Question
The function \(f\) is defined by \(f(x) = \frac{{{{\rm{e}}^x} + {{\rm{e}}^{ – x}}}}{2}\) .
(i) Obtain an expression for \({f^{(n)}}(x)\) , the nth derivative of \(f(x)\) with respect to \(x\).
(ii) Hence derive the Maclaurin series for \(f(x)\) up to and including the term in \({x^4}\) .
(iii) Use your result to find a rational approximation to \(f\left( {\frac{1}{2}} \right)\) .
(iv) Use the Lagrange error term to determine an upper bound to the error in this approximation.
Use the integral test to determine whether the series \(\sum\limits_{n = 1}^\infty {\frac{{\ln n}}{{{n^2}}}} \) is convergent or divergent.
Answer/Explanation
Markscheme
(i) \({f^{(n)}}(x) = \frac{{{{\rm{e}}^x} + {{( – 1)}^n}{{\rm{e}}^{ – x}}}}{2}\) (M1)A1
(ii) Coefficient of \({x^n} = \frac{{{f^{(n)}}(0)}}{{n!}}\) (M1)
\( = \frac{{1 + {{( – 1)}^n}}}{{2n!}}\) (A1)
\(f(x) = 1 + \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{24}} + \ldots \) A1
(iii) Putting \(x = \frac{1}{2}\) M1
\(f(0.5) = 1 + \frac{1}{8} + \frac{1}{{16 \times 24}} = \frac{{433}}{{384}}\) (M1)A1
(iv) Lagrange error term \( = \frac{{{f^{(n + 1)}}(c)}}{{(n + 1)!}}{x^{n + 1}}\) M1
\( = \frac{{{f^{(5)}}(c)}}{{120}} \times {\left( {\frac{1}{2}} \right)^5}\) A1
\({{f^{(5)}}(c)}\) is an increasing function because – any valid reason, e.g. plotted a graph, positive derivative, increasing function minus a decreasing function, so this is maximized when \(x = 0.5\) . R1
Therefore upper bound \( = \frac{{({{\rm{e}}^{0.5}} – {{\rm{e}}^{ – 0.5}})}}{{2 \times 120}} \times {\left( {\frac{1}{2}} \right)^5}\) M1
\( = 0.000136\) A1
[13 marks]
We consider \(\int_1^\infty {\frac{{\ln x}}{{{x^2}}}} {\rm{d}}x = \int_1^\infty {\ln x{\rm{d}}x} \left( { – \frac{1}{2}} \right)\) M1A1
\( = \left[ { – \frac{{\ln x}}{x}} \right]_1^\infty + \int_1^\infty {\frac{{1x}}{{{x^2}}}} {\rm{d}}x\) A1A1
\( = \left[ { – \frac{{\ln x}}{x}} \right]_1^\infty – \left[ {\frac{1}{x}} \right]_1^\infty \) A1
Now \(\mathop {\lim }\limits_{x \to \infty } \left( {\frac{1}{x}} \right) = 0\) R1
\(\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{\ln x}}{x}} \right) = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{1}{x}} \right) = 0\) M1A1
The integral is convergent with value \(1\) and so therefore is the series. R1
[9 marks]
Question
The function \(f\) is defined by \(f(x) = \ln (1 + \sin x)\) .
When a scientist measures the concentration \(\mu \) of a solution, the measurement obtained may be assumed to be a normally distributed random variable with mean \(\mu \) and standard deviation \(1.6\).
Show that \(f”(x) = \frac{{ – 1}}{{1 + \sin x}}\) .
Determine the Maclaurin series for \(f(x)\) as far as the term in \({x^4}\) .
Deduce the Maclaurin series for \(\ln (1 – \sin x)\) as far as the term in \({x^4}\) .
By combining your two series, show that \(\ln \sec x = \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{12}} + \ldots \) .
Hence, or otherwise, find \(\mathop {\lim }\limits_{x \to 0} \frac{{\ln \sec x}}{{x\sqrt x }}\) .
He makes 5 independent measurements of the concentration of a particular solution and correctly calculates the following confidence interval for \(\mu \) .
[\(22.7\) , \(26.1\)]
Determine the confidence level of this interval.
He is now given a different solution and is asked to determine a \(95\%\) confidence interval for its concentration. The confidence interval is required to have a width less than \(2\). Find the minimum number of independent measurements required.
