1 Find the exact value of \(\int _ { 2 } ^ { \frac { 7 } { 2 } } \frac { 1 } { \sqrt { 4 x - x ^ { 2 } - 1 } } \mathrm {~d} x\).
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Question 1:
Answer Marks
Guidance
Answer Marks
Guidance
\(-x^2 + 4x - 1 = 3 - (x-2)^2\) B1 Completes the square.
\(\int_2^{\frac{7}{2}} \frac{1}{\sqrt{3-(x-2)^2}}\, dx = \left[\sin^{-1}\left(\frac{x-2}{\sqrt{3}}\right)\right]_2^{\frac{7}{2}}\) M1 A1 Applies formula. \(\left[\sin^{-1}\left(\frac{x-2}{\sqrt{3}}\right)\right]_2^{\frac{7}{2}} = \left[-\cos^{-1}\left(\frac{x-2}{\sqrt{3}}\right)\right]_2^{\frac{7}{2}}\)
\(\sin^{-1}\!\left(\tfrac{1}{2}\sqrt{3}\right) - \sin^{-1}(0) = \tfrac{1}{3}\pi\) M1 A1 Uses limits. Answer must be in radians.
Total: 5
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## Question 1:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-x^2 + 4x - 1 = 3 - (x-2)^2$ | **B1** | Completes the square. |
| $\int_2^{\frac{7}{2}} \frac{1}{\sqrt{3-(x-2)^2}}\, dx = \left[\sin^{-1}\left(\frac{x-2}{\sqrt{3}}\right)\right]_2^{\frac{7}{2}}$ | **M1 A1** | Applies formula. $\left[\sin^{-1}\left(\frac{x-2}{\sqrt{3}}\right)\right]_2^{\frac{7}{2}} = \left[-\cos^{-1}\left(\frac{x-2}{\sqrt{3}}\right)\right]_2^{\frac{7}{2}}$ |
| $\sin^{-1}\!\left(\tfrac{1}{2}\sqrt{3}\right) - \sin^{-1}(0) = \tfrac{1}{3}\pi$ | **M1 A1** | Uses limits. Answer must be in radians. |
| | **Total: 5** | |
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1 Find the exact value of $\int _ { 2 } ^ { \frac { 7 } { 2 } } \frac { 1 } { \sqrt { 4 x - x ^ { 2 } - 1 } } \mathrm {~d} x$.\\
\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q1 [5]}}