IB Math Analysis & Approaches Questionbank-Topic: SL 2.5 Composite functions SL Paper 1

Question

The graph of y = f (x) for -4 ≤ x ≤ 6 is shown in the following diagram.

 
 (a)        Write down the value of

 (i)       f (2) ;

(ii)      ( f o f )(2) .                                                                                                                                                             [2]

 (b)        Let g(x) = \(\frac{1}{2} f (x) +1\) for -4 ≤ x ≤ 6 . On the axes above, sketch the graph of g .                   [3]

Answer/Explanation

Ans:

(a) (i) f(2) = 6

(ii) (fof)2=− 2 [2 marks]

(b)

 

Question

Let \(f(x) = \ln (x + 5) + \ln 2\) , for \(x > – 5\) .

Find \({f^{ – 1}}(x)\) .

[4]
a.

Let \(g(x) = {{\rm{e}}^x}\) .

Find \((g \circ f)(x)\) , giving your answer in the form \(ax + b\) , where \(a,b \in \mathbb{Z}\) .

[3]
b.
Answer/Explanation

Markscheme

METHOD 1

\(\ln (x + 5) + \ln 2 = \ln (2(x + 5))\) \(( = \ln (2x + 10))\)     (A1)

interchanging x and y (seen anywhere)     (M1)

e.g. \(x = \ln (2y + 10)\)

evidence of correct manipulation     (A1)

e.g. \({{\rm{e}}^x} = 2y + 10\)

\({f^{ – 1}}(x) = \frac{{{{\rm{e}}^x} – 10}}{2}\)    A1     N2

METHOD 2

\(y = \ln (x + 5) + \ln 2\)

\(y – \ln 2 = ln(x + 5)\)     (A1)

evidence of correct manipulation     (A1)

e.g. \({{\rm{e}}^{y – \ln 2}} = x + 5\)

interchanging x and y (seen anywhere)     (M1)

e.g. \({{\rm{e}}^{x – \ln 2}} = y + 5\)

\({f^{ – 1}}(x) = {{\rm{e}}^{x – \ln 2}} – 5\)     A1     N2

[4 marks]

a.

METHOD 1

evidence of composition in correct order     (M1)

e.g. \((g \circ f)(x) = g(\ln (x + 5) + \ln 2)\)

\( = {{\rm{e}}^{\ln (2(x + 5))}} = 2(x + 5)\)

\((g \circ f)(x) = 2x + 10\)     A1A1     N2

METHOD 2

evidence of composition in correct order     (M1)

e.g. \((g \circ f)(x) = {{\rm{e}}^{\ln (x + 5) + \ln 2}}\)

\( = {{\rm{e}}^{\ln (x + 5)}} \times {{\rm{e}}^{\ln 2}} = (x + 5)2\)

\((g \circ f)(x) = 2x + 10\)     A1A1     N2

[3 marks]

b.

Question

Let \(f(x) = {x^2}\) and \(g(x) = 2x – 3\) .

Find \({g^{ – 1}}(x)\) .

[2]
a.

Find \((f \circ g)(4)\) .

[3]
b.
Answer/Explanation

Markscheme

for interchanging x and y (may be done later)     (M1)

e.g. \(x = 2y – 3\)

\({g^{ – 1}}(x) = \frac{{x + 3}}{2}\) (accept \(y = \frac{{x + 3}}{2},\frac{{x + 3}}{2}\) )     A1     N2

[2 marks]

a.

METHOD 1

\(g(4) = 5\)     (A1)

evidence of composition of functions     (M1)

\(f(5) = 25\)     A1 N3

METHOD 2

\(f \circ g(x) = {(2x – 3)^2}\)     (M1)

\(f \circ g(4) = {(2 \times 4 – 3)^2}\)     (A1)

= 25     A1     N3

[3 marks]

b.

Question

Let \(f(x) = 2{x^3} + 3\) and \(g(x) = {{\rm{e}}^{3x}} – 2\) .

(i)     Find \(g(0)\) .

(ii)    Find \((f \circ g)(0)\) .

[5]
a.

Find \({f^{ – 1}}(x)\) .

