Consider the vectors a = 3i + 2j b = -i + 4j - HSC - SSCE Mathematics Extension 1 - Question 11 - 2024 - Paper 1

To find the dot product a · b, we use the following formula:

$a · b = (3i + 2j) · (-i + 4j)$

Calculating this, we get:

$a · b = 3(-1) + 2(4) = -3 + 8 = 5$

Therefore, the result is:

$a · b = 5$

To solve the inequality, we rewrite it as:

$x² - 8x - 9 = 0$

Factoring gives:

$(x - 9)(x + 1) = 0$

The critical points are x = 9 and x = -1. We analyze the intervals:

1. For $(- ext{∞}, -1)$: Choose x = -2, yielding positive.
2. For $(-1, 9)$: Choose x = 0, yielding negative.
3. For $(9, ext{∞})$: Choose x = 10, yielding positive.

Thus, the solution is:

$-1 ≤ x ≤ 9$

First, we perform the substitution:

Let $u = x - 1$, thus $x = u + 1$. Now, we need to adjust the limits and the dx:

$dx = du$

The integral becomes:

∫√(x - 1) dx & = ∫√(u) du\ & = ∫u^{1/2} du\ & = rac{u^{3/2}}{3/2} + C \ & = rac{2}{3}u^{3/2} + C\ & = rac{2}{3}(x - 1)^{3/2} + C. d \ d d d d d d d d d d d d d d\ \ Thus, the result is: $$∫√(x - 1) dx = rac{2}{3}(x - 1)^{3/2} + C$$

We separate variables to solve the differential equation:

rac{dy}{y} = x dx

Integrating both sides:

ext{ln}|y| & = rac{x^2}{2} + C \ ext{y} & = e^{rac{x^2}{2} + C}\ & = e^{C} e^{rac{x^2}{2}}\ & = e^{rac{x^2}{2}} ext{ (let e}^{C} = k ext{, a constant)}. Thus, the final result is: $$y = ke^{rac{x^2}{2}}$$

To differentiate, we apply the chain rule:

f'(x) = rac{1}{ ext{√}(1 - (x^5)^2)} imes (5x^4)

This simplifies to:

f'(x) = rac{5x^4}{ ext{√}(1 - x^{10})}.

Given the volume of a sphere:

V = rac{4}{3}πr³

Taking the derivative with respect to time t:

rac{dV}{dt} = 4πr² rac{dr}{dt}

We know (rac{dV}{dt} = 10 \text{ cm}^3/s),

thus,

10 = 4πr² rac{dr}{dt}

Rearranging gives:

rac{dr}{dt} = rac{10}{4πr²} = rac{5}{2πr²}

This shows the required result.

To find the area of the region R bounded by the curves, we set up the integral:

∫_{0}^{rac{ ext{π}}{2}} (x - sin(x)) dx\ ext{Area }=\ A = egin{aligned} ext{ Area (under } sin(x)): A_s = ∫_0^{rac{ ext{π}}{2}} sin(x)dx = [ - cos(x) ]_0^{rac{ ext{π}}{2}} = -(0 - (-1)) = 1.\ ext{Area (under the line):} A_l = rac{1}{2} (base)(height) ext{ (triangle)}\ = rac{1}{2}(rac{ ext{π}}{2})(rac{ ext{π}}{2}) \ = rac{π²}{8}. ext{So the area} = rac{π²}{8} - 1 = rac{π²}{8} - 1\ ext{Thus area R is: } \ A =rac{π}{8} - 1. d ext{ Add the areas: }\ 1 - rac{ ext{π}^2}{8} \\ Therefore, area R = rac{π}{8} - 1. \ }} d ext{Resulting area R is: } \ A total = rac{π}{8} - 1. Final Area R result: $$A = rac{π}{8} - 1$$.