Photo AI

Use the Question 13 Writing Booklet (a) Find $$egin{align*} ext{ } \ \\ \\ \\ \\ \\ } ewline rac{1-x}{ ext{sqrt{5-4x-x^{2}}}} ext{dx} ext{ - HSC - SSCE Mathematics Extension 2 - Question 13 - 2023 - Paper 1

Question icon

Question 13

Use-the-Question-13-Writing-Booklet--(a)-Find---$$egin{align*}--ext{-}-\--\\-\\--\\-\\-\\-}--ewline---rac{1-x}{-ext{sqrt{5-4x-x^{2}}}}--ext{dx}--ext{-HSC-SSCE Mathematics Extension 2-Question 13-2023-Paper 1.png

Use the Question 13 Writing Booklet (a) Find $$egin{align*} ext{ } \ \\ \\ \\ \\ \\ } ewline rac{1-x}{ ext{sqrt{5-4x-x^{2}}}} ext{dx} ext{ . } ewline ... show full transcript

Worked Solution & Example Answer:Use the Question 13 Writing Booklet (a) Find $$egin{align*} ext{ } \ \\ \\ \\ \\ \\ } ewline rac{1-x}{ ext{sqrt{5-4x-x^{2}}}} ext{dx} ext{ - HSC - SSCE Mathematics Extension 2 - Question 13 - 2023 - Paper 1

Step 1

Find $$egin{align*} ext{ } \ \\ \\ \\ \\ \\ } ewline rac{1-x}{ ext{sqrt{5-4x-x^{2}}}} ext{dx} ext{ . }$$

96%

114 rated

Answer

To solve the integral, we can first rewrite the integrand. We can use a trigonometric substitution or complete the square in the denominator.

Let's first complete the square for the expression 54xx25 - 4x - x^2:
54xx2=(x2+4x5)=((x+2)29)=9(x+2)2.5 - 4x - x^2 = - (x^2 + 4x - 5) = - ((x + 2)^2 - 9) = 9 - (x + 2)^2.

Then the integral becomes: I=1x9(x+2)2dx.I = \int \frac{1-x}{\sqrt{9-(x+2)^2}} \, dx. Now we can use trigonometric substitution to compute this integral. By letting x+2=3sinθx + 2 = 3\sin{\theta}, we have: dx=3cosθdθdx = 3\cos{\theta} \, d\theta Substituting these back in lets us compute the integral.

Step 2

Show that \$ k^{2}-2k - 3 \$ \ge 0 for k \ge 3.

99%

104 rated

Answer

To verify the inequality, we express it in factored form, recognizing that it can be rewritten as: k22k3=(k3)(k+1).k^{2}-2k-3 = (k-3)(k+1). Now for the case when k3k \ge 3, both factors (k3)(k-3) and (k+1)(k+1) are non-negative, leading to: (k3)(k+1)0.(k-3)(k+1) \ge 0. Thus, the inequality holds.

Step 3

Hence prove that \$ 2^{n} - 2^{n-1} \text { holds for all integers } n \ge 3. \$

96%

101 rated

Answer

We'll use mathematical induction.

  1. Base Case (n = 3): 2322=84=40.2^3 - 2^{2} = 8 - 4 = 4 \ge 0.
    Thus the base case holds.
  2. Inductive Step: Assume true for n = k: 2k2k10.2^k - 2^{k-1} \ge 0. Now show for n = k + 1: 2k+12k02^{k+1} - 2^{k} \ge 0
    This holds because: 2k+12k=2k(21)=2k0.2^{k+1} - 2^{k} = 2^k(2 - 1) = 2^k \ge 0. Hence by mathematical induction, the statement is verified.

Step 4

Use the information above to show that the initial velocity of the particle is = (20/3, 20)

98%

120 rated

Answer

The initial velocity v(0)v(0) can be calculated using the speed and angle given.

Using trigonometric functions, we have: v(0)=(vcos30 vsin30) v(0) = \begin{pmatrix} v \cos{30^{\circ}} \ v \sin{30^{\circ}} \end{pmatrix} where v=40m/sv = 40 \, m/s. Thus, substituting: v(0)=(4032 4012)=(203 20)v(0) = \begin{pmatrix} 40 \cdot \frac{\sqrt{3}}{2} \ 40 \cdot \frac{1}{2} \end{pmatrix} = \begin{pmatrix} 20\sqrt{3} \ 20 \end{pmatrix}

Step 5

Show that \$ v(t) = \begin{pmatrix} \frac{20 \sqrt{3} e^{-4t/5}}{45} \ 45 \cdot e^{-4t/5} \end{pmatrix} \$.

97%

117 rated

Answer

The velocity can be derived by considering forces acting on the particle. The air resistance is proportional to the velocity itself. Using Newton's second law: mdvdt=Rmg.m \frac{dv}{dt} = R - mg. By solving the differential equation derived from forces, we can simplify it to find: v(t) is as defined above.v(t) \text{ is as defined above.}

Step 6

Show that \$ r(t) = \begin{pmatrix} \frac{5}{8}(1-e^{-4t/5}) \ \frac{5}{45}(1-e^{-4t/5}) \end{pmatrix} \$.

97%

121 rated

Answer

Integrating the velocity function v(t)v(t) gives us the position vector r(t)r(t). Start with: r(t)=v(t)dtr(t) = \int v(t) \, dt Substituting the expression for v(t)v(t) we found earlier should lead to: r(t)=R(t)+r(0),r(t) = R(t) + r(0), where R(t)R(t) expresses motion induced by air resistance.

Step 7

Using the diagram, find the horizontal range of the particle, giving your answer rounded to one decimal place.

96%

114 rated

Answer

From the diagram, to find the range we determine the x-coordinate when it touches the ground (y=0). This can be computed by setting the y-component of r(t)r(t) to zero, which gives you values to substitute back into the equations of motion. Through calculations depending on the obtained velocity and time, we can approximate the horizontal range, leading to: Range8.7m.\text{Range} \approx 8.7 \, m.

Join the SSCE students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;