OCR MEI Further Pure Core 2023 June — Question 15 5 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2023
SessionJune
Marks5
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Mark schemeDownload PDF ↗
TopicIntegration using inverse trig and hyperbolic functions
TypeStandard integral of 1/√(a²-x²)
DifficultyStandard +0.3 This is a standard further maths integration question requiring completing the square to transform the denominator into the form √(a²-(x-b)²), then applying the arcsin formula. While it requires multiple steps (completing the square, recognizing the standard form, applying limits), it's a textbook application of a known technique with no novel insight needed. Slightly easier than average A-level difficulty due to its routine nature, though the further maths context and multi-step process keep it close to baseline.
Spec4.08h Integration: inverse trig/hyperbolic substitutions
15 In this question you must show detailed reasoning. Evaluate \(\int _ { 1 } ^ { 2 } \frac { 1 } { \sqrt { 1 + 2 x - x ^ { 2 } } } d x\), giving your answer in terms of \(\pi\).
Question 15:
AnswerMarks
15DR
1+2𝑥−𝑥(cid:2870) = −(𝑥(cid:2870)−2𝑥)+1 = −(𝑥−1)(cid:2870)+2
𝑥−1 (cid:2870)
(cid:3428)arcsin (cid:3432)
𝑘
(cid:2869)
𝑥−1 (cid:2870)
= (cid:3428)arcsin (cid:3432)
√2
(cid:2869)
1 𝜋
= arcsin(cid:3436) (cid:3440)−arcsin0 =
AnswerMarks
√2 4M1
A1
M1
A1
AnswerMarks
A13.1a
1.1
2.1
2.1
AnswerMarks
2.2acompleting the square
substitution or evaluation of both limits must be seen
Alternative method
AnswerMarks Guidance
1+2𝑥−𝑥(cid:2870) = −(𝑥(cid:2870)−2𝑥)+1 = −(𝑥−1)(cid:2870)+2M1 completing the square
A1
Let 𝑢 = 𝑥−1
1 1
(cid:3505) d𝑢
√2−𝑢2
AnswerMarks Guidance
0M1 M1
𝑢 (cid:2869)
= (cid:3428)arcsin (cid:3432)
√2
AnswerMarks
(cid:2868)A1
1 𝜋
= arcsin(cid:3436) (cid:3440)−arcsin0 =
AnswerMarks Guidance
√2 4A1 substitution or evaluation of both limits must be seen
[5]
AnswerMarks
(cid:2870)×(cid:2872)(cid:2878)(cid:2869)×(cid:2869)(cid:2878)(cid:2868)×(cid:2870)
Distance from point to plane =
√(cid:2870)(cid:3118)(cid:2878)(cid:2869)(cid:3118)(cid:2878)(cid:2870)(cid:3118)
= 3 units
Line is 𝐫 = 3𝐢+𝐣−5𝐤+𝝀(2𝐢+𝑏𝐣+3𝐤)
𝑨(cid:4652)(cid:4652)(cid:4652)(cid:4652)(cid:4652)𝑷(cid:4652)⃗ = 𝐢+5𝐤
𝑨(cid:4652)(cid:4652)(cid:4652)(cid:4652)(cid:4652)𝑷(cid:4652)⃗×𝒅= (𝐢+5𝐤)×(2𝐢+𝑏𝐣+3𝐤)
= −5𝑏𝐢+7𝐣+𝑏𝐤
dist from P to line
AnswerMarks
−5𝑏𝐢+7𝐣+𝑏𝐤√26𝑏(cid:2870)+49
=
AnswerMarks
𝒅√13+𝑏(cid:2870)
(cid:2870)(cid:2874)(cid:3029)(cid:3118)(cid:2878)(cid:2872)(cid:2877)
so =9  26𝑏(cid:2870)+49= 117+9𝑏(cid:2870)
(cid:2869)(cid:2871)(cid:2878)(cid:3029)(cid:3118)
AnswerMarks
 𝑏 = 2M1*
A1
M1
A1
M1
A1
M1*
A1
M1dep
AnswerMarks
A13.1a
1.1
3.1a
2.1
2.1
2.1
3.1a
1.1
2.1
AnswerMarks
3.2aor −𝑨(cid:4652)(cid:4652)(cid:4652)(cid:4652)(cid:4652)𝑷(cid:4652)⃗. Could be seen as part of distance calculation.
