A-Level Maths OCR A H240
Paper 1 - Jun 2018 H240/01
Time allowed: 2 h 0 min.
The total mark for this paper is 100.
The marks for each question are shown in brackets [ ].
Find the equation of the straight line which passes through the point (2, 8) and is perpendicular to AB.
Give your answer in the form \(ax + by = c \), where \(a\), \(b\) and \(c\) are constants.
\(m = \dfrac{17-5}{4-1} (= 4) \)
\(m_{perp} = -\dfrac{1}{4} \)
\(y-8 = -\dfrac{1}{4}(x-2)\)
\(x + 4y = 34 \)
\[\displaystyle\int_0^2 e^{x^2} dx \]
giving your answer correct to 3 significant figures.
(ii) Explain how the trapezium rule could be used to obtain a more accurate estimate.
(i) \(0.5 × 0.5 [(e^0 + e^{2^2}) + 2(e^{0.5^2} + e^{1^2} + e^{1.5^2})] \)
\(= 20.6\)
(ii) Use more trapezia, of a narrower width, over the same interval
Find the two real roots of the equation \(x^4 - 5 = 4x^2 \).
Give the roots in an exact form.
\((x^2 - 5)(x^2 + 1) = 0\)
\(x^2 = 5\)
\(x^2 \geq 0\), so \(x^2 + 1 = 0\) has no real solutions
\(x = \pm \sqrt{5}\)
If \(n\) is even then \(n\) can be written as \(2m\).
\(n^3 + 3n - 1 = 8m^3 + 6m - 1 \)
\(= 2(4m^3 + 3m) - 1\)
For all \(m\), \(2(4m^3+ 3m)\) is even, hence \(2(4m^3 + 3m ) - 1\) is odd.
If \(n\) is odd then \(n\) can be written as \(2m+ 1\).
\(n^3+3n–1 = 8m^3+ 12m^2+ 6m+ 1 + 6m+ 3 -1\)
\(= 8m^3+ 12m^2+ 12m+ 3\)
\(= 2(4m^3+ 6m^2+ 6m) + 3 \)
For all \(m\), \(2(4m^3+ 6m^2+ 6m)\) is even, hence \(2(4m^3+ 6m^2+ 6m) + 3\) is odd.
(i) Find the centre and radius of the circle.
(ii) Find the coordinates of any points where the line \(y = 2x - 3\) meets the circle \(x^2 + y^2 + 6x - 2y - 10 = 0 \)
(iii) State what can be deduced from the answer to part (ii) about the line \(y = 2x - 3\) and the circle \(x^2 + y^2 + 6x - 2y - 10 = 0 \)
(i) \((x+ 3)^2-9 + (y-1)^2-1 -10 = 0\)
\((x+ 3)^2+ (y-1)^2= 20\)
centre is \((-3, 1)\)
radius is \(2\sqrt{5}\) or \(\sqrt{20}\)
(ii) \(x^2 + (2x-3)^2 + 6x - 2(2x-3) - 10 = 0\)
OR
\((x+3)^2 + (2x-4)^2 = 20\)
\(x^2 - 2x + 1 = 0\)
\(x = 1\)
\((1,-1)\)
(iii) The line is a tangent to the circle at \((1,-1)\)
(i) Given that \((x-3)\) is a factor of \(f(x)\), express \(f(x)\) in a fully factorised form.
(ii) Sketch the graph of \(y = f(x)\), indicating the coordinates of any points of intersection with the axes.
(iii) Solve the inequality \(f(x) < 0 \), giving your answer in set notation.
(iv) The graph of \(y = f(x)\) is transformed by a stretch parallel to the x-axis, scale factor \(\frac{1}{2}\).
Find the equation of the transformed graph.
Polynomials Transformations of graphs
(i) \(f(x) = (x-3)(2x^2-x-1)\)
\(f(x) = (x-3)(2x+1)(x-1)\)
(ii) Sketch of positive cubic
(–0.5, 0), (1, 0), (3, 0), (0, 3)
(iii) \(\{x:x < -0.5\}∪\{x:1 < x < 3\}\)
(iv) \(y= 2(2x)^3-7(2x)^2+ 2(2x) + 3\)
\(= 16x^3-28x^2+ 4x+ 3\)
OR
\(y= (2x-3)(4x+ 1)(2x-1)\)
(i) Chris sets himself a target of completing a half marathon in less than 120 minutes. Show that this model predicts that Chris will achieve his target on his thirteenth half marathon.
