Question
Let y = \(\frac{Inx}{x^{4}}\) for x > 0.
(a) Show that \(\frac{dy}{dx}= \frac{1-4Inx}{x^{5}}\)
Consider the function defined by f (x) \(\frac{Inx}{x^{4}}\) = for x> 0 and its graph y = f (x) .
(b) The graph of f has a horizontal tangent at point P. Find the coordinates of P. [5]
(c) Given that f ” (x) = \(\frac{20Lnx-9}{x^{6}}\) show that P is a local maximum point. [3]
(d) Solve f (x) > 0 for x > 0. [2]
(e) Sketch the graph of f , showing clearly the value of the x-intercept and the approximate position of point P. [3]
Answer/Explanation
Ans
Question
Consider \(f(x) = \frac{1}{3}{x^3} + 2{x^2} – 5x\) . Part of the graph of f is shown below. There is a maximum point at M, and a point of inflexion at N.
Find \(f'(x)\) .
Find the x-coordinate of M.
Find the x-coordinate of N.
The line L is the tangent to the curve of f at \((3{\text{, }}12)\). Find the equation of L in the form \(y = ax + b\) .
Answer/Explanation
Markscheme
\(f'(x) = {x^2} + 4x – 5\) A1A1A1 N3
[3 marks]
evidence of attempting to solve \(f'(x) = 0\) (M1)
evidence of correct working A1
e.g. \((x + 5)(x – 1)\) , \(\frac{{ – 4 \pm \sqrt {16 + 20} }}{2}\) , sketch
\(x = – 5\), \(x = 1\) (A1)
so \(x = – 5\) A1 N2
[4 marks]
METHOD 1
\(f”(x) = 2x + 4\) (may be seen later) A1
evidence of setting second derivative = 0 (M1)
e.g. \(2x + 4 = 0\)
\(x = – 2\) A1 N2
METHOD 2
evidence of use of symmetry (M1)
e.g. midpoint of max/min, reference to shape of cubic
correct calculation A1
e.g. \(\frac{{ – 5 + 1}}{2}\)
\(x = – 2\) A1 N2
[3 marks]
attempting to find the value of the derivative when \(x = 3\) (M1)
\(f'(3) = 16\) A1
valid approach to finding the equation of a line M1
e.g. \(y – 12 = 16(x – 3)\) , \(12 = 16 \times 3 + b\)
\(y = 16x – 36\) A1 N2
[4 marks]
Question
A rectangle is inscribed in a circle of radius 3 cm and centre O, as shown below.
The point P(x , y) is a vertex of the rectangle and also lies on the circle. The angle between (OP) and the x-axis is \(\theta \) radians, where \(0 \le \theta \le \frac{\pi }{2}\) .
Write down an expression in terms of \(\theta \) for
(i) \(x\) ;
(ii) \(y\) .
Let the area of the rectangle be A.
Show that \(A = 18\sin 2\theta \) .
(i) Find \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }}\) .
(ii) Hence, find the exact value of \(\theta \) which maximizes the area of the rectangle.
(iii) Use the second derivative to justify that this value of \(\theta \) does give a maximum.
Answer/Explanation
Markscheme
(i) \(x = 3\cos \theta \) A1 N1
(ii) \(y = 3\sin \theta \) A1 N1
[2 marks]
finding area (M1)
e.g. \(A = 2x \times 2y\) , \(A = 8 \times \frac{1}{2}bh\)
substituting A1
e.g. \(A = 4 \times 3\sin \theta \times 3\cos \theta \) , \(8 \times \frac{1}{2} \times 3\cos \theta \times 3\sin \theta \)
\(A = 18(2\sin \theta \cos \theta )\) A1
\(A = 18\sin 2\theta \) AG N0
[3 marks]
(i) \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }} = 36\cos 2\theta \) A2 N2
(ii) for setting derivative equal to 0 (M1)
e.g. \(36\cos 2\theta = 0\) , \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }} = 0\)
\(2\theta = \frac{\pi }{2}\) (A1)
\(\theta = \frac{\pi }{4}\) A1 N2
(iii) valid reason (seen anywhere) R1
e.g. at \(\frac{\pi }{4}\), \(\frac{{{{\rm{d}}^2}A}}{{{\rm{d}}{\theta ^2}}} < 0\) ; maximum when \(f”(x) < 0\)
finding second derivative \(\frac{{{{\rm{d}}^2}A}}{{{\rm{d}}{\theta ^2}}} = – 72\sin 2\theta \) A1
evidence of substituting \(\frac{\pi }{4}\) M1
e.g. \( – 72\sin \left( {2 \times \frac{\pi }{4}} \right)\) , \( – 72\sin \left( {\frac{\pi }{2}} \right)\) , \( – 72\)
\(\theta = \frac{\pi }{4}\) produces the maximum area AG N0
[8 marks]
Question
Let \(f(x) = {{\rm{e}}^{ – 3x}}\) and \(g(x) = \sin \left( {x – \frac{\pi }{3}} \right)\) .
Write down
(i) \(f'(x)\) ;
(ii) \(g'(x)\) .
Let \(h(x) = {{\rm{e}}^{ – 3x}}\sin \left( {x – \frac{\pi }{3}} \right)\) . Find the exact value of \(h’\left( {\frac{\pi }{3}} \right)\) .
Answer/Explanation
Markscheme
(i) \( – 3{{\rm{e}}^{ – 3x}}\) A1 N1
(ii) \(\cos \left( {x – \frac{\pi }{3}} \right)\) A1 N1
[4 marks]
evidence of choosing product rule (M1)
e.g. \(uv’ + vu’\)
correct expression A1
e.g. \( – 3{{\rm{e}}^{ – 3x}}\sin \left( {x – \frac{\pi }{3}} \right) + {{\rm{e}}^{ – 3x}}\cos \left( {x – \frac{\pi }{3}} \right)\)
complete correct substitution of \(x = \frac{\pi }{3}\) (A1)
e.g. \( – 3{{\rm{e}}^{ – 3\frac{\pi }{3}}}\sin \left( {\frac{\pi }{3} – \frac{\pi }{3}} \right) + {{\rm{e}}^{ – 3\frac{\pi }{3}}}\cos \left( {\frac{\pi }{3} – \frac{\pi }{3}} \right)\)
\(h’\left( {\frac{\pi }{3}} \right) = {{\rm{e}}^{ – \pi }}\) A1 N3
[4 marks]
Question
Consider \(f(x) = {x^2} + \frac{p}{x}\) , \(x \ne 0\) , where p is a constant.
Find \(f'(x)\) .
There is a minimum value of \(f(x)\) when \(x = – 2\) . Find the value of \(p\) .
Answer/Explanation
Markscheme
\(f'(x) = 2x – \frac{p}{{{x^2}}}\) A1A1 N2
Note: Award A1 for \(2x\) , A1 for \( – \frac{p}{{{x^2}}}\) .
[2 marks]
evidence of equating derivative to 0 (seen anywhere) (M1)
evidence of finding \(f'( – 2)\) (seen anywhere) (M1)
correct equation A1
e.g. \( – 4 – \frac{p}{4} = 0\) , \( – 16 – p = 0\)
\(p = – 16\) A1 N3
[4 marks]
Question
Let \(f(x) = 3 + \frac{{20}}{{{x^2} – 4}}\) , for \(x \ne \pm 2\) . The graph of f is given below.
