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]