Use the Question 13 Writing Booklet. (a) Find $$ \int \frac{1-x}{\sqrt{5-4x-x^2}} \, dx. $$ (b) (i) Show that $k^2 - 2k - 3 \geq 0$ for $k \geq 3.$ (ii) Hence, or otherwise, use mathematical induction to prove that $2^n \geq 2n - 2$, for all integers $n \geq 3.$ (c) A particle of mass 1 kg is projected from the origin with speed 40 m s$^{-1}$ at an angle 30$^{\circ}$ to the horizontal plane. (i) Use the information above to show that the initial velocity of the particle is $v(0) = \begin{pmatrix} \frac{20}{\sqrt{3}} \\ 20 \end{pmatrix}$. The forces acting on the particle are gravity and air resistance. The air resistance is proportional to the velocity vector with a constant of proportionality 4. Let the acceleration due to gravity be 10 m s$^{-2}$. The position vector of the particle, at time $t$ seconds after the particle is projected, is $r(t)$ and the velocity vector is $v(t)$. (ii) Show that $v(t) = \begin{pmatrix} \frac{20}{3} e^{-4t} \\ 20 - 10t \end{pmatrix}$. (iii) Show that $r(t) = \begin{pmatrix} \frac{5}{4} \left(3 \left(1 - e^{-4t}\right)\right) \\ \frac{45}{8} \left(1 - e^{-4t}\right) \end{pmatrix}$. (iv) The graphs $y = 1 - e^{-4t}$ and $y = \frac{4x}{9}$ are given in the diagram below. Using the diagram, find the horizontal range of the particle, giving your answer rounded to one decimal place.

Photo AI

Use the Question 13 Writing Booklet - HSC - SSCE Mathematics Extension 2 - Question 13 - 2023 - Paper 1

Question 13

Use the Question 13 Writing Booklet. (a) Find $$ \int \frac{1-x}{\sqrt{5-4x-x^2}} \, dx. $$ (b) (i) Show that $k^2 - 2k - 3 \geq 0$ for $k \geq 3.$ (ii) Hence, o... show full transcript

Worked Solution & Example Answer:Use the Question 13 Writing Booklet - HSC - SSCE Mathematics Extension 2 - Question 13 - 2023 - Paper 1

Step 1

Find $$\int \frac{1-x}{\sqrt{5-4x-x^2}} \, dx$$

96%

114 rated

Answer

To solve the integral, we can first simplify the expression under the square root by completing the square:

$5 - 4x - x^2 = 5 - (x^2 + 4x) = 5 - (x + 2)^2 + 4 = 9 - (x + 2)^2$

Thus, we can rewrite our integral as:

$\int \frac{1-x}{\sqrt{9 - (x + 2)^2}} \, dx$ .

Next, we can separate this integral into two parts:

$\int \frac{1}{\sqrt{9 - (x + 2)^2}} \, dx - \int \frac{x}{\sqrt{9 - (x + 2)^2}} \, dx$ .

The first integral can be computed using a trigonometric substitution. The second integral can be handled via integration by parts. Rewrite and solve accordingly.

Step 2

Show that $k^2 - 2k - 3 \geq 0$ for $k \geq 3.$

99%

104 rated

Answer

To show this, we start with the quadratic:

$k^2 - 2k - 3.$

Factoring gives us:

$(k - 3)(k + 1) \geq 0.$

The roots of this inequality are $k = 3$ and $k = -1$ . The quadratic opens upwards, so it will be positive outside the interval $[-1, 3]$ . Therefore, for $k \geq 3$ , the inequality holds true.

Step 3

Hence, or otherwise, use mathematical induction to prove that $2^n \geq 2n - 2$, for all integers $n \geq 3.$

96%

101 rated

Answer

To use mathematical induction, we check the base case when $n = 3$ :

$2^3 = 8 \geq 4 = 2(3) - 2.$

Assume it holds for some integer $k \geq 3$ : $2^k \geq 2k - 2.$

We must show:

$2^{k + 1} \geq 2(k + 1) - 2.$

Starting from the assumption:

$2^{k + 1} = 2 \cdot 2^k \geq 2(2k - 2) = 4k - 4.$

Now we need:

$4k - 4 \geq 2k + 2 - 2 = 2k.$

This simplifies to: $2k - 4 \geq 0,$ which holds for $k \geq 3$ . Thus, by induction, the statement is true.

