Specialist Maths — Question 19

QCAA 2022, Paper 1 · 7 marks

Q19 · 2022 · Technology-freeComplex unfamiliar7 marks

QUESTION 19 (7 marks)

The function

f(x)

passes through the origin. The gradient function of

f(x)

is defined as

g(x)=e^x\sin^{-1}(e^x)

. Determine

f(x)

WORKED SOLUTION

7 marks

ANSWER

f(x)=e^x\sin^{-1}(e^x)+\sqrt{1-e^{2x}}-\frac{\pi}{2}

Worked solution

Since

f'(x)=e^x\sin^{-1}(e^x)

f(x)=\int e^x\sin^{-1}(e^x)\,dx.

Using integration by parts with

u=\sin^{-1}(e^x)

and

dv=e^x dx

gives

f(x)=e^x\sin^{-1}(e^x)-\int\frac{e^{2x}}{\sqrt{1-e^{2x}}}\,dx.

With

u=1-e^{2x}

, the remaining integral gives

f(x)=e^x\sin^{-1}(e^x)+\sqrt{1-e^{2x}}+c.

Using

f(0)=0

gives

c=-\frac{\pi}{2}

, hence

f(x)=e^x\sin^{-1}(e^x)+\sqrt{1-e^{2x}}-\frac{\pi}{2}.

correctly determines the expressions required in preparation for the use of the integration by parts rule

[1 mark]

applies the integration by parts rule and simplifies an integrand

[1 mark]

determines a suitable substitution variable in preparation for using a substitution method of integration

[1 mark]

expresses an integrand in terms of the substitution variable

[1 mark]

determines a general solution for

f(x)

[1 mark]

determines a value of the constant of integration and communicates a solution

[1 mark]

shows logical organisation communicating key steps up to the stage where an integration method using substitution is considered

[1 mark]

One QCAA sample method typeset for web; criterion wording is adapted from the official marking guide.

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