21M.1.SL.TZ1.5

$f\text{'}\left(x\right)=-2\left(x-h\right)$          A1

 

[1 mark]

$g\text{'}\left(x\right)={\mathrm{e}}^{x-2}$  OR  $g\text{'}\left(3\right)={\mathrm{e}}^{3-2}$ (may be seen anywhere)          A1

 

Note: The derivative of $g$ must be explicitly seen, either in terms of $x$ or $3$.

 

recognizing $f\text{'}\left(3\right)=g\text{'}\left(3\right)$          (M1)

$-2\left(3-h\right)={\mathrm{e}}^{3-2} \left(=e\right)$

$-6+2h=\text{e}$  OR  $3-h=-\frac{\text{e}}{2}$          A1

 

Note: The final A1 is dependent on one of the previous marks being awarded.

 

$h=\frac{\text{e}+6}{2}$          AG

 

[3 marks]

$f\left(3\right)=g\left(3\right)$          (M1)

$-{\left(3-h\right)}^2+2k={\text{e}}^{3-2}+k$

correct equation in $k$

 

EITHER

$-{\left(3-\frac{\text{e}+6}{2}\right)}^2+2k={\text{e}}^{3-2}+k$          A1

$k=\text{e}+{\left(\frac{6-\text{e}-6}{2}\right)}^2 \left(=\text{e}+{\left(\frac{-\text{e}}{2}\right)}^2\right)$          A1

 

OR

$k=\text{e}+{\left(3-\frac{\text{e}+6}{2}\right)}^2$          A1

$k=\text{e}+9-3\text{e}-18+\frac{{\text{e}}^2+12\text{e}+36}{4}$          A1

 

THEN

$k=\text{e}+\frac{{\text{e}}^2}{4}$          AG

 

[3 marks]

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