[3]
b.
Answer/Explanation

Markscheme

(i) \(g(0) = {{\rm{e}}^0} – 2\)     (A1)

\( = – 1\)     A1     N2

(ii) METHOD 1

substituting answer from (i)     (M1)

e.g. \((f \circ g)(0) = f( – 1)\)

correct substitution \(f( – 1) = 2{( – 1)^3} + 3\)     (A1)

\(f( – 1) = 1\)     A1     N3

METHOD 2

attempt to find \((f \circ g)(x)\)     (M1)

e.g. \((f \circ g)(x) = f({{\rm{e}}^{3x}} – 2)\) \( = 2{({{\rm{e}}^{3x}} – 2)^3} + 3\)

correct expression for \((f \circ g)(x)\)     (A1)

e.g. \(2{({{\rm{e}}^{3x}} – 2)^3} + 3\)

\((f \circ g)(0) = 1\)     A1     N3

[5 marks]

a.

interchanging x and y (seen anywhere)     (M1)

e.g. \(x = 2{y^3} + 3\)

attempt to solve     (M1)

e.g. \({y^3} = \frac{{x – 3}}{2}\)

\({f^{ – 1}}(x) = \sqrt[3]{{\frac{{x – 3}}{2}}}\)     A1     N3

[3 marks]

b.

Question

Let \(f(x) = lo{g_3}\sqrt x \) , for \(x > 0\) .

Show that \({f^{ – 1}}(x) = {3^{2x}}\) .

[2]
a.

Write down the range of \({f^{ – 1}}\) .

[1]
b.

Let \(g(x) = {\log _3}x\) , for \(x > 0\) .

Find the value of \(({f^{ – 1}} \circ g)(2)\) , giving your answer as an integer.

[4]
c.
Answer/Explanation

Markscheme

interchanging x and y (seen anywhere)     (M1)

e.g. \(x = \log \sqrt y \) (accept any base)

evidence of correct manipulation     A1

e.g. \(3^x = \sqrt y \) , \({3^y} = {x^{\frac{1}{2}}}\) , \(x = \frac{1}{2}{\log _3}y\) , \(2y = {\log _3}x\)

\({f^{ – 1}}(x) = {3^{2x}}\)     AG     N0 

[2 marks]

a.

\(y > 0\) , \({f^{ – 1}}(x) > 0\)     A1     N1

[1 mark]

b.

METHOD 1

finding \(g(2) = lo{g_3}2\) (seen anywhere)     A1

attempt to substitute     (M1)

e.g. \(({f^{ – 1}} \circ g)(2) = {3^{2\log {_3}2}}\)

evidence of using log or index rule     (A1)

e.g. \(({f^{ – 1}} \circ g)(2) = {3^{\log {_3}4}}\) , \({3^{{{\log }_3}2^2}}\)

\(({f^{ – 1}} \circ g)(2) = 4\)     A1     N1

METHOD 2

attempt to form composite (in any order)     (M1)

e.g. \(({f^{ – 1}} \circ g)(x) = {3^{2{{\log }_3}x}}\)

evidence of using log or index rule     (A1)

e.g. \(({f^{ – 1}} \circ g)(x) = {3^{{{\log }_3}{x^2}}}\) , \({3^{{{\log }_3}{x^{}}}}^2\)

\(({f^{ – 1}} \circ g)(x) = {x^2}\)     A1

\(({f^{ – 1}} \circ g)(2) = 4\)     A1     N1

[4 marks]

c.

Question

Let \(f(x) = \cos 2x\) and \(g(x) = 2{x^2} – 1\) .

Find \(f\left( {\frac{\pi }{2}} \right)\) .

[2]
a.

Find \((g \circ f)\left( {\frac{\pi }{2}} \right)\) .

[2]
b.

Given that \((g \circ f)(x)\) can be written as \(\cos (kx)\) , find the value of k, \(k \in \mathbb{Z}\) .

[3]
c.
Answer/Explanation

Markscheme

\(f\left( {\frac{\pi }{2}} \right) = \cos \pi \)     (A1)

\( = – 1\)     A1     N2

[2 marks]

a.

\((g \circ f)\left( {\frac{\pi }{2}} \right) = g( – 1)\) \(( = 2{( – 1)^2} – 1)\)    (A1)

\(= 1\)     A1     N2

[2 marks]

b.