(ii) After twelve months Chris has spent a total of 2974 minutes, to the nearest minute, running half marathons. Use this model to find how many half marathons he has run.
(iii) Give two reasons why this model may not be appropriate when predicting the time for a half marathon.
Geometric sequences and series
(i) \(r = 0.98\)
\(u_{12}= 150 × 0.98^{11} \) or \(u_{13}= 150 × 0.98^{12} \)
\(u_{12} = 120.1\) or \(u_{13} = 117.7\)
\(u_{12} > 120\) and \(u_{13} < 120\), hence the thirteenth half marathon
(ii) \(\dfrac{150(1-0.98^n)}{1-0.98} = 2974\)
\(0.98^n = 0.6035\)
\(n= 25\) hence Chris has run 25 half marathons
(iii) Does not take into account possible variations in conditions
Assumes that times will continue to improve
(ii) The expansion of \(\dfrac{a+bx}{\sqrt{4-x}}\) is \(16-x ... \) Find the values of the constants \(a\) and \(b\).
(i) \((4-x)^{-\frac{1}{2}} = 4^{-\frac{1}{2}}(1-\frac{1}{4}x)^{-\frac{1}{2}} = \frac{1}{2}(1-\frac{1}{4}x)^{-\frac{1}{2}}\)
\((1-\frac{1}{2}x)^{-\frac{1}{2}} = 1 + (-\frac{1}{2})(-\frac{1}{4}x) + (\frac{1}{2})(-\frac{1}{2})(-\frac{3}{2})(-\frac{1}{4}x)^2 \)
\(= 1 + \frac{1}{8}x + \frac{3}{128}x^2\)
\((4-x)^{-\frac{1}{2}} = \frac{1}{2} + \frac{1}{16}x + \frac{3}{256}x^2\)
(ii) \(\frac{1}{2}a = 16 \) hence \(a= 32\)
\(2 + \frac{1}{2}b = -1\)
OR
\(\frac{1}{16}a + \frac{1}{2}b = -1\)
\(b = -6 \)
(i) Given that the range of \(f\) is \(f(x) \leq 19 \), find the value of \(c\).
(ii) Given instead that \(ff(2) = 8\), find the possible values of \(c\).
Composite functions Functions, mappings, domain and range
(i) \(f(x) = c+ 16 -(x-4)^2\)
\(c+ 16 = 19\)
\(c= 3\)
(ii) \(f(2) = c+ 12\)
\(f(c+ 12) = c+ 8(c+ 12) -(c+ 12)^2\)
\(-48 -15c-c^2= 8\)
\(c^2+ 15c+ 56 = 0\)
\(c = -7, c = -8\)
(i) Find \(\dfrac{dy}{dx}\) in terms of \(t\), giving your answer in its simplest form.
(ii) Explain why the curve has no stationary points.
(iii) By considering \(x+y\), or otherwise, find a cartesian equation of the curve, giving your answer in a form not involving fractions or brackets.
(i) \(\dfrac{dx}{dt} = 1-2t^{-2}\)
\(\dfrac{dy}{dt} = 1+2t^{-2}\)
\(\dfrac{dy}{dx} = \dfrac{1+2t^{-2}}{1-2t^{-2}} = \dfrac{\dfrac{t^2+2}{t^2}}{\dfrac{t^2-2}{t^2}}\)
\(\dfrac{dy}{dx} = \dfrac{t^2+2}{t^2-2}\)
(ii) \(\dfrac{dy}{dx} = 0 \implies t^2 + 2 = 0\)
\(t^2 \geq 0\), hence \(t^2+2=0\) has no solutions, hence curve has no stationary points
(iii) \(x + y = 2t\) hence \(t = \frac{1}{2}(x+y)\)
\(x = \frac{1}{2}(x+y) + \dfrac{2}{\frac{1}{2}(x+y)}\)
\(2x(x+ y) = (x+ y)^2+ 8\)
\(2x^2+ 2xy= x^2+ 2xy+ y^2+ 8\)
\(x^2-y^2= 8\)
(i) Find the time taken for the mass to decrease to half of its original value.
A second substance is also decaying exponentially. Initially its mass was 400 grams and, after 10 minutes, its mass was 320 grams.
(ii) Find the time at which both substances are decaying at the same rate.
