2020 H2 Mathematics Paper 1 Question 9

Graphs and Transformations

Answers

The lines

{y=2x}

and

{y=-\frac{1}{2}x}

are perpendicular

{\left( 0, \frac{9}{2} \right]}

{\left( \frac{\pi}{2} - \frac{4}{3} \ln 2 \right) \textrm{ units}^2}

The line

{y=2x}

has gradient

{2}

and is at an angle of

{\tan^{-1}(2)}

anti-clockwise with respect to the positive

{x\textrm{-}}

axis.

The line

{y=-\frac{1}{2}x}

has gradient

{-\frac{1}{2}}

and is at an angle of

{\tan^{-1}\left(-\frac{1}{2}\right)}

anti-clockwise with respect to the positive

{x\textrm{-}}

axis.

The two lines are perpendicular since

{m_1m_2 = -1.}

Hence

\tan^{-1}(2)- \tan^{-1}\left( -{\textstyle\frac{1}{2}} \right) = {\textstyle\frac{1}{2}}\pi \; \blacksquare

\begin{gather*} \frac{1}{x^2+1} = \frac{k}{3x+4} \\ kx^2 + k = 3x + 4 \\ kx^2 - 3x + k - 4 = 0 \end{gather*}

For

{C_1}

and

{C_2}

to intersect,

\begin{align*} 3^2 - 4(k)(k-4) &\geq 0 \\ 9 - 4k^2 + 16k &\geq 0 \\ 4 k^2 - 16 k - 9 &\leq 0 \\ (2 k + 1)(2 k - 9) &\leq 0 \end{align*}

- \frac{1}{2} \leq k \leq \frac{9}{2}

Since

{k>0,}

0 < k \leq \frac{9}{2}

Set of values of

{k:}

\left( 0, \frac{9}{2} \right] \; \blacksquare

\begin{align*} & \textrm{Area of region} \\ & = \int_{-\frac{1}{2}}^2 \left( \frac{1}{x^2 + 1} - \frac{2}{3 x + 4} \right) \; \mathrm{d}x \\ & = \left[ \tan^{-1} x - \frac{2}{3} \ln \left| 3 x + 4\right| \right]_{-\frac{1}{2}}^2 \\ & = \tan^{-1}(2)- \tan^{-1}\left( -{\textstyle\frac{1}{2}} \right) - \frac{2}{3}\ln 10 + \frac{2}{3} \ln \frac{5}{2} \\ & = \frac{\pi}{2} - \frac{2}{3} \ln 4 \\ & = \left( \frac{\pi}{2} - \frac{4}{3} \ln 2 \right) \textrm{ units}^2 \; \blacksquare \end{align*}