2 Find the exact value of \(\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 3 + 4 x - 4 x ^ { 2 } } } \mathrm {~d} x\).
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Question 2:
Answer Marks
Guidance
Answer Marks
Guidance
\(3 + 4x - 4x^2 = 4 - (2x-1)^2\) M1
Complete the square
EITHER Solution 1: \(2x - 1 = 2\sin\theta\); \(\sqrt{4 - 4\sin^2\theta} = 2\cos\theta\)(M1)
Use appropriate substitution
Integral \(= \int \frac{\cos\theta}{2\cos\theta}d\theta = \frac{1}{2}\sin^{-1}\!\left(\frac{2x-1}{2}\right)\) M1A1
Integrate
OR Solution 2: \(\int \frac{1}{\sqrt{1-\left(x - \frac{1}{2}\right)^2}}\,dx\)(M1)
Use formula
\(= \frac{1}{2}\sin^{-1}\!\left(x - \frac{1}{2}\right)\) M1A1
\(\frac{1}{2}\!\left(\sin^{-1}(0.5) - \sin^{-1}(-0.5)\right)\) M1
Use limits
\(= \dfrac{\pi}{6}\) A1
CAO; NOTE: answer of \(\frac{\pi}{6}\) without working scores 0 (as per front cover instructions)
6
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# Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3 + 4x - 4x^2 = 4 - (2x-1)^2$ | M1 | Complete the square |
| **EITHER** Solution 1: $2x - 1 = 2\sin\theta$; $\sqrt{4 - 4\sin^2\theta} = 2\cos\theta$ | (M1) | Use appropriate substitution |
| Integral $= \int \frac{\cos\theta}{2\cos\theta}d\theta = \frac{1}{2}\sin^{-1}\!\left(\frac{2x-1}{2}\right)$ | M1A1 | Integrate |
| **OR** Solution 2: $\int \frac{1}{\sqrt{1-\left(x - \frac{1}{2}\right)^2}}\,dx$ | (M1) | Use formula |
| $= \frac{1}{2}\sin^{-1}\!\left(x - \frac{1}{2}\right)$ | M1A1 | |
| $\frac{1}{2}\!\left(\sin^{-1}(0.5) - \sin^{-1}(-0.5)\right)$ | M1 | Use limits |
| $= \dfrac{\pi}{6}$ | A1 | CAO; NOTE: answer of $\frac{\pi}{6}$ without working scores 0 (as per front cover instructions) |
| | **6** | |
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2 Find the exact value of $\int _ { 0 } ^ { 1 } \frac { 1 } { \sqrt { 3 + 4 x - 4 x ^ { 2 } } } \mathrm {~d} x$.\\
\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q2 [6]}}