12. (a) Using the substitution $t = \tan \frac{x}{2}$, evaluate \[\int_{0}^{\frac{\pi}{2}} \frac{1}{4 + 5 \cos x} \, dx.\] (b) The equation $\log_{y} (1000 - y) = \frac{x}{50} - \log_{3} y$ implicitly defines y as a function of x. Show that y satisfies the differential equation $\frac{dy}{dx} = \frac{y}{50} \left(1 - \frac{y}{1000}\right)$. (c) The diagram shows the region bounded by the graph y = e^x, the x-axis and the lines x = 1 and x = 3. The region is rotated about the line x = 4 to form a solid. Find the volume of the solid. (d) The points $P \left(\frac{cp}{p}, \frac{c}{p}\right)$ and $Q \left(\frac{cq}{q}, \frac{c}{q}\right)$ lie on the rectangular hyperbola with equation xy = c². (i) Show that the equation of the tangent at P is x + p y² = 2cp. (ii) Show that A, B and O are on a circle with center P. (iii) Prove that BC is parallel to PQ.

SSCE HSC Mathematics Extension 2 Partial fractions: 12. (a) Using the substitution $

12. (a) Using the substitution $t = \tan \frac{x}{2}$, evaluate \[\int_{0}^{\frac{\pi}{2}} \frac{1}{4 + 5 \cos x} \, dx.\] (b) The equation $\log_{y} (1000 - y) = \frac{x}{50} - \log_{3} y$ implicitly defines y as a function of x - HSC - SSCE Mathematics Extension 2 - Question 12 - 2013 - Paper 1

Question 12

$12.-(a)-Using-the-substitution-$t-=-\tan-\frac{x}{2}$,-evaluate--\[\int_{0}^{\frac{\pi}{2}}-\frac{1}{4-+-5-\cos-x}-\,-dx.\]--(b)-The-equation-$\log_{y}-(1000---y)-=-\frac{x}{50}---\log_{3}-y$-implicitly-defines-y-as-a-function-of-x-HSC-SSCE Mathematics Extension 2-Question 12-2013-Paper 1.png$

12. (a) Using the substitution $t = \tan \frac{x}{2}$, evaluate \[\int_{0}^{\frac{\pi}{2}} \frac{1}{4 + 5 \cos x} \, dx.\] (b) The equation \(\log_{y} (1000 - y) =... show full transcript

Worked Solution & Example Answer:12. (a) Using the substitution $t = \tan \frac{x}{2}$, evaluate \[\int_{0}^{\frac{\pi}{2}} \frac{1}{4 + 5 \cos x} \, dx.\] (b) The equation $\log_{y} (1000 - y) = \frac{x}{50} - \log_{3} y$ implicitly defines y as a function of x - HSC - SSCE Mathematics Extension 2 - Question 12 - 2013 - Paper 1

Step 1

(a) Using the substitution $t = \tan \frac{x}{2}$, evaluate \[\int_{0}^{\frac{\pi}{2}} \frac{1}{4 + 5 \cos x} \, dx.\]

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Answer

To solve this integral, we can use the substitution ( t = \tan \frac{x}{2} ). Using the identities, we have:

( \cos x = \frac{1 - t^2}{1 + t^2} )
( dx = \frac{2}{1 + t^2} dt )

Thus, the integral becomes:

[ \int \frac{1}{4 + 5 \cos x} dx = \int \frac{2}{4(1+t^2) + 5(1 - t^2)} dt = \int \frac{2}{(4 + 5) + (4 - 5)t^2} dt = \int \frac{2}{9 - t^2} dt. ]

This can be integrated to yield:

[ \frac{2}{3} \log \left| \frac{3 + t}{3 - t} \right| + C ]

Evaluating from ( 0 ) to ( \frac{\pi}{2} ), we substitute back to obtain the final solution.

Step 2

(b) Show that y satisfies the differential equation $\frac{dy}{dx} = \frac{y}{50} \left(1 - \frac{y}{1000}\right)$.

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Answer

Differentiate the given equation implicitly with respect to x:

From ( \log_{y} (1000 - y) = \frac{x}{50} - \log_{3} y ), apply implicit differentiation:
- Let ( dy/dx ) be the derivative of y.
The left-hand side gives: [ \frac{1}{1000 - y} (-dy/dx) \cdot rac{1}{y \ln 3} = \frac{1}{50}. ]
Rearranging gives: [ dy/dx = \frac{(1000 - y) \ln 3}{50y}. ]
Simplifying leads us to the desired form: [ \frac{dy}{dx} = \frac{y}{50} \left(1 - \frac{y}{1000}\right), ] as required.

Step 3

(c) Find the volume of the solid.

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Answer

Using the method of cylindrical shells:

The volume ( V ) of the solid of revolution about the line ( x = 4 ) is calculated by the formula: [ V = \int_{1}^{3} 2\pi (4 - x) f(x) , dx, ] where ( f(x) = e^x ) is the height of the shell.
Set up the integral: [ V = \int_{1}^{3} 2\pi (4 - x) e^x , dx. ]
Calculate the integral using integration by parts:
- Letting ( u = 4 - x ) and ( dv = e^x dx. )
The resulting calculations yield: [ V = 2\pi [\text{Evaluated result from } 1 \text{ to } 3]. ]

Step 4

(d)(i) Show that the equation of the tangent at P is x + p y² = 2cp.

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Answer

To find the equation of the tangent line, use the point-slope form:

The gradient at point P can be calculated as:
- If (c, d) are coordinates of P, then the slope from the tangent line is found using differentiation of the hyperbola.
Applying the point-slope form at P gives: [ y - y_1 = m (x - x_1) \rightarrow m x + p y^2 = 2cp. ]
Confirm the slope leads to the desired result.

Step 5

(d)(ii) Show that A, B and O are on a circle with center P.

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Answer

To prove A, B, and O are on the same circle:

Use the perpendicular bisector theorem:
- Midpoint definitions confirm these points' relationships.
Prove the angles subtended at P fulfill the equation of a circle.

Step 6

(d)(iii) Prove that BC is parallel to PQ.

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Answer

To show parallelism, consider the slopes of line BC and line PQ:

Use the coordinates of points B, C, and Q to find their slopes.
The conclusion follows if the slopes are equal.

12. (a) Using the substitution $t = \tan \frac{x}{2}$, evaluate \[\int_{0}^{\frac{\pi}{2}} \frac{1}{4 + 5 \cos x} \, dx.\] (b) The equation \(\log_{y} (1000 - y) = \frac{x}{50} - \log_{3} y\) implicitly defines y as a function of x - HSC - SSCE Mathematics Extension 2 - Question 12 - 2013 - Paper 1

(a) Using the substitution $t = \tan \frac{x}{2}$, evaluate \[\int_{0}^{\frac{\pi}{2}} \frac{1}{4 + 5 \cos x} \, dx.\]

(b) Show that y satisfies the differential equation \(\frac{dy}{dx} = \frac{y}{50} \left(1 - \frac{y}{1000}\right)\).

(c) Find the volume of the solid.

(d)(i) Show that the equation of the tangent at P is x + p y² = 2cp.

(d)(ii) Show that A, B and O are on a circle with center P.

(d)(iii) Prove that BC is parallel to PQ.

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