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2020 HSC Mathematics Extension 1

Question 13(c):
Inverse trigonometric functions

7 marksinverse trigonometric functionsdependency-safe group

Suppose and .

The graph of is given (two branches on and , vertical asymptote at ).

(i) Show that .

(ii) Using part (i), or otherwise, show that .

Graph of y equals g of x, as supplied with the original question.
Graph of y equals g of x, as supplied with the original question.

How to recognise this question

Question type: inverse trigonometric functions

  • The command and notation point to inverse trigonometric functions.
  • The word “hence” means the later part is intended to reuse the earlier result.

How to handle it: use the restricted domain and range that make the inverse function single-valued.

Watch out: Do not restart the second part with unrelated numbers or discard the result proved in the first part.

Step-by-step answer

  1. (i) Differentiate : .
  2. Since , , so .
  3. Differentiate by the quotient rule: .
  4. Hence .
  5. NESA (i): 4 marks; partial for one correct derivative.
  6. (ii) Then on each connected component of the domain , so is constant on each component.
  7. On , test : and , so the constant is .
  8. On , test : and , so the constant is there too.
  9. Therefore on the domain.
  10. NESA (ii): 3 marks for both components; partial for one component or for observing the difference is constant.

Common mistake: choosing a familiar formula before checking that its conditions, signs, interval and units match this question.

The paper records 7 total marks for this group. Where the official marking guide provides useful partial-credit criteria, those criteria are stated in the steps above.

Try your hand at it again by answering these questions

Question 1

Suppose and for .

(i) Show that .

(ii) Using part (i), or otherwise, show that .

Graph of $g(x)=\sqrt{1-x^2}$ on $[-1,1]$.
Graph of on .
See the step-by-step answer guide →

Question 2

Suppose and for .

(i) Show that .

(ii) Using part (i), or otherwise, show that .

Graph of $g(x)=\dfrac{x}{\sqrt{1-x^2}}$ on $(-1,1)$.
Graph of on .
See the step-by-step answer guide →

Question 3

Suppose and for .

(i) Show that .

(ii) Using part (i), or otherwise, show that .

Graph of $g(x)=\sqrt{1+x^2}$ for all real $x$.
Graph of for all real .
See the step-by-step answer guide →