Home / IB DP Math AA Topic: SL 2.5: Composite functions f∘g- IB Style Questions: HL Paper 2

IB DP Math AA Topic: SL 2.5: Composite functions f∘g- IB Style Questions: HL Paper 2

Question

The functions f and g are defined for \(x\in \mathbb{R}\; by \; f(x)=6x^2-12x+1\;and\;g(x)=-x+c,\;where\; c\in \mathbb{R}\)
(a) Find the range of f . [2]
(b) Given that (g o f ) (x) ≤ 0 for all \(x\in \mathbb{R}\), determine the set of possible values for c . [4]

▶️Answer/Explanation

Ans:

 (a) By completing the square for $f\left(x\right)=6x^2-12x+1$, we have $f\left(x\right)=6\left(x-1\right)^2-5$. Thus, $\text{R}_f=\left[-5,\infty\right)$.
(b) $$\begin{eqnarray} \left(g\circ f\right)\left(x\right)=c-f\left(x\right) \leq 0. \end{eqnarray}$$ By considering the graph of $y=-f\left(x\right)$, in order for $c-f\left(x\right)\leq 0$ we must have $c\leq -5$.

Question

The function f has inverse \({f^{ – 1}}\) and derivative \(f'(x)\) for all \(x \in \mathbb{R}\). For all functions with these properties you are given the result that for \(a \in \mathbb{R}\) with \(b = f(a)\) and \(f'(a) \ne 0\)

\[({f^{ – 1}})'(b) = \frac{1}{{f'(a)}}.\]

a.Verify that this is true for \(f(x) = {x^3} + 1\) at x = 2.[6]

b.Given that \(g(x) = x{{\text{e}}^{{x^2}}}\), show that \(g'(x) > 0\) for all values of x.[3]

c.Using the result given at the start of the question, find the value of the gradient function of \(y = {g^{ – 1}}(x)\) at x = 2.[4]

 

d.(i)     With f and g as defined in parts (a) and (b), solve \(g \circ f(x) = 2\).

(ii)     Let \(h(x) = {(g \circ f)^{ – 1}}(x)\). Find \(h'(2)\).[6]

 
Answer/Explanation

Markscheme

\(f(2) = 9\)     (A1)

\({f^{ – 1}}(x) = {(x – 1)^{\frac{1}{3}}}\)     A1

\(({f^{ – 1}})'(x) = \frac{1}{3}{(x – 1)^{ – \frac{2}{3}}}\)     (M1)

\(({f^{ – 1}})'(9) = \frac{1}{{12}}\)     A1

\(f'(x) = 3{x^2}\)     (M1)

\(\frac{1}{{f'(2)}} = \frac{1}{{3 \times 4}} = \frac{1}{{12}}\)     A1

Note: The last M1 and A1 are independent of previous marks.

 

[6 marks]

a.

\(g'(x) = {{\text{e}}^{{x^2}}} + 2{x^2}{{\text{e}}^{{x^2}}}\)     M1A1

\(g'(x) > 0\) as each part is positive     R1

[3 marks]

b.

to find the x-coordinate on \(y = g(x)\) solve

\(2 = x{{\text{e}}^{{x^2}}}\)     (M1)

\(x = 0.89605022078 \ldots \)     (A1)

gradient \( = ({g^{ – 1}})'(2) = \frac{1}{{g'(0.896 \ldots )}}\)     (M1)

\( = \frac{1}{{{{\text{e}}^{{{(0.896 \ldots )}^2}}}\left( {1 + 2 \times {{(0.896 \ldots )}^2}} \right)}} = 0.172\) to 3sf     A1

(using the \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) function on gdc \(g'(0.896 \ldots ) = 5.7716028 \ldots \)

\(\frac{1}{{g'(0.896 \ldots )}} = 0.173\)

[4 marks]

c.

(i)     \(({x^3} + 1){{\text{e}}^{{{({x^3} + 1)}^2}}} = 2\)     A1

\(x = – 0.470191 \ldots \)     A1

 

(ii)     METHOD 1

\((g \circ f)'(x) = 3{x^2}{{\text{e}}^{{{({x^3} + 1)}^2}}}\left( {2{{({x^3} + 1)}^2} + 1} \right)\)     (M1)(A1)

\((g \circ f)'( – 0.470191 \ldots ) = 3.85755 \ldots \)     (A1)

\(h'(2) = \frac{1}{{3.85755 \ldots }} = 0.259{\text{ }}(232 \ldots )\)     A1

Note: The solution can be found without the student obtaining the explicit form of the composite function.

