Sketch the graph of $y=f\left(x\right)$ for $-\frac{5\pi }{8}⩽x⩽\frac{\pi }{8}$.

With reference to your graph, explain why $f$ is a function on the given domain.

Explain why $f$ has no inverse on the given domain.

Explain why $f$ is not a function for $-\frac{3\pi }{4}⩽x⩽\frac{\pi }{4}$.

Show that $g\left(t\right)={\left(\frac{1+t}{1-t}\right)}^2$.

Sketch the graph of $y=g\left(t\right)$ for t ≤ 0. Give the coordinates of any intercepts and the equations of any asymptotes.

Find $\alpha$ and β in terms of $k$.

Show that $\alpha$ + β < −2.

     A1A1

A1 for correct concavity, many to one graph, symmetrical about the midpoint of the domain and with two axes intercepts.

Note: Axes intercepts and scales not required.

A1 for correct domain

[2 marks]

for each value of $x$ there is a unique value of $f\left(x\right)$      A1

Note: Accept “passes the vertical line test” or equivalent.

[1 mark]

no inverse because the function fails the horizontal line test or equivalent      R1

Note: No FT if the graph is in degrees (one-to-one).

[1 mark]

the expression is not valid at either of $x=\frac{\pi }{4}\left(\text{or}-\frac{3\pi }{4}\right)$       R1

[1 mark]

METHOD 1

$f\left(x\right)=\frac{\text{tan}\left(x+\frac{\pi }{4}\right)}{\text{tan}\left(\frac{\pi }{4}-x\right)}$     M1

$=\frac{\frac{\text{tan}x+\text{tan}\frac{\pi }{4}}{1-\text{tan}x\text{tan}\frac{\pi }{4}}}{\frac{\text{tan}\frac{\pi }{4}-\text{tan}x}{1+\text{tan}\frac{\pi }{4}\text{tan}x}}$      M1A1

$={\left(\frac{1+t}{1-t}\right)}^2$      AG

METHOD 2

$f\left(x\right)=\text{tan}\left(x+\frac{\pi }{4}\right)\text{tan}\left(\frac{\pi }{2}-\frac{\pi }{4}+x\right)$      (M1)

$=\text{ta}{\text{n}}^2\left(x+\frac{\pi }{4}\right)$     A1

$g\left(t\right)={\left(\frac{\text{tan}x+\text{tan}\frac{\pi }{4}}{1-\text{tan}x\text{tan}\frac{\pi }{4}}\right)}^2$     A1

$={\left(\frac{1+t}{1-t}\right)}^2$      AG

[3 marks]

for t ≤ 0, correct concavity with two axes intercepts and with asymptote $y$ = 1      A1

t intercept at (−1, 0)      A1

$y$ intercept at (0, 1)       A1

[3 marks]

METHOD 1

$\alpha$, β satisfy $\frac{{\left(1+t\right)}^2}{{\left(1-t\right)}^2}=k$     M1

$1+t^2+2t=k\left(1+t^2-2t\right)$     A1

$\left(k-1\right)t^2-2\left(k+1\right)t+\left(k-1\right)=0$     A1

attempt at using quadratic formula      M1

$\alpha$, β $=\frac{k+1\pm 2\sqrt{k}}{k-1}$ or equivalent     A1

METHOD 2

$\alpha$, β satisfy $\frac{1+t}{1-t}=(\pm )\sqrt{k}$      M1

$t+\sqrt{k}t=\sqrt{k}-1$      M1

$t=\frac{\sqrt{k}-1}{\sqrt{k}+1}$ (or equivalent)      A1

$t-\sqrt{k}t=-\left(\sqrt{k}+1\right)$     M1

$t=\frac{\sqrt{k}+1}{\sqrt{k}-1}$ (or equivalent)       A1

so for eg, $\alpha =\frac{\sqrt{k}-1}{\sqrt{k}+1}$, β $=\frac{\sqrt{k}+1}{\sqrt{k}-1}$

[5 marks]

$\alpha$ + β $=2\frac{\left(k+1\right)}{\left(k-1\right)}\left(=-2\frac{\left(1+k\right)}{\left(1-k\right)}\right)$     A1

since $1+k>1-k$     R1

$\alpha$ + β < −2     AG

Note: Accept a valid graphical reasoning.

[2 marks]