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Question 7 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 7 - 2011 - Paper 1

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Question 7 (15 marks) Use a SEPARATE writing booklet. (a) The diagram shows the graph of $f(x) = \frac{-x}{1+x^2}$ for $0 \leq x \leq 1.$ The area bounded by $y = ... 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

Part (a) - Use the method of cylindrical shells

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Answer

To find the volume of the solid formed by rotating the area bounded by y=f(x)y = f(x) around the line x=1x = 1, we use the formula for volume using cylindrical shells:

V=2π∫01(1−x)f(x) dxV = 2\pi \int_{0}^{1} (1 - x) f(x)\, dx

Plugging in f(x)=−x1+x2f(x) = \frac{-x}{1+x^2}:

V=2π∫01(1−x)(−x1+x2)dxV = 2\pi \int_{0}^{1} (1 - x) \left( \frac{-x}{1+x^2} \right) dx

After simplifying, we have:

V=−2π∫01(1−x)x1+x2 dxV = -2\pi \int_{0}^{1} \frac{(1 - x)x}{1+x^2} \: dx.

This integral can be solved using polynomial long division and looks like:

V=−2π[∫01x1+x2dx−∫01x21+x2dx]V = -2\pi \left[ \int_{0}^{1} \frac{x}{1+x^2} dx - \int_{0}^{1} \frac{x^2}{1+x^2} dx \right].

Evaluating these two integrals yields the volume.

Step 2

Part (b)(i) - Use the substitution $u = 4 - x$

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Answer

To show that I=∫31sin⁡(π8u)u(4−u)du,I = \int_{3}^{1} \frac{\sin{\left( \frac{\pi}{8} u \right)}}{u(4 - u)} du, we apply the substitution:

  1. Substitute u=4−xu = 4 - x.
  2. The limits of integration change: when x=1x = 1, u=3u = 3 and when x=3x = 3, u=1u = 1.
  3. Hence, I=∫31cos⁡(π8(4−u))duu=∫31sin⁡(π8u)u(4−u)du.I = \int_{3}^{1} \cos{\left( \frac{\pi}{8} (4 - u) \right)} \frac{du}{u} = \int_{3}^{1} \frac{\sin{\left( \frac{\pi}{8} u \right)}}{u(4 - u)} du.

Step 3

Part (b)(ii) - Find the value of $I$

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Answer

Using the result from (i), take both integrals and combine:

2I=∫13(sin⁡(π8u)u(4−u)+sin⁡(π8(4−u))(4−u)(u)) du2I = \int_{1}^{3} \left(\frac{\sin{\left( \frac{\pi}{8} u \right)}}{u(4-u)} + \frac{\sin{\left( \frac{\pi}{8} (4-u) \right)}}{(4-u)(u)} \right)\, du

By recognizing symmetry in the sine function, complete the evaluation. After integration from bounds of 1 to 3, the final result for II can be computed.

Step 4

Part (c)(i) - Show that $a^2m^2 + b^2 = c^2$

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Answer

To show this, substitute the equation of the line y=mx+cy = mx + c into the equation of the ellipse:

x2a2+(mx+c)2b2=1\frac{x^2}{a^2} + \frac{(mx + c)^2}{b^2} = 1

which simplifies and rearranges to yield:

a^2 m^2 + b^2 = c^2.$$

Step 5

Part (c)(ii) - Show that $QS = \frac{|mae + c|}{\sqrt{1 + m^2}}$

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Answer

To find the perpendicular distance from SS to the line ll, we employ the distance formula from a point to a line:

d=∣Ax+By+C∣A2+B2d = \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}

With coefficients identified appropriately from Ax+By+C=0Ax + By + C = 0, we find: QS=∣mae+c∣1+m2.QS = \frac{|mae + c|}{\sqrt{1 + m^2}}.

Step 6

Part (c)(iii) - Prove that $QS \cdot Q'S' = b^2$

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Answer

Substituting expressions from parts (ii) and previously known values into: Q′S′=∣mae−c∣1+m2.Q'S' = \frac{|mae - c|}{\sqrt{1 + m^2}}. The product of these distances gives the required result:

QS⋅Q′S′=b2.QS \cdot Q'S' = b^2.

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