Find \[ \int \cos^2(3x) \, dx - HSC - SSCE Mathematics Extension 1 - Question 12 - 2018 - Paper 1

Using the relationship between height ( h ) and the angle ( \theta ):

[ h = 20 , \sin \theta. ]

Differentiating with respect to ( \theta ):

[ \frac{dh}{d\theta} = 20 , \cos \theta. ]

Given that the height of the top of the carriage is 15 m above the diameter, we set:

[ h = 20 + 15 = 35. ]

From ( h = 20 , \sin \theta ):

[ 35 = 20 , \sin \theta ] [ \sin \theta = \frac{35}{20} = 1.75 ] \text{ (not possible)} \text{But when } \theta = \arcsin(1) = \frac{\pi}{2}, \text{ which is maximum height. }

The desired height for the top is at maximum, and speed is given by:

[ \frac{dh}{dt} = \frac{dh}{d\theta} \cdot \frac{d\theta}{dt}. ]

Substituting values,

[ \frac{dh}{dt} = 20 \cos(\theta) \cdot 1.5 , \text{ (converting radians to meters/min)}. ]

Compute the final value for the specific height.

To find the derivative of ( f(x) = \sin^{-1}x + \cos^{-1}x ), we apply the chain rule:

[ f^{\prime}(x) = \frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x^2}} = 0. ]

Since ( f^{\prime}(x) = 0 ), ( f(x) ) is constant. Therefore,

At ( x = 0 ):

[ f(0) = \sin^{-1}(0) + \cos^{-1}(0) = 0 + \frac{\pi}{2} = \frac{\pi}{2}. ]

Thus, ( f(x) = \frac{\pi}{2} ) for all x in its domain.

The function is a constant function equal to ( \frac{\pi}{2} ). Thus, the sketch would be a horizontal line at ( y = \frac{\pi}{2} ) for ( -1 \leq x \leq 1 ).