OCR MEI Further Pure Core 2023 June — Question 9 6 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2023
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIndefinite & Definite Integrals
TypeMean value of function
DifficultyStandard +0.3 This is a straightforward application of mean value integration with a standard trigonometric integral. Students must set up the mean value formula over the given period and integrate sinΒ²(nt) using the double angle identity, which is a routine Further Maths technique. The question is slightly above average difficulty due to the applied context and needing to recall the mean value formula, but the integration itself is standard.
Spec4.08f Integrate using partial fractions

9 In an electrical circuit, the alternating current \(I\) amps is given by \(\mathbf { I } =\) asinnt, where \(t\) is the time in seconds and \(a\) and \(n\) are positive constants. The RMS value of the current, in amps, is defined to be the square root of the mean value of \(I ^ { 2 }\) over one complete period of \(\frac { 2 \pi } { n }\) seconds. Show that the RMS value of the current is \(\frac { a } { \sqrt { 2 } }\) amps.

Question 9:
AnswerMarks
9(cid:2870)(cid:3095) (cid:2870)(cid:3095)
(cid:3041) 1 (cid:3041)
π‘˜(cid:3505) sin(cid:2870)𝑛𝑑d𝑑 = π‘˜(cid:3505) (1βˆ’cos2𝑛𝑑)d𝑑
2
(cid:2868) (cid:2868)
2πœ‹
1 1 𝑛
= π‘˜(cid:4680)π‘‘βˆ’ sin2𝑛𝑑(cid:4681)
2 2𝑛
0
1 2πœ‹ πœ‹π‘Ž(cid:2870)
= π‘Ž(cid:2870) =
2 𝑛 𝑛
a2 2 a2
mean value of I2 = ο€½
n n 2
(cid:3028)
οƒž RMS value =
AnswerMarks
√(cid:2870)M1*
A1
M1dep
A1
M1
AnswerMarks
A13.1a
1.1
1.1
1.1
1.1
AnswerMarks
2.2ause of double angle formula for sin(cid:2870)𝑛𝑑
1
(cid:3428)π‘‘βˆ’ sin2𝑛𝑑(cid:3432)
2𝑛
www
(cid:2870)(cid:3095)
for dividing by
(cid:3041)
www AG
Alternative method
(cid:2870)(cid:3095) (cid:2870)(cid:3095)
1 (cid:3041) 𝑛 (cid:3041)
π‘˜(cid:3505) sin(cid:2870)𝑛𝑑d𝑑 = π‘˜(cid:3505) sin(cid:2870)𝑛𝑑d𝑑
2πœ‹ 2πœ‹
(cid:4672) (cid:4673) (cid:2868) (cid:2868)
AnswerMarks Guidance
𝑛M1 1.1
for dividing their integral by
(cid:3041)
2πœ‹
1 𝑛 𝑛
= Γ— π‘˜(cid:3505) (1βˆ’cos2𝑛𝑑)d𝑑
2 2πœ‹
AnswerMarks
0use of double angle formula for sin(cid:2870)𝑛𝑑
M1*3.1a
A11.1
2πœ‹
𝑛 1 𝑛
= π‘˜(cid:4680)π‘‘βˆ’ sin2𝑛𝑑(cid:4681)
4πœ‹ 2𝑛
AnswerMarks Guidance
0M1dep 1.1
(cid:3428)π‘‘βˆ’ sin2𝑛𝑑(cid:3432)
2𝑛
AnswerMarks
M1dep1.1
π‘Ž(cid:2870)𝑛 2πœ‹ π‘Ž(cid:2870)
= (cid:3436) (cid:3440)=
AnswerMarks Guidance
4πœ‹ 𝑛 2A1 1.1
(cid:3028)
οƒž RMS value =
AnswerMarks Guidance
√(cid:2870)A1 2.2a
[6]
M1
1.1
(cid:2870)(cid:3095)
for dividing their integral by
(cid:3041)
use of double angle formula for sin(cid:2870)𝑛𝑑
Question 9:
9 | (cid:2870)(cid:3095) (cid:2870)(cid:3095)
(cid:3041) 1 (cid:3041)
π‘˜(cid:3505) sin(cid:2870)𝑛𝑑d𝑑 = π‘˜(cid:3505) (1βˆ’cos2𝑛𝑑)d𝑑
2
(cid:2868) (cid:2868)
2πœ‹
1 1 𝑛
= π‘˜(cid:4680)π‘‘βˆ’ sin2𝑛𝑑(cid:4681)
2 2𝑛
0
1 2πœ‹ πœ‹π‘Ž(cid:2870)
= π‘Ž(cid:2870) =
2 𝑛 𝑛
a2 2 a2
mean value of I2 = ο€½
n n 2
(cid:3028)
οƒž RMS value =
√(cid:2870) | M1*
A1
M1dep
A1
M1
A1 | 3.1a
1.1
1.1
1.1
1.1
2.2a | use of double angle formula for sin(cid:2870)𝑛𝑑
1
(cid:3428)π‘‘βˆ’ sin2𝑛𝑑(cid:3432)
2𝑛
www
(cid:2870)(cid:3095)
for dividing by
(cid:3041)
www AG
Alternative method
(cid:2870)(cid:3095) (cid:2870)(cid:3095)
1 (cid:3041) 𝑛 (cid:3041)
π‘˜(cid:3505) sin(cid:2870)𝑛𝑑d𝑑 = π‘˜(cid:3505) sin(cid:2870)𝑛𝑑d𝑑
2πœ‹ 2πœ‹
(cid:4672) (cid:4673) (cid:2868) (cid:2868)
𝑛 | M1 | 1.1 | (cid:2870)(cid:3095)
for dividing their integral by
(cid:3041)
2πœ‹
1 𝑛 𝑛
= Γ— π‘˜(cid:3505) (1βˆ’cos2𝑛𝑑)d𝑑
2 2πœ‹
0 | use of double angle formula for sin(cid:2870)𝑛𝑑
M1* | 3.1a
A1 | 1.1
2πœ‹
𝑛 1 𝑛
= π‘˜(cid:4680)π‘‘βˆ’ sin2𝑛𝑑(cid:4681)
4πœ‹ 2𝑛
0 | M1dep | 1.1 | 1
(cid:3428)π‘‘βˆ’ sin2𝑛𝑑(cid:3432)
2𝑛
M1dep | 1.1
π‘Ž(cid:2870)𝑛 2πœ‹ π‘Ž(cid:2870)
= (cid:3436) (cid:3440)=
4πœ‹ 𝑛 2 | A1 | 1.1 | www | www
(cid:3028)
οƒž RMS value =
√(cid:2870) | A1 | 2.2a | www AG
[6]
M1
1.1
(cid:2870)(cid:3095)
for dividing their integral by
(cid:3041)
use of double angle formula for sin(cid:2870)𝑛𝑑
9 In an electrical circuit, the alternating current $I$ amps is given by $\mathbf { I } =$ asinnt, where $t$ is the time in seconds and $a$ and $n$ are positive constants. The RMS value of the current, in amps, is defined to be the square root of the mean value of $I ^ { 2 }$ over one complete period of $\frac { 2 \pi } { n }$ seconds.

Show that the RMS value of the current is $\frac { a } { \sqrt { 2 } }$ amps.

\hfill \mbox{\textit{OCR MEI Further Pure Core 2023 Q9 [6]}}