| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2015 |
| Session | June |
| Marks | 22 |
| Topic | Volumes of Revolution |
| Type | Surface area of revolution: Cartesian curve |
| Difficulty | Challenging +1.8 This is a challenging Further Maths question requiring integration by parts to derive a reduction formula, a non-trivial substitution for I_0, and application to surface area of revolution. The multi-step proof in (i)(b) and the substitution in (i)(c) require significant technical skill and insight beyond routine A-level integration, though the structure guides students through the problem systematically. |
| Spec | 4.08d Volumes of revolution: about x and y axes4.08f Integrate using partial fractions8.06a Reduction formulae: establish, use, and evaluate recursively8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks |
|---|---|
| 12 (i) | (a) I1 = ∫ 2 x 1+2x2 dx = 1 6 ( 1+2x2 ) 2 3 2 = 1 3 3 |
| Answer | Marks |
|---|---|
| 2 | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Page 10 | Mark Scheme | Syllabus |
| Cambridge Pre-U – May/June 2015 | 9795 | 01 |
| (ii) | ALTERNATIVE (partial) |
| Answer | Marks |
|---|---|
| 2 2I =secθ tanθ +ln | secθ +tanθ |
| Answer | Marks |
|---|---|
| 8 2 8 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Page 11 | Mark Scheme | Syllabus |
| Cambridge Pre-U – May/June 2015 | 9795 | 01 |
| Answer | Marks |
|---|---|
| (iii) | a 1 |
| Answer | Marks |
|---|---|
| leading to … Answer Given | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Page 12 | Mark Scheme | Syllabus |
| Cambridge Pre-U – May/June 2015 | 9795 | 01 |
| Answer | Marks |
|---|---|
| 4 | M1 A1 |
Question 12:
--- 12 (i) ---
12 (i) | (a) I1 = ∫ 2 x 1+2x2 dx = 1 6 ( 1+2x2 ) 2 3 2 = 1 3 3
0 0
2
(b) In = ∫xn − 1. x 1+2x2 dx Correct splitting and use of parts
0
3
= xn − 1. 1 6 ( 1+2x2 )3 2 2 – ∫ 2 (n−1)xn − 2. 1 6 ( 1+2x2 )2 dx
0 0
= 2n − 1 . 27 – 0 – 1(n−1)∫ 2 xn− 2. ( 1+2x2 ) 1+2x2 dx
6 6
0
= 2n − 1 . 27 – 1(n−1){2I +I }
6 6 n n − 2
⇒ 6 In = 27×2n − 1−2(n−1)I −(n−1)I
n n − 2
⇒ (2n + 4) In = 27×2n − 1−(n−1)I Answer Given
n − 2
2
(c) I0 = ∫ 1+2x2 dx
0
Let x 2 =sinhθ ( 2 dx=coshθ dθ , 1+2x2 =coshθ )
1
full substitution I0 = ∫cosh2θ dθ (Ignore limits for now)
2
1
∫( )
trig. identity = 1+cosh2θ dθ
2 2
1 1
= θ + sinh2θ ft ∫n. of a + b cosh2θ only
2 2 2
(0, 2) → (0, sinh – 1 2 2) Limits properly dealt with
1
= sinh−12 2 +6 2
2 2
sinh – 1 2 2 = ln 3+2 2 to at least here = ln 1+ 2 2 = 2 ln 1+ 2
( )
I0 = 3+ 1 ln1+ 2 legitimately Answer Given
2 | M1 A1
A1
[3]
M1
A1 A1
M1
A1
A1
[6]
M1
M1 A1
M1
A1
M1
M1
A1
[8]
Page 10 | Mark Scheme | Syllabus | Paper
Cambridge Pre-U – May/June 2015 | 9795 | 01
(ii) | ALTERNATIVE (partial)
Let x 2 =tanθ 2dx=sec2θ dθ
1
full substitution I0 = ∫sec3θ dθ (Ignore limits for now)
