OCR MEI Further Pure Core 2019 June — Question 13 11 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2019
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHyperbolic functions
TypeDifferentiate inverse hyperbolic functions
DifficultyStandard +0.8 This is a Further Maths question requiring proof of a derivative from logarithmic form, integration by parts, and understanding of domain restrictions. Part (a) requires manipulating the logarithmic definition and differentiating; part (b) needs integration by parts with the derived result; part (c) tests conceptual understanding. While systematic, it demands multiple sophisticated techniques and careful algebraic manipulation beyond standard A-level.
Spec4.07f Inverse hyperbolic: logarithmic forms4.08h Integration: inverse trig/hyperbolic substitutions
13
  1. Using the logarithmic form of \(\operatorname { arcosh } x\), prove that the derivative of \(\operatorname { arcosh } x\) is \(\frac { 1 } { \sqrt { x ^ { 2 } - 1 } }\).
  2. Hence find \(\int _ { 1 } ^ { 2 } \operatorname { arcosh } x \mathrm {~d} x\), giving your answer in exact logarithmic form.
  3. Ali tries to evaluate \(\int _ { 0 } ^ { 1 } \operatorname { arcosh } x \mathrm {~d} x\) using his calculator, and gets an 'error'. Explain why.
Question 13:
AnswerMarks Guidance
13(a) yarcoshxln(x x21)
(lnu)
AnswerMarks
du uor ey = x + (x2  1)
dy 1x(x21)1/2
AnswerMarks Guidance
dx (x(x21)1/2)M1
A11.1b chain rule on (x2  1)
correct expressionchain rule on (x2  1) ey dy/dx = … M1
= 1 + x(x21)1/2 M1
[substituting for ey]
AnswerMarks
1.1bcorrect expression
(x21)1/2x
AnswerMarks Guidance
(x(x21)1/2)(x21)1/2A1 2.1
or
(x(x21)1/2)
2.1
1
 *
AnswerMarks Guidance
(x2 1)1/2A1cao
[5]2.2a NB AG
13(b) let u = arcosh x, u = 1/(x2  1), v = 1, v = x
 2 arcoshxdx xarcoshx 2 2 x dx
AnswerMarks Guidance
1 1 1 x21 2 arcoshxdx xarcoshx 2 2 x dx
1 1 1 x21A1 1.1b
M11.1b
substitution or inspection
  xarcoshx x2 1 
AnswerMarks Guidance
 A1 A1
 dx x21
x21
= 2arcosh 2  arcosh 1  3
AnswerMarks Guidance
= 2ln(2 + 3)  3A1cao oe e.g. ln(7 + 43)  3
Alternative solution
Let x = cosh u, dx = sinh u du
AnswerMarks
 arcoshxdxusinhuduM1
M1A1integration by parts
 coshudu
AnswerMarks
ucoshuintegration by parts
 ucoshusinhu arcosh2
AnswerMarks Guidance
arcosh1A1 limits not needed
= 2ln(2 +3)  3A1cao oe e.g. ln(7 + 43)  3
[5]
AnswerMarks Guidance
13(c) arcosh x does not exist for x < 1
[1]2.4 or (x2  1) = (1) not real,
so ln(x+(x21)) is not realaccept other valid
arguments
dy 1x(x21)1/2
dx (x(x21)1/2)
M1
A1
ey dy/dx = … M1
= 1 + x(x21)1/2 M1
[substituting for ey]
A1
A1cao
[5]
M1
M1A1
Question 13:
13 | (a) | yarcoshxln(x x21) | M1 | d 1
(lnu)
du u | or ey = x + (x2  1)
dy 1x(x21)1/2

dx (x(x21)1/2) | M1
A1 | 1.1b | chain rule on (x2  1)
correct expression | chain rule on (x2  1) | ey dy/dx = … M1
= 1 + x(x21)1/2 M1
[substituting for ey]
1.1b | correct expression
(x21)1/2x

(x(x21)1/2)(x21)1/2 | A1 | 2.1 | [(x21)1/2x](x21)1/2
or
(x(x21)1/2)
2.1
1
 *
(x2 1)1/2 | A1cao
[5] | 2.2a | NB AG | NB AG
13 | (b) | let u = arcosh x, u = 1/(x2  1), v = 1, v = x | M1 | 3.1a | integration by parts
 2 arcoshxdx xarcoshx 2 2 x dx
1 1 1 x21 |  2 arcoshxdx xarcoshx 2 2 x dx
1 1 1 x21 | A1 | 1.1b | ignore limits
M1 | 1.1b
substitution or inspection
  xarcoshx x2 1 
  | A1 | A1 | 1.1b | 1.1b | x
 dx x21
x21
= 2arcosh 2  arcosh 1  3
= 2ln(2 + 3)  3 | A1cao | oe e.g. ln(7 + 43)  3 | isw, not ln 1
Alternative solution
Let x = cosh u, dx = sinh u du
 arcoshxdxusinhudu | M1
M1A1 | integration by parts
 coshudu
ucoshu | integration by parts
 ucoshusinhu arcosh2
arcosh1 | A1 | limits not needed
= 2ln(2 +3)  3 | A1cao | oe e.g. ln(7 + 43)  3 | [isw] , not ln 1
[5]
13 | (c) | arcosh x does not exist for x < 1 | B1
[1] | 2.4 | or (x2  1) = (1) not real,
so ln(x+(x21)) is not real | accept other valid
arguments
dy 1x(x21)1/2

dx (x(x21)1/2)
M1
A1
ey dy/dx = … M1
= 1 + x(x21)1/2 M1
[substituting for ey]
A1
A1cao
[5]
M1
M1A1
13
\begin{enumerate}[label=(\alph*)]
\item Using the logarithmic form of $\operatorname { arcosh } x$, prove that the derivative of $\operatorname { arcosh } x$ is $\frac { 1 } { \sqrt { x ^ { 2 } - 1 } }$.
\item Hence find $\int _ { 1 } ^ { 2 } \operatorname { arcosh } x \mathrm {~d} x$, giving your answer in exact logarithmic form.
\item Ali tries to evaluate $\int _ { 0 } ^ { 1 } \operatorname { arcosh } x \mathrm {~d} x$ using his calculator, and gets an 'error'. Explain why.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2019 Q13 [11]}}
This paper (17 questions)
View full paper
Q1 4 Q2 3 Q3 7 Q4 3 Q5 5 Q6 4 Q7 8 Q8 8 Q9 7 Q10 8 Q11 12 Q12 9 Q13 11 Q14 13 Q15 8 Q16 12 Q17 22