 

METHOD 2

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

\(h'(x) = ({f^{ – 1}})’\left( {{g^{ – 1}}(x)} \right) \times ({g^{ – 1}})'(x)\)     M1

\( = \frac{1}{3}{\left( {{g^{ – 1}}(x) – 1} \right)^{ – \frac{2}{3}}} \times ({g^{ – 1}})'(x)\)     M1

\(h'(2) = \frac{1}{3}{\left( {{g^{ – 1}}(2) – 1} \right)^{ – \frac{2}{3}}} \times ({g^{ – 1}})'(2)\)

\( = \frac{1}{3}{(0.89605 \ldots  – 1)^{ – \frac{2}{3}}} \times 0.171933 \ldots \)

\( = 0.259{\text{ }}(232 \ldots )\)     A1     N4

[6 marks]

 

Question

The function f has inverse \({f^{ – 1}}\) and derivative \(f'(x)\) for all \(x \in \mathbb{R}\). For all functions with these properties you are given the result that for \(a \in \mathbb{R}\) with \(b = f(a)\) and \(f'(a) \ne 0\)

\[({f^{ – 1}})'(b) = \frac{1}{{f'(a)}}.\]

a.Verify that this is true for \(f(x) = {x^3} + 1\) at x = 2.[6]

b.Given that \(g(x) = x{{\text{e}}^{{x^2}}}\), show that \(g'(x) > 0\) for all values of x.[3]

c.Using the result given at the start of the question, find the value of the gradient function of \(y = {g^{ – 1}}(x)\) at x = 2.[4]

 

d.(i)     With f and g as defined in parts (a) and (b), solve \(g \circ f(x) = 2\).

(ii)     Let \(h(x) = {(g \circ f)^{ – 1}}(x)\). Find \(h'(2)\).[6]

 
Answer/Explanation

Markscheme

\(f(2) = 9\)     (A1)

\({f^{ – 1}}(x) = {(x – 1)^{\frac{1}{3}}}\)     A1

\(({f^{ – 1}})'(x) = \frac{1}{3}{(x – 1)^{ – \frac{2}{3}}}\)     (M1)

\(({f^{ – 1}})'(9) = \frac{1}{{12}}\)     A1

\(f'(x) = 3{x^2}\)     (M1)

\(\frac{1}{{f'(2)}} = \frac{1}{{3 \times 4}} = \frac{1}{{12}}\)     A1

Note: The last M1 and A1 are independent of previous marks.

 

[6 marks]

a.

\(g'(x) = {{\text{e}}^{{x^2}}} + 2{x^2}{{\text{e}}^{{x^2}}}\)     M1A1

\(g'(x) > 0\) as each part is positive     R1

[3 marks]

b.

to find the x-coordinate on \(y = g(x)\) solve

\(2 = x{{\text{e}}^{{x^2}}}\)     (M1)

\(x = 0.89605022078 \ldots \)     (A1)

gradient \( = ({g^{ – 1}})'(2) = \frac{1}{{g'(0.896 \ldots )}}\)     (M1)

\( = \frac{1}{{{{\text{e}}^{{{(0.896 \ldots )}^2}}}\left( {1 + 2 \times {{(0.896 \ldots )}^2}} \right)}} = 0.172\) to 3sf     A1

(using the \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) function on gdc \(g'(0.896 \ldots ) = 5.7716028 \ldots \)

\(\frac{1}{{g'(0.896 \ldots )}} = 0.173\)

[4 marks]

c.

(i)     \(({x^3} + 1){{\text{e}}^{{{({x^3} + 1)}^2}}} = 2\)     A1

\(x = – 0.470191 \ldots \)     A1

 

(ii)     METHOD 1

\((g \circ f)'(x) = 3{x^2}{{\text{e}}^{{{({x^3} + 1)}^2}}}\left( {2{{({x^3} + 1)}^2} + 1} \right)\)     (M1)(A1)

\((g \circ f)'( – 0.470191 \ldots ) = 3.85755 \ldots \)     (A1)

\(h'(2) = \frac{1}{{3.85755 \ldots }} = 0.259{\text{ }}(232 \ldots )\)     A1

Note: The solution can be found without the student obtaining the explicit form of the composite function.