\((g \circ f)(x) = 2{(\cos (2x))^2} – 1\) \(( = 2{\cos ^2}(2x) – 1)\)     A1

evidence of \(2{\cos ^2}\theta – 1 = \cos 2\theta \) (seen anywhere)     (M1)

\((g \circ f)(x) = \cos 4x\)

\(k = 4\)     A1     N2

[3 marks]

c.

Question

Let \(f(x) = {x^2} + 4\) and \(g(x) = x – 1\) .

Find \((f \circ g)(x)\) .

[2]
a.

The vector \(\left( {\begin{array}{*{20}{c}}
3\\
{ – 1}
\end{array}} \right)\) translates the graph of \((f \circ g)\) to the graph of h .

Find the coordinates of the vertex of the graph of h .

[3]
b.

The vector \(\left( {\begin{array}{*{20}{c}}
3\\
{ – 1}
\end{array}} \right)\) translates the graph of \((f \circ g)\) to the graph of h .

Show that \(h(x) = {x^2} – 8x + 19\) .

[2]
c.

The vector \(\left( {\begin{array}{*{20}{c}}
3\\
{ – 1}
\end{array}} \right)\) translates the graph of \((f \circ g)\) to the graph of h .

The line \(y = 2x – 6\) is a tangent to the graph of h at the point P. Find the x-coordinate of P.

[5]
d.
Answer/Explanation

Markscheme

attempt to form composition (in any order)     (M1)

\((f \circ g)(x) = {(x – 1)^2} + 4\)    \(({x^2} – 2x + 5)\)     A1     N2

[2 marks]

a.

METHOD 1

vertex of \(f \circ g\) at (1, 4)     (A1)

evidence of appropriate approach     (M1)

e.g. adding \(\left( {\begin{array}{*{20}{c}}
3\\
{ – 1}
\end{array}} \right)\) to the coordinates of the vertex of \(f \circ g\)

vertex of h at (4, 3)     A1     N3

METHOD 2

attempt to find \(h(x)\)     (M1)

e.g. \({((x – 3) – 1)^2} + 4 – 1\) , \(h(x) = (f \circ g)(x – 3) – 1\)

\(h(x) = {(x – 4)^2} + 3\)     (A1)

vertex of h at (4, 3)     A1     N3

[3 marks]

b.

evidence of appropriate approach     (M1)

e.g. \({(x – 4)^2} + 3\) ,\({(x – 3)^2} – 2(x – 3) + 5 – 1\)

simplifying     A1

e.g. \(h(x) = {x^2} – 8x + 16 + 3\) , \({x^2} – 6x + 9 – 2x + 6 + 4\)

\(h(x) = {x^2} – 8x + 19\)     AG     N0

[2 marks]

c.

METHOD 1

equating functions to find intersection point     (M1)

e.g. \({x^2} – 8x + 19 = 2x – 6\) , \(y = h(x)\)

\({x^2} – 10x + 25 + 0\)     A1

evidence of appropriate approach to solve     (M1)

e.g. factorizing, quadratic formula

appropriate working     A1

e.g. \({(x – 5)^2} = 0\)

\(x = 5\)  \((p = 5)\)     A1     N3

METHOD 2

attempt to find \(h'(x)\)     (M1)

\(h(x) = 2x – 8\)     A1

recognizing that the gradient of the tangent is the derivative     (M1)

e.g. gradient at \(p = 2\)

\(2x – 8 = 2\)  \((2x = 10)\)     A1

\(x = 5\)     A1     N3

[5 marks]

d.

Question

Let \(f(x) = 7 – 2x\) and \(g(x) = x + 3\) .

Find \((g \circ f)(x)\) .

[2]
a.

Write down \({g^{ – 1}}(x)\) .

[1]
b.

Find \((f \circ {g^{ – 1}})(5)\) .

[2]
c.
Answer/Explanation

Markscheme

attempt to form composite     (M1)

e.g. \(g(7 – 2x)\) , \(7 – 2x + 3\)

\((g \circ f)(x) = 10 – 2x\)     A1     N2

[2 marks]

a.

