19N.1.SL.TZ0.S_10

pestleMathematicsAASLPaper 119N· sl-2-2-functions-notation-domain-range-and-inverse-as-reflectionsource ↗

Let $g\left( x \right) = {p^x} + q$ , for $x{\text{, }}p{\text{, }}q \in \mathbb{R}{\text{, }}p > 1$ . The point ${\text{A}}\left( {0{\text{, }}a} \right)$ lies on the graph of $g$ .

Let $f\left( x \right) = {g^{ - 1}}\left( x \right)$ . The point ${\text{B}}$ lies on the graph of $f$ and is the reflection of point ${\text{A}}$ in the line $y = x$ .

The line ${L_1}$ is tangent to the graph of $f$ at ${\text{B}}$ .

Write down the coordinates of ${\text{B}}$ .

[2]

Given that $f'\left( a \right) = \frac{1}{{{\text{ln}}\,p}}$ , find the equation of ${L_1}$ in terms of $x$ , $p$ and $q$ .

[5]

The line ${L_2}$ is tangent to the graph of $g$ at ${\text{A}}$ and has equation $y = \left( {{\text{ln}}\,p} \right)x + q + 1$ .

The line ${L_2}$ passes through the point $\left( { - 2{\text{, }} - 2} \right)$ .

The gradient of the normal to $g$ at ${\text{A}}$ is $\frac{1}{{{\text{ln}}\left( {\frac{1}{3}} \right)}}$ .

Find the equation of ${L_1}$ in terms of $x$ .

[7]

Markscheme / solution

${\text{B}}\left( {a{\text{, }}0} \right)$ (accept ${\text{B}}\left( {q + 1{\text{, }}0} \right)$ ) A2 N2

[2 marks]

Note: There are many approaches to this part, and the steps may be done in any order. Please check working and award marks in line with the markscheme, noting that candidates may work with the equation of the line before finding $a$ .

FINDING $a$

valid attempt to find an expression for $a$ in terms of $q$ (M1)

$g\left( 0 \right) = a{\text{, }}\,{p^0} + q = a$

$a = q + 1$ (A1)

FINDING THE EQUATION OF ${L_1}$

EITHER

attempt to substitute tangent gradient and coordinates into equation of straight line (M1)

eg $y - 0 = f'\left( a \right)\left( {x - a} \right){\text{, }}\,y = f'\left( a \right)\left( {x - \left( {q + 1} \right)} \right)$

correct equation in terms of $a$ and $p$ (A1)

eg $y - 0 = \frac{1}{{{\text{ln}}\left( p \right)}}\left( {x - a} \right)$

attempt to substitute tangent gradient and coordinates to find $b$

eg $0 = \frac{1}{{{\text{ln}}\left( p \right)}}\left( a \right) + b$

$b = \frac{{ - a}}{{{\text{ln}}\left( p \right)}}$ (A1)

THEN (must be in terms of both $p$ and $q$ )

$y = \frac{1}{{{\text{ln}}\,p}}\left( {x - q - 1} \right){\text{, }}\,y = \frac{1}{{{\text{ln}}\,p}}x - \frac{{q + 1}}{{{\text{ln}}\,p}}$ A1 N3

Note: Award A0 for final answers in the form ${L_1} = \frac{1}{{{\text{ln}}\,p}}\left( {x - q - 1} \right)$

[5 marks]

Note: There are many approaches to this part, and the steps may be done in any order. Please check working and award marks in line with the markscheme, noting that candidates may find $q$ in terms of $p$ before finding a value for $p$ .

FINDING $p$

valid approach to find the gradient of the tangent (M1)

eg ${m_1}{m_2} = - 1{\text{, }}\,\, - \frac{1}{{\frac{1}{{{\text{ln}}\left( {\frac{1}{3}} \right)}}}}{\text{, }}\,\, - {\text{ln}}\left( {\frac{1}{3}} \right){\text{, }}\,\, - \frac{1}{{{\text{ln}}\,p}} = \frac{1}{{{\text{ln}}\left( {\frac{1}{3}} \right)}}$

correct application of log rule (seen anywhere) (A1)

eg ${\text{ln}}{\left( {\frac{1}{3}} \right)^{ - 1}}{\text{, }}\,\, - \left( {{\text{ln}}\left( 1 \right) - {\text{ln}}\left( 3 \right)} \right)$

correct equation (seen anywhere) A1

eg ${\text{ln}}\,p = {\text{ln}}\,3{\text{, }}\,\,p = 3$

FINDING $q$

correct substitution of $\left( { - 2{\text{, }} - 2} \right)$ into ${L_2}$ equation (A1)

eg $- 2 = \left( {{\text{ln}}\,p} \right)\left( { - 2} \right) + q + 1$

$q = 2\,{\text{ln}}\,p - 3{\text{, }}\,\,q = 2\,{\text{ln}}\,3 - 3$ (seen anywhere) A1

FINDING ${L_1}$

correct substitution of their $p$ and $q$ into their ${L_1}$ (A1)

eg $y = \frac{1}{{{\text{ln}}\,3}}\left( {x - \left( {2\,{\text{ln}}\,3 - 3} \right) - 1} \right)$

$y = \frac{1}{{{\text{ln}}\,3}}\left( {x - 2\,{\text{ln}}\,3 + 2} \right){\text{, }}\,\,y = \frac{1}{{{\text{ln}}\,3}}x - \frac{{2\,{\text{ln}}\,3 - 2}}{{{\text{ln}}\,3}}$ A1 N2

Note: Award A0 for final answers in the form ${L_1} = \frac{1}{{{\text{ln}}\,3}}\left( {x - 2\,{\text{ln}}\,3 + 2} \right)$ .

[7 marks]

Examiners’ report

[N/A]