\)    A1     N4

[5 marks]

c.

(i)     attempt to use a trig ratio     M1

eg\(\,\,\,\,\,\)\(\sin \theta  = \frac{{{\text{DE}}}}{{{\text{CD}}}},{\text{ }}\left| {\overrightarrow {{\text{CE}}} } \right| = \left| {\overrightarrow {{\text{CD}}} } \right|\cos \theta \)

attempt to express \(\overrightarrow {{\text{CD}}} \) in terms of \(\overrightarrow {{\text{OC}}} \)     M1

eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{OC}}}  + \overrightarrow {{\text{CD}}}  = \overrightarrow {{\text{OD}}} ,{\text{ OC}} + {\text{CD}} = {\text{OD}}\)

correct working     A1

eg\(\,\,\,\,\,\)\(\left| {k\overrightarrow {{\text{OC}}}  – \overrightarrow {{\text{OC}}} } \right|\sin \theta \)

\(\left| {\overrightarrow {{\text{DE}}} } \right| = (k – 1)\left| {\overrightarrow {{\text{OC}}} } \right|\sin \theta \)     AG     N0

(ii)     valid approach involving the segment DE     (M1)

eg\(\,\,\,\,\,\)recognizing \(\left| {\overrightarrow {{\text{DE}}} } \right| < 3,{\text{ DE}} = 3\)

correct working (accept equation)     (A1)

eg\(\,\,\,\,\,\)\((k – 1)(\sqrt {13} )\sin 0.271 < 3,{\text{ }}k – 1 = 3.11324\)

\(1 < k < 4.11{\text{ }}({\text{accept }}k < 4.11{\text{ but not }}k = 4.11)\)    A1     N2

[6 marks]

d.

Question

Let \(f(x) = \frac{1}{{x – 1}} + 2\), for \(x > 1\).

Let \(g(x) = a{e^{ – x}} + b\), for \(x \geqslant 1\). The graphs of \(f\) and \(g\) have the same horizontal asymptote.

Write down the equation of the horizontal asymptote of the graph of \(f\).

[2]
a.

Find \(f'(x)\).

[2]
b.

Write down the value of \(b\).

[2]
c.

Given that \(g'(1) =  – e\), find the value of \(a\).

[4]
d.

There is a value of \(x\), for \(1 < x < 4\), for which the graphs of \(f\) and \(g\) have the same gradient. Find this gradient.

[4]
e.
Answer/Explanation

Markscheme

\(y = 2\) (correct equation only)     A2     N2

[2 marks]

a.

valid approach     (M1)

eg\(\,\,\,\,\,\)\({(x – 1)^{ – 1}} + 2,{\text{ }}f'(x) = \frac{{0(x – 1) – 1}}{{{{(x – 1)}^2}}}\)

\( – {(x – 1)^{ – 2}},{\text{ }}f'(x) = \frac{{ – 1}}{{{{(x – 1)}^2}}}\)    A1     N2

[2 marks]

b.

correct equation for the asymptote of \(g\)

eg\(\,\,\,\,\,\)\(y = b\)     (A1)

\(b = 2\)     A1     N2

[2 marks]

c.

correct derivative of g (seen anywhere)     (A2)

eg\(\,\,\,\,\,\)\(g'(x) =  – a{{\text{e}}^{ – x}}\)

correct equation     (A1)

eg\(\,\,\,\,\,\)\( – {\text{e}} =  – a{{\text{e}}^{ – 1}}\)

7.38905

\(a = {{\text{e}}^2}{\text{ }}({\text{exact}}),{\text{ }}7.39\)     A1     N2

[4 marks]

d.

attempt to equate their derivatives     (M1)

eg\(\,\,\,\,\,\)\(f'(x) = g'(x),{\text{ }}\frac{{ – 1}}{{{{(x – 1)}^2}}} =  – a{{\text{e}}^{ – x}}\)

valid attempt to solve their equation     (M1)

eg\(\,\,\,\,\,\)correct value outside the domain of \(f\) such as 0.522 or 4.51,

M16/5/MATME/SP2/ENG/TZ2/09.e/M

correct solution (may be seen in sketch)     (A1)

eg\(\,\,\,\,\,\)\(x = 2,{\text{ }}(2,{\text{ }} – 1)\)

gradient is \( – 1\)     A1     N3

[4 marks]

e.

