Question
Let \(f(x) = {{\rm{e}}^x}\cos x\) . Find the gradient of the normal to the curve of f at \(x = \pi \) .
Answer/Explanation
Markscheme
evidence of choosing the product rule    (M1)
\(f'(x) = {{\rm{e}}^x} \times ( – \sin x) + \cos x \times {{\rm{e}}^x}\) \(( = {{\rm{e}}^x}\cos x – {{\rm{e}}^x}\sin x)\)Â Â Â Â A1A1Â
substituting \(\pi \)Â Â Â Â Â (M1)
e.g. \(f'(\pi ) = {{\rm{e}}^\pi }\cos \pi – {{\rm{e}}^\pi }\sin \pi \) , \({{\rm{e}}^\pi }( – 1 – 0)\) , \( – {{\rm{e}}^\pi }\)
taking negative reciprocal     (M1)
e.g. \( – \frac{1}{{f'(\pi )}}\)
gradient is \(\frac{1}{{{{\rm{e}}^\pi }}}\)Â Â Â Â A1Â Â Â Â N3Â
[6 marks]
Question
Let \(f(x) = {{\rm{e}}^{ – 3x}}\) and \(g(x) = \sin \left( {x – \frac{\pi }{3}} \right)\) .
Write down
(i) Â Â \(f'(x)\) ;
(ii)Â Â Â \(g'(x)\) .
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) \( – 3{{\rm{e}}^{ – 3x}}\)Â Â Â Â Â A1 Â Â N1
(ii) \(\cos \left( {x – \frac{\pi }{3}} \right)\)Â Â Â Â Â A1Â Â Â Â 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 \(x = \frac{\pi }{3}\)Â Â Â Â Â (A1)
e.g. \( – 3{{\rm{e}}^{ – 3\frac{\pi }{3}}}\sin \left( {\frac{\pi }{3} – \frac{\pi }{3}} \right) + {{\rm{e}}^{ – 3\frac{\pi }{3}}}\cos \left( {\frac{\pi }{3} – \frac{\pi }{3}} \right)\)ï€ ïƒ§       
\(h’\left( {\frac{\pi }{3}} \right) = {{\rm{e}}^{ – \pi }}\)Â Â Â Â A1Â Â Â Â N3
[4 marks]
Question
Let \(f(x) = \frac{{\cos x}}{{\sin x}}\) , for \(\sin x \ne 0\) .
In the following table, \(f’\left( {\frac{\pi }{2}} \right) = p\)Â and \(f”\left( {\frac{\pi }{2}} \right) = q\)Â . The table also gives approximate values of \(f'(x)\) and \(f”(x)\) near \(x = \frac{\pi }{2}\) .
Use the quotient rule to show that \(f'(x) = \frac{{ – 1}}{{{{\sin }^2}x}}\) .
Find \(f”(x)\) .
Find the value of p and of q.
Use information from the table to explain why there is a point of inflexion on the graph of f where \(x = \frac{\pi }{2}\) .
Answer/Explanation
Markscheme
\(\frac{{\rm{d}}}{{{\rm{d}}x}}\sin x = \cos x\) , \(\frac{{\rm{d}}}{{{\rm{d}}x}}\cos x = – \sin x\)Â (seen anywhere)Â Â Â Â (A1)(A1)
evidence of using the quotient rule    M1
correct substitution    A1
e.g. \(\frac{{\sin x( – \sin x) – \cos x(\cos x)}}{{{{\sin }^2}x}}\)Â , \(\frac{{ – {{\sin }^2}x – {{\cos }^2}x}}{{{{\sin }^2}x}}\)
\(f'(x) = \frac{{ – ({{\sin }^2}x + {{\cos }^2}x)}}{{{{\sin }^2}x}}\)Â Â Â Â A1
\(f'(x) = \frac{{ – 1}}{{{{\sin }^2}x}}\)Â Â Â Â AG Â Â Â N0
[5 marks]
METHOD 1
appropriate approach    (M1)
e.g. \(f'(x) = – {(\sin x)^{ – 2}}\)
\(f”(x) = 2({\sin ^{ – 3}}x)(\cos x)\)Â \(\left( { = \frac{{2\cos x}}{{{{\sin }^3}x}}} \right)\)Â Â Â Â Â A1A1 Â Â N3
Note: Award A1 for \(2{\sin ^{ – 3}}x\)Â , A1 for \(\cos x\)Â .