2
use of ∫n. by parts on ∫secθsec2θ dθ and use of tan2θ =sec2θ −1
2 2I =secθ tanθ +ln|secθ +tanθ |
0
Further progress (limits, etc.) essentially impossible
ALTERNATIVE
I0 = ∫ 2 1+2x2 ×1dx = x 1+2x2 −∫1 ( 1+2x2 )−1 2 . 4x . xdx ∫n. by parts
2
0
( )
2x2 +1 −1
= x 1+2x2 −∫ dx
1+2x2
1
= x 1+2x2 −I +∫ dx
0
1+2x2
1
⇒ 2I = x 1+2x2 +∫ dx
0
1+2x2
1 1 ( )
⇒ I = x 1+2x2 + sinh−1 x 2 (from MF20)
0 2 2
1
Use of limits (0, 2) ⇒ I =3+ sinh−1(2 2 )
0
2 2
sinh – 1 2 2= ln 3+2 2 = ln 1+ 2 2 = 2 ln 1+ 2
( )
I0 = 3+ 1 ln 1+ 2 legitimately Answer Given
2
dy 2 x2
y= 1 x2 ⇒ = x 2 and S = 2π∫ . 1+2x2 dx
2 dx
2
0
( )
= π 2 I
2
use of R.F. for n = 2: 8 I2 = 54−I
0
( )
use of their (i) result = 51− 1 ln 1+ 2
2
⇒ S = π 2 51− 1 ln 1+ 2 or π 51 2 −ln 1+ 2
8 2 8 | M1
M1 A1
M1
A1
M1 A1
M1
M1
M1 A1
M1
A1
[8]
M1
A1
M1
M1
A1
[5]
Page 11 | Mark Scheme | Syllabus | Paper
Cambridge Pre-U – May/June 2015 | 9795 | 01
13 (i)
(ii)
(iii) | a 1
tan A = , tan B =
1 a
1 π
⇒ tan−1a+tan−1 = A+B=
a 2
( ) ( )
( ) tan tan−1a ±tan tan−1b a±b
tan tan−1a±tan−1b = ( ) ( )=
1mtan tan−1a.tan tan−1b 1mab
1 1 2
tan−1 −tan−1 =tan−1 noted at any stage
n−1 n+1 n2
∑ ∞ tan−1 2 =tan−1 ( 2 ) +tan−1 1 + ∑ ∞ tan−1 2 Splitting off 1st 2 terms
n = 1 n2 2 n = 3 n2
π ∞ 2
= + ∑ tan−1 using (i)’s result
2 n = 3 n2
use of difference method (finite or infinite series)
∞ 2 π
⇒ ∑ tan−1 = +
n = 1 n2 2
( )
tan−11 +tan−11+tan−11 +tan−11+tan−11+tan−11...
2 3 4 5 6 7
( )
− tan−11+tan−11+tan−11+tan−11+...
4 5 6 7
Cancelling of terms made clear
π ( )
= + tan−11+tan−11
2 2 3
1 +1 π
and tan−11 +tan−11 =tan−1 2 3 =tan−11= using (ii)’s result
2 3 1−1.1 4
2 3
leading to … Answer Given | B1
[1]
M1 A1
[2]
M1 A1
B1
M1
M1
A1
B1
[7]
Page 12 | Mark Scheme | Syllabus | Paper
Cambridge Pre-U – May/June 2015 | 9795 | 01
ALTERNATIVE
1 1 2
tan−1 −tan−1 =tan−1 noted at any stage
n−1 n+1 n2
Use of difference method (finite or infinite series)
∑ N tan−1 2 = ( tan−11+tan−11+tan−11+tan−11+... )
n = 1 n2 0 2 3
( )
− tan−11+tan−11+...
2 3
Cancelling of remaining terms made clear
π π
tan−1∞+tan−11 = +
2 4
3π
= Given Answer
4 | M1 A1
M1
A1
M1
A1
A1
[7]Let $I_n = \int_0^2 x^n \sqrt{1 + 2x^2} \, \text{d}x$ for $n = 0, 1, 2, 3, \ldots$.
\begin{enumerate}[label=(\roman*)]
\item \begin{enumerate}[label=(\alph*)]
\item Evaluate $I_1$. [3]
\item Prove that, for $n \geqslant 2$,
$$(2n + 4)I_n = 27 \times 2^{n-1} - (n - 1)I_{n-2}.$$ [6]
\item Using a suitable substitution, or otherwise, show that
$$I_0 = 3 + \frac{1}{\sqrt{2}} \ln(1 + \sqrt{2}).$$ [8]
\end{enumerate}
\item The curve $y = \frac{1}{\sqrt{2}} x^2$, between $x = 0$ and $x = 2$, is rotated through $2\pi$ radians about the $x$-axis to form a surface with area $S$. Find the exact value of $S$. [5]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2015 Q12 [22]}}