Question

Let \(f(x) = {x^2} + 2x + 1\) and \(g(x) = x – 5\), for \(x \in \mathbb{R}\).

Find \(f(8)\).

[2]
a.

Find \((g \circ f)(x)\).

[2]
b.

Solve \((g \circ f)(x) = 0\).

[3]
c.
Answer/Explanation

Markscheme

attempt to substitute \(x = 8\)     (M1)

eg\(\,\,\,\,\,\)\({8^2} + 2 \times 8 + 1\)

\(f(8) = 81\)    A1     N2

[2 marks]

a.

attempt to form composition (in any order)     (M1)

eg\(\,\,\,\,\,\)\(f(x – 5),{\text{ }}g\left( {f(x)} \right),{\text{ }}\left( {{x^2} + 2x + 1} \right) – 5\)

\((g \circ f)(x) = {x^2} + 2x – 4\)     A1     N2

[2 marks]

b.

valid approach     (M1)

eg     \(x = \frac{{ – 2 \pm \sqrt {20} }}{2}\), N16/5/MATME/SP2/ENG/TZ0/01.c/M

\(1.23606,{\text{ }} – 3.23606\)

\(x = 1.24,{\text{ }}x =  – 3.24\)     A1A1     N3

[3 marks]

c.

Question

Let \(f(x) = x{{\text{e}}^{ – x}}\) and \(g(x) =  – 3f(x) + 1\).

The graphs of \(f\) and \(g\) intersect at \(x = p\) and \(x = q\), where \(p < q\).

Find the value of \(p\) and of \(q\).

[3]
a.

Hence, find the area of the region enclosed by the graphs of \(f\) and \(g\).

[3]
b.
Answer/Explanation

Markscheme

valid attempt to find the intersection     (M1)

eg\(\,\,\,\,\,\)\(f = g\), sketch, one correct answer

\(p = 0.357402,{\text{ }}q = 2.15329\)

\(p = 0.357,{\text{ }}q = 2.15\)     A1A1     N3

[3 marks]

a.

attempt to set up an integral involving subtraction (in any order)     (M1)

eg\(\,\,\,\,\,\)\(\int_p^q {\left[ {f(x) – g(x)} \right]{\text{d}}x,{\text{ }}} \int_p^q {f(x){\text{d}}x – } \int_p^q {g(x){\text{d}}x} \)

0.537667

\({\text{area}} = 0.538\)     A2     N3

[3 marks]

b.

Question

Let \(f(x) = \ln x\) and \(g(x) = 3 + \ln \left( {\frac{x}{2}} \right)\), for \(x > 0\).

The graph of \(g\) can be obtained from the graph of \(f\) by two transformations:

\[\begin{array}{*{20}{l}} {{\text{a horizontal stretch of scale factor }}q{\text{ followed by}}} \\ {{\text{a translation of }}\left( {\begin{array}{*{20}{c}} h \\ k \end{array}} \right).} \end{array}\]

Let \(h(x) = g(x) \times \cos (0.1x)\), for \(0 < x < 4\). The following diagram shows the graph of \(h\) and the line \(y = x\).

M17/5/MATME/SP2/ENG/TZ1/10.b.c

The graph of \(h\) intersects the graph of \({h^{ – 1}}\) at two points. These points have \(x\) coordinates 0.111 and 3.31 correct to three significant figures.

Write down the value of \(q\);

[1]
a.i.

Write down the value of \(h\);

[1]
a.ii.

Write down the value of \(k\).

[1]
a.iii.

Find \(\int_{0.111}^{3.31} {\left( {h(x) – x} \right){\text{d}}x} \).

[2]
b.i.

Hence, find the area of the region enclosed by the graphs of \(h\) and \({h^{ – 1}}\).

[3]
b.ii.

