OCR Further Pure Core 1 2019 June — Question 6 6 marks

Exam BoardOCR
ModuleFurther Pure Core 1 (Further Pure Core 1)
Year2019
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration using inverse trig and hyperbolic functions
TypeIntegration by parts with inverse trig
DifficultyChallenging +1.2 Part (a) requires chain rule differentiation of an inverse trig function with a square root composition—straightforward for Further Maths students. Part (b) is more substantial: students must recognize that the integrand relates to the derivative from part (a), perform algebraic manipulation to match forms, then apply integration by parts or substitution. The 'show that' format and exact answer requirement add modest challenge, but this is a standard Further Pure technique with clear signposting from part (a).
Spec4.08h Integration: inverse trig/hyperbolic substitutions
6 You are given that \(y = \tan ^ { - 1 } \sqrt { 2 x }\).
  1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
  2. Show that \(\int _ { \frac { 1 } { 6 } } ^ { \frac { 1 } { 2 } } \frac { \sqrt { x } } { \left( x + 2 x ^ { 2 } \right) } \mathrm { d } x = k \pi\) where \(k\) is a number to be determined in exact form.
Question 6:
AnswerMarks Guidance
6(a) 1 d ( )
= × 2x
( )2 dx
1+ 2x
oe
1 2 1 1
= × = ×
AnswerMarks
1+2x 2 x 1+2x 2xM1
A1
AnswerMarks
[2]1.1a
1.1Attempt to differentiate
using chain rulei.e. product of 2 terms
Alternatively:
tan y = 2x
dy 2 ( )dy 2
⇒sec2 y = ⇒ 1+tan2 x = M1
dx 2 x dx 2 x
dy 1 2
⇒ = × A1
AnswerMarks Guidance
dx 1+2x 2 x[4] Make a substitution
6(b) 1 1
2 1 2 1
∫ dx= 2∫ dx
(1+2x) x (1+2x) 2x
1 1
6 6
1  1 
= 2  tan−1 2x   2 = 2tan−11−tan−1 
1  3
6
π π 2
= 2 −  = π
 4 6  12
2
So k =
AnswerMarks
12M1
A1
M1
AnswerMarks
A13.1a
1.1
1.1
AnswerMarks
1.1Get into form of (a).
Ignore limits
Correct form
Use (a) and correct limits
in correct order.
oe
Alternatively:
Let u= x M1
1
du= dx⇒dx=2 xdu=2udu
2 x
1 x=1 x=1
2 x 2 u 2 1
∫ dx= ∫ 2udu=2 ∫ du A1
( x+2x2) ( u2+2u4) ( 1+2u2)
1 x=1 x=1
6 6 6
2x=1
= 2tan−1u 2 M1
  x=1
6
= 2  tan−1 2x  x x = =1 1 2 = 2    tan−11−tan−1 1 3    = 2    π 4 − π 6   
6
π 2
= A1
AnswerMarks Guidance
12[4] Make a substitution
Get into correct form
Use standard result with
correct limits in correct
order
Question 6:
6 | (a) | 1 d ( )
= × 2x
( )2 dx
1+ 2x
oe
1 2 1 1
= × = ×
1+2x 2 x 1+2x 2x | M1
A1
[2] | 1.1a
1.1 | Attempt to differentiate
using chain rule | i.e. product of 2 terms
Alternatively:
tan y = 2x
dy 2 ( )dy 2
⇒sec2 y = ⇒ 1+tan2 x = M1
dx 2 x dx 2 x
dy 1 2
⇒ = × A1
dx 1+2x 2 x | [4] | Make a substitution
6 | (b) | 1 1
2 1 2 1
∫ dx= 2∫ dx
(1+2x) x (1+2x) 2x
1 1
6 6
1  1 
= 2  tan−1 2x   2 = 2tan−11−tan−1 
1  3
6
π π 2
= 2 −  = π
 4 6  12
2
So k =
12 | M1
A1
M1
A1 | 3.1a
1.1
1.1
1.1 | Get into form of (a).
Ignore limits
Correct form
Use (a) and correct limits
in correct order.
oe
Alternatively:
Let u= x M1
1
du= dx⇒dx=2 xdu=2udu
2 x
1 x=1 x=1
2 x 2 u 2 1
∫ dx= ∫ 2udu=2 ∫ du A1
( x+2x2) ( u2+2u4) ( 1+2u2)
1 x=1 x=1
6 6 6
2x=1
= 2tan−1u 2 M1
  x=1
6
= 2  tan−1 2x  x x = =1 1 2 = 2    tan−11−tan−1 1 3    = 2    π 4 − π 6   
6
π 2
= A1
12 | [4] | Make a substitution
Get into correct form
Use standard result with
correct limits in correct
order
6 You are given that $y = \tan ^ { - 1 } \sqrt { 2 x }$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Show that $\int _ { \frac { 1 } { 6 } } ^ { \frac { 1 } { 2 } } \frac { \sqrt { x } } { \left( x + 2 x ^ { 2 } \right) } \mathrm { d } x = k \pi$ where $k$ is a number to be determined in exact form.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 1 2019 Q6 [6]}}
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