💡Okay, so I have this equation: sin α + 2i cos α = 2i. Hmm, I need to find the range of values for α. Alright, let me think about how to approach this.First, I remember that complex numbers can be expressed in the form a + ib, where a is the real part and b is the imaginary part. So in this equation, sin α is the real part, and 2 cos α is the coefficient of the imaginary part. On the right side of the equation, we have 2i, which can be written as 0 + 2i. That means the real part is 0 and the imaginary part is 2.So, if I equate the real parts and the imaginary parts separately, I should get two equations:1. Real part: sin α = 02. Imaginary part: 2 cos α = 2Let me write that down:1. sin α = 02. 2 cos α = 2Starting with the first equation, sin α = 0. I know that sine of an angle is zero at integer multiples of π. So, α = kπ, where k is any integer. That gives me possible solutions like 0, π, 2π, 3π, and so on, as well as negative multiples like -π, -2π, etc.Now, moving on to the second equation: 2 cos α = 2. If I divide both sides by 2, I get cos α = 1. The cosine of an angle is 1 at even multiples of π. So, α = 2kπ, where k is any integer. That gives me solutions like 0, 2π, 4π, etc., and the negative counterparts like -2π, -4π, etc.Wait a minute, so from the first equation, α can be any integer multiple of π, but from the second equation, α has to be an even multiple of π. That means the solutions have to satisfy both conditions. So, the values of α must be the ones that are even multiples of π because those are the ones that make both sin α = 0 and cos α = 1.Let me double-check that. If α = 0, then sin 0 = 0 and cos 0 = 1, so plugging into the original equation: 0 + 2i*1 = 2i, which matches. If α = π, then sin π = 0, but cos π = -1, so plugging in: 0 + 2i*(-1) = -2i, which is not equal to 2i. So α = π doesn't work. Similarly, α = 2π: sin 2π = 0, cos 2π = 1, so 0 + 2i*1 = 2i, which works. So it seems like only the even multiples of π satisfy both equations.Therefore, the range of α is all real numbers such that α = 2kπ, where k is any integer. Looking back at the options provided:A: α = kπ, k ∈ ℤB: α = (kπ)/2, k ∈ ℤC: α = 2kπ, k ∈ ℤD: α = 2kπ + π/2, k ∈ ℤSo, option C matches my conclusion. Let me just make sure I didn't make any mistakes. I equated the real and imaginary parts correctly, solved both equations, and checked that only the even multiples of π satisfy both conditions. Yeah, that seems right.I think I'm confident now that the correct answer is option C.**Final Answer**boxed{C}

💡Okay, so I have this problem where I need to prove that for a triangle with sides (a), (b), and (c), and inradius (r), the following inequality holds:[frac{1}{a^2} + frac{1}{b^2} + frac{1}{c^2} leq frac{1}{4r^2}]Hmm, let me start by recalling some basic properties of triangles and inradius. The inradius (r) of a triangle is related to its area (A) and semi-perimeter (S) by the formula:[A = r cdot S]Where (S = frac{a + b + c}{2}). Also, Heron's formula tells us that the area (A) can be expressed as:[A = sqrt{S(S - a)(S - b)(S - c)}]So, combining these two expressions for the area, we have:[sqrt{S(S - a)(S - b)(S - c)} = r cdot S]Squaring both sides gives:[S(S - a)(S - b)(S - c) = r^2 S^2]Simplifying, we can divide both sides by (S) (assuming (S neq 0), which it isn't for a valid triangle):[(S - a)(S - b)(S - c) = r^2 S]I'm not sure if this is immediately helpful, but maybe I can express (r) in terms of (a), (b), and (c). Let me think.Another approach might be to use substitution. Let me denote (x = S - a), (y = S - b), and (z = S - c). Then, (x), (y), and (z) are all positive numbers since they represent the distances from the sides to the inradius. Also, we have:[a = y + z, quad b = z + x, quad c = x + y]So, substituting these into the inequality, we get:[frac{1}{(y + z)^2} + frac{1}{(z + x)^2} + frac{1}{(x + y)^2} leq frac{1}{4r^2}]But from earlier, we have:[(S - a)(S - b)(S - c) = r^2 S]Substituting (x), (y), (z) into this, we get:[xyz = r^2 S]But (S = x + y + z), so:[xyz = r^2 (x + y + z)]Therefore, (r^2 = frac{xyz}{x + y + z}). So, substituting back into the inequality, we have:[frac{1}{(y + z)^2} + frac{1}{(z + x)^2} + frac{1}{(x + y)^2} leq frac{1}{4 cdot frac{xyz}{x + y + z}} = frac{x + y + z}{4xyz}]So, the inequality we need to prove becomes:[frac{1}{(y + z)^2} + frac{1}{(z + x)^2} + frac{1}{(x + y)^2} leq frac{x + y + z}{4xyz}]Hmm, okay, so maybe I can manipulate the left-hand side to see if it's less than or equal to the right-hand side.Let me consider each term on the left-hand side. For example, take (frac{1}{(y + z)^2}). I wonder if I can find an upper bound for this term in terms of (x), (y), and (z).I recall that for positive real numbers, the arithmetic mean is greater than or equal to the harmonic mean. Maybe I can use that here. Specifically, for two positive numbers (y) and (z), we have:[frac{y + z}{2} geq frac{2}{frac{1}{y} + frac{1}{z}}]But I'm not sure if this directly helps. Alternatively, I know that ((y + z)^2 geq 4yz) by the AM-GM inequality, since:[frac{y + z}{2} geq sqrt{yz} implies y + z geq 2sqrt{yz} implies (y + z)^2 geq 4yz]Taking reciprocals (since both sides are positive), we get:[frac{1}{(y + z)^2} leq frac{1}{4yz}]Ah, that's useful! So, applying this to each term on the left-hand side, we have:[frac{1}{(y + z)^2} leq frac{1}{4yz}, quad frac{1}{(z + x)^2} leq frac{1}{4zx}, quad frac{1}{(x + y)^2} leq frac{1}{4xy}]Adding these three inequalities together, we get:[frac{1}{(y + z)^2} + frac{1}{(z + x)^2} + frac{1}{(x + y)^2} leq frac{1}{4yz} + frac{1}{4zx} + frac{1}{4xy}]Factor out (frac{1}{4}):[frac{1}{(y + z)^2} + frac{1}{(z + x)^2} + frac{1}{(x + y)^2} leq frac{1}{4} left( frac{1}{yz} + frac{1}{zx} + frac{1}{xy} right )]Now, let's look at the right-hand side of the inequality we need to prove:[frac{x + y + z}{4xyz}]So, we have:[frac{1}{4} left( frac{1}{yz} + frac{1}{zx} + frac{1}{xy} right ) = frac{1}{4} left( frac{x + y + z}{xyz} right ) = frac{x + y + z}{4xyz}]Therefore, combining the two results, we have:[frac{1}{(y + z)^2} + frac{1}{(z + x)^2} + frac{1}{(x + y)^2} leq frac{x + y + z}{4xyz}]Which is exactly what we needed to prove. So, substituting back, we have:[frac{1}{a^2} + frac{1}{b^2} + frac{1}{c^2} leq frac{1}{4r^2}]Therefore, the inequality holds.**Final Answer**boxed{frac{1}{a^2}+frac{1}{b^2}+frac{1}{c^2}le frac{1}{4r^2}}

💡Okay, so I have this problem about a sequence where the first term is 'a' and the second term is 'b'. The recurrence relation is given by 2a_{n+2} = a_{n+1} + a_n. And it's also given that the limit of the sum of the first n terms as n approaches infinity is 4. I need to find the values of 'a' and 'b'.First, let me try to understand the recurrence relation. It says that twice the term two places ahead is equal to the sum of the next term and the current term. So, rearranging that, I can write a_{n+2} = (a_{n+1} + a_n)/2. That seems like each term is the average of the two previous terms. Interesting.Since it's a linear recurrence relation, maybe I can solve it using characteristic equations or something like that. But I'm not too familiar with that method. Maybe I can find a pattern or express the terms in terms of 'a' and 'b'.Let me write out the first few terms to see if I can spot a pattern.Given:a1 = aa2 = ba3 = (a2 + a1)/2 = (b + a)/2a4 = (a3 + a2)/2 = [(b + a)/2 + b]/2 = (a + 3b)/4a5 = (a4 + a3)/2 = [(a + 3b)/4 + (a + b)/2]/2 = [(a + 3b) + 2(a + b)]/8 = (3a + 5b)/8a6 = (a5 + a4)/2 = [(3a + 5b)/8 + (a + 3b)/4]/2 = [(3a + 5b) + 2(a + 3b)]/16 = (5a + 11b)/16Hmm, I see the coefficients of 'a' and 'b' are following some pattern. Let me list them:n | a_n---|---1 | a2 | b3 | (a + b)/24 | (a + 3b)/45 | (3a + 5b)/86 | (5a + 11b)/16Looking at the coefficients of 'a': 1, 0, 1/2, 1/4, 3/8, 5/16...Coefficients of 'b': 0, 1, 1/2, 3/4, 5/8, 11/16...It seems like the coefficients are following a pattern similar to