| Marks available | 5 |
| Reference code | 10M.2.hl.TZ0.6 |
Question
The diagram shows a sketch of the graph of \(y = {x^{ – 4}}\) for \(x > 0\) .
By considering this sketch, show that, for \(n \in {\mathbb{Z}^ + }\) ,\[\sum\limits_{r = n + 1}^\infty {\frac{1}{{{r^4}}}} < \int_n^\infty {\frac{{{\rm{d}}x}}{{{x^4}}}} < \sum\limits_{r = n}^\infty {\frac{1}{{{r^4}}}} .\]
Let \(S = \sum\limits_{r = 1}^\infty {\frac{1}{{{r^4}}}} \) .
Use the result in (a) to show that, for \(n \ge 2\) , the value of \(S\) lies between
\(\sum\limits_{r = 1}^{n – 1} {\frac{1}{{{r^4}}}} + \frac{1}{{3{n^3}}}\) and \(\sum\limits_{r = 1}^n {\frac{1}{{{r^4}}}} + \frac{1}{{3{n^3}}}\) .
(i) Show that, by taking \(n = 8\) , the value of \(S\) can be deduced correct to three decimal places and state this value.
(ii) The exact value of \(S\) is known to be \(\frac{{{\pi ^4}}}{N}\)where \(N \in {\mathbb{Z}^ + }\) . Determine the value of \(N\) .
Now let \(T = \sum\limits_{r = 1}^\infty {\frac{{{{( – 1)}^{r + 1}}}}{{{r^4}}}} \) .
Find the value of \(T\) correct to three decimal places.
Answer/Explanation
Markscheme
(M1)
total area of “upper” rectangles
\( = \frac{1}{{{n^4}}} \times 1 + \frac{1}{{{{(n + 1)}^4}}} \times 1 + \frac{1}{{{{(n + 2)}^4}}} \times 1 + \ldots = \sum\limits_{r = n}^\infty {\frac{1}{{{r^4}}}} \) M1A1
total area of “lower” rectangles
\( = \frac{1}{{{{(n + 1)}^4}}} \times 1 + \frac{1}{{{{(n + 2)}^4}}} \times 1 + \frac{1}{{{{(n + 3)}^4}}} \times 1 + \ldots = \sum\limits_{r = n + 1}^\infty {\frac{1}{{{r^4}}}} \) A1
the total area under the curve from \(x = n\) to infinity lies between these two sums hence \(\sum\limits_{r = n + 1}^\infty {\frac{1}{{{r^4}}}} < \int_n^\infty {\frac{{{\rm{d}}x}}{{{x^4}}}} < \sum\limits_{r = n}^\infty {\frac{1}{{{r^4}}}} \) R1AG
[5 marks]
first evaluate the integral
\(\int_n^\infty {\frac{{{\rm{d}}x}}{{{x^4}}}} = – \left[ {\frac{1}{{3{x^3}}}} \right]_n^\infty = \frac{1}{{3{n^3}}}\) M1A1
it follows that
\(\sum\limits_{r = n + 1}^\infty {\frac{1}{{{r^4}}}} < \frac{1}{{3{n^3}}}\) A1
adding \(\sum\limits_{r = 1}^n {\frac{1}{{{r^4}}}} \) to both sides, M1
\(S < \sum\limits_{r = 1}^n {\frac{1}{{{r^4}}}} + \frac{1}{{3{n^3}}}\) A1
similarly,
\(\sum\limits_{r = n}^\infty {\frac{1}{{{r^4}}}} > \frac{1}{{3{n^3}}}\) A1
adding \(\sum\limits_{r = 1}^{n – 1} {\frac{1}{{{r^4}}}} \) to both sides, M1
\(S > \sum\limits_{r = 1}^{n – 1} {\frac{1}{{{r^4}}}} + \frac{1}{{3{n^3}}}\) A1
hence the value of \(S\) lies between
\(\sum\limits_{r = 1}^{n – 1} {\frac{1}{{{r^4}}}} + \frac{1}{{3{n^3}}}\) and \(\sum\limits_{r = 1}^n {\frac{1}{{{r^4}}}} + \frac{1}{{3{n^3}}}\) AG
[8 marks]
(i) putting \(n = 8\) , we find that
\(S < 1.08243 \ldots \) and \(S > 1.08219 \ldots \) A1A1
it follows that \(S = 1.082\) to 3 decimal places A1
(ii) substituting this value of \(S\),
\(N \approx \frac{{{\pi ^4}}}{{1.082}} \approx 90.0268\) M1A1
\(N = 90\) A1
[6 marks]
EITHER
successive partial sums are
1 M1
0.9375
0.9498…
0.9459…
0.9475…
0.9467…
0.9471… A1
it follows that correct to 3 decimal places \(T = 0.947\) A1
OR
\(T = S – \frac{2}{{16}}S\) M1A1
using part (c)(i) or \(0.94703…\) using the sum given in part (c)(ii) \(0.9471…\)
it follows that \(T = 0.947\) correct to 3 decimal places A1
[3 marks]
| Marks available | 5 |
| Reference code | 12M.2.hl.TZ0.3 |
Question
(i) Show that \(\frac{{\rm{d}}}{{{\rm{d}}\theta }}(\sec \theta \tan \theta + \ln (\sec \theta + \tan \theta )) = 2{\sec ^3}\theta \) .