(i) When \(t= 0\), \(M= 300\)
\(300e^{-0.05t} = 150\)
\(e^{-0.05t} = 0.5\)
\(-0.05t = \ln{0.5}\)
\(t= 13.9\) (minutes)
(ii) \(M_2 = 400e^{kt}\)
\(320 = 400e^{10k}\)
\(k= 0.1\ln{0.8}\)
\(M_2 = 400e^{-0.0223t}\)
Substance 1: \(\dfrac{dM_1}{dt} = -15e^{-0.05t}\)
Substance 2: \(\dfrac{dM_2}{dt} = -8.93e^{-0.0223t}\)
\(-15e^{-0.05t} = -8.93e^{-0.0223t}\)
\(e^{0.0277t} = 1.681\)
\(0.0277t= 0.519\)
time = 18.75 minutes
The diagram shows the curve \(y = \dfrac{4\cos{2x}}{3-\sin{2x}}\), for \(x \geq 0\), and the normal to the curve at the point \((\frac{1}{4}π, 0)\). Show that the exact area of the shaded region enclosed by the curve, the normal to the curve and the y-axis is \(\ln{\dfrac{9}{4} + \dfrac{1}{128}π^2}\)
\(\dfrac{dy}{dx} = \dfrac{(-8\sin{2x})(3-\sin{2x})-(4\cos{2x})(-2\cos{2x})}{(3-\sin{2x})^2}\)
EITHER
when \(x = \frac{1}{4}π\), gradient \(= \dfrac{-16-0}{4} = -4 \)
OR
\(\dfrac{(-8\sin{\frac{π}{2}})(3-\sin{\frac{π}{2}})-(4\cos{\frac{π}{2}})(-2\cos{\frac{π}{2}})}{(3-\sin{\frac{π}{2}})^2} = -4\)
gradient of normal is \(\frac{1}{4}\)
area of triangle is \(\dfrac{1}{2}×\dfrac{1}{16}π×\dfrac{1}{4}π = \dfrac{1}{128}π^2\)
\(\displaystyle\int \dfrac{4\cos{2x}}{3-\sin{2x}} dx = -2\ln{|3-\sin{2x}|}\)
\(\displaystyle\int^{\frac{1}{4}π}_0 \dfrac{4\cos{2x}}{3-\sin{2x}} dx = (-2\ln{2}) - (-2\ln{3})\)
\(2\ln{3} -2 \ln{2} = \ln{9} -\ln{4} = \ln{\frac{9}{4}} \)
OR
\(2\ln{3} -2 \ln{2} = 2\ln{\frac{3}{2}} = \ln{\frac{9}{4}} \)
hence total area is \(\ln{\dfrac{9}{4}} + \dfrac{1}{128}π^2 \)
(i) A scientist assumes that the rate of increase of the number of insects is inversely proportional to the number of insects present at time \(t\).
(a) Write down a differential equation to model this situation.
(b) Solve this differential equation to find \(N\) in terms of \(t\).
(ii) In a revised model it is assumed that \(\dfrac{dN}{dt} = \dfrac{N^2}{3988e^{0.2t}} \).
Solve this differential equation to find \(N\) in terms of \(t\).
(iii) Compare the long-term behaviour of the two models.
Differential equations Solve differential equations
(i) (a) \(\dfrac{dN}{dt} = \dfrac{k}{N}\)
(b) \(ft = \int N dN\)
\(kt = \frac{1}{2} N^2 + c\)
\(0 = 80000+c \implies c = -80000 \)
\(k = 96800-80000=16800\)
\(N = \sqrt{33600t+160000}\)
(ii) \(\int 3988N^{-2} dN = \int e^{-0.2t} dt\)
\(-3988N^{-1} = -5e^{-0.2t} + c\)
OR
\(-N^{-1} = -\dfrac{5}{3988}e^{-0.2t} + c\)
\(-9.97 = -5 + c \implies c = -4.97\)
OR
\(-\dfrac{1}{400} = -\dfrac{5}{3988} + c \implies c = -\dfrac{497}{398800}\)
\(\dfrac{3988}{N} = 5e^{-0.2t} + 4.97\)
OR
\(\dfrac{1}{N} = \dfrac{5}{3988}e^{-0.2t} + \dfrac{497}{398800}\)
\(N = \dfrac{3988}{5e^{-0.2t} + 4.97}\)
(iii) Model in (i) predicts that population will continue to increase
Model in (ii) predicts that population will tend towards a limit of 802