| Date | May 2018 | Marks available | 3 | Reference code | 18M.1.hl.TZ0.11 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Show that | Question number | 11 | Adapted from | N/A |
Question
Given that \(y\) is a function of \(x\), the function \(z\) is given by \(z = \frac{{y - x}}{{y + x}}\), where \(x \in \mathbb{R},\,\,x \ne 3,\,\,y + x \ne 0\).
Show that \(\frac{{{\text{d}}z}}{{{\text{d}}x}} = \frac{2}{{{{\left( {y + x} \right)}^2}}}\left( {x\frac{{{\text{d}}y}}{{{\text{d}}x}} - y} \right)\).
Show that the differential equation \(f\left( x \right)\left( {x\frac{{{\text{d}}y}}{{{\text{d}}x}} - y} \right) = {y^2} - {x^2}\) can be written as \(f\left( x \right)\frac{{{\text{d}}z}}{{{\text{d}}x}} = 2z\).
Hence show that the solution to the differential equation \(\left( {x - 3} \right)\left( {x\frac{{{\text{d}}y}}{{{\text{d}}x}} - y} \right) = {y^2} - {x^2}\) given that \(x = 4\) when \(y = 5\) is \(\frac{{y - x}}{{y + x}} = {\left( {\frac{{x - 3}}{3}} \right)^2}\).
Markscheme
\(z = \frac{{y - x}}{{y + x}}\)
\( \Rightarrow \frac{{{\text{d}}z}}{{{\text{d}}x}} = \frac{{\left( {y + x} \right)\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}} - 1} \right) - \left( {y - x} \right)\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}} + 1} \right)}}{{{{\left( {y + x} \right)}^2}}}\) M1A1
\( \Rightarrow \frac{{{\text{d}}z}}{{{\text{d}}x}} = \frac{{y\frac{{{\text{d}}y}}{{{\text{d}}x}} + x\frac{{{\text{d}}y}}{{{\text{d}}x}} - y - x - y\frac{{{\text{d}}y}}{{{\text{d}}x}} + x\frac{{{\text{d}}y}}{{{\text{d}}x}} - y + x}}{{{{\left( {y + x} \right)}^2}}}\) A1
\( \Rightarrow \frac{{{\text{d}}z}}{{{\text{d}}x}} = \frac{2}{{{{\left( {y + x} \right)}^2}}}\left( {x\frac{{{\text{d}}y}}{{{\text{d}}x}} - y} \right)\) AG
[3 marks]
\(f\left( x \right)\left( {\frac{{{{\left( {y + x} \right)}^2}}}{2}} \right)\frac{{{\text{d}}z}}{{{\text{d}}x}} = {y^2} - {x^2}\) (M1)
\(f\left( x \right)\frac{{{\text{d}}z}}{{{\text{d}}x}} = 2\frac{{\left( {y - x} \right)\left( {y + x} \right)}}{{{{\left( {y + x} \right)}^2}}}\) A1
\(f\left( x \right)\frac{{{\text{d}}z}}{{{\text{d}}x}} = 2\frac{{\left( {y - x} \right)}}{{\left( {y + x} \right)}} = 2z\) AG
[2 marks]
METHOD 1
\(f\left( x \right)\frac{{{\text{d}}z}}{{{\text{d}}x}} = 2z\)
\(\frac{1}{z}\frac{{{\text{d}}z}}{{{\text{d}}x}} = \frac{2}{{f\left( x \right)}}\)
\(\frac{1}{z}\frac{{{\text{d}}z}}{{{\text{d}}x}} = \frac{2}{{x - 3}}\) M1A1
EITHER
\( \Rightarrow {\text{ln}}\,z = 2\,{\text{ln}}\left( {x - 3} \right) + c\) A1
when \(y = 5,\,x = 4 \Rightarrow z = \frac{1}{9}\) M1
\( \Rightarrow c = {\text{ln}}\frac{1}{9}\) A1
\( \Rightarrow {\text{ln}}\,z = 2\,{\text{ln}}\left( {x - 3} \right) + {\text{ln}}\frac{1}{9}\)
\( \Rightarrow {\text{ln}}\,z = {\text{ln}}{\left( {x - 3} \right)^2} - {\text{ln}}\,9\) A1
\( \Rightarrow {\text{ln}}\,z = {\text{ln}}{\left( {\frac{{x - 3}}{3}} \right)^2}\) A1
\( \Rightarrow z = {\left( {\frac{{x - 3}}{3}} \right)^2}\)
OR
\( \Rightarrow {\text{ln}}\,z = 2\,{\text{ln}}\left( {x - 3} \right) + {\text{ln}}\,c\) A1
\(z = c{\left( {x - 3} \right)^2}\) M1A1
when \(y = 5,\,x = 4 \Rightarrow z = \frac{1}{9}\) M1
\( \Rightarrow c = \frac{1}{9}\) A1
THEN
\( \Rightarrow \frac{{y - x}}{{y + x}} = {\left( {\frac{{x - 3}}{3}} \right)^2}\) AG
METHOD 2
\(f\left( x \right)\frac{{{\text{d}}z}}{{{\text{d}}x}} = 2z\)
\(f\left( x \right)\frac{{{\text{d}}z}}{{{\text{d}}x}} - 2z = 0\)
\(\frac{{{\text{d}}z}}{{{\text{d}}x}} - \frac{{2z}}{{x - 3}} = 0\) M1
integrating factor is \({e^{\int {\frac{{ - 2}}{{x - 3}}} {\text{d}}x}}\) A1
\({e^{\int {\frac{{ - 2}}{{x - 3}}} {\text{d}}x}} = {e^{ - 2\,{\text{ln}}\left( {x - 3} \right)}}\)
\( = \frac{1}{{{{\left( {x - 3} \right)}^2}}}\) A1
hence \(\frac{{\text{d}}}{{{\text{d}}x}}\left[ {\frac{z}{{{{\left( {x - 3} \right)}^2}}}} \right] = 0\) M1
\(z = A{\left( {x - 3} \right)^2}\) A1
when when \(y = 5,\,x = 4 \Rightarrow z = \frac{1}{9}\) M1
\( \Rightarrow A = \frac{1}{9}\) A1
\( \Rightarrow \frac{{y - x}}{{y + x}} = {\left( {\frac{{x - 3}}{3}} \right)^2}\) AG
[7 marks]