The y-intercept is at the point A.
(i) Find the coordinates of A.
(ii) Show that \(f'(x) = 0\) at A.
The second derivative \(f”(x) = \frac{{40(3{x^2} + 4)}}{{{{({x^2} – 4)}^3}}}\) . Use this to
(i) justify that the graph of f has a local maximum at A;
(ii) explain why the graph of f does not have a point of inflexion.
Describe the behaviour of the graph of \(f\) for large \(|x|\) .
Write down the range of \(f\) .
Answer/Explanation
Markscheme
(i) coordinates of A are \((0{\text{, }} – 2)\) A1A1 N2
(ii) derivative of \({x^2} – 4 = 2x\) (seen anywhere) (A1)
evidence of correct approach (M1)
e.g. quotient rule, chain rule
finding \(f'(x)\) A2
e.g. \(f'(x) = 20 \times ( – 1) \times {({x^2} – 4)^{ – 2}} \times (2x)\) , \(\frac{{({x^2} – 4)(0) – (20)(2x)}}{{{{({x^2} – 4)}^2}}}\)
substituting \(x = 0\) into \(f'(x)\) (do not accept solving \(f'(x) = 0\) ) M1
at A \(f'(x) = 0\) AG N0
[7 marks]
(i) reference to \(f'(x) = 0\) (seen anywhere) (R1)
reference to \(f”(0)\) is negative (seen anywhere) R1
evidence of substituting \(x = 0\) into \(f”(x)\) M1
finding \(f”(0) = \frac{{40 \times 4}}{{{{( – 4)}^3}}}\) \(\left( { = – \frac{5}{2}} \right)\) A1
then the graph must have a local maximum AG
(ii) reference to \(f”(x) = 0\) at point of inflexion (R1)
recognizing that the second derivative is never 0 A1 N2
e.g. \(40(3{x^2} + 4) \ne 0\) , \(3{x^2} + 4 \ne 0\) , \({x^2} \ne – \frac{4}{3}\) , the numerator is always positive
Note: Do not accept the use of the first derivative in part (b).
[6 marks]
correct (informal) statement, including reference to approaching \(y = 3\) A1 N1
e.g. getting closer to the line \(y = 3\) , horizontal asymptote at \(y = 3\)
[1 mark]
correct inequalities, \(y \le – 2\) , \(y > 3\) , FT from (a)(i) and (c) A1A1 N2
[2 marks]
Question
Let \(f(x) = \frac{1}{2}{x^3} – {x^2} – 3x\) . Part of the graph of f is shown below.
There is a maximum point at A and a minimum point at B(3, − 9) .
Find the coordinates of A.
Write down the coordinates of
(i) the image of B after reflection in the y-axis;
(ii) the image of B after translation by the vector \(\left( {\begin{array}{*{20}{c}}
{ – 2}\\
5
\end{array}} \right)\) ;
(iii) the image of B after reflection in the x-axis followed by a horizontal stretch with scale factor \(\frac{1}{2}\) .
Answer/Explanation
Markscheme
\(f(x) = {x^2} – 2x – 3\) A1A1A1
evidence of solving \(f'(x) = 0\) (M1)
e.g. \({x^2} – 2x – 3 = 0\)
evidence of correct working A1
e.g. \((x + 1)(x – 3)\) , \(\frac{{2 \pm \sqrt {16} }}{2}\)
\(x = – 1\) (ignore \(x = 3\) ) (A1)
evidence of substituting their negative x-value into \(f(x)\) (M1)
e.g. \(\frac{1}{3}{( – 1)^3} – {( – 1)^2} – 3( – 1)\) , \( – \frac{1}{3} – 1 + 3\)
\(y = \frac{5}{3}\) A1
coordinates are \(\left( { – 1,\frac{5}{3}} \right)\) N3
[8 marks]
(i) \(( – 3{\text{, }} – 9)\) A1 N1
(ii) \((1{\text{, }} – 4)\) A1A1 N2
(iii) reflection gives \((3{\text{, }}9)\) (A1)
stretch gives \(\left( {\frac{3}{2}{\text{, }}9} \right)\) A1A1 N3
[6 marks]
Question
Let \(f(x) = k{x^4}\) . The point \({\text{P}}(1{\text{, }}k)\) lies on the curve of f . At P, the normal to the curve is parallel to \(y = – \frac{1}{8}x\) . Find the value of k.
Answer/Explanation
Markscheme
gradient of tangent \(= 8\) (seen anywhere) (A1)
\(f'(x) = 4k{x^3}\) (seen anywhere) A1
recognizing the gradient of the tangent is the derivative (M1)
setting the derivative equal to 8 (A1)
e.g. \(4k{x^3} = 8\) , \(k{x^3} = 2\)
substituting \(x = 1\) (seen anywhere) (M1)
\(k = 2\) A1 N4
[6 marks]
Question
Let \(f(x) = {x^3}\). The following diagram shows part of the graph of f .
The point \({\rm{P}}(a,f(a))\) , where \(a > 0\) , lies on the graph of f . The tangent at P crosses the x-axis at the point \({\rm{Q}}\left( {\frac{2}{3},0} \right)\) . This tangent intersects the graph of f at the point R(−2, −8) .
The equation of the tangent at P is \(y = 3x – 2\) . Let T be the region enclosed by the graph of f , the tangent [PR] and the line \(x = k\) , between \(x = – 2\) and \(x = k\) where \( – 2 < k < 1\) . This is shown in the diagram below.
(i) Show that the gradient of [PQ] is \(\frac{{{a^3}}}{{a – \frac{2}{3}}}\) .
(ii) Find \(f'(a)\) .
(iii) Hence show that \(a = 1\) .
Given that the area of T is \(2k + 4\) , show that k satisfies the equation \({k^4} – 6{k^2} + 8 = 0\) .