METHOD 2
derivative of \({\sin ^2}x = 2\sin x\cos x\)Â (seen anywhere)Â Â Â Â A1
evidence of choosing quotient rule    (M1)
e.g. \(u = – 1\) , Â \(v = {\sin ^2}x\) , \(f” = \frac{{{{\sin }^2}x \times 0 – ( – 1)2\sin x\cos x}}{{{{({{\sin }^2}x)}^2}}}\)
\(f”(x) = \frac{{2\sin x\cos x}}{{{{({{\sin }^2}x)}^2}}}\)Â \(\left( { = \frac{{2\cos x}}{{{{\sin }^3}x}}} \right)\)Â Â Â Â Â A1Â Â Â Â N3
[3 marks]
evidence of substituting \(\frac{\pi }{2}\)Â Â Â Â Â M1
e.g. \(\frac{{ – 1}}{{{{\sin }^2}\frac{\pi }{2}}}\) , \(\frac{{2\cos \frac{\pi }{2}}}{{{{\sin }^3}\frac{\pi }{2}}}\)
\(p = – 1\) , Â \(q = 0\)Â Â Â A1A1 Â Â N1N1
[3 marks]
second derivative is zero, second derivative changes sign    R1R1    N2
[2 marks]
Question
Let \(g(x) = \frac{{\ln x}}{{{x^2}}}\) , for \(x > 0\) .
Use the quotient rule to show that \(g'(x) = \frac{{1 – 2\ln x}}{{{x^3}}}\) .
The graph of g has a maximum point at A. Find the x-coordinate of A.
Answer/Explanation
Markscheme
\(\frac{{\rm{d}}}{{{\rm{d}}x}}\ln x = \frac{1}{x}\) , \(\frac{{\rm{d}}}{{{\rm{d}}x}}{x^2} = 2x\)Â (seen anywhere)Â Â Â Â A1A1
attempt to substitute into the quotient rule (do not accept product rule)Â Â Â Â M1
e.g. \(\frac{{{x^2}\left( {\frac{1}{x}} \right) – 2x\ln x}}{{{x^4}}}\)
correct manipulation that clearly leads to result    A1
e.g. \(\frac{{x – 2x\ln x}}{{{x^4}}}\) , \(\frac{{x(1 – 2\ln x)}}{{{x^4}}}\) , \(\frac{x}{{{x^4}}}\) , \(\frac{{2x\ln x}}{{{x^4}}}\)
\(g'(x) = \frac{{1 – 2\ln x}}{{{x^3}}}\)Â Â Â Â AGÂ Â Â Â N0
[4 marks]
evidence of setting the derivative equal to zero    (M1)
e.g. \(g'(x) = 0\) , \(1 – 2\ln x = 0\)
\(\ln x = \frac{1}{2}\)Â Â Â Â A1
\(x = {{\rm{e}}^{\frac{1}{2}}}\)Â Â Â Â Â A1Â Â Â Â N2
[3 marks]
Question
Let \(h(x) = \frac{{6x}}{{\cos x}}\) . Find \(h'(0)\) .
Answer/Explanation
Markscheme
METHOD 1 (quotient)
derivative of numerator is 6Â Â Â Â (A1)
derivative of denominator is \( – \sin x\)Â Â Â Â (A1)
attempt to substitute into quotient rule    (M1)
correct substitution    A1
e.g. \(\frac{{(\cos x)(6) – (6x)( – \sin x)}}{{{{(\cos x)}^2}}}\)
substituting \(x = 0\)Â Â Â Â Â (A1)
e.g. \(\frac{{(\cos 0)(6) – (6 \times 0)( – \sin 0)}}{{{{(\cos 0)}^2}}}\)
\(h'(0) = 6\)Â Â Â Â Â A1Â Â Â Â N2
METHOD 2 (product)
\(h(x) = 6x \times {(\cos x)^{ – 1}}\)
derivative of 6x is 6Â Â Â Â (A1)
derivative of \({(\cos x)^{ – 1}}\) is \(( – {(\cos x)^{ – 2}}( – \sin x))\)Â Â Â Â (A1)
attempt to substitute into product rule    (M1)
correct substitution    A1
e.g. \((6x)( – {(\cos x)^{ – 2}}( – \sin x)) + (6){(\cos x)^{ – 1}}\)
substituting \(x = 0\)Â Â Â Â (A1)
e.g. \((6 \times 0)( – {(\cos 0)^{ – 2}}( – \sin 0)) + (6){(\cos 0)^{ – 1}}\)
\(h'(0) = 6\)Â Â Â Â Â A1Â Â Â Â N2
[6 marks]
Question
Let \(f(x) = \frac{x}{{ – 2{x^2} + 5x – 2}}\) for \( – 2 \le x \le 4\) , \(x \ne \frac{1}{2}\) , \(x \ne 2\) . The graph of \(f\) is given below.