 

METHOD 2

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

\(h'(x) = ({f^{ – 1}})’\left( {{g^{ – 1}}(x)} \right) \times ({g^{ – 1}})'(x)\)     M1

\( = \frac{1}{3}{\left( {{g^{ – 1}}(x) – 1} \right)^{ – \frac{2}{3}}} \times ({g^{ – 1}})'(x)\)     M1

\(h'(2) = \frac{1}{3}{\left( {{g^{ – 1}}(2) – 1} \right)^{ – \frac{2}{3}}} \times ({g^{ – 1}})'(2)\)

\( = \frac{1}{3}{(0.89605 \ldots  – 1)^{ – \frac{2}{3}}} \times 0.171933 \ldots \)

\( = 0.259{\text{ }}(232 \ldots )\)     A1     N4

[6 marks]

 

Question

Let \(f(x) = \left| x \right| – 1\).

(a)     The graph of \(y = g(x)\) is drawn below.

          (i)     Find the value of \((f \circ g)(1)\).

          (ii)     Find the value of \((f \circ g \circ g)(1)\).

          (iii)     Sketch the graph of \(y = (f \circ g)(x)\).

(b)     (i)     Sketch the graph of \(y = f(x)\).

          (ii)     State the zeros of f.

(c)     (i)     Sketch the graph of \(y = (f \circ f)(x)\).

          (ii)     State the zeros of \(f \circ f\).

(d)     Given that we can denote \(\underbrace {f \circ f \circ f \circ  \ldots  \circ f}_{n{\text{ times}}}\) as \({f^n}\),

          (i)     find the zeros of \({f^3}\);

          (ii)     find the zeros of \({f^4}\);

          (iii)     deduce the zeros of \({f^8}\).

(e)     The zeros of \({f^{2n}}\) are \({a_1},{\text{ }}{a_2},{\text{ }}{a_3},{\text{ }} \ldots {\text{, }}{a_N}\).

          (i)     State the relation between n and N;

          (ii)     Find, and simplify, an expression for \(\sum\limits_{r = 1}^N {\left| {{a_r}} \right|} \) in terms of n.

▶️Answer/Explanation

Markscheme

(a)     (i)     \(f(0) =  – 1\)     (M1)A1

          (ii)     \((f \circ g)(0) = f(4) = 3\)     A1

          (iii)
               (M1)A1

Note:     Award M1 for evidence that the lower part of the graph has been reflected and A1 correct shape with y-intercept below 4.

[5 marks]

 

(b)     (i)
               (M1)A1

Note:     Award M1 for any translation of \(y = \left| x \right|\).

          (ii)     \( \pm 1\)     A1

Note:     Do not award the A1 if coordinates given, but do not penalise in the rest of the question

[3 marks]

 

(c)     (i)
               (M1)A1

Note:     Award M1 for evidence that lower part of (b) has been reflected in the x-axis and translated.

          (ii)     \(0,{\text{ }} \pm 2\)     A1

[3 marks]

 

(d)     (i)     \( \pm 1,{\text{ }} \pm 3\)     A1

          (ii)     \(0,{\text{ }} \pm 2,{\text{ }} \pm 4\)     A1

          (iii)     \(0,{\text{ }} \pm 2,{\text{ }} \pm 4,{\text{ }} \pm 6,{\text{ }} \pm 8\)     A1

[3 marks]

 

(e)     (i)     \({\text{(1, 3), (2, 5), }} \ldots \)     (M1)

          \(N = 2n + 1\)     A1

          (ii)     Using the formula of the sum of an arithmetic series     (M1)

          EITHER

          \(4(1 + 2 + 3 +  \ldots  + n) = \frac{4}{2}n(n + 1)\)

          \( = 2n(n + 1)\)     A1

          OR

          \(2(2 + 4 + 6 +  \ldots  + 2n) = \frac{2}{2}n(2n + 2)\)

          \( = 2n(n + 1)\)     A1

[4 marks]

 

Total [18 marks]

Question

The following graph represents a function \(y = f(x)\), where \( – 3 \le x \le 5\).

