OCR Further Pure Core 2 2023 June — Question 3 6 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2023
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHyperbolic functions
TypeDifferentiate inverse hyperbolic functions
DifficultyStandard +0.3 Part (a) is a standard bookwork proof requiring the chain rule and knowledge of hyperbolic identities (cosh²u - sinh²u = 1). Part (b) applies this result with chain rule to find a derivative at a specific point, then uses it to find a normal line equation—straightforward application once the derivative is known. This is typical Further Maths content but follows standard procedures without requiring novel insight.
Spec4.07e Inverse hyperbolic: definitions, domains, ranges4.08g Derivatives: inverse trig and hyperbolic functions
3
  1. Show that \(\frac { \mathrm { d } } { \mathrm { d } u } \left( \sinh ^ { - 1 } u \right) = \frac { 1 } { \sqrt { u ^ { 2 } + 1 } }\).
  2. Find the equation of the normal to the graph of \(\mathrm { y } = \sinh ^ { - 1 } 2 \mathrm { x }\) at the point where \(x = \sqrt { 6 }\). Give your answer in the form \(\mathrm { y } = \mathrm { mx } + \mathrm { c }\) where \(m\) and \(c\) are given in exact, non-hyperbolic form.
Question 3:
AnswerMarks Guidance
3(a) y = sinh–1u =>sinhy = u
d y d y 1
 c o s h y = 1  =
d u d u c o s h y
d y 1 1
 = =
AnswerMarks Guidance
d u  1 + s i n h 2 y  2 u + 1M1 2.1
differentiating and using
AnswerMarks
cosh2y – sinh2y = 1Condone missing ± for M1
du
Accept at this stage
dy
But gradient of y = sinh–1u is never negative
d y 1
so =
AnswerMarks Guidance
d u u 2 + 1A1 2.4
“always positive”)poor notation can be recovered
Alternative method:M1 AG. Using the definition of
sinh–1 in logarithmic form and
attempting to differentiate using
the chain rule on ln function.
( )
y = s i n h − 1 u = ln u + u 2 + 1
12
1 ( ) −
1 +  2 u 2 u + 1
d y 2
 =
d u u + u 2 + 1
1
( )
u2 +1 2 +u
1 1
= = =
( u2 +1 ) 1 2  u+ ( u2 +1 ) 1 2   ( u2 +1 ) 1 2 u2 +1
 
AnswerMarks Guidance
 A1 AG so some intermediate
working must be seen. www
M1
AG. Using the definition of
sinh–1 in logarithmic form and
attempting to differentiate using
the chain rule on ln function.
AG so some intermediate
working must be seen. www
AnswerMarks Guidance
Alternative method 2:M1 Correctly integrates RHS
1
I =  d u
2 u + 1
u = s i n h v  d u = c o s h v d v
1
 I =  c o s h v d v
s i n 2 h v + 1
c o s h v
=  d v =  1 d v = v ( + c )
c o s h 2 v
= s i n h − 1 u ( + ) c
D i f f e r e n t i a t i n g b o t h s i d e s w r t u g i v e s
1 = d ( s i n h − 1 u )
AnswerMarks
u 2 + 1 d uA1
[2]
M1
Correctly integrates RHS
Condone omission of c
AnswerMarks
(b)y=sinh −12x
dy 2 2
 = =
AnswerMarks Guidance
dx (2x)2 +1 4x2 +1M1 1.1
or the formula booklet1
Giving
1
+ x 2
4
AnswerMarks Guidance
x = 6  y = s i n h − 1 2 6 ( = ln ( 5 + 2 6 ) )M1 1.1
equation to find the y coordinate
of the point
d y 2
x = 6  =
d x 5
x = 6
5
 m = −
AnswerMarks Guidance
2M1 1.1
theirgradient and taking
negative reciprocal
( ) 5 ( )
y − ln 5 + 2 6 = − x − 6
2
5 ( ) 5 6
 y = − x + ln 5 + 2 6 +
AnswerMarks Guidance
2 2A1 1.1
[4]
Question 3:
3 | (a) | y = sinh–1u =>sinhy = u
d y d y 1
 c o s h y = 1  =
d u d u c o s h y
d y 1 1
 = =
d u  1 + s i n h 2 y  2 u + 1 | M1 | 2.1 | AG. Taking sinh of both sides,
differentiating and using
cosh2y – sinh2y = 1 | Condone missing ± for M1
du
Accept at this stage
dy
But gradient of y = sinh–1u is never negative
d y 1
so =
d u u 2 + 1 | A1 | 2.4 | AG so reason required (accept
“always positive”) | poor notation can be recovered
Alternative method: | M1 | AG. Using the definition of
sinh–1 in logarithmic form and
attempting to differentiate using
the chain rule on ln function.
( )
y = s i n h − 1 u = ln u + u 2 + 1
12
1 ( ) −
1 +  2 u 2 u + 1
d y 2
 =
d u u + u 2 + 1
1
( )
u2 +1 2 +u
1 1
= = =
( u2 +1 ) 1 2  u+ ( u2 +1 ) 1 2   ( u2 +1 ) 1 2 u2 +1
 
  | A1 | AG so some intermediate
working must be seen. www
M1
AG. Using the definition of
sinh–1 in logarithmic form and
attempting to differentiate using
the chain rule on ln function.
AG so some intermediate
working must be seen. www
Alternative method 2: | M1 | Correctly integrates RHS | Condone omission of c
1
I =  d u
2 u + 1
u = s i n h v  d u = c o s h v d v
1
 I =  c o s h v d v
s i n 2 h v + 1
c o s h v
=  d v =  1 d v = v ( + c )
c o s h 2 v
= s i n h − 1 u ( + ) c
D i f f e r e n t i a t i n g b o t h s i d e s w r t u g i v e s
1 = d ( s i n h − 1 u )
u 2 + 1 d u | A1
[2]
M1
Correctly integrates RHS
Condone omission of c
(b) | y=sinh −12x
dy 2 2
 = =
dx (2x)2 +1 4x2 +1 | M1 | 1.1 | Differentiating using chain rule
or the formula booklet | 1
Giving
1
+ x 2
4
x = 6  y = s i n h − 1 2 6 ( = ln ( 5 + 2 6 ) ) | M1 | 1.1 | Substituting the x-value into the
equation to find the y coordinate
of the point
d y 2
x = 6  =
d x 5
x = 6
5
 m = −
2 | M1 | 1.1 | Substituting the x-value into
theirgradient and taking
negative reciprocal
( ) 5 ( )
y − ln 5 + 2 6 = − x − 6
2
5 ( ) 5 6
 y = − x + ln 5 + 2 6 +
2 2 | A1 | 1.1 | oe in correct form
[4]
3
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac { \mathrm { d } } { \mathrm { d } u } \left( \sinh ^ { - 1 } u \right) = \frac { 1 } { \sqrt { u ^ { 2 } + 1 } }$.
\item Find the equation of the normal to the graph of $\mathrm { y } = \sinh ^ { - 1 } 2 \mathrm { x }$ at the point where $x = \sqrt { 6 }$. Give your answer in the form $\mathrm { y } = \mathrm { mx } + \mathrm { c }$ where $m$ and $c$ are given in exact, non-hyperbolic form.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 2 2023 Q3 [6]}}
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