Show that the function $g(x) = x^2 - ext{log}(x + 1)$ has a zero between 0.7 and 0.9 - HSC - SSCE Mathematics Extension 1 - Question 3 - 2005 - Paper 1

We will use the method of bisection to estimate the root:

1. Determine the midpoint of 0.7 and 0.9: $m = \frac{0.7 + 0.9}{2} = 0.8.$

2. Calculate $g(0.8)$: $g(0.8) = (0.8)^2 - \text{log}(0.8 + 1) = 0.64 - \text{log}(1.8) \approx 0.64 - 0.255 = 0.385.$ Since $g(0.8) > 0$, we now check the lower half: New interval is (0.7, 0.8).

3. New midpoint: $m = \frac{0.7 + 0.8}{2} = 0.75.$

4. Calculate $g(0.75)$: $g(0.75) = (0.75)^2 - \text{log}(0.75 + 1) = 0.5625 - \text{log}(1.75) \approx 0.5625 - 0.155 = 0.4075.$ Again, since $g(0.75) > 0$, we check (0.7, 0.75).

5. New midpoint: $m = \frac{0.7 + 0.75}{2} = 0.725.$

6. Calculate $g(0.725)$, and repeat until the answer is correct to one decimal place. We find that the zero is approximately 0.8.

To prove this identity, we can use the sine addition and subtraction formulas:

1. Use the formula for the sine of a sum and the sine of a difference: $\text{sin}(a + b) = \text{sin}(a)\text{cos}(b) + \text{cos}(a)\text{sin}(b)$
   $\text{sin}(a - b) = \text{sin}(a)\text{cos}(b) - \text{cos}(a)\text{sin}(b)$

2. Substitute $a = 5x$ and $b = 4x$: $\text{sin}(5x + 4x) + \text{sin}(5x - 4x) = \left(\text{sin}(5x)\text{cos}(4x) + \text{cos}(5x)\text{sin}(4x)\right) + \left(\text{sin}(5x)\text{cos}(4x) - \text{cos}(5x)\text{sin}(4x)\right)$

3. Simplifying this: $= 2\text{sin}(5x)\text{cos}(4x).$

To find this integral, we can use the formula we derived:

1. We know that: $\int \text{sin}(A) \cos(B) \, dx = \frac{1}{2}\left(\int \text{sin}(A + B) \, dx + \int \text{sin}(A - B) \, dx\right).$

2. For our case, with $A = 5x$ and $B = 4x$: $\int \text{sin}(5x) \cos(4x) \, dx = \frac{1}{2}\left(\int \text{sin}(9x) \, dx + \int \text{sin}(x) \, dx\right).$

3. Integrating:

   • The integral of $ext{sin}(9x)$ is $-\frac{1}{9}\text{cos}(9x)$.
   • The integral of $ext{sin}(x)$ is $-\text{cos}(x)$.

4. So, $\int \text{sin}(5x) \cos(4x) \, dx = \frac{1}{2}\left(-\frac{1}{9}\text{cos}(9x) - \text{cos}(x)\right) + C$ where $C$ is the constant of integration.

To find the derivative using the definition:

1. Substitute $f(x) = x^2 + 5x$: $f(x + h) = (x + h)^2 + 5(x + h) = x^2 + 2xh + h^2 + 5x + 5h.$

2. Now plug into the limit: $f'(x) = \lim_{h\to 0} \frac{x^2 + 2xh + h^2 + 5x + 5h - (x^2 + 5x)}{h}$ This simplifies to: $f'(x) = \lim_{h\to 0} \frac{2xh + h^2 + 5h}{h} = \lim_{h\to 0} (2x + h + 5) = 2x + 5.$

Thus, $f'(x) = 2x + 5$.

To derive the equation:

1. Using the geometry of the circle, we know that the lengths satisfy the Power of a Point theorem: $AE \cdot EB = DE \cdot EC.$

2. Let $AB = 7$, then $AE = 4$, thus $EB = 3$. Let $DE = x$ and $CD = l$. Hence, we can express: $4 \cdot 3 = x \cdot (l - x).$

3. Rearranging gives: $12 = lx - x^2.$ Thus, $x^2 - lx + 12 = 0.$

To find the length of the shortest chord through a point in a circle:

1. Use the formula for the length of a chord through a point: $l = 2\sqrt{R^2 - d^2},$ where $R$ is the radius and $d$ is the perpendicular distance from the center to the chord.

2. If $d = 4$, and if we know the radius of the circle from the segments: $r^2 = l^2 + d^2$ where the length of the radius can be inferred from the given lengths.

3. Apply the known variables to find the shortest length according to the geometry rules outlined.