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

To differentiate ( \tan^{-1}(x^2) ), we apply the chain rule:

Let ( y = \tan^{-1}(x^2) ).

Then, the derivative is:

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

Calculating further:

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

To solve the inequality ( \frac{2x}{x + 1} > 1 ), we first subtract 1 from both sides:

$\frac{2x}{x + 1} - 1 > 0$

This gives:

$\frac{2x - (x + 1)}{x + 1} > 0$

which simplifies to:

$\frac{x - 1}{x + 1} > 0$

To find the critical points, set the numerator equal to zero:

1. ( x - 1 = 0 ) gives ( x = 1 )
2. ( x + 1 = 0 ) gives ( x = -1 )

Next, we will test the intervals (-∞, -1), (-1, 1), and (1, ∞):

• For ( x < -1 ), both numerator and denominator are negative, so the quotient is positive.
• For ( -1 < x < 1 ), the numerator is negative and denominator is positive, so the quotient is negative.
• For ( x > 1 ), both are positive, so the quotient is positive.

Thus, the solution is ( x < -1 ) or ( x > 1 ).

To sketch the graph of ( y = 2 \cos^{-1}(x) ), we identify the domain and range:

• The domain of ( \cos^{-1}(x) ) is ([-1, 1]).
• The range of ( \cos^{-1}(x) ) is ([0, \pi]).

Thus, the range of ( y = 2 \cos^{-1}(x) ) will be ([0, 2\pi]).

Next, we can plot key points:

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

The graph will start at ( ( -1, 2\pi ) ), peak at ( ( 0, \pi ) ), and end at ( ( 1, 0 ) ), with a downward concave shape.

To evaluate the integral, we will perform a substitution: Let ( x = u^2 - 1 ), then ( dx = 2u , du ).

Next, we adjust the limits of integration:

• When ( x = 0 ), ( u = \sqrt{1} = 1 ).
• When ( x = 3 ), ( u = \sqrt{4} = 2 ).

Now, we can rewrite the integral:

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

This equals: $= 2 \left[ \frac{u^3}{3} - u \right]_1^2$

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

Simplifying: $= 2 \left[ \frac{8}{3} - \frac{1}{3} - 1 \right] = 2 \left[ \frac{7}{3} - 1 \right] = 2 \left[ \frac{7 - 3}{3} \right] = 2 \cdot \frac{4}{3} = \frac{8}{3}$

To evaluate the integral ( \int \sin^2 x \cos x , dx ), we can use a substitution: Let ( u = \sin x ), so that ( du = \cos x , dx ).

Rewriting the integral in terms of u, we have:

$\int u^2 \, du$

Now, integrate:

$= \frac{u^3}{3} + C = \frac{\sin^3 x}{3} + C$

Let ( p = \frac{1}{5} ) be the probability of producing red flowers. We are looking for an expression of the probability of exactly k successes in n trials:

Using the binomial probability formula:

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

For our case where ( n = 8 ) and ( k = 3 ):

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

Using a similar approach:

• The probability of not producing red flowers is ( 1 - p = \frac{4}{5} ).
• For none of the eight (k = 0):

The expression will be:

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

We can find this probability by using the complement rule:

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

From our previous part:

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