Question 7 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 7 - 2011 - Paper 1
Question 7
Question 7 (15 marks) Use a SEPARATE writing booklet.
(a) The diagram shows the graph of $y = f(x) = \frac{-x}{1+x^2}$ for $0 \leq x \leq 1$.
The area bounded by $... show full transcript
Worked Solution & Example Answer:Question 7 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 7 - 2011 - Paper 1
Step 1
a) Use the method of cylindrical shells to find the volume of the solid.
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Answer
To find the volume of the solid formed by rotating the area bounded by the curve y=f(x), the line x=1, and the x-axis about the line x=1, we use the method of cylindrical shells.
The volume V can be expressed as:
V=2π∫01(1−x)f(x)dx,
where the shell radius is (1−x) and shell height is f(x)=1+x2−x.
Substituting for f(x) gives:
= -2\pi \int_{0}^{1} \frac{x(1-x)}{1+x^2} \, dx.$$
This integral can be simplified and evaluated to obtain the final volume.
Step 2
b)(i) Use the substitution $u = 4 - x$ to show that $I = \int_{1}^{3} \frac{\sin(\frac{\pi}{8} u)}{u(4-u)} \, du.$
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Answer
To show this, we perform the substitution:
Let u=4−x, which implies du=−dx.
When x=1, u=3 and when x=3, u=1.
Thus the integral transforms as follows:
= \int_{1}^{3} \cos\left(\frac{\pi}{8}(4-u)\right)\, du.$$
Now using the cosine identity, we change $\cos\left(\frac{\pi}{8}(4-u)\right)$ and simplify it to ultimately show:
$$I = \int_{1}^{3} \frac{\sin(\frac{\pi}{8} u)}{u(4-u)} \, du.$$
Step 3
b)(ii) Hence, find the value of $I$.
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Answer
Having established the new form of I in part (i), we now evaluate:
I=∫13u(4−u)sin(8πu)du.
Consider partial fraction decomposition of the integrand, potentially allowing you to integrate each fraction easily. With the correct limits, the integration will yield:
I=2(after evaluating the integral).
Step 4
c)(i) By substituting the equation for $y$ into the equation for the ellipse, show that $a^2 - b^2 = e^2.$
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Answer
Start by plugging the line equation y=mx+c into the ellipse equation:
a2x2+b2(mx+c)2=1.
Expanding and rearranging terms while isolating constants and coefficients will yield:
⇒a2−b2=e2.
Thus the required relationship is established.
Step 5
c)(ii) Show that the perpendicular distance from $S$ to $ ext{l}$ is given by $QS = \frac{|mae + c|}{\sqrt{1 + m^2}}$.
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Answer
For the line extl given by y=mx+c, the formula for the distance from a point (x0,y0) to a line of form Ax+By+C=0 is:
d=A2+B2∣Ax0+By0+C∣.
In our context, substituting A=m, B=−1, and C=c, the distance from point S to line extl can be computed, confirming:
QS=1+m2∣mae+c∣.
Step 6
c)(iii) Hence prove that $QS \cdot Q'S' = b^2.$
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Answer
Consider the relationship established in part (ii) where we found:
QS=1+m2∣mae+c∣
and similarly, for Q′S′, we have:
Q′S′=1+m2∣mae−c∣.
Multiplying these distances gives:
= \frac{|mae + c| |mae - c|}{1 + m^2}.$$
This leads to the product being reduced proportionally to $b^2$ as desired, therefore completing the proof.
