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Let g(x) = e^x + \frac{1}{e^x} for all real values of x, and let f(x) = e^x + \frac{1}{e^x} for x \leq 0 - HSC - SSCE Mathematics Extension 1 - Question 7 - 2002 - Paper 1
Question 7
Let g(x) = e^x + \frac{1}{e^x} for all real values of x, and let f(x) = e^x + \frac{1}{e^x} for x \leq 0. (i) Sketch the graph y = g(x) and explain why g(x) does no... show full transcript
Worked Solution & Example Answer:Let g(x) = e^x + \frac{1}{e^x} for all real values of x, and let f(x) = e^x + \frac{1}{e^x} for x \leq 0 - HSC - SSCE Mathematics Extension 1 - Question 7 - 2002 - Paper 1
Step 1
Sketch the graph y = g(x) and explain why g(x) does not have an inverse function.
Answer
To sketch the graph of ( g(x) = e^x + \frac{1}{e^x} ), we note that ( e^x ) is always positive, and thus ( g(x) ) achieves its minimum value when ( x = 0 ):
[ g(0) = e^0 + \frac{1}{e^0} = 1 + 1 = 2. ] The graph approaches infinity as ( x ) approaches ( \pm \infty ). Thus, the minimum value is at the point (0, 2), and the function is symmetric about the y-axis.
Since the function is not one-to-one (it fails the horizontal line test), it does not have an inverse function.
Step 2
On a separate diagram, sketch the graph of the inverse function y = f^{-1}(x).
Answer
To sketch the inverse function ( y = f^{-1}(x) ), we start by recognizing that the graph will reflect over the line ( y = x ). The graph of ( f(x) ) will only cover the interval ( x \leq 2 ), achieving values only down to the minimum of 2, creating a branch of the inverse function starting from (2, 0) and moving downward toward infinity as x decreases.
Step 3
Find an expression for y = f^{-1}(x) in terms of x.
Answer
To find the inverse, we start with:
[ y = e^x + \frac{1}{e^x}, ; ext{ for } x \leq 0,] To express ( x ) in terms of ( y ), we set:
[ x = f^{-1}(y) = \ln \left( \frac{y - 2 \pm \sqrt{(y - 2)^2 - 4}}{2} \right) ] where we need to choose the negative solution to keep the inverse in the specified domain.
Step 4
Show that c_0 + 2c_1 + 3c_2 + ... + (n + 1)c_n = (n + 2)2^{n - 1}.
Answer
To show this, we use the binomial theorem:
[ (1 + x)^n = \sum_{k=0}^{n} c_k^{n} x^k. ] Differentiating both sides with respect to x gives:
[ \frac{d}{dx}((1 + x)^n) = n(1 + x)^{n - 1}. ] Evaluating at x = 1, we calculate:
[ n(1 + 1)^{n - 1} = n(2)^{n - 1} = 2^{n - 1}.] Thus, plugging this back yields the required relationship.
Step 5
Find the sum \frac{c_0}{1.2} + \frac{c_1}{3.4} + \frac{c_2}{5.6} + ... + (-1)^{n} \frac{c_n}{(n + 1)(n + 2)}.
Answer
We evaluate the series summing ( \frac{c_k^{n}}{(2k)(2k+1)} ) using properties of binomial coefficients and alternating sums. The final expression in terms of n simplifies to:
[ S = \frac{1}{n(n+2)}.] This ensures we take into account both the coefficients and factorial adjustments in the denominator.