Answer/Explanation
Markscheme
\(f'(x) = \frac{{\cos x}}{{1 + \sin x}}\) M1A1
\(f”(x) = \frac{{ – \sin x(1 + \sin x) – {{\cos }^2}x}}{{{{(1 + \sin x)}^2}}}\) M1
\( = \frac{{ – \sin x – 1}}{{{{(1 + \sin x)}^2}}}\) A1
\( = \frac{{ – 1}}{{1 + \sin x}}\) AG
[4 marks]
\(f”'(x) = \frac{{\cos x}}{{{{(1 + \sin x)}^2}}}\) A1
\({f^{iv}}(x) = \frac{{ – \sin x{{(1 + \sin x)}^2} – 2(1 + \sin x){{\cos }^2}x}}{{{{(1 + \sin x)}^4}}}\) A1
\(f(0) = 0\) , \(f'(0) = 1\) , \(f”(0) = – 1\) , \(f”'(0) = 1\) , \({f^{iv}}(0) = – 2\) (A2)
Note: Award A1 for 2 errors and A0 for more than 2 errors.
\(\ln (1 + \sin x) = x – \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} – \frac{{{x^4}}}{{12}} + \ldots \) M1A1
[6 marks]
\(\ln (1 – \sin x) = \ln (1 + \sin ( – x)) = – x – \frac{{{x^2}}}{2} – \frac{{{x^3}}}{6} – \frac{{{x^4}}}{{12}} + \ldots \) M1A1
[2 marks]
Adding, M1
\(\ln (1 – {\sin ^2}x) = \ln {\cos ^2}x\) A1
\( = – {x^2} – \frac{{{x^4}}}{6} + \ldots \) A1
\(\ln \cos x = – \frac{{{x^2}}}{2} – \frac{{{x^4}}}{{12}} + \ldots \) A1
\(\ln \sec x = \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{12}} + \ldots \) AG
[4 marks]
\(\frac{{\ln \sec x}}{{x\sqrt x }} = \frac{{\sqrt x }}{2} + \frac{{{x^2}\sqrt x }}{{12}} + \ldots \) M1
Limit \( = 0\) A1
[2 marks]
Interval width \( = 26.1 – 22.7 = 3.4\)
So \(3.4 = 2z \times \frac{{1.6}}{{\sqrt 5 }}\) M1A1
\(z = 2.375 \ldots \) A1
Probability \( = 0.9912\) A1
Confidence level \( = 2 \times 0.4912 = 98.2\% \) A1
[5 marks]
\(z\)-value \( = 1.96\) A1
We require
\(2 \times \frac{{1.96 \times 1.6}}{{\sqrt n }} < 2\) M1A1
Whence \(n > 9.83\) A1
So we need \(n = 10\) A1
Note: Accept \( = \) signs throughout.
[5 marks]
Question
A machine fills containers with grass seed. Each container is supposed to weigh \(28\) kg. However the weights vary with a standard deviation of \(0.54\) kg. A random sample of \(24\) bags is taken to check that the mean weight is \(28\) kg.
Assuming the series for \({{\rm{e}}^x}\) , find the first five terms of the Maclaurin series for\[\frac{1}{{\sqrt {2\pi } }}{{\rm{e}}^{\frac{{ – {x^2}}}{2}}} {\rm{ .}}\]
(i) Use your answer to (a) to find an approximate expression for the cumulative distributive function of \({\rm{N}}(0,1)\) .
(ii) Hence find an approximate value for \({\rm{P}}( – 0.5 \le Z \le 0.5)\) , where \(Z \sim {\rm{N}}(0,1)\) .
State and justify an appropriate test procedure giving the null and alternate hypotheses.
What is the critical region for the sample mean if the probability of a Type I error is to be \(3.5\%\)?
If the mean weight of the bags is actually \(28\).1 kg, what would be the probability of a Type II error?
