Question 1 (12 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 1 - 2011 - Paper 1

To differentiate the function $y = \frac{\sin^2{x}}{x}$ with respect to x, we will use the quotient rule. The quotient rule states:

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

Where:

• $u = \sin^2{x}$
• $v = x$

First, calculate the derivatives:

• $u' = 2\sin{x}\cos{x} = \sin{2x}$
• $v' = 1$

Applying the quotient rule gives us:

$\frac{d}{dx}\left( \frac{\sin^2{x}}{x} \right) = \frac{(\sin{2x})(x) - (\sin^2{x})(1)}{x^2} = \frac{x\sin{2x} - \sin^2{x}}{x^2}.$

To solve the inequality $\frac{4 - x}{x} < 1$, we can start by rearranging the terms:

1. Move 1 to the left side: $\frac{4 - x}{x} - 1 < 0$

2. Combine the fractions: $\frac{4 - x - x}{x} < 0$ Simplifying gives: $\frac{4 - 2x}{x} < 0$

3. To solve this inequality, we find the zeros of the numerator and check the sign of the expression. Setting the numerator and denominator to zero:

   • Numerator: $4 - 2x = 0 \Rightarrow x = 2$
   • Denominator: $x = 0$

4. Test the intervals determined by these critical points (x = 0 and x = 2):

   • Interval: (–∞, 0)
   • Interval: (0, 2)
   • Interval: (2, +∞)

5. Check signs for each interval to find where the entire expression is negative:

   • For x < 0, the expression is positive.
   • For 0 < x < 2, the expression is negative (solution).
   • For x > 2, the expression is positive.

Thus, the solution is: $0 < x < 2.$

We start with the integral: $\int_{1}^{4} \frac{\sqrt{x}}{\sqrt{x}} dx$

1. Simplifying the integrand gives us: $\int_{1}^{4} 1 \, dx = [x]_{1}^{4} = 4 - 1 = 3.$

Therefore, the result of the integral is 3.

To find the exact value of $\cos(–1/2)$, we can use the property of cosine that states: $\cos(–x) = \cos(x).$

Thus, $\cos(–1/2) = \cos(1/2).$

This angle is not commonly found in standard trigonometric tables. Therefore, we express it in terms of radians or keep it as is, yielding: $\cos(–1/2) = \cos(1/2).$

To find the range of the function $f(x) = |\ln(x^2 + e)|$, we first analyze the domain of the function:

1. The term $x^2 + e$ is always positive for all real numbers x as $e$ (approximately 2.718) is positive.

2. The natural logarithm $\ln(x^2 + e)$ can take any real value depending on the value of $x^2 + e$.

3. As $x^2$ approaches 0 (when x = 0), we have: $\ln(e) = 1$. As $x^2$ approaches infinity, $\ln(x^2 + e)$ also approaches infinity.

4. Thus, $\ln(x^2 + e)$ can vary from 1 to +∞. The absolute value denotes that the function will take only non-negative values.

Finally, thus the range of the function is: $y \geq 1$.