OCR MEI Further Pure Core 2021 November — Question 16 14 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2021
SessionNovember
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHyperbolic functions
TypeIntegrate using hyperbolic substitution
DifficultyChallenging +1.2 This is a standard Further Maths hyperbolic integration question with two parts: (a) requires routine manipulation of exponential definitions of hyperbolic functions, and (b) uses the standard substitution x = 2sinh(u) which is telegraphed by the √(4+x²) form. While requiring multiple steps and knowledge of hyperbolic identities, this follows a well-established template that Further Maths students practice extensively. It's harder than average A-level due to being Further Maths content, but routine within that context.
Spec4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.08h Integration: inverse trig/hyperbolic substitutions
16
  1. Show using exponentials that \(\cosh 2 u = 1 + 2 \sinh ^ { 2 } u\).
  2. Show that \(\int _ { 0 } ^ { 2 } \frac { x ^ { 2 } } { \sqrt { 4 + x ^ { 2 } } } \mathrm {~d} x = 2 \sqrt { 2 } - 2 \ln ( 1 + \sqrt { 2 } )\).
Question 16:
AnswerMarks Guidance
16(a) e 2 u + e − 2 u
L H S =
2
(eu−e −u) 2
RHS=1+2
2
(e2u−2+e −2u)
=1+2
4
(e2u−2+e −2u) e2u+e −2u
=1+ = = LHS
AnswerMarks
2 2B1
B1
B1
B1
AnswerMarks
[4]2.1
2.1
2.1
2.2a
AnswerMarks Guidance
16(b) d x
Let x = 2 s i n h u  = 2 c o s h u
d u
 2 x 2  a rs in h 1 4 s i n h 2 u
d x = 2 c o s h u d u
4 + x 2 4 + 4 s i n h 2 u
0 0
 a rs in h 1 4 s i n h 2 u
= 2 c o s h u d u
2 c o s h u
0
a rs in h 1
=  4 s i n h 2 u d u
AnswerMarks
0M1
A1
M1
AnswerMarks
A13.1a
1.1
2.1
AnswerMarks
2.11 + s i n h 2 u = c o s h 2 u used
( 1 + 2 ) − 1 = 1 = 1 − 2 = 2 − 1
1 + 2 ( 1 + 2 ) ( 1 − 2 )
giving = 12  ( 3 + 2 2 ) − ( 3 − 2 2 )  − 2 ln ( 1 + 2 )
AnswerMarks Guidance
= 2 2 − 2 ln ( 1 + 2 )A1 2.2a
Alternative for last 6 markss i n h u = 12 ( e u − e − u ) used
AG
=  a rs in h 1 ( e u − e − u ) 2 d u
AnswerMarks
0M1
=  a rs in h 1 ( e 2 u − 2 + e − 2 u ) d u
0
a rs in h 1
=  12 e 2 u − 2 u − 12 e − 2 u 
0
A1
AnswerMarks
a r s i n h 1 = ln ( 1 + 2 )B1
M1
e 2 ln (1 + 2 ) = ( 1 + 2 ) 2
M1
e − 2 ln (1 + 2 ) = ( 1 + 2 ) − 2 = ( 2 − 1 ) 2
3 + 2 2 3 − 2 2
giving − − 2 ln ( 1 + 2 )
2 2
= 2 2 − 2 ln ( 1 + 2 )
A1
[10]
s i n h u = 12 ( e u − e − u ) used
Question 16:
16 | (a) | e 2 u + e − 2 u
L H S =
2
(eu−e −u) 2
RHS=1+2
2
(e2u−2+e −2u)
=1+2
4
(e2u−2+e −2u) e2u+e −2u
=1+ = = LHS
2 2 | B1
B1
B1
B1
[4] | 2.1
2.1
2.1
2.2a
16 | (b) | d x
Let x = 2 s i n h u  = 2 c o s h u
d u
 2 x 2  a rs in h 1 4 s i n h 2 u
d x = 2 c o s h u d u
4 + x 2 4 + 4 s i n h 2 u
0 0
 a rs in h 1 4 s i n h 2 u
= 2 c o s h u d u
2 c o s h u
0
a rs in h 1
=  4 s i n h 2 u d u
0 | M1
A1
M1
A1 | 3.1a
1.1
2.1
2.1 | 1 + s i n h 2 u = c o s h 2 u used
( 1 + 2 ) − 1 = 1 = 1 − 2 = 2 − 1
1 + 2 ( 1 + 2 ) ( 1 − 2 )
giving = 12  ( 3 + 2 2 ) − ( 3 − 2 2 )  − 2 ln ( 1 + 2 )
= 2 2 − 2 ln ( 1 + 2 ) | A1 | 2.2a | AG
Alternative for last 6 marks | s i n h u = 12 ( e u − e − u ) used
AG
=  a rs in h 1 ( e u − e − u ) 2 d u
0 | M1
=  a rs in h 1 ( e 2 u − 2 + e − 2 u ) d u
0
a rs in h 1
=  12 e 2 u − 2 u − 12 e − 2 u 
0
A1
a r s i n h 1 = ln ( 1 + 2 ) | B1
M1
e 2 ln (1 + 2 ) = ( 1 + 2 ) 2
M1
e − 2 ln (1 + 2 ) = ( 1 + 2 ) − 2 = ( 2 − 1 ) 2
3 + 2 2 3 − 2 2
giving − − 2 ln ( 1 + 2 )
2 2
= 2 2 − 2 ln ( 1 + 2 )
A1
[10]
s i n h u = 12 ( e u − e − u ) used
16
\begin{enumerate}[label=(\alph*)]
\item Show using exponentials that $\cosh 2 u = 1 + 2 \sinh ^ { 2 } u$.
\item Show that $\int _ { 0 } ^ { 2 } \frac { x ^ { 2 } } { \sqrt { 4 + x ^ { 2 } } } \mathrm {~d} x = 2 \sqrt { 2 } - 2 \ln ( 1 + \sqrt { 2 } )$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2021 Q16 [14]}}
This paper (17 questions)
View full paper
Q1 7 Q2 4 Q3 6 Q4 4 Q5 6 Q6 4 Q7 6 Q8 9 Q9 11 Q10 13 Q11 9 Q12 4 Q13 7 Q14 14 Q15 6 Q16 14 Q17 20