(ii) Hence write down \(\int {{{\sec }^3}\theta {\rm{d}}\theta } \) .
Consider the differential equation \((1 + {x^2})\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + xy = 1 + {x^2}\) given that \(y = 1\) when \(x = 0\) .
(i) Use Euler’s method with a step length of \(0.1\) to find an approximate value for y when \(x = 0.3\) .
(ii) Find an integrating factor for determining the exact solution of the differential equation.
(iii) Find the solution of the equation in the form \(y = f(x)\) .
(iv) To how many significant figures does the approximation found in part (i) agree with the exact value of \(y\) when \(x = 0.3\) ?
Answer/Explanation
Markscheme
(i) \(\frac{{\rm{d}}}{{{\rm{d}}\theta }}(\sec \theta \tan \theta + \ln (\sec \theta + \tan \theta ))\)
\( = {\sec ^3}\theta + \sec \theta {\tan ^2}\theta + \frac{{\sec \theta \tan \theta + {{\sec }^2}\theta }}{{\sec \theta + \tan \theta }}\) M1A1A1
Note: Award M1 for a valid attempt to differentiate either term.
\( = {\sec ^3}\theta + \sec \theta ({\sec ^2}\theta – 1) + \sec \theta \) A1
\( = 2{\sec ^3}\theta \) AG
(ii) \(\int {{{\sec }^3}\theta {\rm{d}}\theta } = \frac{1}{2}(\sec \theta \tan \theta + \ln (\sec \theta + \tan \theta ))( + C)\) A1
[5 marks]
(i) \(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} = 1 – \frac{{xy}}{{1 + {x^2}}}\) A1
Note: Accept tabular values correct to 3 significant figures.
\(y \approx 1.27\) when \(x = 0.3\) A1
(ii) consider the equation in the form
\(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + \frac{{xy}}{{1 + {x^2}}} = 1\) (M1)
the integrating factor I is given by
\(I = \exp \int {\left( {\frac{x}{{1 + {x^2}}}} \right)} {\rm{d}}x\) A1
\( = \exp \left( {\frac{1}{2}\ln (1 + {x^2})} \right)\) A1
\( = \sqrt {1 + {x^2}} \) A1
Note: Accept also the fact that the integrating factor for the original equation is \(\frac{1}{{\sqrt {1 + {x^2}} }}\) .
