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question:the sequence {a_n}, a_{1}=2, a_{n+1}= frac {1+a_{n}}{1a_{n}}, then a_{2012}= （ ） A: 2 B: frac {1}{3} C: frac {1}{2} D: 3
answer:Given a_{1}=-2, a_{n+1}= frac {1+a_{n}}{1-a_{n}}, we find a_{2}=- frac {1}{3}, a_{3}= frac {1}{2}, a_{4}=3, a_{5}=-2, ..., it can be seen that the terms in the sequence {a_n} appear in a cycle of 4. Thus, a_{2012}=a_{4+502×4}=a_{4}=3. Therefore, the correct answer is boxed{D}. By calculating the first few terms of the sequence given its first term and the recursive formula, we can determine the cycle of the sequence, which leads to the answer. This problem tests the concept of sequences and their simple representation, focusing on the functional properties of sequences. Identifying the cycle of the sequence is key to solving this problem, making it a basic question.
question:Triangle PQR has sides ( PQ=26 ), ( PR=28 ), and ( QR=30 ). Points ( S ) and ( T ) lie on ( overline{PQ} ) and ( overline{PR} ), respectively, such that ( overline{ST} ) is parallel to ( overline{QR} ) and contains the center of the inscribed circle of triangle ( PQR ). Find ( ST ) as a fraction ( frac{m}{n} ), where ( m ) and ( n ) are relatively prime positive integers. What is ( m+n )?
answer:Let ( I ) be the incenter of ( triangle PQR ). We know ( I ) is the intersection point of the angle bisectors. Therefore, triangles ( triangle PQS ) and ( triangle PRT ) are proportional to ( triangle PQR ) by a scale factor ( k ). The semiperimeter of ( triangle PQR ) is ( s = frac{26 + 28 + 30}{2} = 42 ). Using Heron’s Formula: [ text{Area} = sqrt{s(s-a)(s-b)(s-c)} = sqrt{42 times (42-26) times (42-28) times (42-30)} = sqrt{42 times 16 times 14 times 12} ] The inradius ( r ) is given by: [ r = frac{text{Area}}{s} = frac{sqrt{42 times 16 times 14 times 12}}{42} ] This leads us to scale ( k ) as: [ k = frac{PQ + PS + ST}{PQ + PR + QR} ] Assuming the geometry remains simple with ( DE ) holding the incenter, we set: [ ST = k times QR = frac{26+28}{26+28+30} times 30 = frac{54}{84} times 30 ] Simplifying the values with common factors: [ ST = frac{54}{84} times 30 = frac{9}{14} times 30 = frac{270}{14} = frac{135}{7} ] Here, ( m = 135 ) and ( n = 7 ), therefore ( m+n = 135 + 7 = boxed{142} ).
question:Prove that if the limit (lim _{n rightarrow infty} x_{n}=x) exists, where (left{x_{n}right}) is a sequence of rational numbers, then the limit (lim _{n rightarrow infty} a^{x_{n}}) exists and this limit depends only on (x). Let (a) be a positive number. Define (a^{x}) for an arbitrary (x) as follows. Let (left{x_{n}right}) be a sequence of rational numbers converging to (x). Set (a^{x}=lim _{n rightarrow infty} a^{x_{n}}).
answer:1. **Given Information:** - Sequence ({x_n}) is a sequence of rational numbers converging to (x). - We need to prove that (lim_{n to infty} a^{x_n}) exists and depends only on (x) if (a > 0). 2. **Approach:** - Assume (a > 1). The case for (0 < a < 1) is handled similarly. - Consider auxiliary sequences ({x_n'}) and ({x_n'')}) converging to (x): - ({x_n'}) is monotonically increasing. - ({x_n''}) is monotonically decreasing. 3. **Monotonicity and Limits:** - Let us address the auxiliary sequence ({x_n'}), wherein each term (x_n') is rational and (x_n' to x). - Based on Exercise 27.1, shown previously, the sequence ({a^{x_n'}}) is monotonically increasing. - Since (a^{x_n'}) is increasing and bounded, it converges to some limit (c'). Hence: [ lim_{n to infty} a^{x_n'} = c' ] 4. **Handling the Monotonically Decreasing Sequence:** - Similarly consider the sequence ({x_n''}) where (x_n'') is monotonically decreasing and converging to (x). - Again, based on Exercise 27.1, the sequence ({a^{x_n''}}) is monotonically decreasing. - Since (a^{x_n''}) is decreasing and bounded, it converges to some limit (c''). Hence: [ lim_{n to infty} a^{x_n''} = c'' ] 5. **Equality of Limits of Auxiliary Sequences:** - Notice that (x_n'' - x_n' to 0) as (n to infty). - Using Exercise 27.5 and the property of limits, we know: [ frac{c''}{c'} = lim_{n to infty} a^{x_n'' - x_n'} = a^0 = 1 ] - Therefore, it follows that: [ c'' = c' ] - Denote this common limit by (c). 6. **Conclusion for the Original Sequence:** - For the original sequence ({x_n}), by choosing appropriate subsequences indexed by (k_n), such that: [ x_{k_n}' < x_n < x_{k_n}'' ] - We have: [ lim_{n to infty} a^{x_n} = c ] 7. **Final Result:** [ boxed{lim_{n to infty} a^{x_n} = a^x} ] - This limit depends solely on (x). Hence, we have shown that the limit exists and is only dependent on (x), as required.
question:Given that the random variable xi follows the normal distribution N(1, sigma^2), and P(xi leq 4) = 0.84, find the probability P(xi leq 2).
answer:We are given that the random variable xi is normally distributed with a mean of 1 and variance sigma^2, xi sim N(1, sigma^2). According to the properties of the normal distribution, the curve is symmetric around its mean (the line x = 1 in this case). Since P(xi leq 4) = 0.84, this means that 84% of the distribution lies to the left of 4. To find P(xi leq -2), we need to consider the symmetry of the normal distribution. If P(xi leq 4) = 0.84, then the probability that xi falls to the right of 4 is: P(xi > 4) = 1 - P(xi leq 4) = 1 - 0.84 = 0.16. Because of the symmetry, the area to the left of -2 is the same as the area to the right of 4 on a normally distributed curve centered at 1. Thus, we have: P(xi leq -2) = P(xi > 4) = boxed{0.16}.