or 𝒅×𝑨(cid:4652)(cid:4652)(cid:4652)(cid:4652)(cid:4652)𝑷(cid:4652)⃗
equating the two distances
Alternative method 1
AnswerMarks
(cid:2870)×(cid:2872)(cid:2878)(cid:2869)×(cid:2869)(cid:2878)(cid:2868)×(cid:2870)
Distance from point to plane =
AnswerMarks Guidance
√(cid:2870)(cid:3118)(cid:2878)(cid:2869)(cid:3118)(cid:2878)(cid:2870)(cid:3118)M1*
= 3 unitsA1
Line is 𝐫 = 3𝐢+𝐣−5𝐤+𝜆(2𝐢+𝑏𝐣+3𝐤)M1
(2𝜆−1)𝐢+𝑏𝜆𝐣+(3𝜆−5)𝐤A1 vector from P to a point on the line
2𝜆−1 2
(cid:3437) 𝑏𝜆 (cid:3441)∙(cid:3437)𝑏(cid:3441)=0
AnswerMarks Guidance
3𝜆−5 3M1* M1*
17
λ =
AnswerMarks Guidance
13+𝑏(cid:2870)A1
dist from P to lineM1 substituting 𝜆 into (cid:3627)𝑨(cid:4652)(cid:4652)(cid:4652)(cid:4652)(cid:4652)𝑷(cid:4652)⃗(cid:3627)
(cid:2870) (cid:2870) (cid:2870)
(cid:3496)(cid:4672)2(cid:4672) (cid:2869)(cid:2875) (cid:4673)−1(cid:4673) +(cid:3436)𝑏(cid:4672) (cid:2869)(cid:2875) (cid:4673)(cid:3440) +(cid:4672)3(cid:4672) (cid:2869)(cid:2875) (cid:4673)−5(cid:4673)
(cid:2869)(cid:2871)(cid:2878)(cid:3029)(cid:3118) (cid:2869)(cid:2871)(cid:2878)(cid:3029)(cid:3118) (cid:2869)(cid:2871)(cid:2878)(cid:3029)(cid:3118)
√26𝑏(cid:2872)+387𝑏(cid:2870)+637
=
AnswerMarks Guidance
13+𝑏(cid:2870)A1 simplified
(cid:2870)(cid:2874)(cid:3029)(cid:3120)(cid:2878)(cid:2871)(cid:2876)(cid:2875)(cid:3029)(cid:3118)(cid:2878)(cid:2874)(cid:2871)(cid:2875)
So = 9
AnswerMarks Guidance
((cid:2869)(cid:2871)(cid:2878)(cid:3029)(cid:3118))(cid:3118)M1dep equating the two distances
17𝑏(cid:2872)+153𝑏(cid:2870)−884 = 0
AnswerMarks
 𝑏 = 2A1
Alternative method 2
AnswerMarks
𝟐×𝟒(cid:2878)𝟏×𝟏(cid:2878)𝟎×𝟐
Distance from point to plane =
AnswerMarks Guidance
(cid:3493)𝟐𝟐(cid:2878)𝟏𝟐(cid:2878)𝟐𝟐M1*
= 3 unitsA1
Line is 𝐫 = 3𝐢+𝐣−5𝐤+𝜆(2𝐢+𝑏𝐣+3𝐤)M1
(2𝜆−1)𝐢+𝑏𝜆𝐣+(3𝜆−5)𝐤A1 vector from P to a point on the line
(cid:3493)(2𝜆−1)(cid:2870)+(𝑏𝜆)(cid:2870)+(3𝜆−5)(cid:2870)M1* finding magnitude of this vector
(cid:3493)(13+𝑏(cid:2870))𝜆(cid:2870)−34𝜆+26A1 A1
(cid:3493)(13+𝑏(cid:2870))𝜆(cid:2870)−34𝜆+26 = 3M1dep setting distances equal
(13+𝑏(cid:2870))𝜆(cid:2870)−34𝜆+17 = 0A1 correct quadratic equation = 0.
