Question 13 (15 marks) Use the Question 13 Writing Booklet. (a) Prove by mathematical induction that, for n ≥ 1, 2 − 6 + 18 − 54 + ⋯ + 2(−3)^{n−1} = \frac{1 − (−3)^{n}}{2}. (b) The diagram shows the graph y = \frac{−x}{x^2 + 1}, for all real x. Consider the function f(x) = \frac{−x}{x^2 + 1}, for x ≥ 1. The function f(x) has an inverse. (Do NOT prove this.) (i) State the domain and range of f^{−1}(x). (ii) Sketch the graph y = f^{−1}(x). (iii) Find an expression for f^{−1}(x).

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Question 13 (15 marks) Use the Question 13 Writing Booklet - HSC - SSCE Mathematics Extension 1 - Question 13 - 2018 - Paper 1

Question 13

Question 13 (15 marks) Use the Question 13 Writing Booklet. (a) Prove by mathematical induction that, for n ≥ 1, 2 − 6 + 18 − 54 + ⋯ + 2(−3)^{n−1} = \frac{1 − (−3)... show full transcript

Worked Solution & Example Answer:Question 13 (15 marks) Use the Question 13 Writing Booklet - HSC - SSCE Mathematics Extension 1 - Question 13 - 2018 - Paper 1

Step 1

Prove by mathematical induction that, for n ≥ 1, 2 − 6 + 18 − 54 + ⋯ + 2(−3)^{n−1} = \frac{1 − (−3)^{n}}{2}.

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Answer

To prove the statement by mathematical induction, we follow these steps:

Base Case (n = 1):
- For n = 1, the left hand side (LHS) is:
- The right hand side (RHS) is:
  
  $RHS = \frac{1 − (−3)^{1}}{2} = \frac{1 + 3}{2} = 2.**$
- Therefore, LHS = RHS, which verifies our base case.
Inductive Step:
- Assume true for n = k:
  
  $LHS = 2 − 6 + 18 − 54 + ⋯ + 2(−3)^{k−1} = \frac{1 − (−3)^{k}}{2}.**$
- We need to show it holds for n = k + 1:
  
  $LHS' = LHS + 2(−3)^{k} = \frac{1 − (−3)^{k}}{2} + 2(−3)^{k} = \frac{1 + 3^{k} − 2(−3)^{k}}{2} = \frac{1 + (−3)^{k} − (−3)^{k+1}}{2} = \frac{1 − (−3)^{k+1}}{2}.**$
- This shows it holds for k + 1. Therefore, by induction, the statement is true for all n ≥ 1.

Step 2

State the domain and range of f^{−1}(x).

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Answer

The function f(x) = \frac{−x}{x^2 + 1} for x ≥ 1 has the following:

Domain: For the inverse function f^{−1}(x), based on the original function behavior, the domain is all real numbers where the function is valid:

$(-\infty, 0)$ .
Range: The range of the inverse function is the output values that the original function can take, which gives:

$[−\frac{1}{2}, +\infty).$

Step 3

Sketch the graph y = f^{−1}(x).

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Answer

To sketch the graph of f^{−1}(x):

Identify key points from the domain and range derived earlier.
Start by plotting points:
- When x approaches −∞, f^{−1}(x) approaches −1.
- At x = 0, y = −1/2.
The graph will start from (−∞, −1) to (0, −1/2).
Plot these points, and connect them smoothly to illustrate the curve.

The sketch should reflect an increasing function beginning from (−∞, −1) to (0, −1/2) showing the range.

Note: use the properties of the original function to guide the shape of the inverse.

Step 4

Find an expression for f^{−1}(x).

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Answer

To find the expression for the inverse function, start with:

Given:

$y = f(x) = \frac{−x}{x^2 + 1}$

Replace f(x) with y:

$y = \frac{−x}{x^2 + 1}$
Solve for x:
$x^2 y + x + y = 0.$$$
Apply the quadratic formula:

$x = \frac{−b \pm \sqrt{b^2 − 4ac}}{2a}$ where a = y, b = 1, and c = y:
This gives:

$x = \frac{−1 \pm \sqrt{1 − 4y^2}}{2y}$
Select the positive root because we're considering x ≥ 1.

Thus, the expression for f^{−1}(x) is:

$f^{−1}(x) = \frac{−1 + \sqrt{1 − 4x^2}}{2x}$ .

Question 13 (15 marks) Use the Question 13 Writing Booklet - HSC - SSCE Mathematics Extension 1 - Question 13 - 2018 - Paper 1

Worked Solution & Example Answer:Question 13 (15 marks) Use the Question 13 Writing Booklet - HSC - SSCE Mathematics Extension 1 - Question 13 - 2018 - Paper 1

Prove by mathematical induction that, for n ≥ 1, 2 − 6 + 18 − 54 + ⋯ + 2(−3)^{n−1} = \frac{1 − (−3)^{n}}{2}.

State the domain and range of f^{−1}(x).

Sketch the graph y = f^{−1}(x).

Find an expression for f^{−1}(x).

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