IB DP Maths Topic 2.7 Use of technology to solve a variety of equations SL Paper 2

 

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Question

Let \(f(x) = 4{\tan ^2}x – 4\sin x\) , \( – \frac{\pi }{3} \le x \le \frac{\pi }{3}\) .

On the grid below, sketch the graph of \(y = f(x)\) .


[3]
a.

Solve the equation \(f(x) = 1\) .

[3]
b.
Answer/Explanation

Markscheme

     A1A1A1     N3

Note: Award A1 for passing through \((0{\text{, }}0)\), A1 for correct shape, A1 for a range of approximately \( – 1\) to 15.

[3 marks]

a.

evidence of attempt to solve \(f(x) = 1\)     (M1)

e.g. line on sketch, using \(\tan x = \frac{{\sin x}}{{\cos x}}\)

\(x = – 0.207\) , \(x = 0.772\)     A1A1     N3

[3 marks]

b.

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

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

The diagram below shows a quadrilateral ABCD with obtuse angles \({\rm{A}}\widehat {\rm{B}}{\rm{C}}\) and \({\rm{A}}\widehat {\rm{D}}{\rm{C}}\).


AB = 5 cm, BC = 4 cm, CD = 4 cm, AD = 4 cm , \({\rm{B}}\widehat {\rm{A}}{\rm{C}} = {30^ \circ }\) , \({\rm{A}}\widehat {\rm{B}}{\rm{C}} = {x^ \circ }\) , \({\rm{A}}\widehat {\rm{D}}{\rm{C}} = {y^ \circ }\) .

Use the cosine rule to show that \({\rm{AC}} = \sqrt {41 – 40\cos x} \) .

[1]
a.

Use the sine rule in triangle ABC to find another expression for AC.

[2]
b.

(i)     Hence, find x, giving your answer to two decimal places.

(ii)    Find AC .

[6]
c.

(i)     Find y.

(ii)    Hence, or otherwise, find the area of triangle ACD.

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

Markscheme

correct substitution     A1

e.g. \(25 + 16 – 40\cos x\) , \({5^2} + {4^2} – 2 \times 4 \times 5\cos x\)

\({\rm{AC}} = \sqrt {41 – 40\cos x} \)     AG

[1 mark]

a.

correct substitution     A1

e.g. \(\frac{{{\rm{AC}}}}{{\sin x}} = \frac{4}{{\sin 30}}\) , \(\frac{1}{2}{\rm{AC}} = 4\sin x\)

\({\rm{AC}} = 8\sin x\) (accept \(\frac{{4\sin x}}{{\sin 30}}\))     A1     N1

[2 marks]

b.

(i) evidence of appropriate approach using AC     M1

e.g. \(8\sin x = \sqrt {41 – 40\cos x} \) , sketch showing intersection

correct solution \(8.682 \ldots \), \(111.317 \ldots \)     (A1)

obtuse value \(111.317 \ldots \)     (A1)

\(x = 111.32\) to 2 dp (do not accept the radian answer 1.94 )     A1     N2

(ii) substituting value of x into either expression for AC     (M1)

e.g. \({\rm{AC}} = 8\sin 111.32\)

\({\rm{AC}} = 7.45\)     A1     N2

[6 marks]

c.

(i) evidence of choosing cosine rule     (M1)

e.g. \(\cos B = \frac{{{a^2} + {c^2} – {b^2}}}{{2ac}}\)

correct substitution     A1

e.g. \(\frac{{{4^2} + {4^2} – {{7.45}^2}}}{{2 \times 4 \times 4}}\) , \({7.45^2} = 32 – 32\cos y\) , \(\cos y = – 0.734 \ldots \)

\(y = 137\)     A1     N2

(ii) correct substitution into area formula     (A1)

e.g. \(\frac{1}{2} \times 4 \times 4 \times \sin 137\) , \(8\sin 137\)

area \(= 5.42\)     A1     N2

[5 marks]

d(i) and (ii).

Question

Let \(f(x) = A{{\rm{e}}^{kx}} + 3\) . Part of the graph of f is shown below.


The y-intercept is at (0, 13) .

Show that \(A = 10\) .

[2]
a.

Given that \(f(15) = 3.49\) (correct to 3 significant figures), find the value of k.

[3]
b.

(i)     Using your value of k , find \(f'(x)\) .

(ii)    Hence, explain why f is a decreasing function.

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

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

Let \(g(x) = – {x^2} + 12x – 24\) .

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

[6]
d.
Answer/Explanation

Markscheme

substituting (0, 13) into function     M1 

e.g. \(13 = A{{\rm{e}}^0} + 3\)

\(13 = A + 3\)     A1

\(A = 10\)     AG     N0

[2 marks]

a.

substituting into \(f(15) = 3.49\)     A1

e.g. \(3.49 = 10{{\rm{e}}^{15k}} + 3\) , \(0.049 = {{\rm{e}}^{15k}}\)

evidence of solving equation     (M1)

e.g. sketch, using \(\ln \)

\(k = – 0.201\) (accept \(\frac{{\ln 0.049}}{{15}}\) )     A1     N2

[3 marks]

b.

