(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 $x^2 = 4y$. The tangent to the parabola at $P$ meets the parabola $x^2 = -4ay$, $a > 0$, at $Q$ and $R$. Let $M$ be the midpoint of $QR$. (i) Show that the $x$ coordinates of $R$ and $Q$ satisfy $$x^2 + 4apx - 4p^2 = 0.$$ (ii) Show that the coordinates of $M$ are $(-2ap, -p^2(2a + 1)).$ (iii) Find the value of $a$ so that the point $M$ always lies on the parabola $x^2 = -4ay$. (c) The concentration of a drug in a body is $F(t)$, where $t$ is the time in hours after the drug is taken. Initially the concentration of the drug is zero. The rate of change of concentration of the drug is given by $$F'(t) = 50e^{-0.5t} - 0.4F(t).$$ (i) By differentiating the product $F(t)e^{0.4t}$ show that $$\frac{d}{dt} [F(t)e^{0.4t}] = 50e^{-0.1t}.$$ (ii) Hence, or otherwise, show that $F(t) = 500(e^{-0.4t} - e^{-0.5t}).$ (iii) The concentration of the drug increases to a maximum. For what value of $t$ does this maximum occur?

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(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

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$(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.png$

(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.

$P(1)$ is true since $8^3 + 6^1 = 512 + 6 = 518 = 7 \times 74$ , which is divisible by 7.

Assume $P(k)$ is true for integer $k$ . That is, we assume:

$8^{2k+1} + 6^{2k-1} = 7m$

for some integer $m$ .

Now consider $P(k + 1)$ :

$8^{2(k+1)+1} + 6^{2(k+1)-1} = 8^{2k+3} + 6^{2k+1}$

This can be expressed as:

$= 8^{2k+1} \cdot 8^2 + 6^{2k-1} \cdot 6^2 = 64 \cdot 8^{2k+1} + 36 \cdot 6^{2k-1}$

We factor out a 7 from each term:

The first term $64 \cdot 8^{2k+1} = 56 \cdot 8^{2k+1} + 8^{2k+1}$ which is divisible by 7, and also the second term is:

$= 36m + 7n$

which is also divisible by 7. Hence, $P(n + 1)$ is true, thus by induction, the statement is proven.

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 equation is $-4ap$ .

Let the $x$ -coordinates of $R$ and $Q$ be the roots of the equation.

By Vieta's formulas, we know that:

The sum of the roots is $2p$
The product of the roots is $-p^2(2a + 1)$ .

Substituting into the tangent equation gives:

$x^2 - (\text{sum}) x + (\text{product}) = 0,$
thus we have:

$x^2 + 4apx - 4p^2 = 0.$

Step 3

Show that the coordinates of $M$ are $(-2ap, -p^2(2a + 1))$

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Answer

From the equation of the tangent:

$y = px - 2ap - p^2(2a + 1).$

Thus, the coordinates of $M$ can be computed from:

The average of the roots gives us:
- $x$ -coordinate = $M_x = -2ap$
The $y$ $y$ -coordinate is the average of the two roots which can be denoted as:
- $M_y = -p^2(2a + 1)$

Therefore, the coordinates of $M$ are:

$M = (-2ap, -p^2(2a + 1)).$

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 find the value of $a$ such that:

Substituting $M(-2ap, -p^2(2a + 1))$ into the parabola's equation gives:

$(-2ap)^2 = -4a(-p^2(2a + 1)).$

This simplifies to:

$4a^2p^2 = 4ap^2(2a + 1).$

We can divide through by $4p^2$ (assuming $p \neq 0$ ):

$a^2 = a(2a + 1).$

Rearranging gives:

$a^2 - 2a^2 - a = 0$ $(1 - 2a)a = 0,$ Thus, $a = 0$ or $a = 0.5$ .

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

We apply the product rule of differentiation:

$\frac{d}{dt}[F(t)e^{0.4t}] = F'(t)e^{0.4t} + F(t)\cdot 0.4e^{0.4t}.$
Substituting for $F'(t)$ gives:

$= 50e^{-0.5t} - 0.4F(t)e^{0.4t} + 0.4F(t)e^{0.4t}$ The terms involving $F(t)$ cancel out, hence:

$= 50e^{-0.1t}.$

Step 6

Hence, or otherwise, show that $F(t) = 500(e^{-0.4t} - e^{-0.5t})$

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Answer

Integrating the expression from part (i):

$F(t)e^{0.4t} = \int 50e^{-0.1t} dt = -500e^{-0.1t} + C.$

Initially, $F(0) = 0$ leads to:

$0 = -500 + C \Rightarrow C = 500.$
Thus, we have: $F(t) = 500(e^{-0.4t} - e^{-0.5t}).$

Step 7

For what value of $t$ does this maximum occur?

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Answer

To find the maximum concentration, we set:

$F'(t) = 0 = 50e^{-0.4t} - 200e^{-0.5t}.$ Thus:

$250e^{-0.4t} = 200e^{-0.5t}.$ This simplifies to:

$\frac{250}{200} = e^{-0.1t}.$ Taking logarithms gives:

$t = \frac{10}{4} \ln\left(\frac{5}{4}\right) = 2.3 \text{ hours}.$

(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

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

Prove by mathematical induction that $8^{2n+1} + 6^{2n-1}$ is divisible by 7

Show that the $x$ coordinates of $R$ and $Q$ satisfy $x^2 + 4apx - 4p^2 = 0$

Show that the coordinates of $M$ are $(-2ap, -p^2(2a + 1))$

Find the value of $a$ so that the point $M$ always lies on the parabola $x^2 = -4ay$

By differentiating the product $F(t)e^{0.4t}$ show that $\frac{d}{dt}[F(t)e^{0.4t}] = 50e^{-0.1t}$

Hence, or otherwise, show that $F(t) = 500(e^{-0.4t} - e^{-0.5t})$

For what value of $t$ does this maximum occur?

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