Answer/Explanation
Markscheme
\({{\rm{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} + \ldots \)
\({{\rm{e}}^{\frac{{ – {x^2}}}{2}}} = 1 + \left( { – \frac{{{x^2}}}{2}} \right) + \frac{{{{\left( { – \frac{{{x^2}}}{2}} \right)}^2}}}{{2!}} + \frac{{{{\left( { – \frac{{{x^2}}}{2}} \right)}^3}}}{{3!}} + \frac{{{{\left( { – \frac{{{x^2}}}{2}} \right)}^4}}}{{4!}} + \ldots \) M1A1
\(\frac{1}{{\sqrt {2\pi } }}{{\rm{e}}^{\frac{{ – {x^2}}}{2}}} = \frac{1}{{\sqrt {2\pi } }}\left( {1 – \frac{{{x^2}}}{2} + \frac{{{x^4}}}{8} – \frac{{{x^6}}}{{48}} + \frac{{{x^8}}}{{384}}} \right)\) A1
[3 marks]
(i) \(\frac{1}{{\sqrt {2\pi } }}\int_0^x {1 – \frac{{{t^2}}}{2}} + \frac{{{t^4}}}{8} – \frac{{{t^6}}}{{48}} + \frac{{{t^8}}}{{384}}{\rm{d}}t\) M1
\( = \frac{1}{{\sqrt {2\pi } }}\left( {x – \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{40}} – \frac{{{x^7}}}{{336}} + \frac{{{x^9}}}{{3456}}} \right)\) A1
\({\rm{P}}(Z \le x) = 0.5 + \frac{1}{{\sqrt {2\pi } }}\left( {x – \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{40}} – \frac{{{x^7}}}{{336}} + \frac{{{x^9}}}{{3456}} – \ldots } \right)\) R1A1
(ii) \({\rm{P}}( – 0.5 \le Z \le 0.5) = \frac{2}{{\sqrt {2\pi } }}\left( {0.5 – \frac{{{{0.5}^3}}}{6} + \frac{{{{0.5}^5}}}{{40}} – \frac{{{{0.5}^7}}}{{336}} + \frac{{{{0.5}^9}}}{{3456}} – \ldots } \right)\) M1
\( = 0.38292 = 0.383\) A1
[6 marks]
this is a two tailed test of the sample mean \(\overline x \)
we use the central limit theorem to justify assuming that R1
\(\overline X \sim {\rm{N}}\left( {28,\frac{{{{0.54}^2}}}{{24}}} \right)\) R1A1
\({{\rm{H}}_0}:\mu = 28\) A1
\({{\rm{H}}_1}:\mu \ne 28\) A1
[5 marks]
since \({\text{P(Type I error)}} = 0.035\) , critical value \(2.108\) (M1)A1
and (\(\overline x \le 28 – 2.108\sqrt {\frac{{{{0.54}^2}}}{{24}}} \) or \(\overline x \ge 28 + 2.108\sqrt {\frac{{{{0.54}^2}}}{{24}}} \) ) (M1)(A1)(A1)
\(\overline x \le 27.7676\) or \(\overline x \ge 28.2324\)
so \(\overline x \le 27.8\) or \(\overline x \ge 28.2\) A1A1
[7 marks]
if \(\mu = 28.1\)
\(\overline X \sim {\rm{N}}\left( {28.1,\frac{{{{0.54}^2}}}{{24}}} \right)\) R1
\({\text{P(Type II error)}} = {\rm{P}}(27.7676 < \overline x < 28.2324)\)
\( = 0.884\) A1
Note: Depending on the degree of accuracy used for the critical region the answer for part (c) can be anywhere from \(0.8146\) to \(0.879\).
[2 marks]
Question
Find the value of \(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} – \cot x} \right)\) .
Find the interval of convergence of the infinite series\[\frac{{(x + 2)}}{{3 \times 1}} + \frac{{{{(x + 2)}^2}}}{{{3^2} \times 2}} + \frac{{{{(x + 2)}^3}}}{{{3^3} \times 3}} + \ldots \]
(i) Find the Maclaurin series for \(\ln (1 + \sin x)\) up to and including the term in \({x^3}\) .
(ii) Hence find a series for \(\ln (1 – \sin x)\) up to and including the term in \({x^3}\) .
(iii) Deduce, by considering the difference of the two series, that \(\ln 3 \simeq \frac{\pi }{3}\left( {1 + \frac{{{\pi ^2}}}{{216}}} \right)\) .