The graph of \(f\) has a local minimum at A(\(1\), \(1\)) and a local maximum at B.
Use the quotient rule to show that \(f'(x) = \frac{{2{x^2} – 2}}{{{{( – 2{x^2} + 5x – 2)}^2}}}\) .
Hence find the coordinates of B.
Given that the line \(y = k\) does not meet the graph of f , find the possible values of k .
Answer/Explanation
Markscheme
correct derivatives applied in quotient rule    (A1)A1A1
\(1\), \( – 4x + 5\)
Note: Award (A1) for 1, A1 for \( – 4x\) and A1 for \(5\), only if it is clear candidates are using the quotient rule.
correct substitution into quotient rule    A1
e.g. \(\frac{{1 \times ( – 2{x^2} + 5x – 2) – x( – 4x + 5)}}{{{{( – 2{x^2} + 5x – 2)}^2}}}\) , \(\frac{{ – 2{x^2} + 5x – 2 – x( – 4x + 5)}}{{{{( – 2{x^2} + 5x – 2)}^2}}}\)
correct working    (A1)
e.g. \(\frac{{ – 2{x^2} + 5x – 2 – ( – 4{x^2} + 5x)}}{{{{( – 2{x^2} + 5x – 2)}^2}}}\)
expression clearly leading to the answer    A1
e.g. \(\frac{{ – 2{x^2} + 5x – 2 + 4{x^2} – 5x}}{{{{( – 2{x^2} + 5x – 2)}^2}}}\)
\(f'(x) = \frac{{2{x^2} – 2}}{{{{( – 2{x^2} + 5x – 2)}^2}}}\)Â Â Â AGÂ Â Â Â N0
[6 marks]
evidence of attempting to solve \(f'(x) = 0\)Â Â Â Â (M1)
e.g. \(2{x^2} – 2 = 0\)
evidence of correct working    A1
e.g. \({x^2} = 1,\frac{{ \pm \sqrt {16} }}{4}{\text{, }}2(x – 1)(x + 1)\)
correct solution to quadratic    (A1)
e.g. \(x = \pm 1\)
correct x-coordinate \(x = – 1\)Â (may be seen in coordinate form \(\left( { – 1,\frac{1}{9}} \right)\) )Â Â Â A1Â Â Â Â N2
attempt to substitute \( – 1\)Â into f (do not accept any other value)Â Â Â Â (M1)
e.g. \(f( – 1) = \frac{{ – 1}}{{ – 2 \times {{( – 1)}^2} + 5 \times ( – 1) – 2}}\)
correct working
e.g. \(\frac{{ – 1}}{{ – 2 – 5 – 2}}\)    A1
correct y-coordinate \(y = \frac{1}{9}\) (may be seen in coordinate form \(\left( { – 1,\frac{1}{9}} \right)\) )Â Â Â A1Â Â Â Â N2
[7 marks]
recognizing values between max and min    (R1)
\(\frac{1}{9} < k < 1\)Â Â Â Â Â A2 Â Â N3
[3 marks]
Examiners report
While most candidates answered part (a) correctly, there were some who did not show quite enough work for a “show that” question. A very small number of candidates did not follow the instruction to use the quotient rule.
In part (b), most candidates knew that they needed to solve the equation \(f'(x) = 0\)Â , and many were successful in answering this question correctly. However, some candidates failed to find both values of x, or made other algebraic errors in their solutions. One common error was to find only one solution for \({x^2} = 1\)Â ; another was to work with the denominator equal to zero, rather than the numerator.
