The point P divides the interval from A(−4, −4) to B(1, 6) internally in the ratio 2:3 - HSC - SSCE Mathematics Extension 1 - Question 11 - 2017 - Paper 1

Let ( y = \tan^{-1}(x^3) ).

Using the chain rule, the derivative is:

To solve the inequality:

1. Multiply both sides by (x + 1) (noting that this changes the inequality if x + 1 < 0): $\ 2x > x + 1.$
2. Rearranging gives us: $\ 2x - x > 1 \implies x > 1.$
3. Consider the case when (x + 1) < 0, so ( x < -1 ): $\ 2x < x + 1 \implies x < 1.$
4. Therefore, the solution is: $\ x > 1 \text{ or } x < -1.$

The graph of ( y = 2 \cos^{-1}(x) ) is between the points x = -1 and x = 1. The range of y is between 0 and 2π. At x = -1, y = 2π and at x = 1, y = 0. The graph is a decreasing function, falling from (0, 2π) to (1, 0), creating a concave shape.

1. Let ( x = u^2 - 1 ) then ( dx = 2u , du. )
2. The limits change: when x = 0, u = 1, and when x = 3, u = 2.
3. Substitute into the integral:

4. Calculate:

Using the substitution ( u = \sin(x), ; du = \cos(x) , dx. )

The integral becomes:

$\int u^2 \, du = \frac{u^3}{3} + C = \frac{\sin^3(x)}{3} + C.$

Let ( p = \frac{1}{5} ) and ( q = \frac{4}{5} ). The required probability is given by the binomial formula:

$P(X = 3) = \binom{8}{3} p^3 q^{5} = \binom{8}{3} \left(\frac{1}{5}\right)^{3} \left(\frac{4}{5}\right)^{5}.$

The probability that none produces red flowers is given by:

$P(X = 0) = \binom{8}{0} p^0 q^{8} = \left(\frac{4}{5}\right)^{8}.$

The probability that at least one seedling produces red flowers is the complement of none:

$P(X \geq 1) = 1 - P(X = 0) = 1 - \left(\frac{4}{5}\right)^{8}.$