\({g^{ – 1}}(x) = x – 3\)     A1     N1

[1 mark]

b.

METHOD 1

valid approach     (M1)

e.g. \({g^{ – 1}}(5)\) , \(2\) , \(f(5)\)

\(f(2) = 3\)     A1     N2

METHOD 2

attempt to form composite of f and \({g^{ – 1}}\)     (M1)

e.g. \((f \circ {g^{ – 1}})(x) = 7 – 2(x – 3)\) , \(13 – 2x\)

\((f \circ {g^{ – 1}})(5) = 3\)     A1     N2

[2 marks]

c.

Question

Let \(f(x) = 2x – 1\) and  \(g(x) = 3{x^2} + 2\) .

Find \({f^{ – 1}}(x)\) . 

[3]
a.

Find \((f \circ g)(1)\) . 

[3]
b.
Answer/Explanation

Markscheme

interchanging x and y (seen anywhere)     (M1)

e.g. \(x = 2y – 1\)

correct manipulation     (A1)

e.g. \(x + 1 = 2y\)

\({f^{ – 1}}(x) = \frac{{x + 1}}{2}\)      A1     N2

[3 marks]

a.

METHOD 1

attempt to find or \(g(1)\) or \(f(1)\)     (M1)

\(g(1) = 5\)     (A1)

\(f(5) = 9\)     A1     N2 

[3 marks]

METHOD 2

attempt to form composite (in any order)     (M1)

e.g. \(2(3{x^2} + 2) – 1\) , \(3{(2x – 1)^2} + 2\)

\((f \circ g)(1) = 2(3 \times {1^2} + 2) – 1\) \(( = 6 \times {1^2} + 3)\)     (A1)

\((f \circ g)(1) = 9\)     A1     N2

[3 marks]

b.

Question

Let \(f(x) = \sqrt {x – 5} \) , for \(x \ge 5\) .

Find \({f^{ – 1}}(2)\) .

[3]
a.

Let \(g\) be a function such that \({g^{ – 1}}\) exists for all real numbers. Given that \(g(30) = 3\) , find \((f \circ {g^{ – 1}})(3)\)  .

[3]
b.
Answer/Explanation

Markscheme

METHOD 1

attempt to set up equation     (M1)

eg   \(2 = \sqrt {y – 5} \) , \(2 = \sqrt {x – 5} \)

correct working     (A1)

eg   \(4 = y – 5\) , \(x = {2^2} + 5\)

\({f^{ – 1}}(2) = 9\)     A1     N2

METHOD 2

interchanging \(x\) and \(y\) (seen anywhere)     (M1)

eg   \(x = \sqrt {y – 5} \)

correct working     (A1)

eg   \({x^2} = y – 5\) , \(y = {x^2} + 5\)

\({f^{ – 1}}(2) = 9\)     A1     N2

[3 marks]

a.

recognizing \({g^{ – 1}}(3) = 30\)     (M1)

eg   \(f(30)\)

correct working     (A1)

eg   \((f \circ {g^{ – 1}})(3) = \sqrt {30 – 5} \) , \(\sqrt {25} \)

\((f \circ {g^{ – 1}})(3) = 5\)     A1     N2

Note: Award A0 for multiple values, eg \( \pm 5\) .

[3 marks]

b.

Question

Let \(f(x) = 4x – 2\) and \(g(x) = – 2{x^2} + 8\) .

Find \({f^{ – 1}}(x)\) .

[3]
a.

Find \((f \circ g)(1)\) .

[3]
b.
Answer/Explanation

Markscheme

interchanging \(x\) and \(y\) (seen anywhere)     (M1)

eg   \(x = 4y – 2\)

evidence of correct manipulation     (A1)

eg   \(x + 2 = 4y\)

\({f^{ – 1}}(x) = \frac{{x + 2}}{4}\) (accept \(y = \frac{{x + 2}}{4}\) , \(\frac{{x + 2}}{4}\) , \({f^{ – 1}}(x) = \frac{1}{4}x + \frac{1}{2}\)     A1     N2

[3 marks]

a.