In part (c), a significant number of candidates seemed to think that the line \(y = k\)Â was a vertical line, and attempted to find the vertical asymptotes. Others tried looking for a horizontal asymptote. Fortunately, there were still a good number of intuitive candidates who recognized the link with the graph and with part (b), and realized that the horizontal line must pass through the space between the given local minimum and the local maximum they had found in part (b).
Question
Let \(f(x) = \frac{{6x}}{{x + 1}}\) , for \(x > 0\) .
Find \(f'(x)\) .
Let \(g(x) = \ln \left( {\frac{{6x}}{{x + 1}}} \right)\) , for \(x > 0\) .
Show that \(g'(x) = \frac{1}{{x(x + 1)}}\) .
Let \(h(x) = \frac{1}{{x(x + 1)}}\) . The area enclosed by the graph of h , the x-axis and the lines \(x = \frac{1}{5}\) and \(x = k\) is \(\ln 4\) . Given that \(k > \frac{1}{5}\) , find the value of k .
Answer/Explanation
Markscheme
METHOD 1
evidence of choosing quotient rule    (M1)
e.g. \(\frac{{u’v – uv’}}{{{v^2}}}\)
evidence of correct differentiation (must be seen in quotient rule)Â Â Â Â (A1)(A1)
e.g. \(\frac{{\rm{d}}}{{{\rm{d}}x}}(6x) = 6\) , \(\frac{{\rm{d}}}{{{\rm{d}}x}}(x + 1) = 1\)
correct substitution into quotient rule    A1
e.g. \(\frac{{(x + 1)6 – 6x}}{{{{(x + 1)}^2}}}\) , \(\frac{{6x + 6 – 6x}}{{{{(x + 1)}^2}}}\)
\(f'(x) = \frac{6}{{{{(x + 1)}^2}}}\)Â Â Â A1Â Â Â Â N4
[5 marks]
METHOD 2
evidence of choosing product rule    (M1)
e.g. \(6x{(x + 1)^{ – 1}}\) , \(uv’ + vu’\)
evidence of correct differentiation (must be seen in product rule)Â Â Â Â (A1)(A1)
e.g. \(\frac{{\rm{d}}}{{{\rm{d}}x}}(6x) = 6\) , \(\frac{{\rm{d}}}{{{\rm{d}}x}}{(x + 1)^{ – 1}} = – 1{(x + 1)^{ – 2}} \times 1\)
correct working    A1
e.g. \(6x \times – {(x + 1)^{ – 2}} + {(x + 1)^{ – 1}} \times 6\) , \(\frac{{ – 6x + 6(x + 1)}}{{{{(x + 1)}^2}}}\)
\(f'(x) = \frac{6}{{{{(x + 1)}^2}}}\)Â Â A1Â Â Â Â N4
[5 marks]
METHOD 1
evidence of choosing chain rule    (M1)
e.g. formula, \(\frac{1}{{\left( {\frac{{6x}}{{x + 1}}} \right)}} \times \left( {\frac{{6x}}{{x + 1}}} \right)\)
correct reciprocal of \(\frac{1}{{\left( {\frac{{6x}}{{x + 1}}} \right)}}\)Â is \(\frac{{x + 1}}{{6x}}\)Â (seen anywhere)Â Â Â Â A1
correct substitution into chain rule    A1
e.g. \(\frac{1}{{\left( {\frac{{6x}}{{x + 1}}} \right)}} \times \frac{6}{{{{(x + 1)}^2}}}\) , \(\left( {\frac{6}{{{{(x + 1)}^2}}}} \right)\left( {\frac{{x + 1}}{{6x}}} \right)\)
working that clearly leads to the answer    A1
e.g. \(\left( {\frac{6}{{(x + 1)}}} \right)\left( {\frac{1}{{6x}}} \right)\) , \(\left( {\frac{1}{{{{(x + 1)}^2}}}} \right)\left( {\frac{{x + 1}}{x}} \right)\) , \(\frac{{6(x + 1)}}{{6x{{(x + 1)}^2}}}\)
