OCR Further Pure Core 2 2021 November — Question 5 8 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2021
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHyperbolic functions
TypeSolve using double angle formulas
DifficultyStandard +0.8 Part (a) requires proving a double angle formula from exponential definitions (standard Further Maths technique). Part (b) involves substituting the identity to create a quadratic in cosh x, then solving cosh x = t for x using inverse hyperbolic functions in logarithmic form. This is a multi-step problem requiring several Further Maths techniques, but follows a predictable pattern once the substitution is made. Slightly above average difficulty due to the Further Maths content and need for exact logarithmic answers.
Spec4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1
5 In this question you must show detailed reasoning.
  1. Using the definition of \(\cosh x\) in terms of exponentials, show that \(\cosh 2 x \equiv 2 \cosh ^ { 2 } x - 1\).
  2. Solve the equation \(\cosh 2 x = 3 \cosh x + 1\), giving all your answers in exact logarithmic form.
Question 5:
AnswerMarks Guidance
5(a) DR
ex +e−x  2
RHS = 2cosh2x – 1 = 2  −1
 2 
e2x +2+e−2x  e2x +2+e−2x
=2 −1=
 4  2
e2x +e−2x
= =cosh2x=LHS
AnswerMarks
2M1
A1
AnswerMarks
[2]2.1
2.1Uses correct exponential form in
an attempt at proof
AnswerMarks
AGProof must be complete
(b)DR
2cosh2x – 1= 3coshx + 1
=> 2cosh2x – 3coshx – 2 = 0
(2coshx + 1)(coshx – 2) = 0
coshx = 2 or –½
coshx ≥ 1 so ≠ –½
( )
x=cosh−12=ln 2+ 3
( )
AnswerMarks
x=ln 2− 3M1
M1
A1
A1
A1
A1
AnswerMarks
[6]3.1a
1.1
1.1
2.3
1.1
AnswerMarks
1.1Use of identity in (a) to leave a
three term quadratic equation in
just coshx
Attempt to solve eg
−(−3)± (−3)2 −4×2×(−2)
2×2
Justification must be seen and
must contain no incorrect
statements
For either correct answer seen
AnswerMarks
Both correct values for x2
 3 9
or 2coshx−  − −2=0
 4 8
Or solves quadratic BC
( )
Or x=−ln 2+ 3
Mark final answer
AnswerMarks Guidance
AnswerMarks AO 
Question 5:
5 | (a) | DR
ex +e−x  2
RHS = 2cosh2x – 1 = 2  −1
 2 
e2x +2+e−2x  e2x +2+e−2x
=2 −1=
 4  2
e2x +e−2x
= =cosh2x=LHS
2 | M1
A1
[2] | 2.1
2.1 | Uses correct exponential form in
an attempt at proof
AG | Proof must be complete
(b) | DR
2cosh2x – 1= 3coshx + 1
=> 2cosh2x – 3coshx – 2 = 0
(2coshx + 1)(coshx – 2) = 0
coshx = 2 or –½
coshx ≥ 1 so ≠ –½
( )
x=cosh−12=ln 2+ 3
( )
x=ln 2− 3 | M1
M1
A1
A1
A1
A1
[6] | 3.1a
1.1
1.1
2.3
1.1
1.1 | Use of identity in (a) to leave a
three term quadratic equation in
just coshx
Attempt to solve eg
−(−3)± (−3)2 −4×2×(−2)
2×2
Justification must be seen and
must contain no incorrect
statements
For either correct answer seen
Both correct values for x | 2
 3 9
or 2coshx−  − −2=0
 4 8
Or solves quadratic BC
( )
Or x=−ln 2+ 3
Mark final answer
Answer | Marks | AO | Guidance
5 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item Using the definition of $\cosh x$ in terms of exponentials, show that $\cosh 2 x \equiv 2 \cosh ^ { 2 } x - 1$.
\item Solve the equation $\cosh 2 x = 3 \cosh x + 1$, giving all your answers in exact logarithmic form.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 2 2021 Q5 [8]}}
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