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) ). To differentiate, we use the chain rule:

( y' = \frac{1}{1 + (x^2)^2} \cdot \frac{d}{dx}(x^2) )

Calculating the derivative: ( = \frac{1}{1 + x^4} \cdot 2x = \frac{2x}{1 + x^4} )

To solve the inequality:

( \frac{2x}{x + 1} > 1 )

First, multiply both sides by (x + 1) (valid since x + 1 is positive if solved separately):

( 2x > x + 1 )

Rearranging gives: ( 2x - x > 1 ) ( x > 1 )

Next, check the restriction:

Make sure ( x + 1 > 0 ) (i.e., x > -1). So the solution is: ( x > 1 )

The function ( y = 2\cos^{-1}(x) ) has a domain of ( x \in [-1, 1] ) and a range of ( y \in [0, 2\pi] ) since the output of ( \cos^{-1}(x) ) goes from 0 to ( \pi ).

For values of x:

• At x = -1, y = 2 ( \cdot \pi ) = 2\pi
• At x = 0, y = 2 ( \cdot \frac{\pi}{2} ) = ( \pi )
• At x = 1, y = 2 ( \cdot 0 ) = 0 The graph is a decreasing curve from (−1, 2π) to (1, 0).

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

• When x = 0, u = 1;
• When x = 3, u = 2.

Also, dx = 2u du. The integral becomes:

$\int_1^{2} \frac{(u^2 - 1)}{\sqrt{u^2}} \cdot 2u \, du$

This simplifies to: $= 2 \int_1^{2} u^2 - 1 \, du$

Continuing to evaluate: $= 2 \left[ \frac{u^3}{3} - u \right]_1^2$

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

We can use substitution: let ( u = \sin x ), thus ( du = \cos x , dx ). The integral becomes:

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

Let p = 1/5 (the probability of producing red flowers) and q = 4/5 (the probability of not producing red flowers). The expression for exactly three seedlings producing red flowers is given by:

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

The probability that none of the eight seedlings produces red flowers (i.e., zero successes) is:

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

Using the complement rule, the probability that at least one of the eight seedlings produces red flowers is:

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