\(g'(x) = \frac{1}{{x(x + 1)}}\)Â Â Â Â AGÂ Â Â Â N0
[4 marks]
METHOD 2
attempt to subtract logs    (M1)
e.g. \(\ln a – \ln b\) , \(\ln 6x – \ln (x + 1)\)
correct derivatives (must be seen in correct expression)Â Â Â Â A1A1
e.g. \(\frac{6}{{6x}} – \frac{1}{{x + 1}}\) , \(\frac{1}{x} – \frac{1}{{x + 1}}\)
working that clearly leads to the answer    A1
e.g. \(\frac{{x + 1 – x}}{{x(x + 1)}}\) , \(\frac{{6x + 6 – 6x}}{{6x(x + 1)}}\) , \(\frac{{6(x + 1 – x)}}{{6x(x + 1)}}\)
\(g'(x) = \frac{1}{{x(x + 1)}}\)Â Â Â Â AGÂ Â Â Â N0
[4 marks]
valid method using integral of h(x) (accept missing/incorrect limits or missing \({\text{d}}x\) )    (M1)
e.g. \({\rm{area}} = \int_{\frac{1}{5}}^k {h(x){\rm{d}}x} \) , \(\int{\left( {\frac{1}{{x(x + 1)}}} \right)} \)Â
recognizing that integral of derivative will give original function    (R1)
e.g. \(\int{\left( {\frac{1}{{x(x + 1)}}} \right)} {\rm{d}}x = \ln \left( {\frac{{6x}}{{x + 1}}} \right)\)
correct substitution and subtraction    A1
e.g. \(\ln \left( {\frac{{6k}}{{k + 1}}} \right) – \ln \left( {\frac{{6 \times \frac{1}{5}}}{{\frac{1}{5} + 1}}} \right)\) , \(\ln \left( {\frac{{6k}}{{k + 1}}} \right) – \ln (1)\)
setting their expression equal to \(\ln 4\) Â Â Â (M1)Â
e.g. \(\ln \left( {\frac{{6k}}{{k + 1}}} \right) – \ln (1) = \ln 4\) , \(\ln \left( {\frac{{6k}}{{k + 1}}} \right) = \ln 4\) , \(\int_{\frac{1}{5}}^k {h(x){\rm{d}}x = \ln 4} \)
correct equation without logs    A1
e.g.\(\frac{{6k}}{{k + 1}} = 4\) , \(6k = 4(k + 1)\)Â
correct working    (A1)
e.g. \(6k = 4k + 4\) , \(2k = 4\)
\(k = 2\)Â Â Â A1 Â Â N4
[7 marks]
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) \(\cos x\)Â , \(2x\)Â Â Â Â (A1)(A1)
\(f'(x) = {x^2}\cos x + 2x\sin x\)Â Â Â Â A1 N4
[4 marks]
substituting \(\frac{\pi }{2}\)Â into their \(f'(x)\)Â Â Â Â Â (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  \(0 + 2\left( {\frac{\pi }{2}} \right) \times 1\)Â
\(f’\left( {\frac{\pi }{2}} \right) = \pi \)Â Â Â Â A1 N2
[3 marks]
Question
Consider the functions \(f(x)\) , \(g(x)\) and \(h(x)\) . The following table gives some values associated with these functions.
The following diagram shows parts of the graphs of \(h\) and \(h”\) .
There is a point of inflexion on the graph of \(h\) at P, when \(x = 3\) .
Given that \(h(x) = f(x) \times g(x)\) ,
Write down the value of \(g(3)\) , of \(f'(3)\) , and of \(h”(2)\) .
Explain why P is a point of inflexion.
find the \(y\)-coordinate of P.
find the equation of the normal to the graph of \(h\) at P.
