2007 H2 Math Paper 1 Answers

{x < - 3, \, -2 < x < -1 \, \textrm{ or } x > 7.}

{fg}

does not exist because

R_g = {\left[ 0, \infty \right]} \not \subseteq {\left( -\infty , 3 \right) \cup \left( 3 , \infty \right)} = D_f.

\displaystyle gf : x \mapsto \left( \frac{1}{ x - 3 } \right)^2, \quad \allowbreak {\textrm{for } x \in \mathbb{R}, x \neq 3.}

f^{-1}(x) = \frac{1}{ x } + 3.

D_{ f^{-1} } = R_f = \left( -\infty, 0 \right) \cup \left( 0, \infty \right).

{z = - 1 + 2 \mathrm{i}.}

{y = \frac{2}{3} + \frac{4}{3} \mathrm{e}^{- \frac{3}{4} x}.}

{y \to \frac{2}{3}.}

{y = 2 + \frac{ 3 }{ x + 2 }}

Transformation 1: Translate by 2 units in the negative

{x}

-axis direction.
Transformation 2: Scale by a factor of

{3}

parallel to the

{y}

-axis.
Transformation 3: Translate by 2 units in the positive

{y}

-axis direction.

{\frac{1}{3} ( 4 \mathbf{i} + 2 \mathbf{j} + 5 \mathbf{k} )}

{\sqrt{35}}

{k = 64.}

{z^6 + 64} = {(z^2 - 2 \sqrt{3} z + 4)}\allowbreak{(z^2 + 2 \sqrt{3} z + 4)}{(z^2 + 4)}.

{\left( \frac{5}{2} , \frac{3}{2} , \frac{11}{2} \right).}

{78.8^{\circ}.}

{\frac{4}{7} \sqrt{14}\textrm{ units}.}

\pi \left( \frac{1}{2} \mathrm{e}^{10} - \frac{1}{2} \mathrm{e}^{8} - 24 \mathrm{e}^{5} + 18 \mathrm{e}^{4} + 183 \right) \allowbreak \textrm{ units}^3.

{\alpha = 0.619, \, \beta = 1.512.}

{0.619 < x < 1.512.}

{0.18\textrm{ units}^2}

The geometric series is convergent because

{-1 < r = \frac{2}{3} < 1.}

{S_\infty = 3 a.}

{\{ n \in \mathbb{Z}: 6 \leq n \leq 13 \}.}

{\frac{2}{5} \textrm{ units}^2.}