Question 3 - A-Level Maths

OCR FP2 2008 June — Question 3 6 marks

Exam Board	OCR
Module	FP2 (Further Pure Mathematics 2)
Year	2008
Session	June
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Integration using inverse trig and hyperbolic functions
Type	Half-angle tangent substitution t = tan(x/2)
Difficulty	Challenging +1.2 This is a standard Further Maths FP2 question testing the half-angle tangent substitution (Weierstrass substitution), which is a bookwork technique. While it requires knowing the substitution formulas (cos x = (1-t²)/(1+t²), dx = 2dt/(1+t²)) and completing an integration that leads to arctan, it's a direct application of a taught method with no novel problem-solving required. The definite integral with exact answer adds minor complexity but follows a standard template for this topic.
Spec	1.08h Integration by substitution

3 By using the substitution $t = \tan \frac { 1 } { 2 } x$, find the exact value of $$\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { 1 } { 2 - \cos x } \mathrm {~d} x$$ giving the answer in terms of $\pi$.

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Answer	Marks
(i) $\frac{d}{dx}(x^3 y) = x^3 \frac{dy}{dx} + 2xy$ or $\frac{d}{dx}(x^2 y) = 2xy \frac{dy}{dx} + y^2$	*B1
Attempt to solve their differentiated equation for $\frac{dy}{dx}$	dep*M1
$\frac{dy}{dx} = \frac{y^2 - 2xy}{x^2 - 2xy}$ only	A1
(ii)(a) Attempt to solve only $y^2 - 2xy = 0$ & derive $y = 2x$	B1
Clear indication why $y = 0$ is not acceptable	B1
(b) Attempt to solve $y = 2x$ simulti with $x^2 - y - xy^2 = 2$	M1
Produce $-2x^3 = 2$ or $y^3 = -8$	A1
$(-1, -2)$ or $x = -1, y = -2$ only	A1

| | |
|---|---|
| (i) $\frac{d}{dx}(x^3 y) = x^3 \frac{dy}{dx} + 2xy$ or $\frac{d}{dx}(x^2 y) = 2xy \frac{dy}{dx} + y^2$ | *B1 |
| Attempt to solve their differentiated equation for $\frac{dy}{dx}$ | dep*M1 |
| $\frac{dy}{dx} = \frac{y^2 - 2xy}{x^2 - 2xy}$ only | A1 |
| (ii)(a) Attempt to solve only $y^2 - 2xy = 0$ & derive $y = 2x$ | B1 |
| Clear indication why $y = 0$ is not acceptable | B1 |
| (b) Attempt to solve $y = 2x$ simulti with $x^2 - y - xy^2 = 2$ | M1 |
| Produce $-2x^3 = 2$ or $y^3 = -8$ | A1 |
| $(-1, -2)$ or $x = -1, y = -2$ only | A1 |

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3 By using the substitution $t = \tan \frac { 1 } { 2 } x$, find the exact value of

$$\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { 1 } { 2 - \cos x } \mathrm {~d} x$$

giving the answer in terms of $\pi$.

\hfill \mbox{\textit{OCR FP2 2008 Q3 [6]}}

This paper (9 questions)

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Q1 5 Q2 5 Q3 6 Q4 8 Q5 8 Q6 7 Q7 10 Q8 11 Q9 12

Answer	Marks
(i) \(\frac{d}{dx}(x^3 y) = x^3 \frac{dy}{dx} + 2xy\) or \(\frac{d}{dx}(x^2 y) = 2xy \frac{dy}{dx} + y^2\)	*B1
Attempt to solve their differentiated equation for \(\frac{dy}{dx}\)	dep*M1
\(\frac{dy}{dx} = \frac{y^2 - 2xy}{x^2 - 2xy}\) only	A1
(ii)(a) Attempt to solve only \(y^2 - 2xy = 0\) & derive \(y = 2x\)	B1
Clear indication why \(y = 0\) is not acceptable	B1
(b) Attempt to solve \(y = 2x\) simulti with \(x^2 - y - xy^2 = 2\)	M1
Produce \(-2x^3 = 2\) or \(y^3 = -8\)	A1
\((-1, -2)\) or \(x = -1, y = -2\) only	A1