Answer/Explanation
Markscheme
\(g(3) = – 18\) , \(f'(3) = 1\) , \(h”(2) = – 6\)Â Â Â Â Â A1A1A1Â Â Â Â N3
[3 marks]
\(h”(3) = 0\)Â Â Â Â (A1)
valid reasoning    R1
eg  \({h”}\) changes sign at \(x = 3\) , change in concavity of \(h\) at \(x = 3\)
so P is a point of inflexion    AG    N0
[2 marks]
writing \(h(3)\)Â as a product of \(f(3)\)Â and \(g(3)\)Â Â Â Â Â A1
eg  \(f(3) \times g(3)\) , \(3 \times ( – 18)\)
\(h(3) = – 54\)Â Â Â Â A1 N1
[2 marks]
recognizing need to find derivative of \(h\)Â Â Â Â (R1)
eg  \({h’}\) , \(h'(3)\)
attempt to use the product rule (do not accept \(h’ = f’ \times g’\)Â )Â Â Â Â (M1)
eg  \(h’ = fg’ + gf’\) ,  \(h'(3) = f(3) \times g'(3) + g(3) \times f'(3)\)
correct substitution    (A1)
eg  \(h'(3) = 3( – 3) + ( – 18) \times 1\)
\(h'(3) = – 27\)Â Â Â A1
attempt to find the gradient of the normal    (M1)
eg  \( – \frac{1}{m}\) , \( – \frac{1}{{27}}x\)Â
attempt to substitute their coordinates and their normal gradient into the equation of a line    (M1)
eg  \( – 54 = \frac{1}{{27}}(3) + b\) , \(0 = \frac{1}{{27}}(3) + b\) , \(y + 54 = 27(x – 3)\) , \(y – 54 = \frac{1}{{27}}(x + 3)\)
correct equation in any form    A1    N4
eg  \(y + 54 = \frac{1}{{27}}(x – 3)\) , \(y = \frac{1}{{27}}x – 54\frac{1}{9}\)
[7 marks]
Question
Let \(f(x) = \frac{{2x}}{{{x^2} + 5}}\).
Use the quotient rule to show that \(f'(x) = \frac{{10 – 2{x^2}}}{{{{({x^2} + 5)}^2}}}\).
Find \(\int {\frac{{2x}}{{{x^2} + 5}}{\text{d}}x} \).
The following diagram shows part of the graph of \(f\).
The shaded region is enclosed by the graph of \(f\), the \(x\)-axis, and the lines \(x = \sqrt 5 \) and \(x = q\). This region has an area of \(\ln 7\). Find the value of \(q\).
Answer/Explanation
Markscheme
derivative of \(2x\) is \(2\) (must be seen in quotient rule) Â Â (A1)
derivative of \({x^2} + 5\) is \(2x\) (must be seen in quotient rule) Â Â (A1)
correct substitution into quotient rule   A1
eg   \(\frac{{({x^2} + 5)(2) – (2x)(2x)}}{{{{({x^2} + 5)}^2}}},{\text{ }}\frac{{2({x^2} + 5) – 4{x^2}}}{{{{({x^2} + 5)}^2}}}\)
correct working which clearly leads to given answer  A1
eg  \(\frac{{2{x^2} + 10 – 4{x^2}}}{{{{({x^2} + 5)}^2}}},{\text{ }}\frac{{2{x^2} + 10 – 4{x^2}}}{{{x^4} + 10{x^2} + 25}}\)
\(f'(x) = \frac{{10 – 2{x^2}}}{{{{({x^2} + 5)}^2}}}\) Â Â AG Â Â N0
[4 marks]
valid approach using substitution or inspection   (M1)
eg   \(u = {x^2} + 5,{\text{ d}}u = 2x{\text{d}}x,{\text{ }}\frac{1}{2}\ln ({x^2} + 5)\)
\(\int {\frac{{2x}}{{{x^2} + 5}}{\text{d}}x = \int {\frac{1}{u}{\text{d}}u} } \) Â Â (A1)
\(\int {\frac{1}{u}{\text{d}}u = \ln u + c} \) Â Â (A1)
\(\ln ({x^2} + 5) + c\) Â Â A1 Â Â N4
[4 marks]
correct expression for area   (A1)
eg   \(\left[ {\ln \left( {{x^2} + 5} \right)} \right]_{\sqrt 5 }^q,{\text{ }}\int\limits_{\sqrt 5 }^q {\frac{{2x}}{{{x^2} + 5}}{\text{d}}x} \)
substituting limits into their integrated function and subtracting (in either order) Â Â (M1)
eg   \(\ln ({q^2} + 5) – \ln \left( {{{\sqrt 5 }^2} + 5} \right)\)
correct working   (A1)
eg   \(\ln \left( {{q^2} + 5} \right) – \ln 10,{\text{ }}\ln \frac{{{q^2} + 5}}{{10}}\)
equating their expression to \(\ln 7\) (seen anywhere) Â Â (M1)
eg   \(\ln \left( {{q^2} + 5} \right) – \ln 10 = \ln 7,{\text{ }}\ln \frac{{{q^2} + 5}}{{10}} = \ln 7,{\text{ }}\ln ({q^2} + 5) = \ln 7 + \ln 10\)
correct equation without logs   (A1)
eg   \(\frac{{{q^2} + 5}}{{10}} = 7,{\text{ }}{q^2} + 5 = 70\)
\({q^2} = 65\) Â Â (A1)
\(q = \sqrt {65} \) Â Â A1 Â Â N3
Â
Note:Â Award A0 for \(q =Â \pm \sqrt {65} \).
