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.
(i) Find \(f'(x)\) .
(ii) Find \(g'(x)\) .
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.
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]
(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]
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]
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 \(y = f(x)\) .
Solve the equation \(f(x) = 1\) .
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]
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]
Question
The following diagram shows two ships A and B. At noon, ship A was 15 km due north of ship B. Ship A was moving south at 15 km h–1 and ship B was moving east at 11 km h–1.
Find the distance between the ships
(i) at 13:00;
(ii) at 14:00.
Let \(s(t)\) be the distance between the ships t hours after noon, for \(0 \le t \le 4\) .
Show that \(s(t) = \sqrt {346{t^2} – 450t + 225} \) .
Sketch the graph of \(s(t)\) .
Due to poor weather, the captain of ship A can only see another ship if they are less than 8 km apart. Explain why the captain cannot see ship B between noon and 16:00.
Answer/Explanation
Markscheme
(i) evidence of valid approach (M1)
e.g. ship A where B was, B \(11{\text{ km}}\) away
\({\text{distance}} = 11\) A1 N2
(ii) evidence of valid approach (M1)
e.g. new diagram, Pythagoras, vectors
\(s = \sqrt {{{15}^2} + {{22}^2}} \) (A1)
\(\sqrt {709} = 26.62705\)
\(s = 26.6\) A1 N2
Note: Award M0A0A0 for using the formula given in part (b).
[5 marks]
evidence of valid approach (M1)
e.g. a table, diagram, formula \(d = r \times t\)
distance ship A travels t hours after noon is \(15(t – 1)\) (A2)
distance ship B travels in t hours after noon is \(11t\) (A1)
evidence of valid approach M1
e.g. \(s(t) = \sqrt {{{\left[ {15(t – 1)} \right]}^2} + {{(11t)}^2}} \)
correct simplification A1
e.g. \(\sqrt {225({t^2} – 2t + 1) + 121{t^2}} \)
\(s(t) = \sqrt {346{t^2} – 450t + 225} \) AG N0
[6 marks]
A1A1A1 N3
Note: Award A1 for shape, A1 for minimum at approximately \((0.7{\text{, }}9)\), A1 for domain.
[3 marks]
evidence of valid approach (M1)
e.g. \(s'(t) = 0\) , find minimum of \(s(t)\) , graph, reference to “more than 8 km”
\(\min = 8.870455 \ldots \) (accept 2 or more sf) A1
since \({s_{\min}} > 8\) , captain cannot see ship B R1 N0
[3 marks]