| Date | May 2013 | Marks available | 6 | Reference code | 13M.2.hl.TZ1.13 |
| Level | HL only | Paper | 2 | Time zone | TZ1 |
| Command term | Verify | Question number | 13 | Adapted from | N/A |
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 \(f'(a) \ne 0\)
\[({f^{ - 1}})'(b) = \frac{1}{{f'(a)}}.\]
Verify that this is true for \(f(x) = {x^3} + 1\) at x = 2.
Given that \(g(x) = x{{\text{e}}^{{x^2}}}\), show that \(g'(x) > 0\) for all values of x.
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.
(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 \(h'(2)\).
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]
\(g'(x) = {{\text{e}}^{{x^2}}} + 2{x^2}{{\text{e}}^{{x^2}}}\) M1A1
\(g'(x) > 0\) as each part is positive R1
[3 marks]
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]
(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]
Examiners report
There were many good attempts at parts (a) and (b), although in (b) many were unable to give a thorough justification.
There were many good attempts at parts (a) and (b), although in (b) many were unable to give a thorough justification.
Few good solutions to parts (c) and (d)(ii) were seen although many were able to answer (d)(i) correctly.
Few good solutions to parts (c) and (d)(ii) were seen although many were able to answer (d)(i) correctly.