METHOD 1

attempt to substitute \(1\) into \(g(x)\)     (M1)

eg   \(g(1) =  – 2 \times {1^2} + 8\)

\(g(1) = 6\)     (A1)

\(f(6) = 22\)     A1     N3

METHOD 2

attempt to form composite function (in any order)     (M1)

eg   \((f \circ g)(x) = 4( – 2{x^2} + 8) – 2\) \(( =  – 8{x^2} + 30)\)

correct substitution

eg   \((f \circ g)(1) = 4( – 2 \times {1^2} + 8) – 2\) , \( – 8 + 30\)

\(f(6) = 22\)     A1     N3

[3 marks]

b.
 

Question

Let \(f(x) = 3x – 2\) and \(g(x) = \frac{5}{{3x}}\), for \(x \ne 0\).

Let \(h(x) = \frac{5}{{x + 2}}\), for \(x \geqslant 0\). The graph of h has a horizontal asymptote at \(y = 0\).

Find \({f^{ – 1}}(x)\).

[2]
a.

Show that \(\left( {g \circ {f^{ – 1}}} \right)(x) = \frac{5}{{x + 2}}\).

[2]
b.

Find the \(y\)-intercept of the graph of \(h\).

[2]
c(i).

Hence, sketch the graph of \(h\).

[3]
c(ii).

For the graph of \({h^{ – 1}}\), write down the \(x\)-intercept;

[1]
d(i).

For the graph of \({h^{ – 1}}\), write down the equation of the vertical asymptote.

[1]
d(ii).

Given that \({h^{ – 1}}(a) = 3\), find the value of \(a\).

[3]
e.
Answer/Explanation

Markscheme

interchanging \(x\) and \(y\)     (M1)

eg     \(x = 3y – 2\)

\({f^{ – 1}}(x) = \frac{{x + 2}}{3}{\text{   }}\left( {{\text{accept }}y = \frac{{x + 2}}{3},{\text{ }}\frac{{x + 2}}{3}} \right)\)     A1     N2

[2 marks]

a.

attempt to form composite (in any order)     (M1)

eg     \(g\left( {\frac{{x + 2}}{3}} \right),{\text{ }}\frac{{\frac{5}{{3x}} + 2}}{3}\)

correct substitution     A1

eg     \(\frac{5}{{3\left( {\frac{{x + 2}}{3}} \right)}}\)

\(\left( {g \circ {f^{ – 1}}} \right)(x) = \frac{5}{{x + 2}}\)     AG     N0

[2 marks]

b.

valid approach     (M1)

eg     \(h(0),{\text{ }}\frac{5}{{0 + 2}}\)

\(y = \frac{5}{2}{\text{   }}\left( {{\text{accept (0, 2.5)}}} \right)\)     A1     N2

[2 marks]

c(i).

     A1A2     N3

Notes:     Award A1 for approximately correct shape (reciprocal, decreasing, concave up).

     Only if this A1 is awarded, award A2 for all the following approximately correct features: y-intercept at \((0, 2.5)\), asymptotic to x-axis, correct domain \(x \geqslant 0\).

     If only two of these features are correct, award A1.

[3 marks]

c(ii).

\(x = \frac{5}{2}{\text{   }}\left( {{\text{accept (2.5, 0)}}} \right)\)     A1     N1

[1 mark]

d(i).

\(x = 0\)   (must be an equation)     A1     N1

[1 mark]

d(ii).

METHOD 1

attempt to substitute \(3\) into \(h\) (seen anywhere)     (M1)

eg     \(h(3),{\text{ }}\frac{5}{{3 + 2}}\)

correct equation     (A1)

eg     \(a = \frac{5}{{3 + 2}},{\text{ }}h(3) = a\)

\(a = 1\)     A1     N2

[3 marks]

METHOD 2

attempt to find inverse (may be seen in (d))     (M1)

eg     \(x = \frac{5}{{y + 2}},{\text{ }}{h^{ – 1}} = \frac{5}{x} – 2,{\text{ }}\frac{5}{x} + 2\)

correct equation, \(\frac{5}{x} – 2 = 3\)     (A1)

\(a = 1\)     A1     N2

[3 marks]

e.

Question

The following diagram shows the graph of a function \(f\).