Answer/Explanation
Markscheme
(i) substitute into gradient \( = \frac{{{y_1} – {y_2}}}{{{x_1} – {x_2}}}\) (M1)
e.g. \(\frac{{f(a) – 0}}{{a – \frac{2}{3}}}\)
substituting \(f(a) = {a^3}\)
e.g. \(\frac{{{a^3} – 0}}{{a – \frac{2}{3}}}\) A1
gradient \(\frac{{{a^3}}}{{a – \frac{2}{3}}}\) AG N0
(ii) correct answer A1 N1
e.g. \(3{a^2}\) , \(f'(a) = 3\) , \(f'(a) = \frac{{{a^3}}}{{a – \frac{2}{3}}}\)
(iii) METHOD 1
evidence of approach (M1)
e.g. \(f'(a) = {\rm{gradient}}\) , \(3{a^2} = \frac{{{a^3}}}{{a – \frac{2}{3}}}\)
simplify A1
e.g. \(3{a^2}\left( {a – \frac{2}{3}} \right) = {a^3}\)
rearrange A1
e.g. \(3{a^3} – 2{a^2} = {a^3}\)
evidence of solving A1
e.g. \(2{a^3} – 2{a^2} = 2{a^2}(a – 1) = 0\)
\(a = 1\) AG N0
METHOD 2
gradient RQ \( = \frac{{ – 8}}{{ – 2 – \frac{2}{3}}}\) A1
simplify A1
e.g. \(\frac{{ – 8}}{{ – \frac{8}{3}}},3\)
evidence of approach (M1)
e.g. \(f'(a) = {\rm{gradient}}\) , \(3{a^2} = \frac{{ – 8}}{{ – 2 – \frac{2}{3}}}\) , \(\frac{{{a^3}}}{{a – \frac{2}{3}}} = 3\)
simplify A1
e.g. \(3{a^2} = 3\) , \({a^2} = 1\)
\(a = 1\) AG N0
[7 marks]
approach to find area of T involving subtraction and integrals (M1)
e.g. \(\int {f – (3x – 2){\rm{d}}x} \) , \(\int_{ – 2}^k {(3x – 2) – \int_{ – 2}^k {{x^3}} } \) , \(\int {({x^3} – 3x + 2)} \)
correct integration with correct signs A1A1A1
e.g. \(\frac{1}{4}{x^4} – \frac{3}{2}{x^2} + 2x\) , \(\frac{3}{2}{x^2} – 2x – \frac{1}{4}{x^4}\)
correct limits \( – 2\) and k (seen anywhere) A1
e.g. \(\int_{ – 2}^k {({x^3} – 3x + 2){\rm{d}}x} \) , \(\left[ {\frac{1}{4}{x^4} – \frac{3}{2}{x^2} + 2x} \right]_{ – 2}^k\)
attempt to substitute k and \( – 2\) (M1)
correct substitution into their integral if 2 or more terms A1
e.g. \(\left( {\frac{1}{4}{k^4} – \frac{3}{2}{k^2} + 2k} \right) – (4 – 6 – 4)\)
setting their integral expression equal to \(2k + 4\) (seen anywhere) (M1)
simplifying A1
e.g. \(\frac{1}{4}{k^4} – \frac{3}{2}{k^2} + 2 = 0\)
\({k^4} – 6{k^2} + 8 = 0\) AG N0
[9 marks]
Question
The following diagram shows part of the graph of a quadratic function f .
The x-intercepts are at \(( – 4{\text{, }}0)\) and \((6{\text{, }}0)\) , and the y-intercept is at \((0{\text{, }}240)\) .
Write down \(f(x)\) in the form \(f(x) = – 10(x – p)(x – q)\) .
Find another expression for \(f(x)\) in the form \(f(x) = – 10{(x – h)^2} + k\) .
Show that \(f(x)\) can also be written in the form \(f(x) = 240 + 20x – 10{x^2}\) .
A particle moves along a straight line so that its velocity, \(v{\text{ m}}{{\text{s}}^{ – 1}}\) , at time t seconds is given by \(v = 240 + 20t – 10{t^2}\) , for \(0 \le t \le 6\) .
(i) Find the value of t when the speed of the particle is greatest.
(ii) Find the acceleration of the particle when its speed is zero.
Answer/Explanation
Markscheme
\(f(x) = – 10(x + 4)(x – 6)\) A1A1 N2
[2 marks]
METHOD 1
attempting to find the x-coordinate of maximum point (M1)
e.g. averaging the x-intercepts, sketch, \(y’ = 0\) , axis of symmetry
attempting to find the y-coordinate of maximum point (M1)
e.g. \(k = – 10(1 + 4)(1 – 6)\)
\(f(x) = – 10{(x – 1)^2} + 250\) A1A1 N4
METHOD 2
attempt to expand \(f(x)\) (M1)
e.g. \( – 10({x^2} – 2x – 24)\)
attempt to complete the square (M1)
e.g. \( – 10({(x – 1)^2} – 1 – 24)\)
\(f(x) = – 10{(x – 1)^2} + 250\) A1A1 N4
[4 marks]
attempt to simplify (M1)
e.g. distributive property, \( – 10(x – 1)(x – 1) + 250\)
correct simplification A1
e.g. \( – 10({x^2} – 6x + 4x – 24)\) , \( – 10({x^2} – 2x + 1) + 250\)
\(f(x) = 240 + 20x – 10{x^2}\) AG N0
[2 marks]
(i) valid approach (M1)
e.g. vertex of parabola, \(v'(t) = 0\)
\(t = 1\) A1 N2
(ii) recognizing \(a(t) = v'(t)\) (M1)
\(a(t) = 20 – 20t\) A1A1
speed is zero \( \Rightarrow t = 6\) (A1)
\(a(6) = – 100\) (\({\text{m}}{{\text{s}}^{ – 2}}\)) A1 N3
[7 marks]
Question
The following diagram shows the graph of \(f(x) = a\sin (b(x – c)) + d\) , for \(2 \le x \le 10\) .
There is a maximum point at P(4, 12) and a minimum point at Q(8, −4) .
Use the graph to write down the value of
(i) a ;
(ii) c ;
(iii) d .
Show that \(b = \frac{\pi }{4}\) .
Find \(f'(x)\) .
At a point R, the gradient is \( – 2\pi \) . Find the x-coordinate of R.
Answer/Explanation
Markscheme
(i) \(a = 8\) A1 N1
(ii) \(c = 2\) A1 N1
(iii) \(d = 4\) A1 N1
[3 marks]
METHOD 1
recognizing that period \( = 8\) (A1)
correct working A1
e.g. \(8 = \frac{{2\pi }}{b}\) , \(b = \frac{{2\pi }}{8}\)
\(b = \frac{\pi }{4}\) AG N0
METHOD 2
attempt to substitute M1
e.g. \(12 = 8\sin (b(4 – 2)) + 4\)
correct working A1
e.g. \(\sin 2b = 1\)
\(b = \frac{\pi }{4}\) AG N0
[2 marks]
evidence of attempt to differentiate or choosing chain rule (M1)
e.g. \(\cos \frac{\pi }{4}(x – 2)\) , \(\frac{\pi }{4} \times 8\)
\(f'(x) = 2\pi \cos \left( {\frac{\pi }{4}(x – 2)} \right)\) (accept \(2\pi \cos \frac{\pi }{4}(x – 2)\) ) A2 N3
[3 marks]
recognizing that gradient is \(f'(x)\) (M1)
e.g. \(f'(x) = m\)
correct equation A1
e.g. \( – 2\pi = 2\pi \cos \left( {\frac{\pi }{4}(x – 2)} \right)\) , \( – 1 = \cos \left( {\frac{\pi }{4}(x – 2)} \right)\)
correct working (A1)
e.g. \({\cos ^{ – 1}}( – 1) = \frac{\pi }{4}(x – 2)\)
using \({\cos ^{ – 1}}( – 1) = \pi \) (seen anywhere) (A1)
e.g. \(\pi = \frac{\pi }{4}(x – 2)\)
simplifying (A1)
e.g. \(4 = (x – 2)\)
\(x = 6\) A1 N4
[6 marks]
Question
Let \(f(x) = {{\rm{e}}^{6x}}\) .
Write down \(f'(x)\) .
The tangent to the graph of f at the point \({\text{P}}(0{\text{, }}b)\) has gradient m .
(i) Show that \(m = 6\) .
(ii) Find b .
Hence, write down the equation of this tangent.
Answer/Explanation
Markscheme
\(f'(x) = 6{{\rm{e}}^{6x}}\) A1 N1
[1 mark]
(i) evidence of valid approach (M1)
e.g. \(f'(0)\) , \(6{{\rm{e}}^{6 \times 0}}\)
correct manipulation A1
e.g. \(6{{\rm{e}}^0}\) , \(6 \times 1\)
\(m = 6\) AG N0
(ii) evidence of finding \(f(0)\) (M1)
e.g. \(y = {{\rm{e}}^{6(0)}}\)
\(b = 1\) A1 N2
[4 marks]
\(y = 6x + 1\) A1 N1
[1 mark]
Question
Consider \(f(x) = {x^2}\sin x\) .
