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 $x^2 = -4y$. 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?

SSCE HSC Mathematics Extension 1 Proving divisibility by induction: a) Prove by mathematical inducti

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-$n-\geq-1$-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 = 4... 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 $n = 1$ , we find:

$8^{2(1)+1} + 6^{2(1)-1} = 8^3 + 6^1 = 512 + 6 = 518,$

which is divisible by 7 since $518 \div 7 = 74.$

Assume $P(k)$ is true for some integer $k$ , i.e.,

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

for some integer $m$ . We need to show $P(k+1)$ :

$8^{2(k+1)+1} + 6^{2(k+1)-1} = 8^{2k+3} + 6^{2k+1} = 8^{2k+1} \cdot 8^2 + 6^{2k-1} \cdot 6^2 = 64(8^{2k+1}) + 36(6^{2k-1}).$

Since both terms are divisible by 7 (by the inductive hypothesis), $P(k+1)$ holds true. Hence, by induction, the statement is established for all integers $n \geq 1$ .

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 $-4ap$ . Since $x$ -coordinate of $M$ is average of the roots, we have:

$M_x = -2ap.$

Using the tangent line at point $P$ , it can be derived that:

$y = p(x - 2p) - p^2.$

Substituting into the parabola’s equation yields:

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

Step 3

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

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Answer

Using the relationship between the roots $R$ and $Q$ , the coordinates of $M$ can be computed. Since $M$ is the midpoint:

$M_x = \frac{R_x + Q_x}{2} = -2ap,$

and substituting gives the $y$ -coordinate:

$M_y = p(x - 2p) - p^2(2a+1).$

Thus, the coordinates of $M$ are verified: $(-2ap, -p^2(2a + 1))$ .

Step 4

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

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Answer

We know that $M = (-2ap, -p^2(2a + 1))$ must satisfy:

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

This leads to:

$4a^2p^2 = 4p^2(2a + 1) \Rightarrow a^2 = 2a + 1 \Rightarrow a^2 - 2a - 1 = 0.$

Solving this using the quadratic formula provides the value of $a$ as:

$a = 1 \pm \sqrt{2}.$

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

Differentiating gives:

$\frac{d}{dt}(F(t)e^{0.4t}) = F'(t)e^{0.4t} + F(t)(0.4)e^{0.4t}.$

Substituting the expression for $F'(t)$ leads to:

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

Therefore,

$\frac{d}{dt}(F(t)e^{0.4t}) = 50e^{-0.1t}.$

Step 6

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

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Answer

Integrating $F'(t)$ , we have:

$F(t) = \int(50e^{-0.5t} - 0.4F(t))dt.$

Solving gives:

$F(t) = 500(e^{-0.4t} - e^{-0.5t}) + C.$

Given $F(0) = 0$ , we find $C = 500.$

Step 7

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

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Setting $F'(t) = 0$ results in:

$500(0.4e^{-0.4t} + 0.5e^{-0.5t}) = 0.$

Solving yields:

$0.4e^{-0.4t} + 0.5e^{-0.5t} = 0.$

This will find the maximum concentration, occurring when $t = 2.23$ 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 $x^2 = -4y$.

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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