Let \(d\) be the vertical distance from a point on the graph of \(h\) to the line \(y = x\). There is a point \({\text{P}}(a,{\text{ }}b)\) on the graph of \(h\) where \(d\) is a maximum.

Find the coordinates of P, where \(0.111 < a < 3.31\).

[7]
c.
Answer/Explanation

Markscheme

\(q = 2\)     A1     N1

Note:     Accept \(q = 1\), \(h = 0\), and \(k = 3 – \ln (2)\), 2.31 as candidate may have rewritten \(g(x)\) as equal to \(3 + \ln (x) – \ln (2)\).

[1 mark]

a.i.

\(h = 0\)     A1     N1

Note:     Accept \(q = 1\), \(h = 0\), and \(k = 3 – \ln (2)\), 2.31 as candidate may have rewritten \(g(x)\) as equal to \(3 + \ln (x) – \ln (2)\).

[1 mark]

a.ii.

\(k = 3\)     A1     N1

Note:     Accept \(q = 1\), \(h = 0\), and \(k = 3 – \ln (2)\), 2.31 as candidate may have rewritten \(g(x)\) as equal to \(3 + \ln (x) – \ln (2)\).

[1 mark]

a.iii.

2.72409

2.72     A2     N2

[2 marks]

b.i.

recognizing area between \(y = x\) and \(h\) equals 2.72     (M1)

eg\(\,\,\,\,\,\)M17/5/MATME/SP2/ENG/TZ1/10.b.ii/M

recognizing graphs of \(h\) and \({h^{ – 1}}\) are reflections of each other in \(y = x\)     (M1)

eg\(\,\,\,\,\,\)area between \(y = x\) and \(h\) equals between \(y = x\) and \({h^{ – 1}}\)

\(2 \times 2.72\int_{0.111}^{3.31} {\left( {x – {h^{ – 1}}(x)} \right){\text{d}}x = 2.72} \)

5.44819

5.45     A1     N3

[??? marks]

b.ii.

valid attempt to find \(d\)     (M1)

eg\(\,\,\,\,\,\)difference in \(y\)-coordinates, \(d = h(x) – x\)

correct expression for \(d\)     (A1)

eg\(\,\,\,\,\,\)\(\left( {\ln \frac{1}{2}x + 3} \right)(\cos 0.1x) – x\)

valid approach to find when \(d\) is a maximum     (M1)

eg\(\,\,\,\,\,\)max on sketch of \(d\), attempt to solve \(d’ = 0\)

0.973679

\(x = 0.974\)     A2     N4 

substituting their \(x\) value into \(h(x)\)     (M1)

2.26938

\(y = 2.27\)     A1     N2

[7 marks]

c.

Question

Consider a geometric sequence where the first term is 768 and the second term is 576.

Find the least value of \(n\) such that the \(n\)th term of the sequence is less than 7.

Answer/Explanation

Markscheme

attempt to find \(r\)     (M1)

eg\(\,\,\,\,\,\)\(\frac{{576}}{{768}},{\text{ }}\frac{{768}}{{576}},{\text{ }}0.75\)

correct expression for \({u_n}\)     (A1)

eg\(\,\,\,\,\,\)\(768{(0.75)^{n – 1}}\)

EITHER (solving inequality)

valid approach (accept equation)     (M1)

eg\(\,\,\,\,\,\)\({u_n} < 7\)

valid approach to find \(n\)     M1

eg\(\,\,\,\,\,\)\(768{(0.75)^{n – 1}} = 7,{\text{ }}n – 1 > {\log _{0.75}}\left( {\frac{7}{{768}}} \right)\), sketch

correct value

eg\(\,\,\,\,\,\)\(n = 17.3301\)     (A1)

\(n = 18\) (must be an integer)     A1     N2

OR (table of values)

valid approach     (M1)

eg\(\,\,\,\,\,\)\({u_n} > 7\), one correct crossover value

both crossover values, \({u_{17}} = 7.69735\) and \({u_{18}} = 5.77301\)     A2

\(n = 18\) (must be an integer)     A1     N2

OR (sketch of functions)

valid approach     M1

eg\(\,\,\,\,\,\)sketch of appropriate functions

valid approach     (M1) 

eg\(\,\,\,\,\,\)finding intersections or roots (depending on function sketched)

correct value

eg\(\,\,\,\,\,\)\(n = 17.3301\)     (A1)

\(n = 18\) (must be an integer)     A1     N2

[6 marks]

Question

Let \(f(x) = {x^2} – 1\) and \(g(x) = {x^2} – 2\), for \(x \in \mathbb{R}\).