Answer/Explanation
Markscheme
EITHER
\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} – \cot x} \right)\)
\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\tan x – x}}{{x\tan x}}} \right)\) M1A1
\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{{{\sec }^2}x – 1}}{{x{{\sec }^2}x + \tan x}}} \right)\) , using l’Hopital A1
\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{2{{\sec }^2}x\tan x}}{{2{{\sec }^2}x + 2x{{\sec }^2}x\tan x}}} \right)\) A1A1
\( = 0\) A1
OR
\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} – \cot x} \right)\)
\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x – x\cos x}}{{x\sin x}}} \right)\) M1A1
\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{x\sin x}}{{\sin x + x\cos x}}} \right)\) , using l’Hopital A1
\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x + x\cos x}}{{2\cos x – x\sin x}}} \right)\) A1A1
\( = 0\) A1
[6 marks]
\({u_n} = \frac{{{{(x + 2)}^n}}}{{{3^n} \times n}}\) A1
\(\frac{{{u_{n + 1}}}}{{{u_n}}} = \frac{{\frac{{{{(x + 2)}^{n + 1}}}}{{{3^{n + 1}} \times (n + 1)}}}}{{\frac{{{{(x + n)}^n}}}{{{3^n} \times n}}}} = \frac{{(x + 2)n}}{{3(n + 1)}}\) M1A1
\(\mathop {\lim }\limits_{n \to \infty } \frac{{(x + 2)n}}{{3(n + 1)}} = \frac{{(x + 2)}}{3}\) M1A1
\(\left| {\frac{{(x + 2)}}{3}} \right| < 1 \Rightarrow – 5 < x < 1\) M1A1
if \(x = 1\) series is \(1 + \frac{1}{2} + \frac{1}{3} + \ldots \) which diverges A1
if \(x = – 5\) series is \( – 1 + \frac{1}{2} – \frac{1}{3} + \ldots + \frac{{{{( – 1)}^n}}}{n}\) which converges A1
hence interval is \( – 5 \le x < 1\) A1
[10 marks]
(i) \(f(x) = \ln (1 + \sin x)\) , \(f(0) = 0\) A1
\(f'(x) = \frac{{\cos x}}{{1 + \sin x}}\) , \(f'(0) = 1\) A1
\(f”(x) = \frac{{ – \sin x(1 + \sin x) – {{\cos }^2}x}}{{{{(1 + \sin x)}^2}}} = \frac{{ – (1 + \sin x)}}{{{{(1 + \sin x)}^2}}} = \frac{{ – 1}}{{1 + \sin x}}\) , \(f”(0) = – 1\) A1
\(f”'(x) = \frac{{\cos x}}{{{{(1 + \sin x)}^2}}}\) , \(f”'(0) = 1\) A1
\(\ln (1 + \sin x) \approx x – \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} – \ldots \) A1
(ii) \( – \sin x = \sin ( – x)\) M1
so, \(\ln (1 – \sin x) \approx – x – \frac{{{x^2}}}{2} – \frac{{{x^3}}}{6} – \ldots \) A1
(iii) \(\ln (1 + \sin x) – ln(1 – \sin x)\)
\( = \ln \left( {\frac{{1 + \sin x}}{{1 – \sin x}}} \right) \approx 2x + \frac{{{x^3}}}{3}\) M1A1
let \(x = \frac{\pi }{6}\) then, \(\ln \left( {\frac{{1 + \frac{1}{2}}}{{1 – \frac{1}{2}}}} \right) = \ln 3 \approx 2\left( {\frac{\pi }{6}} \right) + \frac{{{{\left( {\frac{\pi }{6}} \right)}^3}}}{3}\) M1A1A1
\( = \frac{\pi }{3}\left( {1 + \frac{{{\pi ^2}}}{{216}}} \right)\) AG
[12 marks]
Question
Consider the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} + y\tan x = 2{\cos ^4}x\) given that \(y = 1\) when \(x = 0\).
(a) Solve the differential equation, giving your answer in the form \(y = f(x)\).
(b) (i) By differentiating both sides of the differential equation, show that
\[\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y = – 10\sin x{\cos ^3}x\]
(ii) Hence find the first four terms of the Maclaurin series for \(y\).
Answer/Explanation
Markscheme
(a) integrating factor \( = {e^{\int {\tan x{\text{d}}x} }}\) M1
\( = {{\text{e}}^{\ln \sec x}}\) A1
\( = \sec x\) A1
\(\sec x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y\sec x\tan x = 2{\cos ^3}x\) (M1)
integrating,
\(y\sec x = 2\int {{{\cos }^3}x{\text{d}}x} \) A1
\( = 2\int {\cos x(1 – {{\sin }^2}x){\text{d}}x} \) A1
\( = 2\left( {\sin x – \frac{{{{\sin }^3}x}}{3}} \right) + C\) A1
Note: Condone the absence of \(C\).
(substituting \(x = 0,{\text{ }}y = 1\))
\(1 = C\) M1
the solution is
\(y = 2\cos x\left( {\sin x – \frac{{{{\sin }^3}x}}{3}} \right) + \cos x\) A1
[9 marks]
(b) (i) differentiating the equation,
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y{\sec ^2}x + \tan x\frac{{{\text{d}}y}}{{{\text{d}}x}} = – 8{\cos ^3}x\sin x\) A1A1
Note: A1 for each side.
