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1) Compute the product\( \frac{(1998^2 - 1996^2)(1998^2 - 1995^2) \dotsm (1998^2 - 0^2)}{(1997^2 - 1996^2)(1997^2 - 1995^2) \dotsm (1997^2 - 0^2)}.\)

2) Let \(f(x) = \frac{3}{9^x + 3}.\) Find \(f \left( \frac{1}{1001} \right) + f \left( \frac{2}{1001} \right) + f \left( \frac{3}{1001} \right) + \dots + f \left( \frac{1000}{1001} \right).\)

3) Let \(f(x) = \frac{x + 6}{x}.\) The sequence \( (f_n)\) of functions is defined by \(f_1 = f\) and \(f_n = f \circ f_{n - 1}\)for all \(n \ge 2.\) 

For example,

\(f_2(x) = f(f(x)) = \frac{\frac{x + 6}{x} + 6}{\frac{x + 6}{x}} = \frac{7x + 6}{x + 6}\)

and

\(f_3(x) = f(f_2(x)) = \frac{\frac{7x + 6}{x + 6} + 6}{\frac{7x + 6}{x + 6}} = \frac{13x + 42}{7x + 6}.\)

Let S be the set of all real numbers x such that \(f_n(x) = x\)for some positive integer n. Find the number of elements in S.

Thanks :)

 Jul 8, 2019
 #1
avatar+22892 
+3

1) 

Compute the product

\(\dfrac{(1998^2 - 1996^2)(1998^2 - 1995^2) \dotsm (1998^2 - 0^2)}{(1997^2 - 1996^2)(1997^2 - 1995^2) \dotsm (1997^2 - 0^2)}.\)

 

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{(1998^2 - 1996^2)(1998^2 - 1995^2) \dotsm (1998^2 - 0^2)}{(1997^2 - 1996^2)(1997^2 - 1995^2) \dotsm (1997^2 - 0^2)}} \\\\ &=& \dfrac{(1998 - 1996)(1998 + 1996)(1998 - 1995)(1998 + 1995) \dotsm (1998 - 0)(1998 + 0)} {(1997 - 1996)(1997 + 1996)(1997 - 1995)(1997 + 1995) \dotsm (1997 - 0)(1997 + 0)} \\\\ &=& \dfrac{2\cdot 3994 \cdot 3 \cdot 3993 \dotsm 1998\cdot 1998 } {1\cdot 3993 \cdot 2 \cdot 3992 \dotsm 1997\cdot 1997 } \\\\ &=& \dfrac{2\cdot 3 \dotsm 1998 } {1 \cdot 2 \dotsm 1997 } \times \dfrac{3994 \cdot 3993 \dotsm 1998 } {3993 \cdot 3992 \dotsm 1997 } \\\\ &=& 1998 \times \dfrac{3994 } {1997 } \\\\ &=& 1998 \times \dfrac{2\cdot 1997} {1997 } \\\\ &=& 2\cdot 1998 \\ &=& \mathbf{3996 } \\ \hline \end{array}\)

 

laugh

 Jul 8, 2019
 #2
avatar+7709 
+3

2)

\(f(x) = \dfrac{3}{9^x + 3}\\ f(1-x) = \dfrac{3}{9^{1-x}+3}\\ f(1-x) = \dfrac{3\cdot 9^x}{9+3\cdot 9^x}\\ f(1-x) = \dfrac{9^x}{3+9^x}\\ \therefore f(x) + f(1-x) = 1 \quad\forall x\in \mathbb R\)

\(\quad\text{Answer}\\=\displaystyle \sum_{k=1}^{1000}f\left(\dfrac{k}{1001}\right)\\ = \displaystyle \sum_{k=1}^{500} \left(f\left(\dfrac{k}{1001}\right) + f\left(1-\dfrac{k}{1001}\right)\right)\\ = \displaystyle \sum_{k=1}^{500} 1\\ = 500\)

.
 Jul 8, 2019
 #3
avatar+7709 
+2

3)

\(\text{When }f_1 = x,\\ f_2 = f_1(f_1) = f(x) = f_1(x) = x\\ f_3 = f_2(f_1) = f(x) = x\\ \text{When } f_k = x, \\f_{k+1} = f_k(f_1) = f(x) = f_1 (x) = x\\ \implies \text{When }f_1(x) = x, f_n (x) = x \quad \forall x \in \mathbb Z^+\\ \implies \text{The set }S \text{ only contains the solutions to the equation } f_1(x) = x.\\ \text{When }f_1(x) = x,\\ \dfrac{x+6}{x} = x\\ x^2 - x - 6 = 0\\ (x-3)(x + 2) = 0\\ x = 3 \vee x = -2\\ card(S) = 2\)

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 Jul 8, 2019
edited by MaxWong  Jul 8, 2019

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