Find \({f^{ – 1}}( – 1)\).

[2]
a.

Find \((f \circ f)( – 1)\).

[3]
b.

On the same diagram, sketch the graph of \(y = f( – x)\).

[2]
c.
Answer/Explanation

Markscheme

valid approach     (M1)

eg\(\;\;\;\)horizontal line on graph at \( – 1,{\text{ }}f(a) =  – 1,{\text{ }}( – 1,5)\)

\({f^{ – 1}}( – 1) = 5\)     A1     N2

[2 marks]

a.

attempt to find \(f( – 1)\)     (M1)

eg\(\;\;\;\)line on graph

\(f( – 1) = 2\)     (A1)

\((f \circ f)( – 1) = 1\)     A1     N3

[3 marks]

b.

     A1A1     N2

Note:     The shape must be an approximately correct shape (concave down and increasing). Only if the shape is approximately correct, award the following for points in circles:

A1 for the \(y\)-intercept,

A1 for any two of these points \(( – 5,{\text{ }} – 1),{\text{ }}( – 2,{\text{ }}1),{\text{ }}(1,{\text{ }}2)\).

[2 marks]

Total [7 marks]

c.

Question

Let \(f(x) = {(x – 5)^3}\), for \(x \in \mathbb{R}\).

Find \({f^{ – 1}}(x)\).

[3]
a.

Let \(g\) be a function so that \((f \circ g)(x) = 8{x^6}\). Find \(g(x)\).

[3]
b.
Answer/Explanation

Markscheme

interchanging \(x\) and \(y\) (seen anywhere)     (M1)

eg\(\;\;\;x = {(y – 5)^3}\)

evidence of correct manipulation     (A1)

eg\(\;\;\;y – 5 = \sqrt[3]{x}\)

\({f^{ – 1}}(x) = \sqrt[3]{x} + 5\;\;\;({\text{accept }}5 + {x^{\frac{1}{3}}},{\text{ }}y = 5 + \sqrt[3]{x})\)     A1     N2

Notes:     If working shown, and they do not interchange \(x\) and \(y\), award A1A1M0 for \(\sqrt[3]{y} + 5\).

If no working shown, award N1 for \(\sqrt[3]{y} + 5\).

a.

METHOD 1

attempt to form composite (in any order)     (M1)

eg\(\;\;\;g\left( {{{(x – 5)}^3}} \right),{\text{ }}{\left( {g(x) – 5} \right)^3} = 8{x^6},{\text{ }}f(2{x^2} + 5)\)

correct working     (A1)

eg\(\;\;\;g – 5 = 2{x^2},{\text{ }}{\left( {(2{x^2} + 5) – 5} \right)^3}\)

\(g(x) = 2{x^2} + 5\)     A1     N2

METHOD 2

recognising inverse relationship     (M1)

eg\(\;\;\;{f^{ – 1}}(8{x^6}) = g(x),{\text{ }}{f^{ – 1}}(f \circ g)(x) = {f^{ – 1}}(8{x^6})\)

correct working

eg\(\;\;\;g(x) = \sqrt[3]{{(8{x^6})}} + 5\)     (A1)

\(g(x) = 2{x^2} + 5\)     A1     N2

b.

Question

Let \(f(x) = 8x + 3\) and \(g(x) = 4x\), for \(x \in \mathbb{R}\).

Write down \(g(2)\).

[1]
a.

Find \((f \circ g)(x)\).

[2]
b.

Find \({f^{ – 1}}(x)\).

[2]
c.
Answer/Explanation

Markscheme

\(g(2) = 8\)    A1     N1

[1 mark]

a.

attempt to form composite (in any order)     (M1)

eg\(\,\,\,\,\,\)\(f(4x),{\text{ }}4 \times (8x + 3)\)

\((f \circ g)(x) = 32x + 3\)     A1     N2

[2 marks]

b.

interchanging \(x\) and \(y\) (may be seen at any time)     (M1)

eg\(\,\,\,\,\,\)\(x = 8y + 3\)

\({f^{ – 1}}(x) = \frac{{x – 3}}{8}\,\,\,\,\,\left( {{\text{accept }}\frac{{x – 3}}{8},{\text{ }}y = \frac{{x – 3}}{8}} \right)\)     A1     N2

[2 marks]

c.

