The point P divides the interval joining A(-1,-2) to B(9,3) internally in the ratio 4 : 1 - HSC - SSCE Mathematics Extension 1 - Question 1 - 2011 - Paper 1

To differentiate ( \frac{\sin^2 x}{x} ), we apply the quotient rule, which is given by:

$\frac{d}{dx}\left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$

where ( u = \sin^2 x ) and ( v = x ).

Calculating derivatives:

• ( u' = 2 \sin x \cos x = \sin(2x) )
• ( v' = 1 )

Now, applying the quotient rule:

$\frac{d}{dx}\left( \frac{\sin^2 x}{x} \right) = \frac{\sin(2x) \cdot x - \sin^2 x}{x^2}$

To solve ( \frac{4 - x}{x} < 1 ), we start by simplifying:

$4 - x < x\implies 4 < 2x\implies x > 2$

Since ( x ) cannot be zero (as it is in the denominator), we conclude that the solution is ( x > 2 ).

By substituting ( u = \sqrt{x} ), we find:

• When ( x = 1, u = 1 )
• When ( x = 4, u = 2 )

The integral transforms as follows:

$dx = 2u \, du$

Thus:

$\int_{1}^{4} \frac{\sqrt{x}}{\sqrt{x}} \,dx = \int_{1}^{2} 2u \, du$

Calculating gives:

$= \left[ u^2 \right]_{1}^{2} = 4 - 1 = 3$

So, the final answer is 3.

The angle whose cosine is (-\frac{1}{2}) is ( \frac{2\pi}{3} ) radians or 120 degrees. Thus:

$\cos^{-1}( -\frac{1}{2}) = \frac{2\pi}{3}$

To find the range of ( f(x) = \ln(x^2 + e) ):

1. As ( x^2 + e ) is always greater than or equal to ( e ) (the minimum value occurs when ( x = 0 )), the minimum value of the logarithm is:

$\ln(e) = 1$

2. As ( x ) approaches both positive and negative infinity, ( x^2 + e ) approaches infinity, meaning:

$\ln(x^2 + e) \to \infty$

Thus, the range of ( f(x) ) is ( [1, \infty) ).