Show that \((f \circ g)(x) = {x^4} – 4{x^2} + 3\).

[2]
a.

On the following grid, sketch the graph of \((f \circ g)(x)\), for \(0 \leqslant x \leqslant 2.25\).

M17/5/MATME/SP2/ENG/TZ2/06.b

[3]
b.

The equation \((f \circ g)(x) = k\) has exactly two solutions, for \(0 \leqslant x \leqslant 2.25\). Find the possible values of \(k\).

[3]
c.
Answer/Explanation

Markscheme

attempt to form composite in either order     (M1)

eg\(\,\,\,\,\,\)\(f({x^2} – 2),{\text{ }}{({x^2} – 1)^2} – 2\)

\(({x^4} – 4{x^2} + 4) – 1\)     A1

\((f \circ g)(x) = {x^4} – 4{x^2} + 3\)     AG     N0

[2 marks]

a.

M17/5/MATME/SP2/ENG/TZ2/06.b/M    A1

A1A1     N3

Note:     Award A1 for approximately correct shape which changes from concave down to concave up. Only if this A1 is awarded, award the following:

A1 for left hand endpoint in circle and right hand endpoint in oval,

A1 for minimum in oval.

[3 marks]

b.

evidence of identifying max/min as relevant points     (M1)

eg\(\,\,\,\,\,\)\(x = 0,{\text{ }}1.41421,{\text{ }}y =  – 1,{\text{ }}3\)

correct interval (inclusion/exclusion of endpoints must be correct)     A2     N3

eg\(\,\,\,\,\,\)\( – 1 < k \leqslant 3,{\text{ }}\left] { – 1,{\text{ 3}}} \right],{\text{ }}( – 1,{\text{ }}3]\)

[3 marks]

c.

Question

The following table shows a probability distribution for the random variable \(X\), where \({\text{E}}(X) = 1.2\).

M17/5/MATME/SP2/ENG/TZ2/10

A bag contains white and blue marbles, with at least three of each colour. Three marbles are drawn from the bag, without replacement. The number of blue marbles drawn is given by the random variable \(X\).

A game is played in which three marbles are drawn from the bag of ten marbles, without replacement. A player wins a prize if three white marbles are drawn.

Find \(q\).

[2]
a.i.

Find \(p\).

[2]
a.ii.

Write down the probability of drawing three blue marbles.

[1]
b.i.

Explain why the probability of drawing three white marbles is \(\frac{1}{6}\).

[1]
b.ii.

The bag contains a total of ten marbles of which \(w\) are white. Find \(w\).

[3]
b.iii.

Grant plays the game until he wins two prizes. Find the probability that he wins his second prize on his eighth attempt.

[4]
d.
Answer/Explanation

Markscheme

correct substitution into \({\text{E}}(X)\) formula     (A1)

eg\(\,\,\,\,\,\)\(0(p) + 1(0.5) + 2(0.3) + 3(q) = 1.2\)

\(q = \frac{1}{{30}}\), 0.0333     A1     N2

[2 marks]

a.i.

evidence of summing probabilities to 1     (M1)

eg\(\,\,\,\,\,\)\(p + 0.5 + 0.3 + q = 1\)

\(p = \frac{1}{6},{\text{ }}0.167\)     A1     N2

[2 marks]

a.ii.

