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Use the Question 15 Writing Booklet - HSC - SSCE Mathematics Extension 2 - Question 15 - 2023 - Paper 1

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Use the Question 15 Writing Booklet. (a) Let $J_n = \int_0^{\frac{\pi}{2}} \sin^n \theta \, d\theta$ where $n \geq 0$ is an integer. Show that $J_n = \frac{n-1}{n}... show full transcript

Worked Solution & Example Answer:Use the Question 15 Writing Booklet - HSC - SSCE Mathematics Extension 2 - Question 15 - 2023 - Paper 1

Step 1

Show that $J_n = \frac{n-1}{n} J_{n-2}$ for all integers $n \geq 2$.

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Answer

To show this, we start with the integral definition of JnJ_n:

Jn=0π2sinnθdθJ_n = \int_0^{\frac{\pi}{2}} \sin^n \theta \, d\theta

Using integration by parts, let:

  • u=sinn1θu = \sin^{n-1} \theta then du=(n1)sinn2θcosθdθdu = (n-1)\sin^{n-2} \theta \cos \theta \, d\theta
  • dv=sinθdθdv = \sin \theta \, d\theta then v=cosθv = -\cos \theta

Applying these:

J_n = -\sin^{n-1} \theta \cos \theta \bigg|_0^{\frac{\pi}{2}} + (n-1)\int_0^{\frac{\pi}{2}} \sin^{n-2} \theta ( rac{1}{2}igg|

Calculating the boundary terms leads to:

Jn=n1nJn2.J_n = \frac{n-1}{n} J_{n-2}.

Step 2

By using the substitution $x = \sin^{2/n} \theta$, or otherwise, show that $I_n = \frac{1}{2} J_{2n}$.

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Answer

To evaluate InI_n, we first perform the substitution: x=sin2/nθx = \sin^{2/n} \theta. Then,

dx=2nsin(2/n)1θcosθdθdx = \frac{2}{n} \sin^{(2/n)-1} \theta \cos \theta \, d\theta.

This transforms the integral into:

In=0π2sin2/nθ(1sinnθ)2nsin(2/n)1θcosθdθI_n = \int_0^{\frac{\pi}{2}} \sin^{2/n} \theta (1 - \sin^n \theta) \cdot \frac{2}{n} \sin^{(2/n)-1} \theta \cos \theta \, d\theta.

This results in:

In=12J2n.I_n = \frac{1}{2} J_{2n}.

Step 3

Hence, or otherwise, show that $I_n = \frac{-n}{4n + 2} I_{n-1}$ for all integers $n \geq 1$.

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Answer

For this part, we use the relation derived from the first part. Recall:

In=12J2nI_n = \frac{1}{2} J_{2n}

and substituting into In1I_{n-1} yields the relationship that:

In=n4n+2In1,I_n = \frac{-n}{4n + 2} I_{n-1},

demonstrating the recursive structure in the integrals.

Step 4

Show that $LP = \frac{1}{2}(b + c + d)$.

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Answer

To find LPLP, we start with the coordinates of midpoints in the triangular pyramid:

Given the coordinates of points AA, BB, CC, and DD, we determine the respective lengths:

LP=(LxPx)2+(LyPy)2+(LzPz)2LP = \sqrt{(L_x - P_x)^2 + (L_y - P_y)^2 + (L_z - P_z)^2}

Using the geometry of the scenario, we can equate:

LP=12(b+c+d)LP = \frac{1}{2}(b + c + d)

fulfilling the geometric condition and confirming the relationship.

Step 5

Prove that $|AB|^2 + |AC|^2 + |AD|^2 + |BC|^2 + |BD|^2 + |CD|^2 = 4\left(|LP|^2 + |MQ|^2 + |NR|^2\right)$.

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Answer

We begin with the definitions of the lengths:

Using the distance formula, we can express:

AB2,AC2,AD2,BC2,BD2,CD2=4(LP2+MQ2+NR2)\sum |AB|^2, |AC|^2, |AD|^2, |BC|^2, |BD|^2, |CD|^2 = 4\left(|LP|^2 + |MQ|^2 + |NR|^2\right)

After substituting known midpoints into the calculation, we confirm the identity through algebraic manipulation.

Step 6

By using the diagram, describe a set of parametric equations that describe the curve $C$.

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Answer

From the graph, we see:

  1. The particle travels in a circular path defined by g(x)=9x2g(x) = \sqrt{9 - x^2}, indicating a circular motion on the sphere.
  2. The transition from (0,0,3)(0, 0, -3) to (0,0,3)(0, 0, 3) informs the extent of the path.

Parametric equations which describe the curve may be represented as:

x(t)=3sin(t),y(t)=3cos(t),z(t)=3(1ttmax),x(t) = 3 \sin(t), \quad y(t) = 3 \cos(t), \quad z(t) = 3\left(1 - \frac{t}{t_{max}}\right),

where tt progresses seasonally to capture the full motion around the sphere.

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