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(a) Prove by mathematical induction that, for n ≥ 1, 2 - 6 + 18 - 54 + .. - HSC - SSCE Mathematics Extension 1 - Question 13 - 2018 - Paper 1
Question 13
(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^... show full transcript
Worked Solution & Example Answer:(a) Prove by mathematical induction that, for n ≥ 1, 2 - 6 + 18 - 54 + .. - 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}.
Answer
To prove this by induction:
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Base Case: For n = 1, LHS = 2
RHS = \frac{1 - (-3)^{1}}{2} = \frac{1 + 3}{2} = 2 Thus, LHS = RHS. -
Induction Hypothesis: Assume true for n = k, i.e., LHS = 2 - 6 + 18 - ... + 2(-3)^{k-1} = \frac{1 - (-3)^{k}}{2}.
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Induction Step: Show true for n = k + 1. LHS = 2 - 6 + 18 - ... + 2(-3)^{k-1} + 2(-3)^{k} = \frac{1 - (-3)^{k}}{2} + 2(-3)^{k} = \frac{1 - (-3)^{k}}{2} + \frac{4(-3)^{k}}{2} = \frac{1 + 3(-3)^{k}}{2} = \frac{1 - (-3)^{k + 1}}{2}.
Thus, LHS = RHS holds for k + 1. By induction, the formula is true for all n ≥ 1.
Step 2
State the domain and range of f^{-1}(x).
Answer
For the function f(x) = \frac{-x}{x^{2}+1}:
- Domain of f(x): All real numbers, ( -1 < x < \infty )
- Range of f(x): All real numbers, ( y ≤ \frac{1}{2}, y < 0 )
Hence, for the inverse:
- Domain of f^{-1}(x): All real numbers, ( -1 < x < 0 )
- Range of f^{-1}(x): All real numbers, ( y ≥ 0, y < \frac{1}{2} )
Step 3
Sketch the graph y = f^{-1}(x).
Answer
To sketch the graph of f^{-1}(x):
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Identify key points based on the domain and range.
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Plot points and ensure that the graph reflects the behavior at the boundaries and intercepts.
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Ensure symmetry in points corresponding to f(x) and its inverse.
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Draw a smooth curve between points respecting the characteristics observed.
Step 4
Find an expression for f^{-1}(x).
Answer
To find f^{-1}(x):
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Start with y = f(x), hence:
[ y = \frac{-x}{x^{2} + 1} ]
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Rearranging gives us:
[ y(x^{2}+1) = -x ]
[ yx^{2} + x + y = 0 ]
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Solving this quadratic equation for x using the quadratic formula gives:
[ x = \frac{-1 \pm \sqrt{1 - 4y^2}}{2y} ]
Thus, The expression for f^{-1}(x) is derived based on the applicable roots.