Date | May 2018 | Marks available | 7 | Reference code | 18M.1.hl.TZ1.7 |
Level | HL only | Paper | 1 | Time zone | TZ1 |
Command term | Find | Question number | 7 | Adapted from | N/A |
Question
Let \(y = {\text{arccos}}\left( {\frac{x}{2}} \right)\)
Find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\).
Find \(\int_0^1 {{\text{arccos}}\left( {\frac{x}{2}} \right){\text{d}}x} \).
Markscheme
\(y = {\text{arccos}}\left( {\frac{x}{2}} \right) \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = - \frac{1}{{2\sqrt {1 - {{\left( {\frac{x}{2}} \right)}^2}} }}\left( { = - \frac{1}{{\sqrt {4 - {x^2}} }}} \right)\) M1A1
Note: M1 is for use of the chain rule.
[2 marks]
attempt at integration by parts M1
\(u = {\text{arccos}}\left( {\frac{x}{2}} \right) \Rightarrow \frac{{{\text{d}}u}}{{{\text{d}}x}} = - \frac{1}{{\sqrt {4 - {x^2}} }}\)
\(\frac{{{\text{d}}v}}{{{\text{d}}x}} = 1 \Rightarrow v = x\) (A1)
\(\int_0^1 {{\text{arccos}}\left( {\frac{x}{2}} \right){\text{d}}x} = \left[ {x\,\,{\text{arccos}}\left( {\frac{x}{2}} \right)} \right]_0^1 + \int_0^1 {\frac{1}{{\sqrt {4 - {x^2}} }}} dx\) A1
using integration by substitution or inspection (M1)
\(\left[ {x\,\,{\text{arccos}}\left( {\frac{x}{2}} \right)} \right]_0^1 + \left[ { - {{\left( {4 - {x^2}} \right)}^{\frac{1}{2}}}} \right]_0^1\) A1
Note: Award A1 for \({ - {{\left( {4 - {x^2}} \right)}^{\frac{1}{2}}}}\) or equivalent.
Note: Condone lack of limits to this point.
attempt to substitute limits into their integral M1
\( = \frac{\pi }{3} - \sqrt 3 + 2\) A1
[7 marks]