Find \(f'(x)\) .
Find the gradient of the curve of \(f\) at \(x = \frac{\pi }{2}\) .
Answer/Explanation
Markscheme
evidence of choosing product rule (M1)
eg \(uv’ + vu’\)
correct derivatives (must be seen in the product rule) \(\cos x\) , \(2x\) (A1)(A1)
\(f'(x) = {x^2}\cos x + 2x\sin x\) A1 N4
[4 marks]
substituting \(\frac{\pi }{2}\) into their \(f'(x)\) (M1)
eg \(f’\left( {\frac{\pi }{2}} \right)\) , \({\left( {\frac{\pi }{2}} \right)^2}\cos \left( {\frac{\pi }{2}} \right) + 2\left( {\frac{\pi }{2}} \right)\sin \left( {\frac{\pi }{2}} \right)\)
correct values for both \(\sin \frac{\pi }{2}\) and \(\cos \frac{\pi }{2}\) seen in \(f'(x)\) (A1)
eg \(0 + 2\left( {\frac{\pi }{2}} \right) \times 1\)
\(f’\left( {\frac{\pi }{2}} \right) = \pi \) A1 N2
[3 marks]
Question
Let \(f(x) = \sin x + \frac{1}{2}{x^2} – 2x\) , for \(0 \le x \le \pi \) .
Let \(g\) be a quadratic function such that \(g(0) = 5\) . The line \(x = 2\) is the axis of symmetry of the graph of \(g\) .
The function \(g\) can be expressed in the form \(g(x) = a{(x – h)^2} + 3\) .
Find \(f'(x)\) .
Find \(g(4)\) .
(i) Write down the value of \(h\) .
(ii) Find the value of \(a\) .
Find the value of \(x\) for which the tangent to the graph of \(f\) is parallel to the tangent to the graph of \(g\) .
Answer/Explanation
Markscheme
\(f'(x) = \cos x + x – 2\) A1A1A1 N3
Note: Award A1 for each term.
[3 marks]
recognizing \(g(0) = 5\) gives the point (\(0\), \(5\)) (R1)
recognize symmetry (M1)
eg vertex, sketch
\(g(4) = 5\) A1 N3
[3 marks]
(i) \(h = 2\) A1 N1
(ii) substituting into \(g(x) = a{(x – 2)^2} + 3\) (not the vertex) (M1)
eg \(5 = a{(0 – 2)^2} + 3\) , \(5 = a{(4 – 2)^2} + 3\)
working towards solution (A1)
eg \(5 = 4a + 3\) , \(4a = 2\)
\(a = \frac{1}{2}\) A1 N2
[4 marks]
\(g(x) = \frac{1}{2}{(x – 2)^2} + 3 = \frac{1}{2}{x^2} – 2x + 5\)
correct derivative of \(g\) A1A1
eg \(2 \times \frac{1}{2}(x – 2)\) , \(x – 2\)
evidence of equating both derivatives (M1)
eg \(f’ = g’\)
correct equation (A1)
eg \(\cos x + x – 2 = x – 2\)
working towards a solution (A1)
eg \(\cos x = 0\) , combining like terms
\(x = \frac{\pi }{2}\) A1 N0
Note: Do not award final A1 if additional values are given.
[6 marks]
Question
Let \(f(x) = \frac{{{{(\ln x)}^2}}}{2}\), for \(x > 0\).
Let \(g(x) = \frac{1}{x}\). The following diagram shows parts of the graphs of \(f’\) and g.
The graph of \(f’\) has an x-intercept at \(x = p\).
Show that \(f'(x) = \frac{{\ln x}}{x}\).
There is a minimum on the graph of \(f\). Find the \(x\)-coordinate of this minimum.
Write down the value of \(p\).
The graph of \(g\) intersects the graph of \(f’\) when \(x = q\).
Find the value of \(q\).
The graph of \(g\) intersects the graph of \(f’\) when \(x = q\).
Let \(R\) be the region enclosed by the graph of \(f’\), the graph of \(g\) and the line \(x = p\).
Show that the area of \(R\) is \(\frac{1}{2}\).
Answer/Explanation
Markscheme
METHOD 1
correct use of chain rule A1A1
eg \(\frac{{2\ln x}}{2} \times \frac{1}{x},{\text{ }}\frac{{2\ln x}}{{2x}}\)
Note: Award A1 for \(\frac{{2\ln x}}{{2x}}\), A1 for \( \times \frac{1}{x}\).
\(f'(x) = \frac{{\ln x}}{x}\) AG N0
[2 marks]
METHOD 2
correct substitution into quotient rule, with derivatives seen A1
eg \(\frac{{2 \times 2\ln x \times \frac{1}{x} – 0 \times {{(\ln x)}^2}}}{4}\)
correct working A1
eg \(\frac{{4\ln x \times \frac{1}{x}}}{4}\)
\(f'(x) = \frac{{\ln x}}{x}\) AG N0
[2 marks]
setting derivative \( = 0\) (M1)
eg \(f'(x) = 0,{\text{ }}\frac{{\ln x}}{x} = 0\)
correct working (A1)
eg \(\ln x = 0,{\text{ }}x = {{\text{e}}^0}\)
\(x = 1\) A1 N2
[3 marks]
intercept when \(f'(x) = 0\) (M1)
\(p = 1\) A1 N2
[2 marks]
equating functions (M1)
eg \(f’ = g,{\text{ }}\frac{{\ln x}}{x} = \frac{1}{x}\)
correct working (A1)
eg \(\ln x = 1\)
\(q = {\text{e (accept }}x = {\text{e)}}\) A1 N2
[3 marks]
evidence of integrating and subtracting functions (in any order, seen anywhere) (M1)
eg \(\int_q^e {\left( {\frac{1}{x} – \frac{{\ln x}}{x}} \right){\text{d}}x{\text{, }}\int {f’ – g} } \)
correct integration \(\ln x – \frac{{{{(\ln x)}^2}}}{2}\) A2
substituting limits into their integrated function and subtracting (in any order) (M1)
eg \((\ln {\text{e}} – \ln 1) – \left( {\frac{{{{(\ln {\text{e}})}^2}}}{2} – \frac{{{{(\ln 1)}^2}}}{2}} \right)\)
Note: Do not award M1 if the integrated function has only one term.
correct working A1
eg \((1 – 0) – \left( {\frac{1}{2} – 0} \right),{\text{ }}1 – \frac{1}{2}\)
\({\text{area}} = \frac{1}{2}\) AG N0
Notes: Candidates may work with two separate integrals, and only combine them at the end. Award marks in line with the markscheme.
[5 marks]
Question
Let \(f(x) = p{x^3} + p{x^2} + qx\).
Find \(f'(x)\).
Given that \(f'(x) \geqslant 0\), show that \({p^2} \leqslant 3pq\).
Answer/Explanation
Markscheme
\(f'(x) = 3p{x^2} + 2px + q\) A2 N2
Note: Award A1 if only 1 error.