So (−34)(cid:2870)−4(13+𝑏(cid:2870))(17)= 0M1 setting discriminant equal to 0
 𝑏 = 2A1
Alternative method 3
AnswerMarks
(cid:2870)×(cid:2872)(cid:2878)(cid:2869)×(cid:2869)(cid:2878)(cid:2868)×(cid:2870)
Distance from point to plane =
AnswerMarks Guidance
√(cid:2870)(cid:3118)(cid:2878)(cid:2869)(cid:3118)(cid:2878)(cid:2870)(cid:3118)M1*
= 3 unitsA1
Line is 𝐫 = 3𝐢+𝐣−5𝐤+𝜆(2𝐢+𝑏𝐣+3𝐤)M1
𝑨(cid:4652)(cid:4652)(cid:4652)(cid:4652)(cid:4652)𝑷(cid:4652)⃗ = 𝐢+5𝐤A1 or −𝑨(cid:4652)(cid:4652)(cid:4652)(cid:4652)(cid:4652)𝑷(cid:4652)⃗. Could be seen as part of distance calculation.
1 2 1 2
(cid:3437)0(cid:3441)∙(cid:3437)𝑏(cid:3441)=(cid:3629)(cid:3437)0(cid:3441)(cid:3629)(cid:3629)(cid:3437)𝑏(cid:3441)(cid:3629)cos𝜃
AnswerMarks Guidance
5 3 5 3M1 M1
1×2+5×3 = (cid:3493)1(cid:2870)+5(cid:2870)(cid:3493)2(cid:2870)+𝑏(cid:2870)+3(cid:2870)cos𝜃M1 evaluating scalar product and magnitudes
17
cos𝜃 =
AnswerMarks Guidance
√26√13+𝑏(cid:2870)A1 expression for cos𝜃
17
√26cos𝜃 =
AnswerMarks Guidance
√13+𝑏(cid:2870)M1* expression for the distance from (3, 1, -5) to the foot of the
perpendicular to the line
2
17
26−(cid:4678) (cid:4679) = 32
AnswerMarks Guidance
(cid:3493)13+𝑏2M1dep using their value correctly with Pythagoras oe to lead to a
value for 𝑏using their value correctly with Pythagoras oe to lead to a
value for 𝑏
AnswerMarks
 𝑏 = 2A1
[10]
Question 15:
15 | DR
1+2𝑥−𝑥(cid:2870) = −(𝑥(cid:2870)−2𝑥)+1 = −(𝑥−1)(cid:2870)+2
𝑥−1 (cid:2870)
(cid:3428)arcsin (cid:3432)
𝑘
(cid:2869)
𝑥−1 (cid:2870)
= (cid:3428)arcsin (cid:3432)
√2
(cid:2869)
1 𝜋
= arcsin(cid:3436) (cid:3440)−arcsin0 =
√2 4 | M1
A1
M1
A1
A1 | 3.1a
1.1
2.1
2.1
2.2a | completing the square
substitution or evaluation of both limits must be seen
Alternative method
1+2𝑥−𝑥(cid:2870) = −(𝑥(cid:2870)−2𝑥)+1 = −(𝑥−1)(cid:2870)+2 | M1 | completing the square
A1
Let 𝑢 = 𝑥−1
1 1
(cid:3505) d𝑢
√2−𝑢2
0 | M1 | M1 | complete substitution including limits | complete substitution including limits
𝑢 (cid:2869)
= (cid:3428)arcsin (cid:3432)
√2
(cid:2868) | A1
1 𝜋
= arcsin(cid:3436) (cid:3440)−arcsin0 =
√2 4 | A1 | substitution or evaluation of both limits must be seen
[5]
|(cid:2870)×(cid:2872)(cid:2878)(cid:2869)×(cid:2869)(cid:2878)(cid:2868)×(cid:2870)|
Distance from point to plane =