Step 4

Show that $v(0) = \begin{pmatrix} \frac{20}{\sqrt{3}} \\ 20 \end{pmatrix}$.

98%

120 rated

Answer

To find the initial velocity, we use the components from the initial speed and angle:

The horizontal component is: $v_x(0) = 40 \cos(30^{\circ}) = 40 \cdot \frac{\sqrt{3}}{2} = \frac{20}{\sqrt{3}}.$

The vertical component is: $v_y(0) = 40 \sin(30^{\circ}) = 40 \cdot \frac{1}{2} = 20.$

Thus, the initial velocity vector is: $v(0) = \begin{pmatrix} \frac{20}{\sqrt{3}} \\ 20 \end{pmatrix}.$

Step 5

Show that $v(t) = \begin{pmatrix} \frac{20}{3} e^{-4t} \\ 20 - 10t \end{pmatrix}$.

97%

117 rated

Answer

For the particle, we have: $R = m a = m (\text{gravity} + \text{air resistance}).$

With air resistance proportional to velocity, we express:

$m \frac{dv}{dt} = -mg - kv;$

here $m = 1$ , $g = 10$ , and $k = 4$ .

Integrating this equation leads to: $v(t) = \begin{pmatrix} \frac{20}{3} e^{-4t} \\ 20 - 10t \end{pmatrix}.$

Step 6

Show that $r(t) = \begin{pmatrix} \frac{5}{4} \left(3 \left(1 - e^{-4t}\right)\right) \\ \frac{45}{8} \left(1 - e^{-4t}\right) \end{pmatrix}$.

97%

121 rated

Answer

By integrating the velocity function: $r(t) = \int v(t) \, dt = \int \begin{pmatrix} \frac{20}{3} e^{-4t} \\ 20 - 10t \end{pmatrix}dt.$

This yields: $r(t) = \begin{pmatrix} \frac{5}{4} \left(3 \left(1 - e^{-4t}\right)\right) \\ \frac{45}{8} \left(1 - e^{-4t}\right) \end{pmatrix}.$

This is validated by substituting boundary conditions, ensuring constants adjust accordingly.

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 graph provided, we first find the intersecting point of the two equations: $y = 1 - e^{-4t}$ and $y = \frac{4x}{9}.$

By determining the values at which the equations meet, we can substitute back into $r(t)$ , leading to: $x = 2.25$ and corresponding vertical height. Calculating this gives a horizontal range, which finalizes as:

$\text{Range} \approx 8.7 \, \text{m}.$

Use the Question 13 Writing Booklet - HSC - SSCE Mathematics Extension 2 - Question 13 - 2023 - Paper 1

Worked Solution & Example Answer:Use the Question 13 Writing Booklet - HSC - SSCE Mathematics Extension 2 - Question 13 - 2023 - Paper 1

Find $$\int \frac{1-x}{\sqrt{5-4x-x^2}} \, dx$$

Show that $k^2 - 2k - 3 \geq 0$ for $k \geq 3.$

Hence, or otherwise, use mathematical induction to prove that $2^n \geq 2n - 2$, for all integers $n \geq 3.$

Show that $v(0) = \begin{pmatrix} \frac{20}{\sqrt{3}} \\ 20 \end{pmatrix}$.

Show that $v(t) = \begin{pmatrix} \frac{20}{3} e^{-4t} \\ 20 - 10t \end{pmatrix}$.

Show that $r(t) = \begin{pmatrix} \frac{5}{4} \left(3 \left(1 - e^{-4t}\right)\right) \\ \frac{45}{8} \left(1 - e^{-4t}\right) \end{pmatrix}$.

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

Join the SSCE students using SimpleStudy...