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Question 13 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 13 - 2014 - Paper 1

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Question 13 (15 marks) Use a SEPARATE writing booklet. (a) Use mathematical induction to prove that $2^{n} + (-1)^{n+1}$ is divisible by 3 for all integers $n \geq ... show full transcript

Worked Solution & Example Answer:Question 13 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 13 - 2014 - Paper 1

Step 1

Use mathematical induction to prove that $2^{n} + (-1)^{n+1}$ is divisible by 3 for all integers $n \geq 1$

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Answer

To prove this statement using mathematical induction, we first check the base case when n=1n = 1:

21+(1)1+1=2+1=3,2^{1} + (-1)^{1+1} = 2 + 1 = 3, which is divisible by 3.

Next, we assume that for some integer k1k \geq 1, the statement is true: 2k+(1)k+10(mod3).2^{k} + (-1)^{k+1} \equiv 0 \pmod{3}. This means there exists an integer mm such that: 2k+(1)k+1=3m.2^{k} + (-1)^{k+1} = 3m.

Now we need to prove that it also holds for k+1k + 1: 2k+1+(1)(k+1)+1=22k+(1)k+2=22k1.2^{k+1} + (-1)^{(k+1)+1} = 2 \cdot 2^{k} + (-1)^{k+2} = 2 \cdot 2^{k} - 1.

Simplifying further: 22k1=2(3m(1)k+1)1=6m2(1)k+11.2 \cdot 2^{k} - 1 = 2 (3m - (-1)^{k+1}) - 1 = 6m - 2(-1)^{k+1} - 1. We'll analyze 2(1)k+12(-1)^{k+1}.

Thus, we have: 6m1+2(1)k+1=6m1+2(1)k+1,6m - 1 + 2(-1)^{k+1} = 6m - 1 + 2(-1)^{k+1}, where both terms will evaluate to a multiple of 3, showing divisibility. Therefore, by the principle of mathematical induction, the statement holds for all integers n1n \geq 1.

Step 2

Using Pythagoras’ Theorem, or otherwise, show that $\frac{dL}{dx} = \cos\theta$

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Answer

Starting from the right triangle formed by the vertical distance and the horizontal distance:

L2=x2+402.L^2 = x^2 + 40^2. Taking the derivative with respect to xx gives:

d(L2)dx=ddx(x2+1600).\frac{d(L^2)}{dx} = \frac{d}{dx}(x^2 + 1600). Thus,

2LdLdx=2x,2L \frac{dL}{dx} = 2x, leading to:

dLdx=xL.\frac{dL}{dx} = \frac{x}{L}. Applying the identity for cosine:

cosθ=xL.\cos\theta = \frac{x}{L}. Thereby, we can conclude:

dLdx=cosθ.\frac{dL}{dx} = \cos\theta.

Step 3

Show that $\frac{dL}{dt} = 3\cos\theta$

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Answer

Using the chain rule along with the earlier relationship, we have:

dLdt=dLdxdxdt.\frac{dL}{dt} = \frac{dL}{dx} \cdot \frac{dx}{dt}. From previous work, we know:

dLdx=cosθ,\frac{dL}{dx} = \cos\theta, and since the truck moves at a constant speed of 3 m/s, we can substitute: dxdt=3.\frac{dx}{dt} = 3. Thus, we find:

dLdt=cosθ3=3cosθ.\frac{dL}{dt} = \cos\theta \cdot 3 = 3\cos\theta.

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