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 use the chain rule. Let:

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

Thus,

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

Calculating ( \frac{d}{dx}(x^2) ) gives:

$\frac{d}{dx}(x^2) = 2x$

So,

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

To solve ( \frac{2x}{x + 1} > 1 ), we first rearrange the inequality:

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

This can be rewritten as:

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

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

Next, we find the critical points by setting the numerator and denominator to zero:

1. Numerator: ( x - 1 = 0 ) gives ( x = 1 )
2. Denominator: ( x + 1 = 0 ) gives ( x = -1 )

Now, we test the intervals created by these critical points. The intervals are ((-\infty, -1)\

To evaluate the integral ( \int_0^{3} \frac{x}{\sqrt{x + 1}} dx ), we use the substitution ( x = u^2 - 1 ).

Then, we find:

1. If ( x = 0 ), then ( u^2 = 1 ) so ( u = 1 )
2. If ( x = 3 ), ( u^2 = 4 ) so ( u = 2 )

So, our new limits of integration will be from 1 to 2. We also need the differential:

$dx = 2u \, du$

Now, substituting into the integral:

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

Calculating:

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

$$= 2 \left[ \frac{8}{3} - 2 + 1 - \frac{1}{3} \right] = 2 \left[ \frac{5}{3} \right] = \frac{10}{3}.$

To find ( \int \sin^2(x) \cos(x) dx ), we can use the substitution method:

Let ( u = \sin(x) ) then ( du = \cos(x) dx ). Thus:

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

To find the probability that exactly three out of the eight seedlings produce red flowers, we can use 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:

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

To determine the probability that none of the eight seedlings produces red flowers, we again use the binomial probability formula. Here, we have:

$P(X = 0) = \binom{8}{0} p^0 (1 - p)^{8}$

Since ( \binom{8}{0} = 1 ) and ( p = \frac{1}{5} ):

Thus:

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

To find the probability that at least one of the eight seedlings produces red flowers, we're looking for the complement of none producing red flowers.

Thus:

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

We already calculated ( P(X = 0) ):

Therefore:

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