(i) \(f(x) = 10{{\rm{e}}^{ – 0.201x}} + 3\)

\(f(x) = 10{{\rm{e}}^{ – 0.201x}} \times – 0.201\) \(( = – 2.01{{\rm{e}}^{ – 0.201x}})\)     A1A1A1     N3

Note: Award A1 for \(10{{\rm{e}}^{ – 0.201x}}\) , A1 for \( \times – 0.201\) , A1 for the derivative of 3 is zero.

(ii) valid reason with reference to derivative     R1     N1

e.g. \(f'(x) < 0\) , derivative always negative

(iii) \(y = 3\)     A1     N1

[5 marks]

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

finding limits \(3.8953 \ldots \), \(8.6940 \ldots \) (seen anywhere)     A1A1

evidence of integrating and subtracting functions     (M1)

correct expression     A1

e.g. \(\int_{3.90}^{8.69} {g(x) – f(x){\rm{d}}x} \) , \(\int_{3.90}^{8.69} {\left[ {\left( { – {x^2} + 12x – 24} \right) – \left( {10{{\rm{e}}^{ – 0.201x}} + 3} \right)} \right]} {\rm{d}}x\)

area \(= 19.5\)     A2     N4

[6 marks]

d.

Question

Consider \(f(x) = 2 – {x^2}\) , for \( – 2 \le x \le 2\) and \(g(x) = \sin {{\rm{e}}^x}\) , for \( – 2 \le x \le 2\) . The graph of f is given below.


On the diagram above, sketch the graph of g.

[3]
a.

Solve \(f(x) = g(x)\) .

[2]
b.

Write down the set of values of x such that \(f(x) > g(x)\) .

[2]
c.
Answer/Explanation

Markscheme


     A1A1A1     N3

[3 marks]

a.

\(x = – 1.32\) , \(x = 1.68\) (accept \(x = – 1.41\) , \(x = 1.39\) if working in degrees)     A1A1     N2

[2 marks]

b.

\( – 1.32 < x < 1.68\) (accept \( – 1.41 < x < 1.39\) if working in degrees)     A2     N2

[2 marks]

c.

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

The following diagram shows the graph of \(f(x) = {{\rm{e}}^{ – {x^2}}}\) .


The points A, B, C, D and E lie on the graph of f . Two of these are points of inflexion.

Identify the two points of inflexion.

[2]
a.

(i)     Find \(f'(x)\) .

(ii)    Show that \(f”(x) = (4{x^2} – 2){{\rm{e}}^{ – {x^2}}}\) .

[5]
b(i) and (ii).

Find the x-coordinate of each point of inflexion.

[4]
c.

Use the second derivative to show that one of these points is a point of inflexion.

[4]
d.
Answer/Explanation

Markscheme

B, D     A1A1     N2

[2 marks]

a.

(i) \(f'(x) = – 2x{{\rm{e}}^{ – {x^2}}}\)     A1A1     N2

Note: Award A1 for \({{\rm{e}}^{ – {x^2}}}\) and A1 for \( – 2x\) .

(ii) finding the derivative of \( – 2x\) , i.e. \( – 2\)     (A1)

evidence of choosing the product rule     (M1)

e.g. \( – 2{{\rm{e}}^{ – {x^2}}}\) \( – 2x \times – 2x{{\rm{e}}^{ – {x^2}}}\)

\( – 2{{\rm{e}}^{ – {x^2}}} + 4{x^2}{{\rm{e}}^{ – {x^2}}}\)     A1

\(f”(x) = (4{x^2} – 2){{\rm{e}}^{ – {x^2}}}\)     AG     N0

[5 marks]

b(i) and (ii).

valid reasoning     R1

e.g. \(f”(x) = 0\)

attempting to solve the equation     (M1)

e.g. \((4{x^2} – 2) = 0\) , sketch of \(f”(x)\)

\(p = 0.707\) \(\left( { = \frac{1}{{\sqrt 2 }}} \right)\) , \(q = – 0.707\) \(\left( { = – \frac{1}{{\sqrt 2 }}} \right)\)     A1A1     N3

[4 marks]

c.

evidence of using second derivative to test values on either side of POI     M1

e.g. finding values, reference to graph of \(f”\) , sign table

correct working     A1A1

e.g. finding any two correct values either side of POI,

checking sign of \(f”\) on either side of POI

reference to sign change of \(f”(x)\)     R1     N0

[4 marks]

d.

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

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 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

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.

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.
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