[2 marks]
evidence of discriminant (must be seen explicitly, not in quadratic formula) (M1)
eg \({b^2} – 4ac\)
correct substitution into discriminant (may be seen in inequality) A1
eg \({(2p)^2} – 4 \times 3p \times q,{\text{ }}4{p^2} – 12pq\)
\(f'(x) \geqslant 0\) then \(f’\) has two equal roots or no roots (R1)
recognizing discriminant less or equal than zero R1
eg \(\Delta \leqslant 0,{\text{ }}4{p^2} – 12pq \leqslant 0\)
correct working that clearly leads to the required answer A1
eg \({p^2} – 3pq \leqslant 0,{\text{ }}4{p^2} \leqslant 12pq\)
\({p^2} \leqslant 3pq\) AG N0
[5 marks]
Question
A function \(f\) has its derivative given by \(f'(x) = 3{x^2} – 2kx – 9\), where \(k\) is a constant.
Find \(f”(x)\).
The graph of \(f\) has a point of inflexion when \(x = 1\).
Show that \(k = 3\).
Find \(f'( – 2)\).
Find the equation of the tangent to the curve of \(f\) at \(( – 2,{\text{ }}1)\), giving your answer in the form \(y = ax + b\).
Given that \(f'( – 1) = 0\), explain why the graph of \(f\) has a local maximum when \(x = – 1\).
Answer/Explanation
Markscheme
\(f”(x) = 6x – 2k\) A1A1 N2
[2 marks]
substituting \(x = 1\) into \(f”\) (M1)
eg\(\;\;\;f”(1),{\text{ }}6(1) – 2k\)
recognizing \(f”(x) = 0\;\;\;\)(seen anywhere) M1
correct equation A1
eg\(\;\;\;6 – 2k = 0\)
\(k = 3\) AG N0
[3 marks]
correct substitution into \(f'(x)\) (A1)
eg\(\;\;\;3{( – 2)^2} – 6( – 2) – 9\)
\(f'( – 2) = 15\) A1 N2
[2 marks]
recognizing gradient value (may be seen in equation) M1
eg\(\;\;\;a = 15,{\text{ }}y = 15x + b\)
attempt to substitute \(( – 2,{\text{ }}1)\) into equation of a straight line M1
eg\(\;\;\;1 = 15( – 2) + b,{\text{ }}(y – 1) = m(x + 2),{\text{ }}(y + 2) = 15(x – 1)\)
correct working (A1)
eg\(\;\;\;31 = b,{\text{ }}y = 15x + 30 + 1\)
\(y = 15x + 31\) A1 N2
[4 marks]
METHOD 1 (\({{\text{2}}^{{\text{nd}}}}\) derivative)
recognizing \(f” < 0\;\;\;\)(seen anywhere) R1
substituting \(x = – 1\) into \(f”\) (M1)
eg\(\;\;\;f”( – 1),{\text{ }}6( – 1) – 6\)
\(f”( – 1) = – 12\) A1
therefore the graph of \(f\) has a local maximum when \(x = – 1\) AG N0
METHOD 2 (\({{\text{1}}^{{\text{st}}}}\) derivative)
recognizing change of sign of \(f'(x)\;\;\;\)(seen anywhere) R1
eg\(\;\;\;\)sign chart\(\;\;\;\)
correct value of \(f’\) for \( – 1 < x < 3\) A1
eg\(\;\;\;f'(0) = – 9\)
correct value of \(f’\) for \(x\) value to the left of \( – 1\) A1
eg\(\;\;\;f'( – 2) = 15\)
therefore the graph of \(f\) has a local maximum when \(x = – 1\) AG N0
[3 marks]
Total [14 marks]
Question
Let \(f'(x) = \frac{{6 – 2x}}{{6x – {x^2}}}\), for \(0 < x < 6\).
The graph of \(f\) has a maximum point at P.
The \(y\)-coordinate of P is \(\ln 27\).
Find the \(x\)-coordinate of P.
Find \(f(x)\), expressing your answer as a single logarithm.
The graph of \(f\) is transformed by a vertical stretch with scale factor \(\frac{1}{{\ln 3}}\). The image of P under this transformation has coordinates \((a,{\text{ }}b)\).
Find the value of \(a\) and of \(b\), where \(a,{\text{ }}b \in \mathbb{N}\).
Answer/Explanation
Markscheme
recognizing \(f'(x) = 0\) (M1)
correct working (A1)
eg\(\,\,\,\,\,\)\(6 – 2x = 0\)
\(x = 3\) A1 N2
[3 marks]
evidence of integration (M1)
eg\(\,\,\,\,\,\)\(\int {f’,{\text{ }}\int {\frac{{6 – 2x}}{{6x – {x^2}}}{\text{d}}x} } \)
using substitution (A1)
eg\(\,\,\,\,\,\)\(\int {\frac{1}{u}{\text{d}}u} \) where \(u = 6x – {x^2}\)
correct integral A1
eg\(\,\,\,\,\,\)\(\ln (u) + c,{\text{ }}\ln (6x – {x^2})\)
substituting \((3,{\text{ }}\ln 27)\) into their integrated expression (must have \(c\)) (M1)
eg\(\,\,\,\,\,\)\(\ln (6 \times 3 – {3^2}) + c = \ln 27,{\text{ }}\ln (18 – 9) + \ln k = \ln 27\)
correct working (A1)
eg\(\,\,\,\,\,\)\(c = \ln 27 – \ln 9\)
EITHER
\(c = \ln 3\) (A1)
attempt to substitute their value of \(c\) into \(f(x)\) (M1)
eg\(\,\,\,\,\,\)\(f(x) = \ln (6x – {x^2}) + \ln 3\) A1 N4
OR
attempt to substitute their value of \(c\) into \(f(x)\) (M1)
eg\(\,\,\,\,\,\)\(f(x) = \ln (6x – {x^2}) + \ln 27 – \ln 9\)
correct use of a log law (A1)
eg\(\,\,\,\,\,\)\(f(x) = \ln (6x – {x^2}) + \ln \left( {\frac{{27}}{9}} \right),{\text{ }}f(x) = \ln \left( {27(6x – {x^2})} \right) – \ln 9\)
\(f(x) = \ln \left( {3(6x – {x^2})} \right)\) A1 N4
[8 marks]
\(a = 3\) A1 N1
correct working A1
eg\(\,\,\,\,\,\)\(\frac{{\ln 27}}{{\ln 3}}\)
correct use of log law (A1)
eg\(\,\,\,\,\,\)\(\frac{{3\ln 3}}{{\ln 3}},{\text{ }}{\log _3}27\)
\(b = 3\) A1 N2
[4 marks]
Question
Fred makes an open metal container in the shape of a cuboid, as shown in the following diagram.
The container has height \(x{\text{ m}}\), width \(x{\text{ m}}\) and length \(y{\text{ m}}\). The volume is \(36{\text{ }}{{\text{m}}^3}\).
Let \(A(x)\) be the outside surface area of the container.
Show that \(A(x) = \frac{{108}}{x} + 2{x^2}\).
Find \(A'(x)\).
Given that the outside surface area is a minimum, find the height of the container.
Fred paints the outside of the container. A tin of paint covers a surface area of \({\text{10 }}{{\text{m}}^{\text{2}}}\) and costs $20. Find the total cost of the tins needed to paint the container.
