OCR Further Pure Core 1 Specimen — Question 5 5 marks

Exam BoardOCR
ModuleFurther Pure Core 1 (Further Pure Core 1)
SessionSpecimen
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration using inverse trig and hyperbolic functions
TypeDerivative of inverse hyperbolic function
DifficultyStandard +0.8 This is a Further Maths question requiring knowledge of inverse hyperbolic functions. Part (i) is a standard 'show that' derivative using the chain rule and hyperbolic identities. Part (ii) requires completing the square and recognizing the inverse sinh form, which is non-trivial but follows a standard technique for this topic. The question tests multiple steps and pattern recognition beyond standard A-level, placing it moderately above average difficulty.
Spec4.07d Differentiate/integrate: hyperbolic functions4.07e Inverse hyperbolic: definitions, domains, ranges4.08h Integration: inverse trig/hyperbolic substitutions

5
  1. Show that \(\frac { \mathrm { d } } { \mathrm { d } x } \left( \sinh ^ { - 1 } ( 2 x ) \right) = \frac { 2 } { \sqrt { 4 x ^ { 2 } + 1 } }\).
  2. Find \(\int \frac { 1 } { \sqrt { 2 - 2 x + x ^ { 2 } } } \mathrm {~d} x\).

Question 5:
AnswerMarks Guidance
5(i) d (cid:11) (cid:16)1(cid:11)2x(cid:12) (cid:12)
sinh
dx
p
1
=
x2 (cid:14)(1)2
2 S
1 2 2
(cid:32) (cid:117) (cid:32)
x2 (cid:14)(1)2 2 4x2 (cid:14)1
AnswerMarks
2e
M1
E1
AnswerMarks
[2]c
1.1a
AnswerMarks
2.1Use formulae from formulae booklet
AG At least one intermediate step must
AnswerMarks
be seenOR
M1 sinhy(cid:32)2x
dy
so coshy(cid:32)2
dx
dy 2 2
E1 (cid:32) (cid:32)
dx coshy 4x2 (cid:14)1
AnswerMarks Guidance
5(ii) 2(cid:16)2x(cid:14)x2 (cid:32)1(cid:14)(cid:11)x(cid:16)1(cid:12)2
(cid:180) 1
(cid:159)(cid:181) dx(cid:32)(cid:170)sinh(cid:16)1(cid:11)x(cid:16)1(cid:12)(cid:186)(cid:14)c
(cid:181) (cid:172) (cid:188)
1(cid:14)(cid:11)x(cid:16)1(cid:12)2
AnswerMarks
(cid:182)B1
M1
A1
AnswerMarks
[3]3.1a
1.1
AnswerMarks
1.1Complete square soi
Use correct form
oe

5(i)
n
e
m
i
c

AnswerMarks
5(ii)e
p
S
Question 5:
5 | (i) | d (cid:11) (cid:16)1(cid:11)2x(cid:12) (cid:12)
sinh
dx
p
1
=
x2 (cid:14)(1)2
2 S
1 2 2
(cid:32) (cid:117) (cid:32)
x2 (cid:14)(1)2 2 4x2 (cid:14)1
2 | e
M1
E1
[2] | c
1.1a
2.1 | Use formulae from formulae booklet
AG At least one intermediate step must
be seen | OR
M1 sinhy(cid:32)2x
dy
so coshy(cid:32)2
dx
dy 2 2
E1 (cid:32) (cid:32)
dx coshy 4x2 (cid:14)1
5 | (ii) | 2(cid:16)2x(cid:14)x2 (cid:32)1(cid:14)(cid:11)x(cid:16)1(cid:12)2
(cid:180) 1
(cid:159)(cid:181) dx(cid:32)(cid:170)sinh(cid:16)1(cid:11)x(cid:16)1(cid:12)(cid:186)(cid:14)c
(cid:181) (cid:172) (cid:188)
1(cid:14)(cid:11)x(cid:16)1(cid:12)2
(cid:182) | B1
M1
A1
[3] | 3.1a
1.1
1.1 | Complete square soi
Use correct form
oe
--- 5(i) ---
5(i)
n
e
m
i
c
--- 5(ii) ---
5(ii) | e
p
S
5 (i) Show that $\frac { \mathrm { d } } { \mathrm { d } x } \left( \sinh ^ { - 1 } ( 2 x ) \right) = \frac { 2 } { \sqrt { 4 x ^ { 2 } + 1 } }$.\\
(ii) Find $\int \frac { 1 } { \sqrt { 2 - 2 x + x ^ { 2 } } } \mathrm {~d} x$.

\hfill \mbox{\textit{OCR Further Pure Core 1  Q5 [5]}}