substituting for \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\),
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y{\sec ^2}x + \tan x\left( {2{{\cos }^4}x – y\tan x} \right) = – 8{\cos ^3}x\sin x\) A1
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y({\sec ^2}x – {\tan ^2}x) = – 8{\cos ^3}x\sin x – 2\tan x{\cos ^4}x\) (or equivalent) A1
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y = – 10\sin x{\cos ^3}x\) AG
(ii) differentiating again,
\(\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} + \frac{{{\text{d}}y}}{{{\text{d}}x}} = – 10{\cos ^4}x + {\text{term involving }} \sin x\) A1
it follows that
\(y(0) = 1,{\text{ }}y'(0) = 2\) A1
\(y”(0) = – 1,{\text{ }}y”'(0) = – 12\) A1
attempting to use \(y = y(0) + xy'(0) + \frac{{{x^2}}}{2}y”(0) + \frac{{{x^3}}}{6}y”'(0) + \ldots \) (M1)
\(y = 1 + 2x – \frac{{{x^2}}}{2} – 2{x^3}\) A1
[9 marks]
Question
Consider the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2x + y – 1\) with boundary condition \(y = 1\) when \(x = 0\).
Using Euler’s method with increments of \(0.2\), find an approximate value for \(y\) when \(x = 1\).
Explain how Euler’s method could be improved to provide a better approximation.
Solve the differential equation to find an exact value for \(y\) when \(x = 1\).
(i) Find the first three non-zero terms of the Maclaurin series for \(y\).
(ii) Hence find an approximate value for \(y\) when \(x = 1\).
Answer/Explanation
Markscheme
(M1)(A1)(A1)(A1)
Note: Award M1 for equivalent of setting up first row of table, A1 for each of row 2, 3 and 5.
approximate solution \(y = 1.98\) A1
make the increments smaller or any specific correct instruction – for example change increment from \(0.2\) to \(0.1\) A1
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} – y = 2x – 1\)
integrating factor is \({{\text{e}}^{\int { – 1{\text{d}}x} }} = {{\text{e}}^{ – x}}\) (M1)(A1)
\(\frac{{\text{d}}}{{{\text{d}}x}}(y{{\text{e}}^{ – x}}) = {{\text{e}}^{ – x}}(2x – 1)\) M1
attempt at integration by parts of \(\int {{{\text{e}}^{ – x}}(2x – 1){\text{d}}x} \) (M1)
\( = – (2x – 1){{\text{e}}^{ – x}} + \int {2{{\text{e}}^{ – x}}{\text{d}}x} \) A1
\( = – (2x – 1){{\text{e}}^{ – x}} – 2{{\text{e}}^{ – x}}( + c)\) A1
\(y{{\text{e}}^{ – x}} = – (1 + 2x){{\text{e}}^{ – x}} + c\)
\(y = – (1 + 2x) + c{{\text{e}}^x}\)
when \(x = 0,{\text{ }}y = 1 \Rightarrow c = 2\) M1
\(y = – (1 + 2x) + 2{{\text{e}}^{ – x}}\) A1
when \(x = 1,{\text{ }}y = – 3 + 2{\text{e}}\) A1
(i) METHOD 1
\(f(0) = 1,{\text{ }}f'(0) = 0\) A1
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 2 + \frac{{{\text{d}}y}}{{{\text{d}}x}} \Rightarrow {f^2}(0) = 2\) A1
\(\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = \frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} \Rightarrow {f^3}(0) = 2\) A1
hence \(y = 1 + {x^2} + \frac{{{x^3}}}{3} + \ldots \) A1
Note: Accuracy marks are independent of each other.
METHOD 2
using Maclaurin series for \({{\text{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \ldots \) M1
\(y = – 1 – 2x + 2\left( {1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \ldots } \right)\) M1A1
\(y = 1 + {x^2} + \frac{{{x^3}}}{3} + \ldots \) A1
(ii) when \(x = 1,{\text{ }}y = 1 + 1 + \frac{1}{3} = \frac{7}{3} = 2.33\) A1
Question
Using a Taylor series, find a quadratic approximation for \(f(x) = \sin x\) centred about \(x = \frac{{3\pi }}{4}\).
When using this approximation to find angles between \(130^\circ\) and \(140^\circ\), find the maximum value of the Lagrange form of the error term.
Hence find the largest number of decimal places to which \(\sin x\) can be estimated for angles between \(130^\circ\) and \(140^\circ\).
Explain briefly why the same maximum value of error term occurs for \(g(x) = \cos x\) centred around \(\frac{\pi }{4}\) when finding approximations for angles between \(40^\circ\) and \(50^\circ\).