Answer/Explanation
Markscheme
correct substitution into the formula for volume A1
eg\(\,\,\,\,\,\)\(36 = y \times x \times x\)
valid approach to eliminate \(y\) (may be seen in formula/substitution) M1
eg\(\,\,\,\,\,\)\(y = \frac{{36}}{{{x^2}}},{\text{ }}xy = \frac{{36}}{x}\)
correct expression for surface area A1
eg\(\,\,\,\,\,\)\(xy + xy + xy + {x^2} + {x^2},{\text{ area}} = 3xy + 2{x^2}\)
correct expression in terms of \(x\) only A1
eg\(\,\,\,\,\,\)\(3x\left( {\frac{{36}}{{{x^2}}}} \right) + 2{x^2},{\text{ }}{x^2} + {x^2} + \frac{{36}}{x} + \frac{{36}}{x} + \frac{{36}}{x},{\text{ }}2{x^2} + 3\left( {\frac{{36}}{x}} \right)\)
\(A(x) = \frac{{108}}{x} + 2{x^2}\) AG N0
[4 marks]
\(A'(x) = – \frac{{108}}{{{x^2}}} + 4x,{\text{ }}4x – 108{x^{ – 2}}\) A1A1 N2
Note: Award A1 for each term.
[2 marks]
recognizing that minimum is when \(A'(x) = 0\) (M1)
correct equation (A1)
eg\(\,\,\,\,\,\)\( – \frac{{108}}{{{x^2}}} + 4x = 0,{\text{ }}4x = \frac{{108}}{{{x^2}}}\)
correct simplification (A1)
eg\(\,\,\,\,\,\)\( – 108 + 4{x^3} = 0,{\text{ }}4{x^3} = 108\)
correct working (A1)
eg\(\,\,\,\,\,\)\({x^3} = 27\)
\({\text{height}} = 3{\text{ (m) }}({\text{accept }}x = 3)\) A1 N2
[5 marks]
attempt to find area using their height (M1)
eg\(\,\,\,\,\,\)\(\frac{{108}}{3} + 2{(3)^2},{\text{ }}9 + 9 + 12 + 12 + 12\)
minimum surface area \( = 54{\text{ }}{{\text{m}}^{\text{2}}}\) (may be seen in part (c)) A1
attempt to find the number of tins (M1)
eg\(\,\,\,\,\,\)\(\frac{{54}}{{10}},{\text{ }}5.4\)
6 (tins) (A1)
$120 A1 N3
[5 marks]
Question
Let \(f(x) = \cos x\).
Let \(g(x) = {x^k}\), where \(k \in {\mathbb{Z}^ + }\).
Let \(k = 21\) and \(h(x) = \left( {{f^{(19)}}(x) \times {g^{(19)}}(x)} \right)\).
(i) Find the first four derivatives of \(f(x)\).
(ii) Find \({f^{(19)}}(x)\).
(i) Find the first three derivatives of \(g(x)\).
(ii) Given that \({g^{(19)}}(x) = \frac{{k!}}{{(k – p)!}}({x^{k – 19}})\), find \(p\).
(i) Find \(h'(x)\).
(ii) Hence, show that \(h'(\pi ) = \frac{{ – 21!}}{2}{\pi ^2}\).
Answer/Explanation
Markscheme
(i) \(f'(x) = – \sin x,{\text{ }}f”(x) = – \cos x,{\text{ }}{f^{(3)}}(x) = \sin x,{\text{ }}{f^{(4)}}(x) = \cos x\) A2 N2
(ii) valid approach (M1)
eg\(\,\,\,\,\,\)recognizing that 19 is one less than a multiple of 4, \({f^{(19)}}(x) = {f^{(3)}}(x)\)
\({f^{(19)}}(x) = \sin x\) A1 N2
[4 marks]
(i) \(g'(x) = k{x^{k – 1}}\)
\(g”(x) = k(k – 1){x^{k – 2}},{\text{ }}{g^{(3)}}(x) = k(k – 1)(k – 2){x^{k – 3}}\) A1A1 N2
(ii) METHOD 1
correct working that leads to the correct answer, involving the correct expression for the 19th derivative A2
eg\(\,\,\,\,\,\)\(k(k – 1)(k – 2) \ldots (k – 18) \times \frac{{(k – 19)!}}{{(k – 19)!}},{{\text{ }}_k}{P_{19}}\)
\(p = 19\) (accept \(\frac{{k!}}{{(k – 19)!}}{x^{k – 19}}\)) A1 N1
METHOD 2
correct working involving recognizing patterns in coefficients of first three derivatives (may be seen in part (b)(i)) leading to a general rule for 19th coefficient A2
eg\(\,\,\,\,\,\)\(g” = 2!\left( {\begin{array}{*{20}{c}} k \\ 2 \end{array}} \right),{\text{ }}k(k – 1)(k – 2) = \frac{{k!}}{{(k – 3)!}},{\text{ }}{g^{(3)}}(x){ = _k}{P_3}({x^{k – 3}})\)
\({g^{(19)}}(x) = 19!\left( {\begin{array}{*{20}{c}} k \\ {19} \end{array}} \right),{\text{ }}19! \times \frac{{k!}}{{(k – 19)! \times 19!}},{{\text{ }}_k}{P_{19}}\)
\(p = 19\) (accept \(\frac{{k!}}{{(k – 19)!}}{x^{k – 19}}\)) A1 N1
[5 marks]
(i) valid approach using product rule (M1)
eg\(\,\,\,\,\,\)\(uv’ + vu’,{\text{ }}{f^{(19)}}{g^{(20)}} + {f^{(20)}}{g^{(19)}}\)
correct 20th derivatives (must be seen in product rule) (A1)(A1)
eg\(\,\,\,\,\,\)\({g^{(20)}}(x) = \frac{{21!}}{{(21 – 20)!}}x,{\text{ }}{f^{(20)}}(x) = \cos x\)
\(h'(x) = \sin x(21!x) + \cos x\left( {\frac{{21!}}{2}{x^2}} \right){\text{ }}\left( {{\text{accept }}\sin x\left( {\frac{{21!}}{{1!}}x} \right) + \cos x\left( {\frac{{21!}}{{2!}}{x^2}} \right)} \right)\) A1 N3
(ii) substituting \(x = \pi \) (seen anywhere) (A1)
eg\(\,\,\,\,\,\)\({f^{(19)}}(\pi ){g^{(20)}}(\pi ) + {f^{(20)}}(\pi ){g^{(19)}}(\pi ),{\text{ }}\sin \pi \frac{{21!}}{{1!}}\pi + \cos \pi \frac{{21!}}{{2!}}{\pi ^2}\)
evidence of one correct value for \(\sin \pi \) or \(\cos \pi \) (seen anywhere) (A1)
eg\(\,\,\,\,\,\)\(\sin \pi = 0,{\text{ }}\cos \pi = – 1\)
evidence of correct values substituted into \(h'(\pi )\) A1
eg\(\,\,\,\,\,\)\(21!(\pi )\left( {0 – \frac{\pi }{{2!}}} \right),{\text{ }}21!(\pi )\left( { – \frac{\pi }{2}} \right),{\text{ }}0 + ( – 1)\frac{{21!}}{2}{\pi ^2}\)
Note: If candidates write only the first line followed by the answer, award A1A0A0.
\(\frac{{ – 21!}}{2}{\pi ^2}\) AG N0
[7 marks]
Question
Let \(f(x) = {x^2}\). The following diagram shows part of the graph of \(f\).
