answer:We start by defining the sequence ( s_{n} = a x^{n} + b y^{n} ) for ( n = 1, 2, 3, ldots ). Given the conditions, we have: [ s_1 = a x + b y = 3, ] [ s_2 = a x^2 + b y^2 = 7, ] [ s_3 = a x^3 + b y^3 = 16, ] [ s_4 = a x^4 + b y^4 = 42. ] To find ( s_5 ), we use the recurrence relation derived from the polynomial properties of ( x ) and ( y ). Let us explore the recurrence relation: [ s_{n+2} = a x^{n+2} + b y^{n+2} ] Note that ( ax^{n+2} + by^{n+2} = (ax^{n+1} + by^{n+1})(x + y) - xy (ax^n + by^n) ). Therefore: [ s_{n+2} = (x + y)s_{n+1} - xy s_n ] # Step-by-Step Application of the Recurrence Relation: 1. **Formulate the Given Equations:** From the problem, we need to solve: [ begin{cases} 16 = 7(x + y) - 3xy 42 = 16(x + y) - 7xy end{cases} ] 2. **Solve the System of Equations:** 1. Rearrange the first equation to have: [ 16 = 7(x + y) - 3(xy), ] or equivalently, [ xy = frac{7(x+y) - 16}{3}. ] 2. Similarly, from the second equation: [ 42 = 16(x + y) - 7(xy). ] Substitute the expression from the rearranged first equation ( xy = frac{7(x+y) - 16}{3} ) into the second equation: [ 42 = 16(x+y) - 7 left( frac{7(x+y) - 16}{3} right). ] 3. **Simplify and Solve for ( x+y ) and ( xy ):** Simplifying the equation gives: [ 42 = 16(x+y) - frac{49(x+y) - 112}{3}, ] [ 42 = 16(x+y) - frac{49(x+y)}{3} + frac{112}{3}, ] [ 42 = frac{48(x+y) - 49(x+y)}{3} + frac{112}{3}, ] [ 42 = -frac{1(x+y)}{3} + frac{112}{3}, ] [ 42 = frac{-x+y + 112}{3}, ] draw simplify: [ 126 = -x+y + 112, ] then: [ -xy = 112 - 126, ] Simplify: [ - xy = -14. ] **So:** [ x + y = -14 quad text{and} quad xy = -38. ] # Find ( s_5 ): Finally, using the recurrence relation: [ s_5 = (x+y)s_4 - xy s_3 ] Substitute the known values: [ s_5 = (-14) cdot 42 - (-38) cdot 16 ] Simplify: [ s_5 = -588 + 608 ] [ s_5 = 20 ] Thus, we conclude: [boxed{s_5 = 20}]

answer:1. **Given Sets:** - (A = { (x, y) mid y = ax + 1, x, y in mathbb{R} }) - (B = { (x, y) mid y = |x|, x, y in mathbb{R} }) 2. **Intersection Condition:** - We need to find the values of (a) such that (A cap B) is a singleton set (a set containing exactly one element). 3. **Points of Intersection:** - For (A cap B) to be a singleton, the line (y = ax + 1) must intersect the curve (y = |x|) at exactly one point. 4. **Case Analysis of (y = |x|):** - ( y = |x| ) can be split into two cases: - ( y = x ) when ( x geq 0 ) - ( y = -x ) when ( x < 0 ) 5. **Analyzing each piece of (y = |x|):** - For ( y = x ): [ ax + 1 = x implies (a-1)x + 1 = 0 ] Solving for (x): [ x = - frac{1}{a-1} quad text{provided} quad a neq 1 ] - For ( y = -x ): [ ax + 1 = -x implies (a+1)x + 1 = 0 ] Solving for (x): [ x = - frac{1}{a+1} quad text{provided} quad a neq -1 ] 6. **Ensuring Singleton Intersection:** - For the intersection to be a singleton, both expressions (-frac{1}{a-1}) and (-frac{1}{a+1}) from the two scenarios must yield the same point. - This implies that the line should intersect the curve at just one point, either in the positive (x) case or negative (x) case, but not both. 7. **Examining (a leq -1) and (a geq 1):** - For (a = 1) and (a = -1), the line (y = ax + 1) parallels or coincides with one of the lines defined by (y = |x|), providing two intersection points or none. - Therefore, they fail our condition of being a singleton. - Hence the only valid values of (a) are such that: [ a in (-infty, -1] cup [1, +infty) ] # Conclusion: [ boxed{(-infty, -1] cup [1, +infty)} ]

answer:The function f(x) = log_a{x} is an increasing function on (0, +infty) if and only if a > 1. The function g(x) = (1 - a) cdot a^x is a decreasing function on mathbb{R} if and only if one of the following cases hold: 1. a > 1 and 1 - a < 0, or 2. 0 < a < 1 and 1 - a > 0. Simplifying these conditions, we get a > 1 or 0 < a < 1. Therefore, the statement "the function f(x) = log_a{x} is an increasing function on (0, +infty)" is a sufficient but not necessary condition for the statement "the function g(x) = (1 - a) cdot a^x is a decreasing function on mathbb{R}". Hence, the correct answer is: boxed{text{C}}. This problem primarily tests the understanding of necessary and sufficient conditions and the monotonicity of functions. The key to solving this problem is to find equivalent conditions based on the monotonicity of the functions.

answer:Let ( x = cos 1^circ + i sin 1^circ ). Then, recognizing ( cos theta = frac{x^{theta} + x^{-theta}}{2} ), and that ( sec theta = frac{1}{cos theta} ): [ sec (4k-3)^circ = frac{1}{cos (4k-3)^circ} = frac{2}{x^{4k-3} + x^{-(4k-3)}} ] Our product ( M ) becomes: [ M = prod_{k=1}^{45} sec^2 (4k-3)^circ = prod_{k=1}^{45} left( frac{2}{x^{4k-3} + x^{-(4k-3)}} right)^2 ] Noting that ( x^{4k-3} ) and ( x^{-(4k-3)} ) represent conjugate pairs so their sum is real: Since ( sec theta = sec (360^circ - theta) ), the angles ( (4k-3)^circ ) for ( k = 1 ) to ( 45 ) and their complementary angles cover a complete set of non-repeated angles in one full circle. The product ( sec^2 theta ) for all these angles effectively considers all angles around the circle exactly once due to symmetry. Hence: [ prod_{theta=1}^{360} sec^2 theta = left( frac{1}{cos 1^circ} frac{1}{cos 2^circ} ldots frac{1}{cos 360^circ} right)^2 = 4^{360} ] Since ( sec^2 theta ) for each ( theta ) is considered exactly once in our product, we conclude ( M = 4^{360} ), so ( m = 4 ) and ( n = 360 ). Thus, ( m + n = 4 + 360 = boxed{364} ).