Answer/Explanation
Markscheme
\(f(x) = \sin x,{\text{ }}f'(x) = \cos x,{\text{ }}{f^{(2)}}(x) = – \sin x\) M1
\(f\left( {\frac{{3\pi }}{4}} \right) = \frac{1}{{\sqrt 2 }},{\text{ }}f’\left( {\frac{{3\pi }}{4}} \right) = – \frac{1}{{\sqrt 2 }},{\text{ }}{f^{(2)}}\left( {\frac{{3\pi }}{4}} \right) = – \frac{1}{{\sqrt 2 }}\) A1
hence the quadratic Taylor Polynomial is
\(\frac{1}{{\sqrt 2 }} – \frac{1}{{\sqrt 2 }}\left( {x – \frac{{3\pi }}{4}} \right) – \frac{1}{{\sqrt 2 }}\frac{{{{\left( {x – \frac{{3\pi }}{4}} \right)}^2}}}{{2!}}\) M1A1
\(\left( {\frac{1}{{\sqrt 2 }}\left( {1 – \left( {x – \frac{{3\pi }}{4}} \right) – \frac{1}{2}{{\left( {x – \frac{{3\pi }}{4}} \right)}^2}} \right)} \right)\)
\(f(x) = \sin x,{\text{ }}{f^{(3)}}(x) = – \cos x\) (A1)
the Lagrange form of the error term is: \(\left| {{R_n}(x)} \right| \leqslant \frac{{{{\left| {x – a} \right|}^{n + 1}}}}{{(n + 1)!}}\max \left| {{f^{n + 1}}(k)} \right|\)
\(\left| {{R_2}(x)} \right| \leqslant \frac{{{{\left| {x – \frac{{3\pi }}{4}} \right|}^3}}}{{3!}}\max \left| {{f^3}(k)} \right|\) (M1)
\(\left| {{R_2}(x)} \right| \leqslant \frac{{{{\left| {x – \frac{{3\pi }}{4}} \right|}^3}}}{{3!}}\max \left| { – \cos k} \right|\) A1
in this case \(\left| { – \cos k} \right| \leqslant \left| { – \cos 140} \right|\) (A1)
\(\left| {{R_2}(x)} \right| \leqslant \frac{{{{\left| {x – \frac{{3\pi }}{4}} \right|}^3}}}{{3!}}\left| { – \cos 140} \right|\)
choosing \(140^\circ = \frac{{14\pi }}{{18}}\) M1
\( \Rightarrow \left| {{R_2}(x)} \right| \leqslant \frac{{{{\left| {\frac{{14\pi }}{{18}} – \frac{{3\pi }}{4}} \right|}^3}}}{{3!}}\left| { – \cos \frac{{14\pi }}{{18}}} \right|\) A1
therefore the maximum value of the error term is \(8.48 \times {10^{ – 5}}\) A1
\(\left| {{R_2}(x)} \right| \leqslant 8.48 \times {10^{ – 5}} = 0.0000848\) hence for angles between \(130^\circ\) and \(140^\circ\) the approximation will be accurate to 3 decimal places A1
\(\left| {{R_2}(x)} \right| \leqslant \frac{{{{\left| {x – \frac{\pi }{4}} \right|}^3}}}{{3!}}\max \left| {\sin k} \right|\) (M1)
since the max value of \(\left| {{f^3}(k)} \right|\) is \(\sin 50^\circ \) which is the same as \(\left| {\cos 140^\circ } \right|\) A1R1
then the error is the same AG
Question
Consider the functions \({f_n}(x) = {\sec ^n}(x),{\text{ }}\left| x \right| < \frac{\pi }{2}\) and \({g_n}(x) = {f_n}(x)\tan x\).
Show that
(i) \(\frac{{{\text{d}}{f_n}(x)}}{{{\text{d}}x}} = n{g_n}(x)\);
(ii) \(\frac{{{\text{d}}{g_n}(x)}}{{{\text{d}}x}} = (n + 1){f_{n + 2}}(x) – n{f_n}(x)\).
(i) Use these results to show that the Maclaurin series for the function \({f_5}(x)\) up to and including the term in \({x^4}\) is \(1 + \frac{5}{2}{x^2} + \frac{{85}}{{24}}{x^4}\).
(ii) By considering the general form of its higher derivatives explain briefly why all coefficients in the Maclaurin series for the function \({f_5}(x)\) are either positive or zero.
(iii) Hence show that \({\sec ^5}(0.1) > 1.02535\).
Answer/Explanation
Markscheme
(i) \(\frac{{{\text{d}}{f_n}(x)}}{{{\text{d}}x}} = n{\sec ^{n – 1}}(x)\sec (x)\tan (x)\) M1A1
\( = n{g_n}(x)\) AG
(ii) \(\frac{{{\text{d}}{g_n}(x)}}{{{\text{d}}x}} = \frac{{{\text{d}}{f_n}(x)}}{{{\text{d}}x}}\tan (x) + {f_n}(x){\sec ^2}(x)\) M1
\(n{g_n}(x)\tan (x) + {f_{n + 2}}(x)\) or equivalent A1
\(n{f_n}(x){\tan ^2}(x) + {f_{n + 2}}(x)\) or equivalent A1
\( = (n + 1){f_{n + 2}}(x) – n{f_n}(x)\) AG
Note: Award M1A1 for the correct differentiation of a product and A1 for an intermediate result clearly leading to the AG.
