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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.
(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
Consider the curve \(y = \ln (3x – 1)\) . Let P be the point on the curve where \(x = 2\) .
Write down the gradient of the curve at P.
The normal to the curve at P cuts the x-axis at R. Find the coordinates of R.
Answer/Explanation
Markscheme
gradient is \(0.6\) A2 N2
[2 marks]
at R, \(y = 0\) (seen anywhere) A1
at \(x = 2\) , \(y = \ln 5\) \(( = 1.609 \ldots )\) (A1)
gradient of normal \( = – 1.6666 \ldots \) (A1)
evidence of finding correct equation of normal A1
e.g. \(y = \ln 5 = – \frac{5}{3}(x – 2)\) , \(y = – 1.67x + c\)
\(x = 2.97\) (accept 2.96) A1
coordinates of R are (2.97,0) N3
[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 .
Write down
(i) the amplitude;
(ii) the period;
(iii) the x-intercept that lies between \( – \frac{\pi }{2}\) and 0.
Hence write \(f(x)\) in the form \(p\sin (qx + r)\) .
Write down one value of x such that \(f'(x) = 0\) .
Write down the two values of k for which the equation \(f(x) = k\) has exactly two solutions.
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.
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]
(i) 5 A1 N1
(ii) \(2\pi \) (6.28) A1 N1
(iii) \( – 0.927\) A1 N1
[3 marks]
\(f(x) = 5\sin (x + 0.927)\) (accept \(p = 5\) , \(q = 1\) , \(r = 0.927\) ) A1A1A1 N3
[3 marks]
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]
\(k = – 5\) , \(k = 5\) A1A1 N2
[2 marks]
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]
Question
Let \(f(x) = \cos 2x\) and \(g(x) = \ln (3x – 5)\) .
Find \(f'(x)\) .
Find \(g'(x)\) .
Let \(h(x) = f(x) \times g(x)\) . Find \(h'(x)\) .
Answer/Explanation
Markscheme
(a) \(f'(x) = – \sin 2x \times 2( = – 2\sin 2x)\) A1A1 N2
Note: Award A1 for 2, A1 for \( – \sin 2x\) .
[2 marks]
\(g'(x) = 3 \times \frac{1}{{3x – 5}}\) \(\left( { = \frac{3}{{3x – 5}}} \right)\) A1A1 N2
Note: Award A1 for 3, A1 for \(\frac{1}{{3x – 5}}\) .
[2 marks]
evidence of using product rule (M1)
\(h'(x) = (\cos 2x)\left( {\frac{3}{{3x – 5}}} \right) + \ln (3x – 5)( – 2\sin 2x)\) A1 N2
[2 marks]
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 \(A = 10\) .
Given that \(f(15) = 3.49\) (correct to 3 significant figures), find the value of k.
(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 .
Let \(g(x) = – {x^2} + 12x – 24\) .
Find the area enclosed by the graphs of f and g .
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]
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]
(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]
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]
Question
Let \(f(x) = \sqrt[3]{{{x^4}}} – \frac{1}{2}\).
Find \(f'(x)\).
Find \(\int {f(x){\text{d}}x} \).
Answer/Explanation
Markscheme
expressing \(f\) as \({x^{\frac{4}{3}}}\) (M1)
\(f'(x) = \frac{4}{3}{x^{\frac{1}{3}}}{\text{ }}\left( { = \frac{4}{3}\sqrt[3]{x}} \right)\) A1 N2
[2 marks]
attempt to integrate \({\sqrt[3]{{{x^4}}}}\) (M1)
eg \(\frac{{{x^{\frac{4}{3} + 1}}}}{{\frac{4}{3} + 1}}\)
\(\int {f(x){\text{d}}x = \frac{3}{7}{x^{\frac{7}{3}}} – \frac{x}{2} + c} \) A1A1A1 N4
[4 marks]
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 \(f\).
Find \(f'(x)\).
Write down the value of \(b\).
Given that \(g'(1) = – e\), find the value of \(a\).
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.
Answer/Explanation
Markscheme
\(y = 2\) (correct equation only) A2 N2
[2 marks]
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]
correct equation for the asymptote of \(g\)
eg\(\,\,\,\,\,\)\(y = b\) (A1)
\(b = 2\) A1 N2
[2 marks]
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]
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,
correct solution (may be seen in sketch) (A1)
eg\(\,\,\,\,\,\)\(x = 2,{\text{ }}(2,{\text{ }} – 1)\)
gradient is \( – 1\) A1 N3
[4 marks]
Question
Let \(f(x) = {({x^2} + 3)^7}\). Find the term in \({x^5}\) in the expansion of the derivative, \(f’(x)\).
Answer/Explanation
Markscheme
METHOD 1
derivative of \(f(x)\) A2
\(7{({x^2} + 3)^6}(x2)\)
recognizing need to find \({x^4}\) term in \({({x^2} + 3)^6}\) (seen anywhere) R1
eg\(\,\,\,\,\,\)\(14x{\text{ (term in }}{x^4})\)
valid approach to find the terms in \({({x^2} + 3)^6}\) (M1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 6 \\ r \end{array}} \right){({x^2})^{6 – r}}{(3)^r},{\text{ }}{({x^2})^6}{(3)^0} + {({x^2})^5}{(3)^1} + \ldots \), Pascal’s triangle to 6th row
identifying correct term (may be indicated in expansion) (A1)
eg\(\,\,\,\,\,\)\({\text{5th term, }}r = 2,{\text{ }}\left( {\begin{array}{*{20}{c}} 6 \\ 4 \end{array}} \right),{\text{ }}{({x^2})^2}{(3)^4}\)
correct working (may be seen in expansion) (A1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 6 \\ 4 \end{array}} \right){({x^2})^2}{(3)^4},{\text{ }}15 \times {3^4},{\text{ }}14x \times 15 \times 81{({x^2})^2}\)
\(17010{x^5}\) A1 N3
METHOD 2
recognition of need to find \({x^6}\) in \({({x^2} + 3)^7}\) (seen anywhere) R1
valid approach to find the terms in \({({x^2} + 3)^7}\) (M1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 7 \\ r \end{array}} \right){({x^2})^{7 – r}}{(3)^r},{\text{ }}{({x^2})^7}{(3)^0} + {({x^2})^6}{(3)^1} + \ldots \), Pascal’s triangle to 7th row
identifying correct term (may be indicated in expansion) (A1)
eg\(\,\,\,\,\,\)6th term, \(r = 3,{\text{ }}\left( {\begin{array}{*{20}{c}} 7 \\ 3 \end{array}} \right),{\text{ (}}{{\text{x}}^2}{)^3}{(3)^4}\)
correct working (may be seen in expansion) (A1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 7 \\ 4 \end{array}} \right){{\text{(}}{{\text{x}}^2})^3}{(3)^4},{\text{ }}35 \times {3^4}\)
correct term (A1)
\(2835{x^6}\)
differentiating their term in \({x^6}\) (M1)
eg\(\,\,\,\,\,\)\((2835{x^6})’,{\text{ (6)(2835}}{{\text{x}}^5})\)
\(17010{x^5}\) A1 N3
[7 marks]
Question
Let f(x) = ln x − 5x , for x > 0 .
Find f ’(x).
Find f ”(x).
Solve f ’(x) = f ”(x).
Answer/Explanation
Markscheme
\(f’\left( x \right) = \frac{1}{x} – 5\) A1A1 N2
[2 marks]
f ”(x) = −x−2 A1 N1
[1 mark]
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]