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^2) ). Using the chain rule:

$\frac{dy}{dx} = \frac{1}{1 + (x^2)^2} \cdot \frac{d}{dx}(x^2)$

This simplifies to:

$\frac{dy}{dx} = \frac{2x}{1 + x^4}$

Starting with the inequality:

$\frac{2x}{x + 1} > 1$

Multiply both sides by ( (x + 1)^2 ) (since it is always positive for valid x):

$2x(x + 1) > (x + 1)^2$

This expands to:

$2x^2 + 2x > x^2 + 2x + 1$

Rearranging gives:

$x^2 > 1$

Thus, ( x > 1 \quad \text{or} \quad x < -1 ).

The function ( y = 2 \cos^{-1}(x) ) is defined for ( -1 \leq x , \leq 1 ).

At the boundaries:

• When ( x = -1, y = 2\pi )
• When ( x = 1, y = 0 )

The function decreases from 2π to 0 as x increases from -1 to 1, creating a smooth curve descending.

Using the substitution ( x = u^2 - 1 ), we have:

$dx = 2u \, du$

Changing the limits:

• When ( x = 0, u = 1 )
• When ( x = 3, u = 2 )

The integral becomes:

$\int_1^2 \frac{u^2 - 1}{\sqrt{u^2}} 2u \, du = 2 \int_1^2 u^2 - 1 \, du$

Calculating involves finding the primitives, resulting in:

$= 2 \left[ \frac{u^3}{3} - u \right]_1^2 = 2 \left( \frac{8}{3} - 2 - \frac{1}{3} + 1 \right) = \frac{8}{3}$

Using the substitution ( u = sin x ) which gives ( du = cos x , dx ):

$\int sin^2 x \cos x dx = \int u^2 \, du = \frac{u^3}{3} + C = \frac{sin^3 x}{3} + C$

Using the binomial probability formula:

$P(X = k) = C(n, k) p^k (1 - p)^{n - k}$

For ( n = 8, k = 3, p = \frac{1}{5} ):

$P(X = 3) = C(8, 3) \left( \frac{1}{5} \right)^3 \left( \frac{4}{5} \right)^5$

For none to produce red flowers:

$P(X = 0) = C(8, 0) \left( \frac{1}{5} \right)^0 \left( \frac{4}{5} \right)^8 = \left( \frac{4}{5} \right)^8$

Using the complement rule:

( P(X \geq 1) = 1 - P(X = 0) )

So:

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