The line \(L\) is the tangent to the graph of \(f\) at the point \({\text{A}}( – k,{\text{ }}{k^2})\), and intersects the \(x\)-axis at point B. The point C is \(( – k,{\text{ }}0)\).
The region \(R\) is enclosed by \(L\), the graph of \(f\), and the \(x\)-axis. This is shown in the following diagram.
Write down \(f'(x)\).
Find the gradient of \(L\).
Show that the \(x\)-coordinate of B is \( – \frac{k}{2}\).
Find the area of triangle ABC, giving your answer in terms of \(k\).
Given that the area of triangle ABC is \(p\) times the area of \(R\), find the value of \(p\).
Answer/Explanation
Markscheme
\(f'(x) = 2x\) A1 N1
[1 mark]
attempt to substitute \(x = – k\) into their derivative (M1)
gradient of \(L\) is \( – 2k\) A1 N2
[2 marks]
METHOD 1
attempt to substitute coordinates of A and their gradient into equation of a line (M1)
eg\(\,\,\,\,\,\)\({k^2} = – 2k( – k) + b\)
correct equation of \(L\) in any form (A1)
eg\(\,\,\,\,\,\)\(y – {k^2} = – 2k(x + k),{\text{ }}y = – 2kx – {k^2}\)
valid approach (M1)
eg\(\,\,\,\,\,\)\(y = 0\)
correct substitution into \(L\) equation A1
eg\(\,\,\,\,\,\)\( – {k^2} = – 2kx – 2{k^2},{\text{ }}0 = – 2kx – {k^2}\)
correct working A1
eg\(\,\,\,\,\,\)\(2kx = – {k^2}\)
\(x = – \frac{k}{2}\) AG N0
METHOD 2
valid approach (M1)
eg\(\,\,\,\,\,\)\({\text{gradient}} = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}},{\text{ }} – 2k = \frac{{{\text{rise}}}}{{{\text{run}}}}\)
recognizing \(y = 0\) at B (A1)
attempt to substitute coordinates of A and B into slope formula (M1)
eg\(\,\,\,\,\,\)\(\frac{{{k^2} – 0}}{{ – k – x}},{\text{ }}\frac{{ – {k^2}}}{{x + k}}\)
correct equation A1
eg\(\,\,\,\,\,\)\(\frac{{{k^2} – 0}}{{ – k – x}} = – 2k,{\text{ }}\frac{{ – {k^2}}}{{x + k}} = – 2k,{\text{ }} – {k^2} = – 2k(x + k)\)
correct working A1
eg\(\,\,\,\,\,\)\(2kx = – {k^2}\)
\(x = – \frac{k}{2}\) AG N0
[5 marks]
valid approach to find area of triangle (M1)
eg\(\,\,\,\,\,\)\(\frac{1}{2}({k^2})\left( {\frac{k}{2}} \right)\)
area of \({\text{ABC}} = \frac{{{k^3}}}{4}\) A1 N2
[2 marks]
METHOD 1 (\(\int {f – {\text{triangle}}} \))
valid approach to find area from \( – k\) to 0 (M1)
eg\(\,\,\,\,\,\)\(\int_{ – k}^0 {{x^2}{\text{d}}x,{\text{ }}\int_0^{ – k} f } \)
correct integration (seen anywhere, even if M0 awarded) A1
eg\(\,\,\,\,\,\)\(\frac{{{x^3}}}{3},{\text{ }}\left[ {\frac{1}{3}{x^3}} \right]_{ – k}^0\)
substituting their limits into their integrated function and subtracting (M1)
eg\(\,\,\,\,\,\)\(0 – \frac{{{{( – k)}^3}}}{3}\), area from \( – k\) to 0 is \(\frac{{{k^3}}}{3}\)
Note: Award M0 for substituting into original or differentiated function.
attempt to find area of \(R\) (M1)
eg\(\,\,\,\,\,\)\(\int_{ – k}^0 {f(x){\text{d}}x – {\text{ triangle}}} \)
correct working for \(R\) (A1)
eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{3} – \frac{{{k^3}}}{4},{\text{ }}R = \frac{{{k^3}}}{{12}}\)
correct substitution into \({\text{triangle}} = pR\) (A1)
eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{3} – \frac{{{k^3}}}{4}} \right),{\text{ }}\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{12}}} \right)\)
\(p = 3\) A1 N2
METHOD 2 (\(\int {(f – L)} \))
valid approach to find area of \(R\) (M1)
eg\(\,\,\,\,\,\)\(\int_{ – k}^{ – \frac{k}{2}} {{x^2} – ( – 2kx – {k^2}){\text{d}}x + \int_{ – \frac{k}{2}}^0 {{x^2}{\text{d}}x,{\text{ }}\int_{ – k}^{ – \frac{k}{2}} {(f – L) + \int_{ – \frac{k}{2}}^0 f } } } \)
correct integration (seen anywhere, even if M0 awarded) A2
eg\(\,\,\,\,\,\)\(\frac{{{x^3}}}{3} + k{x^2} + {k^2}x,{\text{ }}\left[ {\frac{{{x^3}}}{3} + k{x^2} + {k^2}x} \right]_{ – k}^{ – \frac{k}{2}} + \left[ {\frac{{{x^3}}}{3}} \right]_{ – \frac{k}{2}}^0\)
substituting their limits into their integrated function and subtracting (M1)
eg\(\,\,\,\,\,\)\(\left( {\frac{{{{\left( { – \frac{k}{2}} \right)}^3}}}{3} + k{{\left( { – \frac{k}{2}} \right)}^2} + {k^2}\left( { – \frac{k}{2}} \right)} \right) – \left( {\frac{{{{( – k)}^3}}}{3} + k{{( – k)}^2} + {k^2}( – k)} \right) + (0) – \left( {\frac{{{{\left( { – \frac{k}{2}} \right)}^3}}}{3}} \right)\)
Note: Award M0 for substituting into original or differentiated function.
correct working for \(R\) (A1)
eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{{24}},{\text{ }} – \frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{4} – \frac{{{k^3}}}{2} + \frac{{{k^3}}}{3} – {k^3} + {k^3} + \frac{{{k^3}}}{{24}},{\text{ }}R = \frac{{{k^3}}}{{12}}\)
correct substitution into \({\text{triangle}} = pR\) (A1)
eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{{24}}} \right),{\text{ }}\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{12}}} \right)\)
\(p = 3\) A1 N2
[7 marks]
Question
Let \(f(x) = {x^2} – x\), for \(x \in \mathbb{R}\). The following diagram shows part of the graph of \(f\).
The graph of \(f\) crosses the \(x\)-axis at the origin and at the point \({\text{P}}(1,{\text{ }}0)\).
The line L is the normal to the graph of f at P.
The line \(L\) intersects the graph of \(f\) at another point Q, as shown in the following diagram.
Show that \(f’(1) = 1\).
Find the equation of \(L\) in the form \(y = ax + b\).
Find the \(x\)-coordinate of Q.
Find the area of the region enclosed by the graph of \(f\) and the line \(L\).
