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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 not... 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
(i) Sketch the graph y = g(x) and explain why g(x) does not have an inverse function.
Answer
To sketch the graph of y = g(x) = e^x + \frac{1}{e^x}, first note that the function is defined for all real numbers.
- Behavior at Zero: At x = 0, we have g(0) = 2.
- Limits: As x approaches negative infinity, e^x approaches 0 and thus g(x) approaches 1. As x approaches positive infinity, g(x) also approaches infinity.
- Monotonicity: The derivative g'(x) = e^x - \frac{1}{e^x} is always positive for x > 0, indicating that g(x) is increasing on this interval. However, for x < 0, g'(x) is negative, indicating that g(x) is decreasing. Since g(x) does not pass the horizontal line test, it does not have an inverse function.
Step 2
(ii) On a separate diagram, sketch the graph of the inverse function y = f^{-1}(x).
Answer
To sketch y = f^{-1}(x), first note that f(x) = g(x) for x \leq 0. The graph will reflect the sketch of g(x) for the domain of g(x).
- Reflecting: Since g(x) is not one-to-one, the inverse function will only be a section of the reflection in the line y = x.
- Graph: The inverse function will start from the point (1,0) and increase monotonically, culminating at (2, 0) with the negative x-axis being mirrored.
Step 3
(iii) Find an expression for y = f^{-1}(x) in terms of x.
Answer
To find y = f^{-1}(x), we set x = g(y) = e^y + \frac{1}{e^y}.
- Rearranging: Multiply through by e^y to form the equation ( x e^y = e^{2y} + 1 ).
- Quadratic: Rearranging gives the quadratic form ( e^{2y} - x e^y + 1 = 0 ).
- Solving: Let ( z = e^y ) which leads us to the quadratic formula: ( z = \frac{x \pm \sqrt{x^2 - 4}}{2}. ) Hence, we can express y in terms of x as: [ y = \ln \left( \frac{x + \sqrt{x^2 - 4}}{2} \right). ]
Step 4
(i) Show that c_0 + 2c_1 + 3c_2 + ... + (n + 1)c_n = (n + 2)2^{n - 1}.
Answer
To prove this statement, consider the binomial expansion ( (1+x)^n ):
- Differentiation: Take the derivative of ( (1+x)^n ): [ \frac{d}{dx}((1+x)^n) = n(1+x)^{n-1}. ]
- Evaluate at 1: Substituting x = 1 gives us ( n(1 + 1)^{n-1} = n \cdot 2^{n-1}. )
- Summation: The sum ( c_0 + 2c_1 + 3c_2 + ... + (n + 1)c_n ) can be derived from applying the derivative and evaluating it as shown, ultimately leading to the conclusion that this equals ( (n + 2)2^{n - 1}. )
Step 5
(ii) Find the sum \frac{c_0}{1.2} + \frac{c_1}{2.3} + \frac{c_2}{3.4} + ... + (-1)^n \frac{c_n}{(n + 1)(n + 2)}.
Answer
To calculate this sum, we first express it in terms of binomial coefficients:
- Expression Reformulation: The sum can be written down as: [ S = \sum_{k=0}^{n} (-1)^k \frac{c_k}{(k + 1)(k + 2)}. ]
- Known Summation: Each term involves binomial coefficients, hence similar techniques can be applied such as finding relationships within the expansion of ( (1 + x)^n ).
- Final Result: The result of this sum uses known identities leading to a final expression in terms of n, ultimately shown as ( \frac{(-1)^n}{(n + 1)(n + 2)}. )