Â
[7 marks]
Question
Let \(f(x) = \sqrt {4x + 5} \), for \(x \geqslant  – 1.25\).
Consider another function \(g\). Let R be a point on the graph of \(g\). The \(x\)-coordinate of R is 1. The equation of the tangent to the graph at R is \(y = 3x + 6\).
Find \(f'(1)\).
Write down \(g'(1)\).
Find \(g(1)\).
Let \(h(x) = f(x) \times g(x)\). Find the equation of the tangent to the graph of \(h\) at the point where \(x = 1\).
Answer/Explanation
Markscheme
choosing chain rule   (M1)
eg\(\,\,\,\,\,\)\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}u}} \times \frac{{{\text{d}}u}}{{{\text{d}}x}},{\text{ }}u = 4x + 5,{\text{ }}u’ = 4\)
correct derivative of \(f\) Â Â A2
eg\(\,\,\,\,\,\)\(\frac{1}{2}{(4x + 5)^{ – \frac{1}{2}}} \times 4,{\text{ }}f'(x) = \frac{2}{{\sqrt {4x + 5} }}\)
\(f'(1) = \frac{2}{3}\) Â Â A1 Â Â N2
[4 marks]
recognize that \(g'(x)\) is the gradient of the tangent   (M1)
eg\(\,\,\,\,\,\)\(g'(x) = m\)
\(g'(1) = 3\) Â Â A1 Â Â N2
[2 marks]
recognize that R is on the tangent   (M1)
eg\(\,\,\,\,\,\)\(g(1) = 3 \times 1 + 6\), sketch
\(g(1) = 9\) Â Â A1 Â Â N2
[2 marks]
\(f(1) = \sqrt {4 + 5} {\text{ }}( = 3)\) (seen anywhere) Â Â A1
\(h(1) = 3 \times 9{\text{ }}( = 27)\) (seen anywhere) Â Â A1
choosing product rule to find \(h'(x)\) Â Â (M1)
eg\(\,\,\,\,\,\)\(uv’ + u’v\)
correct substitution to find \(h'(1)\) Â Â (A1)
eg\(\,\,\,\,\,\)\(f(1) \times g'(1) + f'(1) \times g(1)\)
\(h'(1) = 3 \times 3 + \frac{2}{3} \times 9{\text{ }}( = 15)\) Â Â A1
EITHER
attempt to substitute coordinates (in any order) into the equation of a straight line   (M1)
eg\(\,\,\,\,\,\)\(y – 27 = h'(1)(x – 1),{\text{ }}y – 1 = 15(x – 27)\)
\(y – 27 = 15(x – 1)\) Â Â A1 Â Â N2
OR
attempt to substitute coordinates (in any order) to find the \(y\)-intercept   (M1)
eg\(\,\,\,\,\,\)\(27 = 15 \times 1 + b,{\text{ }}1 = 15 \times 27 + b\)
\(y = 15x + 12\) Â Â A1 Â Â N2
[7 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) Â Â Find the first four derivatives of \(f(x)\).
(ii) Â Â Find \({f^{(19)}}(x)\).
(i) Â Â Find the first three derivatives of \(g(x)\).
(ii) Â Â Given that \({g^{(19)}}(x) = \frac{{k!}}{{(k – p)!}}({x^{k – 19}})\), find \(p\).
(i) Â Â Find \(h'(x)\).
(ii) Â Â Hence, show that \(h'(\pi ) = \frac{{ – 21!}}{2}{\pi ^2}\).
