a) 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, we differentiate:

$\frac{dy}{dx} = \frac{1}{1 + (x^2)^2} \cdot \frac{d}{dx}(x^2) = \frac{1}{1 + x^4} \cdot 2x = \frac{2x}{1 + x^4}$

To solve the inequality:

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

Multiply both sides by ( x + 1 ) (assuming ( x + 1 \neq 0 )):

$2x > x + 1$

Rearranging gives:

$x > 1$.

Thus, the solution is ( x > 1 ).

The function ( y = 2 \cos^{-1}(x) ) is defined for ( x \in [-1, 1] ) and it takes values from ( [0, 2\pi] ). The maximum value occurs at ( x = 0 ) and the minimum values occur at ( x = \pm1 ).

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

The graph will be a decreasing curve from ( (1, 0) ) to ( (-1, 2\pi) ).

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

$dx = 2u \, du$

Change the limits:

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

Thus:

$\int_{0}^{3} \frac{x}{\sqrt{x + 1}} dx = \int_{1}^{2} \frac{u^2 - 1}{u} \cdot 2u \, du = 2 \int_{1}^{2} (u^2 - 1) \, du = 2 \left[ \frac{u^3}{3} - u \right]_{1}^{2}$

Evaluate this gives:

$= 2 \left[ \frac{8}{3} - 2 - (\frac{1}{3} - 1) \right] = \frac{8}{3} - 4 + 2 = \frac{8 - 12 + 6}{3} = \frac{2}{3}$

Using substitution: Let ( u = \sin(x) ), then ( du = \cos(x) dx ).

The integral becomes:

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

Using the binomial probability formula:

$P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k}$

Here, n = 8, k = 3, and p = \frac{1}{5}.

Thus, the expression is:

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

This situation corresponds to k = 0:

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

The probability that at least one produces red flowers is 1 minus the probability that none produces red flowers:

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