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Date May 2013 Marks available 3 Reference code 13M.2.hl.TZ1.13
Level HL only Paper 2 Time zone TZ1
Command term Show that 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.

[6]
a.

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

[3]
b.

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]
c.

(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]
d.

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]

d.

Examiners report

There were many good attempts at parts (a) and (b), although in (b) many were unable to give a thorough justification.

a.

There were many good attempts at parts (a) and (b), although in (b) many were unable to give a thorough justification.

b.

Few good solutions to parts (c) and (d)(ii) were seen although many were able to answer (d)(i) correctly.

c.

Few good solutions to parts (c) and (d)(ii) were seen although many were able to answer (d)(i) correctly.

d.

Syllabus sections

Topic 6 - Core: Calculus » 6.2 » Derivatives of \({x^n}\) , \(\sin x\) , \(\cos x\) , \(\tan x\) , \({{\text{e}}^x}\) and \\(\ln x\) .
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