Answer/Explanation
Markscheme
(i) Â Â \(f'(x) = Â – \sin x,{\text{ }}f”(x) = Â – \cos x,{\text{ }}{f^{(3)}}(x) = \sin x,{\text{ }}{f^{(4)}}(x) = \cos x\)Â Â Â A2 Â Â N2
(ii)   valid approach   (M1)
eg\(\,\,\,\,\,\)recognizing that 19 is one less than a multiple of 4, \({f^{(19)}}(x) = {f^{(3)}}(x)\)
\({f^{(19)}}(x) = \sin x\)Â Â Â A1 Â Â N2
[4 marks]
(i) Â Â Â \(g'(x) = k{x^{k – 1}}\)
\(g”(x) = k(k – 1){x^{k – 2}},{\text{ }}{g^{(3)}}(x) = k(k – 1)(k – 2){x^{k – 3}}\)Â Â Â A1A1 Â Â N2
(ii) Â Â METHOD 1
correct working that leads to the correct answer, involving the correct expression for the 19th derivative   A2
eg\(\,\,\,\,\,\)\(k(k – 1)(k – 2) \ldots (k – 18) \times \frac{{(k – 19)!}}{{(k – 19)!}},{{\text{ }}_k}{P_{19}}\)
\(p = 19\)Â (accept \(\frac{{k!}}{{(k – 19)!}}{x^{k – 19}}\)) Â Â A1 Â Â 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\(\,\,\,\,\,\)\(g” = 2!\left( {\begin{array}{*{20}{c}} k \\ 2 \end{array}} \right),{\text{ }}k(k – 1)(k – 2) = \frac{{k!}}{{(k – 3)!}},{\text{ }}{g^{(3)}}(x){ = _k}{P_3}({x^{k – 3}})\)
\({g^{(19)}}(x) = 19!\left( {\begin{array}{*{20}{c}} k \\ {19} \end{array}} \right),{\text{ }}19! \times \frac{{k!}}{{(k – 19)! \times 19!}},{{\text{ }}_k}{P_{19}}\)
\(p = 19\)Â (accept \(\frac{{k!}}{{(k – 19)!}}{x^{k – 19}}\)) Â Â A1 Â Â 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) Â Â (A1)(A1)
eg\(\,\,\,\,\,\)\({g^{(20)}}(x) = \frac{{21!}}{{(21 – 20)!}}x,{\text{ }}{f^{(20)}}(x) = \cos x\)
\(h'(x) = \sin x(21!x) + \cos x\left( {\frac{{21!}}{2}{x^2}} \right){\text{ }}\left( {{\text{accept }}\sin x\left( {\frac{{21!}}{{1!}}x} \right) + \cos x\left( {\frac{{21!}}{{2!}}{x^2}} \right)} \right)\) Â Â A1 Â Â N3
(ii) Â Â substituting \(x = \pi \) (seen anywhere) Â Â (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 \(\cos \pi \)Â (seen anywhere) Â Â (A1)
eg\(\,\,\,\,\,\)\(\sin \pi  = 0,{\text{ }}\cos \pi  =  – 1\)
evidence of correct values substituted into \(h'(\pi )\)Â Â Â 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.
\(\frac{{ – 21!}}{2}{\pi ^2}\)Â Â Â AG Â Â N0
[7 marks]
Question
The values of the functions \(f\) and \(g\) and their derivatives for \(x = 1\) and \(x = 8\) are shown in the following table.
Let \(h(x) = f(x)g(x)\).
Find \(h(1)\).
Find \(h'(8)\).
Answer/Explanation
Markscheme
expressing \(h(1)\) as a product of \(f(1)\) and \(g(1)\)   (A1)
eg\(\,\,\,\,\,\)\(f(1) \times g(1),{\text{ }}2(9)\)
\(h(1) = 18\) Â Â A1 Â Â N2
[2 marks]
attempt to use product rule (do not accept \(h’ = f’ \times g’\))   (M1)
eg\(\,\,\,\,\,\)\(h’ = fg’ + gf’,{\text{ }}h'(8) = f'(8)g(8) + g’(8)f(8)\)
correct substitution of values into product rule   (A1)Â
eg\(\,\,\,\,\,\)\(h’(8) = 4(5) + 2( – 3),{\text{ }} – 6 + 20\)
\(h’(8) = 14\)   A1 N2
[3 marks]