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

To compute the dot product:

a · b = (3i + 2j) · (-i + 4j) = 3(-1) + 2(4) = -3 + 8 = 5.

First, we solve the equation:
x² − 8x − 9 = 0.
Factoring, we find:
(x - 9)(x + 1) = 0,
thus, roots are x = 9 and x = -1.
The inequality holds between the roots, leading to:
-1 ≤ x ≤ 9.

With the substitution u = x − 1,
we have: x = u + 1,
dx = du.
Thus, the integral becomes:
∫ √u du = (2/3)u^(3/2) + C = (2/3)(x − 1)^(3/2) + C.

This can be solved using separation of variables:

dy/y = x dx.
Integrating both sides: $ext{ln |y|} = \frac{x^2}{2} + C$ Exponentiating yields: y = e^(x²/2 + C) = e^(C)e^(x²/2) = Ce^(x²/2).

Using the chain rule:

f'(x) = 1/√(1 - (x⁵)²) * 5x⁴ = 5x⁴/√(1 - x^{10}).

Given V = (4/3)πr³, we differentiate:
dV/dt = 4πr²(dr/dt).
Setting dV/dt = 10 cm³/s gives us:

10 = 4πr²(dr/dt).

Solving for dr/dt yields:
dr/dt = 10/(4πr²) = 5/(2πr²) cm/s.

To find the area R bounded by y = sin x and y = x from x = 0 to x = π/2:

Area = ∫₀^(π/2) (x - sin x) dx. Calculating gives:

= [x²/2 - cos x]₀^(π/2)
= (π²/8) - (0 - 1)
= (π²/8) + 1.

Thus, area = (π²/8) - 0 = π²/8.