OCR FP2 2008 June — Question 5 8 marks

Exam BoardOCR
ModuleFP2 (Further Pure Mathematics 2)
Year2008
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicReduction Formulae
TypeMethod of differences
DifficultyChallenging +1.2 This is a standard reduction formula question requiring integration of tan^n(x), establishing a recurrence relation using the identity tan²x = sec²x - 1, then applying it iteratively. While it involves multiple steps and Further Maths content, the technique is well-practiced in FP2 and follows a predictable pattern with clear guidance ('considering I_n + I_{n-2}').
Spec1.08i Integration by parts

5 It is given that, for \(n \geqslant 0\), $$I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \tan ^ { n } x \mathrm {~d} x$$
  1. By considering \(I _ { n } + I _ { n - 2 }\), or otherwise, show that, for \(n \geqslant 2\), $$( n - 1 ) \left( I _ { n } + I _ { n - 2 } \right) = 1 .$$
  2. Find \(I _ { 4 }\) in terms of \(\pi\).

AnswerMarks
(a) Obtain expression of form \(\frac{a \tan \alpha}{b + c \tan^2 \alpha}\)M1
State correct \(\frac{2 \tan \alpha}{1 - \tan^2 \alpha}\)A1
Attempt to produce polynomial equation in \(\tan \alpha\)M1
Obtain at least one correct value of \(\tan \alpha\)A1
Obtain \(41.8\)A1
Obtain \(138.2\) and no other values between 0 and 180A1
(b)(i) State \(\frac{2}{6}\)B1
(ii) Attempt use of identity linking \(\cot^2 \beta\) and \(\cosec^2 \beta\)M1
Obtain \(\frac{13}{36}\)A1
| | |
|---|---|
| (a) Obtain expression of form $\frac{a \tan \alpha}{b + c \tan^2 \alpha}$ | M1 |
| State correct $\frac{2 \tan \alpha}{1 - \tan^2 \alpha}$ | A1 |
| Attempt to produce polynomial equation in $\tan \alpha$ | M1 |
| Obtain at least one correct value of $\tan \alpha$ | A1 |
| Obtain $41.8$ | A1 |
| Obtain $138.2$ and no other values between 0 and 180 | A1 |
| (b)(i) State $\frac{2}{6}$ | B1 |
| (ii) Attempt use of identity linking $\cot^2 \beta$ and $\cosec^2 \beta$ | M1 |
| Obtain $\frac{13}{36}$ | A1 |
5 It is given that, for $n \geqslant 0$,

$$I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \tan ^ { n } x \mathrm {~d} x$$

(i) By considering $I _ { n } + I _ { n - 2 }$, or otherwise, show that, for $n \geqslant 2$,

$$( n - 1 ) \left( I _ { n } + I _ { n - 2 } \right) = 1 .$$

(ii) Find $I _ { 4 }$ in terms of $\pi$.

\hfill \mbox{\textit{OCR FP2 2008 Q5 [8]}}