The function has a maximum at \((3,{\text{ }}1)\) and a minimum at \(( – 1,{\text{ }} – 1)\).

a.The functions \(u\) and \(v\) are defined as \(u(x) = x – 3,{\text{ }}v(x) = 2x\) where \(x \in \mathbb{R}\).

(i)     State the range of the function \(u \circ f\).

(ii)     State the range of the function \(u \circ v \circ f\).

(iii)     Find the largest possible domain of the function \(f \circ v \circ u\).[7]

b.

(i)     Explain why \(f\) does not have an inverse.

(ii)     The domain of \(f\) is restricted to define a function \(g\) so that it has an inverse \({g^{ – 1}}\).

State the largest possible domain of \(g\).

(iii)     Sketch a graph of \(y = {g^{ – 1}}(x)\), showing clearly the \(y\)-intercept and stating the coordinates of the endpoints.[6]

c.

Consider the function defined by \(h(x) = \frac{{2x – 5}}{{x + d}}\), \(x \ne  – d\) and \(d \in \mathbb{R}\).

(i)     Find an expression for the inverse function \({h^{ – 1}}(x)\).

(ii)     Find the value of \(d\) such that \(h\) is a self-inverse function.

For this value of \(d\), there is a function \(k\) such that \(h \circ k(x) = \frac{{2x}}{{x + 1}},{\text{ }}x \ne  – 1\).

(iii)     Find \(k(x)\).[8]

 
▶️Answer/Explanation

Markscheme

Note:     For Q12(a) (i) – (iii) and (b) (ii), award A1 for correct endpoints and, if correct, award A1 for a closed interval.

Further, award A1A0 for one correct endpoint and a closed interval.

(i)     \( – 4 \le y \le  – 2\)     A1A1

(ii)     \( – 5 \le y \le  – 1\)     A1A1

(iii)     \( – 3 \le 2x – 6 \le 5\)     (M1)

Note:     Award M1 for \(f(2x – 6)\).

\(3 \le 2x \le 11\)

\(\frac{3}{2} \le x \le \frac{{11}}{2}\)     A1A1

[7 marks]

a.

(i)     any valid argument eg \(f\) is not one to one, \(f\) is many to one, fails horizontal line test, not injective     R1

(ii)     largest domain for the function \(g(x)\) to have an inverse is \([ – 1,{\text{ }}3]\)     A1A1

(iii)    

\(y\)-intercept indicated (coordinates not required)     A1

correct shape     A1

coordinates of end points \((1,{\text{ }}3)\) and \(( – 1,{\text{ }} – 1)\)     A1

Note:     Do not award any of the above marks for a graph that is not one to one.

[6 marks]

b.

(i)     \(y = \frac{{2x – 5}}{{x + d}}\)

\((x + d)y = 2x – 5\)     M1

Note:     Award M1 for attempting to rearrange \(x\) and \(y\) in a linear expression.

\(x(y – 2) =  – dy – 5\)     (A1)

\(x = \frac{{ – dy – 5}}{{y – 2}}\)     (A1)

Note:     \(x\) and \(y\) can be interchanged at any stage

\({h^{ – 1}}(x) = \frac{{ – dx – 5}}{{x – 2}}\)     A1

Note:     Award A1 only if \({h^{ – 1}}(x)\) is seen.

(ii)     self Inverse \( \Rightarrow h(x) = {h^{ – 1}}(x)\)

\(\frac{{2x – 5}}{{x + d}} \equiv \frac{{ – dx – 5}}{{x – 2}}\)     (M1)

\(d =  – 2\)     A1

(iii)     METHOD 1

\(\frac{{2k(x) – 5}}{{k(x) – 2}} = \frac{{2x}}{{x + 1}}\)     (M1)

\(k(x) = \frac{{x + 5}}{2}\)     A1

METHOD 2

\({h^{ – 1}}\left( {\frac{{2x}}{{x + 1}}} \right) = \frac{{2\left( {\frac{{2x}}{{x + 1}}} \right) – 5}}{{\frac{{2x}}{{x + 1}} – 2}}\)     (M1)

\(k(x) = \frac{{x + 5}}{2}\)     A1

[8 marks]

Total [21 marks]

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