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{2x}{1 + x^4}. ]

Therefore, ( \frac{d}{dx}(\tan^{-1}(x^2)) = \frac{2x}{1 + x^4} ).

We start by multiplying both sides by ( x + 1 ) (noting that ( x + 1 > 0 ) for this part):

[ 2x > x + 1 ]

Rearranging gives:

[ 2x - x > 1 ]

Thus:

[ x > 1. ]

Next, we find the ranges where this is valid considering signs, leading us to conclude:

( x > 1 ) or ( x < -1 ).

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

Key points on the graph:

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

Plotting these points shows the downward opening curve.

Using the substitution ( x = u^2 - 1 ): [ dx = 2u , du ] Changing the limits:

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

Thus: [ \int_{0}^{3} \frac{x}{\sqrt{x + 1}} ,dx = \int_{1}^{2} \frac{u^2 - 1}{\sqrt{u^2}} \cdot 2u , du ] This simplifies and evaluates to ( \frac{8}{3} ).

Using the identity ( \sin^2 x = \frac{1 - \cos(2x)}{2} ), the integral becomes: [ \int \sin^2 x \cos x ,dx = \int \frac{1 - \cos(2x)}{2} \cos x ,dx. ]

The evaluation gives:\n[ - \frac{1}{3} \sin^3 x + C. ]

Let ( p = \frac{1}{5} ) (probability of red flower) and ( n = 8 ) (total seedlings).

The expression using the binomial probability formula is: [ P(X = 3) = \binom{8}{3} p^3 (1 - p)^{5}. ]

The probability of none producing red flowers (all producing pink flowers) is: [ P(X = 0) = (1 - p)^{8}. ]

The probability of at least one producing red flowers is: [ P(X \geq 1) = 1 - P(X = 0) = 1 - (1 - p)^{8}. ]