√(cid:2870)(cid:3118)(cid:2878)(cid:2869)(cid:3118)(cid:2878)(cid:2870)(cid:3118)
= 3 units
Line is 𝐫 = 3𝐢+𝐣−5𝐤+𝝀(2𝐢+𝑏𝐣+3𝐤)
𝑨(cid:4652)(cid:4652)(cid:4652)(cid:4652)(cid:4652)𝑷(cid:4652)⃗ = 𝐢+5𝐤
𝑨(cid:4652)(cid:4652)(cid:4652)(cid:4652)(cid:4652)𝑷(cid:4652)⃗×𝒅= (𝐢+5𝐤)×(2𝐢+𝑏𝐣+3𝐤)
= −5𝑏𝐢+7𝐣+𝑏𝐤
dist from P to line
|−5𝑏𝐢+7𝐣+𝑏𝐤| √26𝑏(cid:2870)+49
=
|𝒅| √13+𝑏(cid:2870)
(cid:2870)(cid:2874)(cid:3029)(cid:3118)(cid:2878)(cid:2872)(cid:2877)
so =9  26𝑏(cid:2870)+49= 117+9𝑏(cid:2870)
(cid:2869)(cid:2871)(cid:2878)(cid:3029)(cid:3118)
 𝑏 = 2 | M1*
A1
M1
A1
M1
A1
M1*
A1
M1dep
A1 | 3.1a
1.1
3.1a
2.1
2.1
2.1
3.1a
1.1
2.1
3.2a | or −𝑨(cid:4652)(cid:4652)(cid:4652)(cid:4652)(cid:4652)𝑷(cid:4652)⃗. Could be seen as part of distance calculation.
or 𝒅×𝑨(cid:4652)(cid:4652)(cid:4652)(cid:4652)(cid:4652)𝑷(cid:4652)⃗
equating the two distances
Alternative method 1
|(cid:2870)×(cid:2872)(cid:2878)(cid:2869)×(cid:2869)(cid:2878)(cid:2868)×(cid:2870)|
Distance from point to plane =
√(cid:2870)(cid:3118)(cid:2878)(cid:2869)(cid:3118)(cid:2878)(cid:2870)(cid:3118) | M1*
= 3 units | A1
Line is 𝐫 = 3𝐢+𝐣−5𝐤+𝜆(2𝐢+𝑏𝐣+3𝐤) | M1
(2𝜆−1)𝐢+𝑏𝜆𝐣+(3𝜆−5)𝐤 | A1 | vector from P to a point on the line
2𝜆−1 2
(cid:3437) 𝑏𝜆 (cid:3441)∙(cid:3437)𝑏(cid:3441)=0
3𝜆−5 3 | M1* | M1*
17
λ =
13+𝑏(cid:2870) | A1
dist from P to line | M1 | substituting 𝜆 into (cid:3627)𝑨(cid:4652)(cid:4652)(cid:4652)(cid:4652)(cid:4652)𝑷(cid:4652)⃗(cid:3627)
(cid:2870) (cid:2870) (cid:2870)
(cid:3496)(cid:4672)2(cid:4672) (cid:2869)(cid:2875) (cid:4673)−1(cid:4673) +(cid:3436)𝑏(cid:4672) (cid:2869)(cid:2875) (cid:4673)(cid:3440) +(cid:4672)3(cid:4672) (cid:2869)(cid:2875) (cid:4673)−5(cid:4673)
(cid:2869)(cid:2871)(cid:2878)(cid:3029)(cid:3118) (cid:2869)(cid:2871)(cid:2878)(cid:3029)(cid:3118) (cid:2869)(cid:2871)(cid:2878)(cid:3029)(cid:3118)
√26𝑏(cid:2872)+387𝑏(cid:2870)+637
=
13+𝑏(cid:2870) | A1 | simplified
(cid:2870)(cid:2874)(cid:3029)(cid:3120)(cid:2878)(cid:2871)(cid:2876)(cid:2875)(cid:3029)(cid:3118)(cid:2878)(cid:2874)(cid:2871)(cid:2875)
So = 9
((cid:2869)(cid:2871)(cid:2878)(cid:3029)(cid:3118))(cid:3118) | M1dep | equating the two distances
17𝑏(cid:2872)+153𝑏(cid:2870)−884 = 0
 𝑏 = 2 | A1
Alternative method 2
|𝟐×𝟒(cid:2878)𝟏×𝟏(cid:2878)𝟎×𝟐|
