Show that the tangent to the curve \(y = f(x)\) at the point \((1,{\text{ }}0)\) is normal to the curve \(y = g(x)\) at the point \((1,{\text{ }}0)\).

Find \(g(x)\).

Use Euler’s method with steps of \(0.2\) to estimate \(f(2)\) to \(5\) decimal places.

Explain why \(y = f(x)\) cannot cross the isocline \(x - {y^2} = 0\), for \(x > 1\).

(i)     Sketch the isoclines \(x - {y^2} =  - 2,{\text{ }}0,{\text{ }}1\).

(ii)     On the same set of axes, sketch the graph of \(f\).

gradient of \(f\) at \((1,{\text{ }}0)\) is \(1 - {0^2} = 1\) and the gradient of \(g\) at \((1,{\text{ }}0)\) is \(0 - {1^2} =  - 1\)     A1

so gradient of normal is \(1\)     A1

= Gradient of the tangent of \(f\) at \((1,{\text{ }}0)\)     AG

[2 marks]

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} - y =  - {x^2}\)

integrating factor is \({{\text{e}}^{\int { - 1{\text{d}}x} }} = {{\text{e}}^{ - x}}\)     M1

\(y{{\text{e}}^{ - x}} = \int { - {x^2}{{\text{e}}^{ - x}}{\text{d}}x} \)     A1

\( = {x^2}{{\text{e}}^{ - x}} - \int {2x{{\text{e}}^{ - x}}{\text{d}}x} \)     M1

\( = {x^2}{{\text{e}}^{ - x}} + 2x{{\text{e}}^{ - x}} - \int {2{{\text{e}}^{ - x}}{\text{d}}x} \)

\( = {x^2}{{\text{e}}^{ - x}} + 2x{{\text{e}}^{ - x}} + 2{{\text{e}}^{ - x}} + c\)     A1

Note:     Condone missing \( + c\) at this stage.

\( \Rightarrow g(x) = {x^2} + 2x + 2 + c{{\text{e}}^x}\)

\(g(1) = 0 \Rightarrow c =  - \frac{5}{{\text{e}}}\)     M1

\( \Rightarrow g(x) = {x^2} + 2x + 2 - 5{{\text{e}}^{x - 1}}\)     A1

[6 marks]

use of \({y_{n + 1}} = {y_n} + hf'({x_n},{\text{ }}{y_n})\)     (M1)

\({x_0} = 1,{\text{ }}{y_0} = 0\)

\({x_1} = 1.2,{\text{ }}{y_1} = 0.2\)     A1

\({x_2} = 1.4,{\text{ }}{y_2} = 0.432\)     (M1)(A1)

\({x_3} = 1.6,{\text{ }}{y_3} = 0.67467 \ldots \)

\({x_4} = 1.8,{\text{ }}{y_4} = 0.90363 \ldots \)

\({x_5} = 2,{\text{ }}{y_5} = 1.1003255 \ldots \)

answer \( = 1.10033\)     A1     N3

Note:     Award A0 or N1 if \(1.10\) given as answer.

[5 marks]

at the point \((1,{\text{ }}0)\), the gradient of \(f\) is positive so the graph of \(f\) passes into the first quadrant for \(x > 1\)

in the first quadrant below the curve \(x - {y^2} = 0\) the gradient of \(f\) is positive     R1

the curve \(x - {y^2} = 0\) has positive gradient in the first quadrant     R1

if \(f\) were to reach \(x - {y^2} = 0\) it would have gradient of zero, and therefore would not cross     R1

[3 marks]

(i) and (ii)

     A4

Note:     Award A1 for 3 correct isoclines.

Award A1 for \(f\) not reaching \(x - {y^2} = 0\).

Award A1 for turning point of \(f\) on \(x - {y^2} = 0\).

Award A1 for negative gradient to the left of the turning point.

Note:     Award A1 for correct shape and position if curve drawn without any isoclines.

[4 marks]

Total [20 marks]