(iii) consider the equation in the form
\(\sqrt {1 + {x^2}} \frac{{{\rm{d}}y}}{{{\rm{d}}x}} + \frac{{xy}}{{\sqrt {1 + {x^2}} }} = \sqrt {1 + {x^2}} \) (M1)
integrating,
\(y\sqrt {1 + {x^2}} = \int {\sqrt {1 + {x^2}} {\rm{d}}x} \) A1
to integrate the right hand side, put \(x = \tan \theta \) , \({\rm{d}}x = {\sec ^2}\theta {\rm{d}}\theta \) M1A1
\(\int {\sqrt {1 + {x^2}} } {\rm{d}}x = \int {\sqrt {1 + {{\tan }^2}\theta } } .{\sec ^2}\theta {\rm{d}}\theta \) A1
\( = \int {{{\sec }^3}\theta {\rm{d}}\theta } \) A1
\( = \frac{1}{2}(\sec \theta \tan \theta + \ln (\sec \theta + \tan \theta ))\)
\( = \frac{1}{2}\left( {x\sqrt {1 + {x^2}} + \ln (x + \sqrt {1 + {x^2}} )} \right)\) A1
the solution to the differential equation is therefore
\(y\sqrt {1 + {x^2}} = \frac{1}{2}\left( {x\sqrt {1 + {x^2}} + \ln \left( {x + \sqrt {1 + {x^2}} } \right)} \right) + C\) A1
Note: Do not penalize the omission of C at this stage.
\(y = 1\) when \(x = 0\) gives \(C = 1\) M1A1
the solution is \(y = \frac{1}{{2\sqrt {1 + {x^2}} }}\left( {x\sqrt {1 + {x^2}} + \ln \left( {x + \sqrt {1 + {x^2}} } \right)} \right) + \frac{1}{{\sqrt {1 + {x^2}} }}\) A1
(iv) when \(x = 0.3\) , \(y = 1.249 \ldots \) A1
the approximation is only correct to 1 significant figure A1
[24 marks]
| Marks available | 6 |
| Reference code | 13M.2.hl.TZ0.2 |
Question
(i) Show that the improper integral \(\int_0^\infty {\frac{1}{{{x^2} + 1}}} {\rm{d}}x\) is convergent.
(ii) Use the integral test to deduce that the series \(\sum\limits_{n = 0}^\infty {\frac{1}{{{n^2} + 1}}} \) is convergent, giving reasons why this test can be applied.
(i) Show that the series \(\sum\limits_{n = 0}^\infty {\frac{{{{( – 1)}^n}}}{{{n^2} + 1}}} \) is convergent.
(ii) If the sum of the above series is \(S\), show that \(\frac{3}{5} < S < \frac{2}{3}\) .
For the series \(\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{{n^2} + 1}}} \)
(i) determine the radius of convergence;
(ii) determine the interval of convergence using your answers to (b) and (c).
Answer/Explanation
Markscheme
(i) consider \(\int_0^R {\frac{1}{{{x^2} + 1}}} {\rm{d}}x\) M1
\( = \left[ {\arctan (x)} \right]_0^R = \arctan (R)\) A1
\(\mathop {\lim }\limits_{R \to \infty } \arctan (R) = \frac{\pi }{2}\) (a finite number) R1
hence the improper integral is convergent AG
(ii) the terms of the series are positive A1
the terms are decreasing A1
the terms tend to zero A1
by the integral test, the series converges AG
[6 marks]
(i) the absolute values of the terms are monotonically decreasing A1
to zero A1
the series converges by the alternating series test R1AG
Note: Accept absolute convergence, with reference to part (b)(ii) \( \Rightarrow \) convergence.
(ii) statement that successive partial sums bound the total sum R1
\(S > \frac{1}{1} – \frac{1}{2} + \frac{1}{5} – \frac{1}{{10}} = \frac{3}{5}\) A1
\(S < \frac{1}{1} – \frac{1}{2} + \frac{1}{5} – \frac{1}{{10}} + \frac{1}{{17}} = 0.6588\) A1
\(S < 0.6588 < \frac{2}{3}\) AG
[6 marks]
(i) consider \(\left| {\frac{{\frac{{{x^{n + 1}}}}{{{{(n + 1)}^2} + 1}}}}{{\frac{{{x^n}}}{{{n^2} + 1}}}}} \right|\) M1
\( = \left| {\frac{{x({n^2} + 1)}}{{{{(n + 1)}^2} + 1}}} \right|\) A1
\( \to \left| x \right|\) as \(n \to \infty \) A1
therefore radius of convergence \( = 1\) A1
(ii) interval of convergence \( = \left[ { – 1,1} \right]\) A1A1
Note: A1 for [\( – 1\), and A1 for \(1\)].
