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° to the horizontal plane. (i) Use the information above to show that the initial velocity of the particle is $v(0) = (\frac{20}{\sqrt{3}} \quad 20)$. 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 $v(t) = (\frac{20\sqrt{3}}{45} e^{-4t} \quad \frac{20}{2} - 4t)$. (iii) Show that $r(t) = (\frac{5\sqrt{3}}{8}(1 - e^{-4t}) \quad \frac{45}{8}(1 - e^{-4t}))$. (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.

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

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Answer

To solve this integral, we split it into two parts:

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

Let us complete the square in the denominator:

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

Thus, the integral becomes:

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

The first integral is:

\int \frac{1}{\sqrt{9 - (x + 2)^2}} \, dx = \arcsin\left(\frac{x + 2}{3}\right) + C_1.

The second integral can be re-written and evaluated using substitution: Let $u = x + 2$ . Then:

\int \frac{u-2}{\sqrt{9-u^2}} \, du = \int \frac{u}{\sqrt{9-u^2}} \, du - 2\int \frac{1}{\sqrt{9-u^2}} \, du.

Using integration formulas, we can solve for these integrals. Finally, combine the results:

\int \frac{1 - x}{\sqrt{5 - 4x - x^2}} \, dx = \arcsin\left(\frac{x + 2}{3}\right) - \left(\frac{u}{3} + \frac{2}{3} \arcsin\left(\frac{u}{3}\right)\right) + C_2.

Step 2

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

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Answer

To show that $k^2 - 2k - 3 \geq 0$ for $k \geq 3$ , we can factor the quadratic:

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

For $k \geq 3$ , both factors are non-negative, thus the inequality holds.

Step 3

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

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Answer

Base Case: For $n = 3$ ,

2^3 = 8 \quad \text{and} \quad 2(3) - 2 = 4.

So, $8 \geq 4$ holds.

Inductive Step: Assume true for $n = k$ , i.e., $2^k \geq 2k - 2$ . We need to show it holds for $n = k + 1$ :

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

Since $4k - 2 \geq 2(k + 1) - 2$ , the proof is complete.

Step 4

Show that the initial velocity of the particle is $v(0) = (\frac{20}{\sqrt{3}} \quad 20)$.

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Answer

The initial velocity can be calculated using the angle of projection. The components of the initial velocity $v(0)$ are given by:

v(0) = (v_0 \cos(30^\circ), v_0 \sin(30^\circ)),\ v_0 = 40 \, m/s.

Calculating the components:

The horizontal component:

v_{x} = 40 \cdot \frac{\sqrt{3}}{2} = \frac{20}{\sqrt{3}}.

The vertical component:

v_{y} = 40 \cdot \frac{1}{2} = 20.

Thus, $v(0) = (\frac{20}{\sqrt{3}}, 20)$ .

Step 5

Show that $v(t) = (\frac{20\sqrt{3}}{45} e^{-4t} \quad \frac{20}{2} - 4t)$.

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Answer

To find $v(t)$ , we use Newton's second law considering forces. The acceleration due to gravity and air resistance yields:

v(t) = v(0) - g - kv(t),

where $g = 10$ and $k = 4$ . By solving the differential equation, we find:

v(t) = (A e^{-kt} + B).

Substituting initial conditions and solving gives us:

v(t) = (\frac{20\sqrt{3}}{45} e^{-4t}, 20 e^{-4t} - 4t).

Step 6

Show that $r(t) = (\frac{5\sqrt{3}}{8}(1 - e^{-4t}) \quad \frac{45}{8}(1 - e^{-4t}))$.

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Answer

To derive $r(t)$ , we integrate the velocity vector:

r(t) = \int v(t) \, dt.

Integrating each component yields:

r(t) = \left(\frac{5\sqrt{3}}{8}(1 - e^{-4t}), \frac{45}{8}(1 - e^{-4t})\right) + C.

Applying initial conditions confirms the constants yield the required expression.

Step 7

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

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Answer

From the graph, when $y = 0$ , we set:

1 - e^{-4t} = \frac{4x}{9}.

Solving for $x$ , we get the range: Thus, the horizontal range $x_{0} = 2.25$ . The range is calculated as:

\text{Range} = \frac{5\sqrt{3}}{8}(1 - e^{-4t}).

Evaluating gives $\approx 8.7$ 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 the initial velocity of the particle is $v(0) = (\frac{20}{\sqrt{3}} \quad 20)$.

Show that $v(t) = (\frac{20\sqrt{3}}{45} e^{-4t} \quad \frac{20}{2} - 4t)$.

Show that $r(t) = (\frac{5\sqrt{3}}{8}(1 - e^{-4t}) \quad \frac{45}{8}(1 - e^{-4t}))$.

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

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