Answer/Explanation
Markscheme
\(f’(x) = 2x – 1\) A1A1
correct substitution A1
eg\(\,\,\,\,\,\)\(2(1) – 1,{\text{ }}2 – 1\)
\(f’(1) = 1\) AG N0
[3 marks]
correct approach to find the gradient of the normal (A1)
eg\(\,\,\,\,\,\)\(\frac{{ – 1}}{{f'(1)}},{\text{ }}{m_1}{m_2} = – 1,{\text{ slope}} = – 1\)
attempt to substitute correct normal gradient and coordinates into equation of a line (M1)
eg\(\,\,\,\,\,\)\(y – 0 = – 1(x – 1),{\text{ }}0 = – 1 + b,{\text{ }}b = 1,{\text{ }}L = – x + 1\)
\(y = – x + 1\) A1 N2
[3 marks]
equating expressions (M1)
eg\(\,\,\,\,\,\)\(f(x) = L,{\text{ }} – x + 1 = {x^2} – x\)
correct working (must involve combining terms) (A1)
eg\(\,\,\,\,\,\)\({x^2} – 1 = 0,{\text{ }}{x^2} = 1,{\text{ }}x = 1\)
\(x = – 1\,\,\,\,\,\left( {{\text{accept }}Q( – 1,{\text{ }}2)} \right)\) A2 N3
[4 marks]
valid approach (M1)
eg\(\,\,\,\,\,\)\(\int {L – f,{\text{ }}\int_{ – 1}^1 {(1 – {x^2}){\text{d}}x} } \), splitting area into triangles and integrals
correct integration (A1)(A1)
eg\(\,\,\,\,\,\)\(\left[ {x – \frac{{{x^3}}}{3}} \right]_{ – 1}^1,{\text{ }} – \frac{{{x^3}}}{3} – \frac{{{x^2}}}{2} + \frac{{{x^2}}}{2} + x\)
substituting their limits into their integrated function and subtracting (in any order) (M1)
eg\(\,\,\,\,\,\)\(1 – \frac{1}{3} – \left( { – 1 – \frac{{ – 1}}{3}} \right)\)
Note: Award M0 for substituting into original or differentiated function.
area \( = \frac{4}{3}\) A2 N3
[6 marks]
Question
A closed cylindrical can with radius r centimetres and height h centimetres has a volume of 20\(\pi \) cm3.
The material for the base and top of the can costs 10 cents per cm2 and the material for the curved side costs 8 cents per cm2. The total cost of the material, in cents, is C.
Express h in terms of r.
Show that \(C = 20\pi {r^2} + \frac{{320\pi }}{r}\).
Given that there is a minimum value for C, find this minimum value in terms of \(\pi \).
Answer/Explanation
Markscheme
correct equation for volume (A1)
eg \(\pi {r^2}h = 20\pi \)
\(h = \frac{{20}}{{{r^2}}}\) A1 N2
[2 marks]
attempt to find formula for cost of parts (M1)
eg 10 × two circles, 8 × curved side
correct expression for cost of two circles in terms of r (seen anywhere) A1
eg \(2\pi {r^2} \times 10\)
correct expression for cost of curved side (seen anywhere) (A1)
eg \(2\pi r \times h \times 8\)
correct expression for cost of curved side in terms of r A1
eg \(8 \times 2\pi r \times \frac{{20}}{{{r^2}}},\,\,\frac{{320\pi }}{{{r^2}}}\)
\(C = 20\pi {r^2} + \frac{{320\pi }}{r}\) AG N0
[4 marks]
recognize \(C’ = 0\) at minimum (R1)
eg \(C’ = 0,\,\,\frac{{{\text{d}}C}}{{{\text{d}}r}} = 0\)
correct differentiation (may be seen in equation)
\(C’ = 40\pi r – \frac{{320\pi }}{{{r^2}}}\) A1A1
correct equation A1
eg \(40\pi r – \frac{{320\pi }}{{{r^2}}} = 0,\,\,40\pi r\frac{{320\pi }}{{{r^2}}}\)
correct working (A1)
eg \(40{r^3} = 320,\,\,{r^3} = 8\)
r = 2 (m) A1
attempt to substitute their value of r into C
eg \(20\pi \times 4 + 320 \times \frac{\pi }{2}\) (M1)
correct working
eg \(80\pi + 160\pi \) (A1)
\(240\pi \) (cents) A1 N3
Note: Do not accept 753.6, 753.98 or 754, even if 240\(\pi \) is seen.
[9 marks]
Question
Given that \(f(x) = \frac{1}{x}\) , answer the following.
Find the first four derivatives of \(f(x)\) .
Write an expression for \({f^{(n)}}(x)\) in terms of x and n .
Answer/Explanation
Markscheme
\(f'(x) = – {x^{ – 2}}\) (or \( – \frac{1}{{{x^2}}}\) ) A1 N1
\(f”(x) = 2{x^{ – 3}}\) (or \(\frac{2}{{{x^3}}}\) ) A1 N1
\(f”'(x) = – 6{x^{ – 4}}\) (or \( – \frac{6}{{{x^4}}}\) ) A1 N1
\({f^{(4)}}(x) = 24{x^{ – 5}}\) (or \(\frac{{24}}{{{x^5}}}\) ) A1 N1
[4 marks]
\({f^{(n)}}(x) = \frac{{{{( – 1)}^n}n!}}{{{x^{n + 1}}}}\) or \({( – 1)^n}n!({x^{ – (n + 1)}})\) A1A1A1 N3
[3 marks]
Question
Let \(f(x) = \cos x + \sqrt 3 \sin x\) , \(0 \le x \le 2\pi \) . The following diagram shows the graph of \(f\) .
The \(y\)-intercept is at (\(0\), \(1\)) , there is a minimum point at A (\(p\), \(q\)) and a maximum point at B.
Find \(f'(x)\) .
Hence
(i) show that \(q = – 2\) ;
(ii) verify that A is a minimum point.
Find the maximum value of \(f(x)\) .
The function \(f(x)\) can be written in the form \(r\cos (x – a)\) .
Write down the value of r and of a .
Answer/Explanation
Markscheme
\(f'(x) = – \sin x + \sqrt 3 \cos x\) A1A1 N2
[2 marks]
(i) at A, \(f'(x) = 0\) R1
correct working A1
e.g. \(\sin x = \sqrt 3 \cos x\)
\(\tan x = \sqrt 3 \) A1
\(x = \frac{\pi }{3}\) , \(\frac{{4\pi }}{3}\) A1
attempt to substitute their x into \(f(x)\) M1
e.g. \(\cos \left( {\frac{{4\pi }}{3}} \right) + \sqrt 3 \sin \left( {\frac{{4\pi }}{3}} \right)\)
correct substitution A1
e.g. \( – \frac{1}{2} + \sqrt 3 \left( { – \frac{{\sqrt 3 }}{2}} \right)\)
correct working that clearly leads to \( – 2\) A1
e.g. \( – \frac{1}{2} – \frac{3}{2}\)
\(q = – 2\) AG N0
(ii) correct calculations to find \(f'(x)\) either side of \(x = \frac{{4\pi }}{3}\) A1A1
e.g. \(f'(\pi ) = 0 – \sqrt 3 \) , \(f'(2\pi ) = 0 + \sqrt 3 \)
\(f'(x)\) changes sign from negative to positive R1
so A is a minimum AG N0
[10 marks]
max when \(x = \frac{\pi }{3}\) R1
correctly substituting \(x = \frac{\pi }{3}\) into \(f(x)\) A1
e.g. \(\frac{1}{2} + \sqrt 3 \left( {\frac{{\sqrt 3 }}{2}} \right)\)
max value is 2 A1 N1
[3 marks]
\(r = 2\) , \(a = \frac{\pi }{3}\) A1A1 N2
[2 marks]