[6 marks]
| Marks available | 4 |
| Reference code | 17M.2.hl.TZ0.5 |
Question
Consider the differential equation
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} + y\tan x = 2{\sec ^2}x,{\text{ }}0 \leqslant x < \frac{\pi }{2}\), given that \(y = 1\) when \(x = 0\).
By considering integration as the reverse of differentiation, show that for
\(0 \leqslant x < \frac{\pi }{2}\)
\[\int {\sec x{\text{d}}x = \ln (\sec x + \tan x) + C.} \]
Hence, using integration by parts, show that
\[\int {{{\sec }^3}x{\text{d}}x = \frac{1}{2}\left( {\sec x\tan x + \ln (\sec x + \tan x)} \right) + C.} \]
Find an integrating factor and hence solve the differential equation, giving your answer in the form \(y = f(x)\).
Starting with the differential equation, show that
\[\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y = 2{\sec ^2}x\tan x.\]
Hence, by using your calculator to draw two appropriate graphs or otherwise, find the \(x\)-coordinate of the point of inflection on the graph of \(y = f(x)\).
Answer/Explanation
Markscheme
\(\frac{{\text{d}}}{{{\text{d}}x}}\left( {\ln (\sec x + \tan x)} \right) = \frac{{\sec x\tan x + {{\sec }^2}x}}{{\sec x + \tan x}}\) M1
\( = \sec x\) A1
therefore \(\int {\sec x{\text{d}}x = \ln (\sec x + \tan x) + C} \) AG
[4 marks]
\(\int {{{\sec }^3}x{\text{d}}x = \int {\sec x \times {{\sec }^2}x{\text{d}}x} } \) M1
\( = \sec x\tan x – \int {\sec x{{\tan }^2}x{\text{d}}x} \) A1A1
\( = \sec x\tan x – \int {\sec x({{\sec }^2}x – 1){\text{d}}x} \) A1
\( = \sec x\tan x – \int {{{\sec }^3}x{\text{d}}x + \int {\sec x{\text{d}}x} } \)
\( = \sec x\tan x – \int {{{\sec }^3}x{\text{d}}x + \ln (\sec x + \tan x)} \) A1
\(2\int {{{\sec }^3}x{\text{d}}x = \left( {\sec x\tan x + \ln (\sec x + \tan x)} \right)} \) A1
therefore
\(\int {{{\sec }^3}x{\text{d}}x = \frac{1}{2}\left( {\sec x\tan x + \ln (\sec x + \tan x)} \right) + C} \) AG
[4 marks]
\({\text{int factor}} = {{\text{e}}^{\int {\tan x{\text{d}}x} }}\) (M1)
\( = {{\text{e}}^{\ln \sec x}}\) (A1)
\( = \sec x\) A1
the differential equation can be written as
\(\frac{{\text{d}}}{{{\text{d}}x}}(y\sec x) = 2{\sec ^3}x\) M1A1
integrating,
\(y\sec x = \sec x\tan x + \ln (\sec x + \tan x) + C\) A1
putting \(x = 0,{\text{ }}y = 1,\) M1
\(C = 1\) A1
the solution is \(y = \cos x\left( {\sec x\tan x + \ln (\sec x + \tan x) + 1} \right)\) A1
[??? marks]
differentiating the differential equation,
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + \frac{{{\text{d}}y}}{{{\text{d}}x}}\tan x + y{\sec ^2}x = 4{\sec ^2}x\tan x\) A1A1
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + (2{\sec ^2}x – y\tan x)\tan x + y{\sec ^2}x = 4{\sec ^2}x\tan x\) A1
\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y = 2{\sec ^2}x\tan x\) AG
[??? marks]
at a point of inflection, \(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 0\) so \(y = 2{\sec ^2}x\tan x\) (M1)
therefore the point of inflection can be found as the point of intersection of the graphs of \(y = \cos x\left( {\sec x\tan x + \ln (\sec x + \tan x) + 1} \right)\)
and \(y = 2{\sec ^2}x\tan x\) (M1)
drawing these graphs on the calculator, \(x = 0.605\) A2
[??? marks]