Distance from point to plane =
(cid:3493)𝟐𝟐(cid:2878)𝟏𝟐(cid:2878)𝟐𝟐 | M1*
= 3 units | A1
Line is 𝐫 = 3𝐢+𝐣−5𝐤+𝜆(2𝐢+𝑏𝐣+3𝐤) | M1
(2𝜆−1)𝐢+𝑏𝜆𝐣+(3𝜆−5)𝐤 | A1 | vector from P to a point on the line
(cid:3493)(2𝜆−1)(cid:2870)+(𝑏𝜆)(cid:2870)+(3𝜆−5)(cid:2870) | M1* | finding magnitude of this vector | finding magnitude of this vector
(cid:3493)(13+𝑏(cid:2870))𝜆(cid:2870)−34𝜆+26 | A1 | A1 | expanding and simplifying
(cid:3493)(13+𝑏(cid:2870))𝜆(cid:2870)−34𝜆+26 = 3 | M1dep | setting distances equal
(13+𝑏(cid:2870))𝜆(cid:2870)−34𝜆+17 = 0 | A1 | correct quadratic equation = 0.
So (−34)(cid:2870)−4(13+𝑏(cid:2870))(17)= 0 | M1 | setting discriminant equal to 0
 𝑏 = 2 | A1
Alternative method 3
|(cid:2870)×(cid:2872)(cid:2878)(cid:2869)×(cid:2869)(cid:2878)(cid:2868)×(cid:2870)|
Distance from point to plane =
√(cid:2870)(cid:3118)(cid:2878)(cid:2869)(cid:3118)(cid:2878)(cid:2870)(cid:3118) | M1*
= 3 units | A1
Line is 𝐫 = 3𝐢+𝐣−5𝐤+𝜆(2𝐢+𝑏𝐣+3𝐤) | M1
𝑨(cid:4652)(cid:4652)(cid:4652)(cid:4652)(cid:4652)𝑷(cid:4652)⃗ = 𝐢+5𝐤 | A1 | or −𝑨(cid:4652)(cid:4652)(cid:4652)(cid:4652)(cid:4652)𝑷(cid:4652)⃗. Could be seen as part of distance calculation.
1 2 1 2
(cid:3437)0(cid:3441)∙(cid:3437)𝑏(cid:3441)=(cid:3629)(cid:3437)0(cid:3441)(cid:3629)(cid:3629)(cid:3437)𝑏(cid:3441)(cid:3629)cos𝜃
5 3 5 3 | M1 | M1 | scalar product formula with 𝑨(cid:4652)(cid:4652)(cid:4652)(cid:4652)(cid:4652)𝑷(cid:4652)⃗ ∙𝒅 to find a value for cos𝜃
1×2+5×3 = (cid:3493)1(cid:2870)+5(cid:2870)(cid:3493)2(cid:2870)+𝑏(cid:2870)+3(cid:2870)cos𝜃 | M1 | evaluating scalar product and magnitudes
17
cos𝜃 =
√26√13+𝑏(cid:2870) | A1 | expression for cos𝜃
17
√26cos𝜃 =
√13+𝑏(cid:2870) | M1* | expression for the distance from (3, 1, -5) to the foot of the
perpendicular to the line
2
17
26−(cid:4678) (cid:4679) = 32
(cid:3493)13+𝑏2 | M1dep | using their value correctly with Pythagoras oe to lead to a
value for 𝑏 | using their value correctly with Pythagoras oe to lead to a
value for 𝑏
 𝑏 = 2 | A1
[10]
15 In this question you must show detailed reasoning.
Evaluate $\int _ { 1 } ^ { 2 } \frac { 1 } { \sqrt { 1 + 2 x - x ^ { 2 } } } d x$, giving your answer in terms of $\pi$.

\hfill \mbox{\textit{OCR MEI Further Pure Core 2023 Q15 [5]}}
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