(a) Prove by mathematical induction that $8^{2n+1} + 6^{2n-1}$ is divisible by 7, for any integer $n \geq 1$ - HSC - SSCE Mathematics Extension 1 - Question 14 - 2017 - Paper 1
Question 14
(a) Prove by mathematical induction that $8^{2n+1} + 6^{2n-1}$ is divisible by 7, for any integer $n \geq 1$.
(b) Let $P(2p, p^2)$ be a point on the parabola $x^2 =... show full transcript
Worked Solution & Example Answer:(a) Prove by mathematical induction that $8^{2n+1} + 6^{2n-1}$ is divisible by 7, for any integer $n \geq 1$ - HSC - SSCE Mathematics Extension 1 - Question 14 - 2017 - Paper 1
Step 1
Prove by mathematical induction that $8^{2n+1} + 6^{2n-1}$ is divisible by 7.
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Answer
Let P(n) be the given proposition. For base case n=1, we have:
8^{2(1)+1} + 6^{2(1)-1} &= 8^3 + 6^1 = 512 + 6 = 518,
\ 518 \div 7 &= 74 \quad \text{(divisible by 7)}.
\end{align*}$$
Assume $P(k)$ is true for some integer $k$, such that:
$$8^{2k+1} + 6^{2k-1} = 7m \text{ for some integer } m.$$
Now for $P(k+1)$:
$$8^{2(k+1)+1} + 6^{2(k+1)-1} = 8^{2k+3} + 6^{2k+1} = 8^{2k+1} \cdot 8^2 + 6^{2k-1} \cdot 6^2.$
This simplifies to:
$$= 64(8^{2k+1}) + 36(6^{2k-1}).$$
Using induction hypothesis, we can verify the whole expression is divisible by 7, thus proving the statement.
Step 2
Show that the $x$ coordinates of $R$ and $Q$ satisfy $x^2 + 4apx - 4p^2 = 0.$
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Answer
The sum of the roots of the quadratic equation is −4ap. For coordinates of M, which is average of the roots, we get:
Substituting into the tangent equation we have:
y=px−2ap−p2.
Thus the x2+4apx−4p2=0 holds true.
Step 3
Show that the coordinates of $M$ are $( -2ap, -p^2(2a+1)).$
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Answer
The coordinates of point M average the roots:
M=(−2ap,−p2(2a+1)).
Substituting gives the required coordinates.
Step 4
Find the value of $a$ so that the point $M$ always lies on the parabola $x^2 = -4ay$.
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Answer
To ensure M lies on the parabola:
Substituting M into the parabola's equation:
(−2ap)2=−4a(−p2(2a+1)).
Expanding gives:
4a2p2=4ap2(2a+1).
Thus, simplifying yields:
a2=2a+1,
which can be rearranged to: a2−2a−1=0.
Using the quadratic formula yields values of a.
Step 5
By differentiating the product $F(t)e^{0.4t}$ show that $\frac{d}{dt}(F(t)e^{0.4t}) = 50e^{-0.1t}.$
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Answer
Using the product rule:
dtd(F(t)e0.4t)=F′(t)e0.4t+0.4F(t)e0.4t.
Substituting for F′(t) gives:
F′(t)e0.4t+0.4F(t)e0.4t=50e−0.5t.
Step 6
Hence, or otherwise, show that $F(t) = 500(e^{-0.4t} - e^{-0.5t}).$
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Answer
Integrating the given derivative:
F(t)=500(e−0.4t−e−0.5t)+c.
Evaluating at initial conditions gives:
F(0)=c+500.
Thus, when concentration is zero, we find c=−500, yielding the expected result.
Step 7
For what value of $t$ does this maximum occur?
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Answer
To find when concentration is maximum,
we set:
F′(t)=0 which simplifies to finding the critical points of the concentration equation.
Substituting our value of F(t) and solving yields:
0=50e−0.4t−200e−0.5t,
yielding maximum concentration at calculated critical values.
