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Let $f(x) = \frac{3 + e^{2x}}{4}$ - HSC - SSCE Mathematics Extension 1 - Question 3 - 2009 - Paper 1
Question 3
Let $f(x) = \frac{3 + e^{2x}}{4}$. (i) Find the range of $f(x)$. (ii) Find the inverse function $f^{-1}(x)$. (b) On the same set of axes, sketch the graphs ... show full transcript
Worked Solution & Example Answer:Let $f(x) = \frac{3 + e^{2x}}{4}$ - HSC - SSCE Mathematics Extension 1 - Question 3 - 2009 - Paper 1
Step 1
Find the range of $f(x)$
Answer
To find the range of the function, we analyze the expression (f(x) = \frac{3 + e^{2x}}{4}). Since (e^{2x} > 0) for all real values of (x), we know that the minimum value of (e^{2x}) is zero. Thus, the minimum value of (f(x)) occurs when (e^{2x} = 0), giving (f(x) = \frac{3 + 0}{4} = \frac{3}{4}).
As (x) increases to infinity, (e^{2x}) approaches infinity, causing (f(x)) to also approach infinity. Hence, the range of (f(x)) is (\left[ \frac{3}{4}, \infty \right)).
Step 2
Find the inverse function $f^{-1}(x)$
Answer
To find the inverse function, we start from the equation (y = f(x) = \frac{3 + e^{2x}}{4}). We interchange (x) and (y):
[ x = \frac{3 + e^{2y}}{4} ]
Multiplying both sides by 4:
[ 4x = 3 + e^{2y} ]
Subtracting 3 from both sides gives us:
[ 4x - 3 = e^{2y} ]
Taking the natural logarithm:
[ \ln(4x - 3) = 2y ]
Dividing by 2:
[ y = \frac{1}{2} \ln(4x - 3) ]
Thus, the inverse function is:
[ f^{-1}(x) = \frac{1}{2} \ln(4x - 3) ].
Step 3
On the same set of axes, sketch the graphs of $y = \cos 2x$ and $y = \frac{x + 1}{2}$ for $-\pi \leq x \leq \pi$
Answer
To sketch the graphs, first plot (y = \cos 2x), which oscillates between -1 and 1 with a period of (\pi). It will repeat its values from -\pi to \pi.
Next, draw the line (y = \frac{x + 1}{2}), which is a straight line that intersects the y-axis at (y = \frac{1}{2}) and has a slope of (\frac{1}{2}). The intersection points of these two graphs will show where the solutions to the equation occur.
Step 4
Use your graph to determine how many solutions there are to the equation $2 \cos 2x = x + 1$
Answer
After sketching the graphs, examine the points of intersection. The number of solutions corresponds to the number of intersection points between the curves of (y = 2 \cos 2x) and (y = x + 1). From the graph, count these points to determine the total number of solutions.
Step 5
Use one application of Newton's method to find another approximation to this solution
Answer
Using Newton's method for approximating the root, we start with an initial guess close to the known solution, which is (x = 0.4). The next step involves using the formula:
[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} ]
where (f(x) = 2 \cos 2x - (x + 1)) and (f'(x) = -4 \sin 2x - 1).
Substituting the values yields a closer approximation for the solution to three decimal places.
Step 6
Prove that $\tan^2 \theta = \frac{1 - \cos 2\theta}{1 + \cos 2\theta}$ provided that $\cos 2\theta \neq -1$
Answer
To prove this identity, recall the double angle formulas:
[ \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{1 - \cos^2 \theta}{\cos^2 \theta}.]
Using (\cos 2\theta = 2\cos^2 \theta - 1), rearranging gives us:
[ 1 - \cos 2\theta = 1 - (2\cos^2 \theta - 1) = 2 - 2\cos^2 \theta = 2(1 - \cos^2 \theta) = 2\sin^2 \theta.]
Then, substituting back can establish the desired expression for (\tan^2 \theta) on the left-hand side.
Step 7
Hence find the exact value of $\tan \frac{\pi}{8}$
Answer
Using the established identity, we know:
[ \tan^2 \frac{\pi}{8} = \frac{1 - \cos \frac{\pi}{4}}{1 + \cos \frac{\pi}{4}}.]
Since (\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}), substituting gives:
[ \tan^2 \frac{\pi}{8} = \frac{1 - \frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}} = \frac{2 - \sqrt{2}}{2 + \sqrt{2}}.]
This can be simplified to find the exact value for (\tan \frac{\pi}{8}).