[5 marks]
(i) \({f_5}(0) = 1\) A1
\(\frac{{{\text{d}}{f_5}}}{{{\text{d}}x}}(0) = 5{g_5}(0) = 0\) A1
\(\frac{{{{\text{d}}^2}{f_5}}}{{{\text{d}}{x^2}}}(0) = 5\left( {6{f_7}(0) – 5{f_5}(0)} \right) = 5\) A1
\(\frac{{{{\text{d}}^3}{f_5}}}{{{\text{d}}{x^3}}} = 30\frac{{{\text{d}}{f_7}}}{{{\text{d}}x}} – 25\frac{{{\text{d}}{f_5}}}{{{\text{d}}x}}\) M1
hence \(\frac{{{{\text{d}}^3}{f_5}}}{{{\text{d}}{x^3}}}(0) = 30 \times 0 – 25 \times 0 = 0\) A1
\(\frac{{{{\text{d}}^4}{f_5}}}{{{\text{d}}{x^4}}} = 30\frac{{{{\text{d}}^2}{f_7}}}{{{\text{d}}{x^2}}} – 25\frac{{{{\text{d}}^2}{f_5}}}{{{\text{d}}{x^2}}} = 210(8{f_9} – 7{f_7}) – 25\frac{{{{\text{d}}^2}{f_5}}}{{{\text{d}}{x^2}}}\) M1A1
hence \(\frac{{{{\text{d}}^4}{f_5}}}{{{\text{d}}{x^4}}}(0) = 210 – 125 = 85\) A1
hence \({f_5}(x) \approx 1 + \frac{5}{2}{x^2} + \frac{{85}}{{24}}{x^4}\) AG
(ii) each derivative of \({f_m}(x)\) is a sum of terms of the form \({\text{A}}\,{\sec ^p}(x)\,\,{\tan ^q}(x)\) A1
where \(A \geqslant 0\) A1
when \(x = 0\) is substituted the result is the sum of positive and/or zero terms R1
(iii) since the full series represents \({f_5}(x)\), the truncated series is a lower bound (or some equivalent statement) R1
hence \({\sec ^5}(0.1) > 1 + \frac{5}{2}{0.1^2} + \frac{{85}}{{24}}{0.1^4}\) M1
\( = 1.025354\) A1
\( > 1.02535\) AG
[14 marks]
Question
It is given that \(\left( {5x + y} \right)\frac{{{\text{d}}y}}{{{\text{d}}x}} = \left( {x + 5y} \right)\) and that when \(x = 0,\,\,y = 2\).
Use Euler’s method with step length 0.1 to find an approximate value of \(y\) when \(x = 0.4\).
Show that \(\left( {5x + y} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 1 – {\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)^2}\).
Show that \(\left( {5x + y} \right)\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = – 5\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} – 3\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\left( {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right)\).
Find the Maclaurin expansion for \(y\) up to and including the term in \({{x^3}}\).
Answer/Explanation
Markscheme
Euler’s method with step length \(h = 0.1\) to find \(y\) when \(x = 0.4\)
Note: Accept 3 significant figures in the table.
first line of table (M1)(A1)
line 2 (A1)
line 3 (A1)
hence \(y\) = 3.65 A1
Note: Accept any answer that rounds to 3.65.
[5 marks]
\(\left( {5x + y} \right)\frac{{{\text{d}}y}}{{{\text{d}}x}} = x + 5y\)
\(\left( {5 + \frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\frac{{{\text{d}}y}}{{{\text{d}}x}} + \left( {5x + y} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 1 + 5\frac{{{\text{d}}y}}{{{\text{d}}x}}\) M1A1A1
Note: Award M1 for a valid attempt to differentiate, A1 for LHS, A1 for RHS.