Question

Let \(f(x) = 6x\sqrt {1 – {x^2}} \), for \( – 1 \leqslant x \leqslant 1\), and \(g(x) = \cos (x)\), for \(0 \leqslant x \leqslant \pi \).

Let \(h(x) = (f \circ g)(x)\).

Write \(h(x)\) in the form \(a\sin (bx)\), where \(a,{\text{ }}b \in \mathbb{Z}\).

[5]
a.

Hence find the range of \(h\).

[2]
b.
Answer/Explanation

Markscheme

attempt to form composite in any order     (M1)

eg\(\,\,\,\,\,\)\(f\left( {g(x)} \right),{\text{ }}\cos \left( {6x\sqrt {1 – {x^2}} } \right)\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(6\cos x\sqrt {1 – {{\cos }^2}x} \)

correct application of Pythagorean identity (do not accept \({\sin ^2}x + {\cos ^2}x = 1\))     (A1)

eg\(\,\,\,\,\,\)\({\sin ^2}x = 1 – {\cos ^2}x,{\text{ }}6\cos x\sin x,{\text{ }}6\cos x \left| \sin x\right|\)

valid approach (do not accept \(2\sin x\cos x = \sin 2x\))     (M1)

eg\(\,\,\,\,\,\)\(3(2\cos x\sin x)\)

\(h(x) = 3\sin 2x\)    A1     N3

[5 marks]

a.

valid approach     (M1)

eg\(\,\,\,\,\,\)amplitude \( = 3\), sketch with max and min \(y\)-values labelled, \( – 3 < y < 3\)

correct range     A1     N2

eg\(\,\,\,\,\,\)\( – 3 \leqslant y \leqslant 3\), \([ – 3,{\text{ }}3]\) from \( – 3\) to 3

Note:     Do not award A1 for \( – 3 < y < 3\) or for “between \( – 3\) and 3”.

[2 marks]

b.

Question

Let \(f(x) = 5x\) and \(g(x) = {x^2} + 1\), for \(x \in \mathbb{R}\).

Find \({f^{ – 1}}(x)\).

[2]
a.

Find \((f \circ g)(7)\).

[3]
b.
Answer/Explanation

Markscheme

interchanging \(x\) and \(x\)     (M1)

eg\(\,\,\,\,\,\)\(x = 5y\)

\({f^{ – 1}}\left( x \right) = \frac{x}{5}\)     A1     N2

[2 marks]

a.

METHOD 1

attempt to substitute 7 into \(g(x)\) or \(f(x)\)     (M1)

eg\(\,\,\,\,\,\)\({7^2} + 1,{\text{ }}5 \times 7\)

\(g(7) = 50\)     (A1)

\(f\left( {50} \right) = 250\)     A1     N2

METHOD 2

attempt to form composite function (in any order)     (M1)

eg\(\,\,\,\,\,\)\(5({x^2} + 1),{\text{ }}{(5x)^2} + 1\)

correct substitution     (A1)

eg\(\,\,\,\,\,\)\(5 \times ({7^2} + 1)\)

\((f \circ g)(7) = 250\)     A1     N2

[3 marks]

b.

Examiners report

[N/A]

a.

[N/A]

b.

Question

Let \(f(x) = 1 + {{\text{e}}^{ – x}}\) and \(g(x) = 2x + b\), for \(x \in \mathbb{R}\), where \(b\) is a constant.

Find \((g \circ f)(x)\).

[2]
a.

Given that \(\mathop {\lim }\limits_{x \to  + \infty } (g \circ f)(x) =  – 3\), find the value of \(b\).

