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

$y' = \frac{1}{1 + (x^2)^2} \cdot (2x) = \frac{2x}{1 + x^4}.$

Hence, the derivative is ( \frac{2x}{1 + x^4} ).

To solve the inequality ( \frac{2x}{x+1} > 1 ), we start by multiplying both sides by ( x + 1 ) (noting that this is valid for ( x + 1 > 0 )):

$2x > x + 1$

Simplifying gives:

$x > 1.$

Now we need to check the case when ( x + 1 < 0 ) which gives the inequality:

$2x < x + 1 \implies x < 1.$

In conclusion, ( x > 1 ) is valid for the initial assumption, hence the solution is ( x > 1 ).

The function ( y = 2 \cos^{-1}(x) ) has a range of ( [0, 2\pi] ) and is defined for ( x ) in the interval ( [-1, 1] ). The graph starts from ( ( -1, 2\pi) ) and goes to ( (1, 0) ), decreasing monotonically. A sketch of this curve reflects these bounds.

To evaluate the integral, we can use the substitution ( x = u^2 - 1 ). Then, ( dx = 2u du ). Changing the limits accordingly:

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

Substituting gives us:

$\int_{1}^{2} \frac{u^2 - 1}{\sqrt{u^2}} \cdot 2u \, du = 2 \int_{1}^{2} \frac{u^3 - u}{u} \ du = 2 \int_{1}^{2} (u^2 - 1) \, du.$

Calculating the integral:

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

Thus, the evaluated integral is ( \frac{8}{3} ).

To solve ( \int \sin^3 x \cos x , dx. ), we can use substitution:

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

Thus, the integral becomes:

$\int u^3 \, du = \frac{u^4}{4} + C = \frac{\sin^4 x}{4} + C.$

The required integral is ( \frac{\sin^4 x}{4} + C. )

The probability of exactly ( k ) successes in ( n ) independent Bernoulli trials can be given by:

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

where ( p = \frac{1}{5} ), ( n = 8 ), ( k = 3 ). The expression is:

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

The probability that none of the seedlings produces red flowers is:

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

To find the probability that at least one seedling produces red flowers, we can use the complement rule:

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