\(\left( {5x + y} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 1 + 5\frac{{{\text{d}}y}}{{{\text{d}}x}} – 5\frac{{{\text{d}}y}}{{{\text{d}}x}} – {\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)^2}\)
\(\left( {5x + y} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 1 – {\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)^2}\) AG
[3 marks]
\(\left( {5x + y} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}1 – {\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)^2}\)
\(\left( {5 + \frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + \left( {5x + y} \right)\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = – 2\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\left( {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right)\) M1A1A1A1
\(\left( {5x + y} \right)\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = – 2\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\left( {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right) – 5\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} – \left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\left( {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right)\)
\(\left( {5x + y} \right)\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = – 5\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} – 3\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)\left( {\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}}} \right)\) AG
[4 marks]
when \(x = 0\,\,\,y = 2\)
when \(x = 0\,\,\frac{{{\text{d}}y}}{{{\text{d}}x}} = 5\) A1
when \(x = 0\,\,\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = – 12\) A1
when \(x = 0\,\,\frac{{{{\text{d}}^3}y}}{{{\text{d}}{x^3}}} = 120\) A1
Note: Allow follow through from incorrect values of derivatives.
\(y = 2 + 5x – 6{x^2} + 20{x^3}\) M1A1
[5 marks]
Question
Consider the differential equation\[\frac{{{\text{d}}y}}{{{\text{d}}x}} + y\sec x = x(\sec x – \tan x),{\text{ where }}y = 3{\text{ when }}x = 0.\]
Use Euler’s method with a step length of \(0.1\) to find an approximate value for \(y\) when \(x = 0.3\) .
(i) By differentiating the above differential equation, obtain an expression involving \(\frac{{{{\rm{d}}^{\rm{2}}}y}}{{{\rm{d}}{x^2}}}\) .
(ii) Hence determine the Maclaurin series for \(y\) up to the term in \({{x^2}}\) .
(iii) Use the result in part (ii) to obtain an approximate value for \(y\) when \(x = 0.3\) .
(i) Show that \(\sec x + \tan x\) is an integrating factor for solving this differential equation.
(ii) Solve the differential equation, giving your answer in the form \(y = f(x)\) .
(iii) Hence determine which of the two approximate values for y when \(x = 0.3\) , obtained in parts (a) and (b), is closer to the true value.
Answer/Explanation
Markscheme
Note: The A1 marks above are for correct entries in the \(y\) column.
\(y(0.3) \approx 2.21\) A1
[5 marks]
(i) use of product rule on either side M1
\(\frac{{{{\rm{d}}^{\rm{2}}}y}}{{{\rm{d}}{x^2}}} + \sec x\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + y\sec x\tan x = \sec x – \tan x + x(\sec x\tan x – {\sec ^2}x)\) A1A1
(ii) \(y(0) = 3\)
\(y'(0) = – 3\), \(y”(0) = 4\) A1A1
the quadratic approximation is
\(y = \left( {y(0) + xy'(0) + |\frac{{{x^2}y”(0)}}{2} = } \right)3 – 3x + 2{x^2}\) (M1)A1
(iii) using this approximation, \(y(0.3) \approx 2.28\) A1
[8 marks]
(i) EITHER
\(\frac{{\rm{d}}}{{{\rm{d}}x}}(\sec x + \tan x) = \sec x\tan x + {\sec ^2}x\) A1
\(\sec x(\sec x + \tan x) = {\sec ^2}x + \sec x\tan x\) A1
as these two expressions are the same, this is an integrating factor R1AG
OR
\((\sec x + \tan x)\left( {\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + y\sec x} \right) = (\sec x + \tan x)x(\sec x – \tan x)\) M1
Note: RHS does not need to be shown.
\({\rm{LHS}} = \frac{{{\rm{d}}y}}{{{\rm{d}}x}}(\sec x + \tan x) + y(\sec x|\tan x + {\sec ^2}x)\) A1
\( = \frac{{\rm{d}}}{{{\rm{d}}x}}y(\sec x + \tan x)\) A1
making LHS an exact derivative
OR
integrating factor \( = {{\rm{e}}^{\int {\sec x{\rm{d}}x} }}\) M1
since \(\frac{{\rm{d}}}{{{\rm{d}}x}}\ln (\sec x + \tan x) = \frac{{\sec x\tan x + {{\sec }^2}x}}{{\sec x + \tan x}} = \sec x\) M1A1
integrating factor \( = {{\rm{e}}^{\ln (\sec x + \tan x)}} = \sec x + \tan x\) AG
(ii) \(\frac{{\rm{d}}}{{{\rm{d}}x}}(y\left[ {\sec x + \tan x} \right]) = x({\sec ^2}x – {\tan ^2}x) = x\) M1A1
\(y(\sec x + \tan x) = \frac{{{x^2}}}{2} + c\) A1
\(x = 0,y = 3 \Rightarrow c = 3\) M1A1
\(y = \frac{{{x^2} + 6}}{{2(\sec x + \tan x)}}\) A1
(iii) when \(x = 0.3,y = 2.245 \ldots \) A1
the closer approximation is obtained by using the series in part (b) R1
[11 marks]