\({\text{P (3 blue)}} = \frac{1}{{30}},{\text{ }}0.0333\)     A1     N1

[1 mark]

b.i.

valid reasoning     R1

eg\(\,\,\,\,\,\)\({\text{P (3 white)}} = {\text{P(0 blue)}}\)

\({\text{P(3 white)}} = \frac{1}{6}\)     AG     N0

[1 mark]

b.ii.

valid method     (M1)

eg\(\,\,\,\,\,\)\({\text{P(3 white)}} = \frac{w}{{10}} \times \frac{{w – 1}}{9} \times \frac{{w – 2}}{8},{\text{ }}\frac{{_w{C_3}}}{{_{10}{C_3}}}\)

correct equation     A1

eg\(\,\,\,\,\,\)\(\frac{w}{{10}} \times \frac{{w – 1}}{9} \times \frac{{w – 2}}{8} = \frac{1}{6},{\text{ }}\frac{{_w{C_3}}}{{_{10}{C_3}}} = 0.167\)

\(w = 6\)     A1     N2

[3 marks]

b.iii.

recognizing one prize in first seven attempts     (M1)

eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 7 \\ 1 \end{array}} \right),{\text{ }}{\left( {\frac{1}{6}} \right)^1}{\left( {\frac{5}{6}} \right)^6}\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 7 \\ 1 \end{array}} \right){\left( {\frac{1}{6}} \right)^1}{\left( {\frac{5}{6}} \right)^6},{\text{ }}0.390714\)

correct approach     (A1)

eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 7 \\ 1 \end{array}} \right){\left( {\frac{1}{6}} \right)^1}{\left( {\frac{5}{6}} \right)^6} \times \frac{1}{6}\)

0.065119

0.0651     A1     N2

[4 marks]

d.

Question

The following table shows a probability distribution for the random variable \(X\), where \({\text{E}}(X) = 1.2\).

M17/5/MATME/SP2/ENG/TZ2/10

A bag contains white and blue marbles, with at least three of each colour. Three marbles are drawn from the bag, without replacement. The number of blue marbles drawn is given by the random variable \(X\).

A game is played in which three marbles are drawn from the bag of ten marbles, without replacement. A player wins a prize if three white marbles are drawn.

Jill plays the game nine times. Find the probability that she wins exactly two prizes.

Answer/Explanation

Markscheme

valid approach     (M1)

eg\(\,\,\,\,\,\)\({\text{B}}(n,{\text{ }}p),{\text{ }}\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right){p^r}{q^{n – r}},{\text{ }}{(0.167)^2}{(0.833)^7},{\text{ }}\left( {\begin{array}{*{20}{c}} 9 \\ 2 \end{array}} \right)\)

0.279081

0.279     A1     N2

[2 marks]

Question

Let \(f(x) = 6 – \ln ({x^2} + 2)\), for \(x \in \mathbb{R}\). The graph of \(f\) passes through the point \((p,{\text{ }}4)\), where \(p > 0\).

Find the value of \(p\).

[2]
a.

The following diagram shows part of the graph of \(f\).

N17/5/MATME/SP2/ENG/TZ0/05.b

The region enclosed by the graph of \(f\), the \(x\)-axis and the lines \(x =  – p\) and \(x = p\) is rotated 360° about the \(x\)-axis. Find the volume of the solid formed.

[3]
b.
Answer/Explanation

Markscheme

valid approach     (M1)

eg\(\,\,\,\,\,\)\(f(p) = 4\), intersection with \(y = 4,{\text{ }} \pm 2.32\)

2.32143

\(p = \sqrt {{{\text{e}}^2} – 2} \) (exact), 2.32     A1     N2

[2 marks]

a.

attempt to substitute either their limits or the function into volume formula (must involve \({f^2}\), accept reversed limits and absence of \(\pi \) and/or \({\text{d}}x\), but do not accept any other errors)     (M1)

eg\(\,\,\,\,\,\)\(\int_{ – 2.32}^{2.32} {{f^2},{\text{ }}\pi \int {{{\left( {6 – \ln ({x^2} + 2)} \right)}^2}{\text{d}}x,{\text{ 105.675}}} } \)

331.989

\({\text{volume}} = 332\)     A2     N3

[3 marks]

b.

Question

Let f(x) = ln x − 5x , for x > 0 .

Find f ’(x).

[2]
a.

Find f ”(x).

[1]
b.

Solve f ’(x) = f ”(x).