[4]
b.
Answer/Explanation

Markscheme

attempt to form composite     (M1)

eg\(\,\,\,\,\,\)\(g(1 + {{\text{e}}^{ – x}})\)

correct function     A1     N2

eg\(\,\,\,\,\,\)\((g \circ f)(x) = 2 + b + 2{{\text{e}}^{ – x}},{\text{ }}2(1 + {{\text{e}}^{ – x}}) + b\)

[2 marks]

a.

evidence of \(\mathop {\lim }\limits_{x \to \infty } (2 + b + 2{{\text{e}}^{ – x}}) = 2 + b + \mathop {\lim }\limits_{x \to \infty } (2{{\text{e}}^{ – x}})\)     (M1)

eg\(\,\,\,\,\,\)\(2 + b + 2{{\text{e}}^{ – \infty }}\), graph with horizontal asymptote when \(x \to \infty \)

Note:     Award M0 if candidate clearly has incorrect limit, such as \(x \to 0,{\text{ }}{{\text{e}}^\infty },{\text{ }}2{{\text{e}}^0}\).

evidence that \({{\text{e}}^{ – x}} \to 0\) (seen anywhere)     (A1)

eg\(\,\,\,\,\,\)\(\mathop {\lim }\limits_{x \to \infty } ({{\text{e}}^{ – x}}) = 0,{\text{ }}1 + {{\text{e}}^{ – x}} \to 1,{\text{ }}2(1) + b =  – 3,{\text{ }}{{\text{e}}^{{\text{large negative number}}}} \to 0\), graph of \(y = {{\text{e}}^{ – x}}\) or

\(y = 2{{\text{e}}^{ – x}}\) with asymptote \(y = 0\), graph of composite function with asymptote \(y =  – 3\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(2 + b =  – 3\)

\(b =  – 5\)     A1     N2

[4 marks]

b.

Examiners report

[N/A]

a.

[N/A]

b.

Question

Consider a function \(f\). The line L1 with equation \(y = 3x + 1\) is a tangent to the graph of \(f\) when \(x = 2\)

Let \(g\left( x \right) = f\left( {{x^2} + 1} \right)\) and P be the point on the graph of \(g\) where \(x = 1\).

Write down \(f’\left( 2 \right)\).

[2]
a.i.

Find \(f\left( 2 \right)\).

[2]
a.ii.

Show that the graph of g has a gradient of 6 at P.

[5]
b.

Let L2 be the tangent to the graph of g at P. L1 intersects L2 at the point Q.

Find the y-coordinate of Q.

[7]
c.
Answer/Explanation

Markscheme

recognize that \(f’\left( x \right)\) is the gradient of the tangent at \(x\)     (M1)

eg   \(f’\left( x \right) = m\)

\(f’\left( 2 \right) = 3\)  (accept m = 3)     A1 N2

[2 marks]

a.i.

recognize that \(f\left( 2 \right) = y\left( 2 \right)\)     (M1)

eg  \(f\left( 2 \right) = 3 \times 2 + 1\)

\(f\left( 2 \right) = 7\)     A1 N2

[2 marks]

a.ii.

recognize that the gradient of the graph of g is \(g’\left( x \right)\)      (M1)

choosing chain rule to find \(g’\left( x \right)\)      (M1)

eg  \(\frac{{{\text{d}}y}}{{{\text{d}}u}} \times \frac{{{\text{d}}u}}{{{\text{d}}x}},\,\,u = {x^2} + 1,\,\,u’ = 2x\)

\(g’\left( x \right) = f’\left( {{x^2} + 1} \right) \times 2x\)     A2

\(g’\left( 1 \right) = 3 \times 2\)     A1

\(g’\left( 1 \right) = 6\)     AG N0

[5 marks]

b.

 at Q, L1L2 (seen anywhere)      (M1)

recognize that the gradient of L2 is g’(1)  (seen anywhere)     (M1)
eg  m = 6

finding g (1)  (seen anywhere)      (A1)
eg  \(g\left( 1 \right) = f\left( 2 \right),\,\,g\left( 1 \right) = 7\)

attempt to substitute gradient and/or coordinates into equation of a straight line      M1
eg  \(y – g\left( 1 \right) = 6\left( {x – 1} \right),\,\,y – 1 = g’\left( 1 \right)\left( {x – 7} \right),\,\,7 = 6\left( 1 \right) + {\text{b}}\)

correct equation for L2 

eg  \(y – 7 = 6\left( {x – 1} \right),\,\,y = 6x + 1\)     A1

correct working to find Q       (A1)
eg   same y-intercept, \(3x = 0\)

\(y = 1\)     A1 N2

[7 marks]

c.

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