[2]
c.
Answer/Explanation

Markscheme

\(f’\left( x \right) = \frac{1}{x} – 5\)     A1A1 N2

[2 marks]

a.

f ”(x) = −x−2      A1 N1

[1 mark]

b.

METHOD 1 (using GDC)

valid approach      (M1)

eg 

0.558257

x = 0.558       A1 N2

Note: Do not award A1 if additional answers given.

METHOD 2 (analytical)

attempt to solve their equation f '(x) = f ”(x)  (do not accept \(\frac{1}{x} – 5 =  – \frac{1}{{{x^2}}}\))      (M1)

eg  \(5{x^2} – x – 1 = 0,\,\,\frac{{1 \pm \sqrt {21} }}{{10}},\,\,\frac{1}{x} = \frac{{ – 1 \pm \sqrt {21} }}{2},\,\, – 0.358\)

0.558257

x = 0.558       A1 N2

Note: Do not award A1 if additional answers given.

[2 marks]

c.

Question

Let \(f\left( x \right) = {{\text{e}}^{2\,{\text{sin}}\left( {\frac{{\pi x}}{2}} \right)}}\), for x > 0.

The k th maximum point on the graph of f has x-coordinate xk where \(k \in {\mathbb{Z}^ + }\).

Given that xk + 1 = xk + a, find a.

[4]
a.

Hence find the value of n such that \(\sum\limits_{k = 1}^n {{x_k} = 861} \).

[4]
b.
Answer/Explanation

Markscheme

valid approach to find maxima     (M1)

eg  one correct value of xk, sketch of f

any two correct consecutive values of xk      (A1)(A1)

eg  x1 = 1, x2 = 5

a = 4      A1 N3

[4 marks]

a.

recognizing the sequence x1,  x2,  x3, …, xn is arithmetic  (M1)

eg  d = 4

correct expression for sum       (A1)

eg  \(\frac{n}{2}\left( {2\left( 1 \right) + 4\left( {n – 1} \right)} \right)\)

valid attempt to solve for n      (M1)

eg  graph, 2n2n − 861 = 0

n = 21       A1 N2

[4 marks]

b.

Question

A rock falls off the top of a cliff. Let \(h\) be its height above ground in metres, after \(t\) seconds.

The table below gives values of \(h\) and \(t\) .

Jane thinks that the function \(f(t) = – 0.25{t^3} – 2.32{t^2} + 1.93t + 106\) is a suitable model for the data. Use Jane’s model to

(i)     write down the height of the cliff;

(ii)    find the height of the rock after 4.5 seconds;

(iii)   find after how many seconds the height of the rock is \(30{\text{ m}}\).

[5]
a(i), (ii) and (iii).

Kevin thinks that the function \(g(t) = – 5.2{t^2} + 9.5t + 100\) is a better model for the data. Use Kevin’s model to find when the rock hits the ground.

[3]
b.

(i)     On graph paper, using a scale of 1 cm to 1 second, and 1 cm to 10 m, plot the data given in the table.

(ii)    By comparing the graphs of f and g with the plotted data, explain which function is a better model for the height of the falling rock.

[6]
c(i) and (ii).
Answer/Explanation

Markscheme

(i) \(106{\text{ m}}\)     A1     N1

(ii) substitute \(t = 4.5\)     M1

\(h = 44.9{\text{ m}}\)     A1     N2

(iii) set up suitable equation     M1

e.g. \(f(t) = 30\)

\(t = 4.91\)     A1     N1

[5 marks]

a(i), (ii) and (iii).

recognizing that height is 0     A1

set up suitable equation     M1

e.g. \(g(t) = 0\)

\(t = 5.39{\text{ secs}}\)     A1     N2

[3 marks]

b.


     A1A2     N3

Note: Award A1 for correct scales on axes, A2 for 5 correct points, A1 for 3 or 4 correct points.

 

(ii) Jane’s function, with 2 valid reasons     A1R1R1     N3

e.g. Jane’s passes very close to all the points, Kevin’s has the rock clearly going up initially – not possible if rock falls

Note: Although Jane’s also goes up initially, it only goes up very slightly, and so is the better model.

[6 marks]

c(i) and (ii).
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