Fibonacci numbers but scaled by powers of 1/2. Maybe I can express a_n as a combination of 'a' and 'b' with coefficients involving Fibonacci numbers and powers of 1/2.Alternatively, maybe I can express the difference between consecutive terms. Let's try that.Let me define d_n = a_{n+1} - a_n.Then, from the recurrence relation:a_{n+2} = (a_{n+1} + a_n)/2So, a_{n+2} - a_{n+1} = (a_{n+1} + a_n)/2 - a_{n+1} = (-a_{n+1} + a_n)/2 = (-1/2)(a_{n+1} - a_n) = (-1/2)d_nTherefore, d_{n+1} = (-1/2)d_nThis is a geometric sequence for the differences d_n. So, d_n = d_1 * (-1/2)^{n-1}Given that d_1 = a2 - a1 = b - aTherefore, d_n = (b - a)(-1/2)^{n-1}So, the difference between consecutive terms is a geometric sequence with ratio -1/2.Now, to find a_n, we can sum up the differences:a_n = a1 + sum_{k=1}^{n-1} d_k = a + sum_{k=1}^{n-1} (b - a)(-1/2)^{k-1}This is a geometric series. The sum of the first m terms of a geometric series with first term c and ratio r is c*(1 - r^m)/(1 - r)So, sum_{k=1}^{n-1} (b - a)(-1/2)^{k-1} = (b - a)*(1 - (-1/2)^{n-1}) / (1 - (-1/2)) = (b - a)*(1 - (-1/2)^{n-1}) / (3/2) = (2/3)(b - a)*(1 - (-1/2)^{n-1})Therefore, a_n = a + (2/3)(b - a)*(1 - (-1/2)^{n-1})Simplify that:a_n = a + (2/3)(b - a) - (2/3)(b - a)(-1/2)^{n-1}Which can be written as:a_n = [a + (2/3)(b - a)] + [ - (2/3)(b - a)(-1/2)^{n-1} ]Simplify the first part:a + (2/3)(b - a) = (3a + 2b - 2a)/3 = (a + 2b)/3So, a_n = (a + 2b)/3 - (2/3)(b - a)(-1/2)^{n-1}Now, the problem is about the sum S_n = a1 + a2 + ... + a_n, and the limit as n approaches infinity of S_n is 4.So, let's compute S_n.S_n = sum_{k=1}^n a_k = sum_{k=1}^n [ (a + 2b)/3 - (2/3)(b - a)(-1/2)^{k-1} ]We can split this sum into two parts:S_n = sum_{k=1}^n (a + 2b)/3 - sum_{k=1}^n (2/3)(b - a)(-1/2)^{k-1}Compute each sum separately.First sum: sum_{k=1}^n (a + 2b)/3 = n*(a + 2b)/3Second sum: (2/3)(b - a) * sum_{k=1}^n (-1/2)^{k-1}Again, this is a geometric series with first term 1 and ratio -1/2. The sum of the first n terms is [1 - (-1/2)^n] / [1 - (-1/2)] = [1 - (-1/2)^n] / (3/2) = (2/3)[1 - (-1/2)^n]Therefore, the second sum becomes:(2/3)(b - a) * (2/3)[1 - (-1/2)^n] = (4/9)(b - a)[1 - (-1/2)^n]Putting it all together:S_n = n*(a + 2b)/3 - (4/9)(b - a)[1 - (-1/2)^n]Now, take the limit as n approaches infinity.First, consider the term n*(a + 2b)/3. For the limit to be finite (since it's given as 4), the coefficient of n must be zero. Otherwise, as n grows, S_n would go to infinity or negative infinity.So, set the coefficient of n to zero:(a + 2b)/3 = 0 => a + 2b = 0 => a = -2bNow, substitute a = -2b into the expression for S_n.First, compute (a + 2b)/3 = 0, as we already set.Then, compute (b - a):b - a = b - (-2b) = 3bSo, the second term becomes:(4/9)(3b)[1 - (-1/2)^n] = (12/9)b[1 - (-1/2)^n] = (4/3)b[1 - (-1/2)^n]Therefore, S_n = 0 - (4/3)b[1 - (-1/2)^n] = - (4/3)b + (4/3)b*(-1/2)^nNow, take the limit as n approaches infinity.The term (4/3)b*(-1/2)^n approaches zero because |(-1/2)| < 1, so it goes to zero.Therefore, lim_{n->infty} S_n = - (4/3)b = 4So, - (4/3)b = 4 => b = 4 * (-3/4) = -3Now, since a = -2b, and b = -3, then a = -2*(-3) = 6So, the values are a = 6 and b = -3.Let me double-check my steps to make sure I didn't make any mistakes.1. I started by writing out the recurrence relation and expressing a_{n+2} in terms of a_{n+1} and a_n.2. Then, I defined the difference d_n = a_{n+1} - a_n and found that it's a geometric sequence with ratio -1/2.3. I summed the differences to express a_n in terms of a and b, which gave me an expression involving a geometric series.4. Then, I computed the sum S_n by summing a_n from k=1 to n, splitting it into two sums.5. I recognized that for the limit to be finite, the coefficient of n must be zero, leading to a = -2b.6. Substituted a = -2b into the expression for S_n and found that b = -3, hence a = 6.Everything seems to check out. I think the values are correct.

💡Okay, so I have this equation: ( x^3 - 2a x^2 + (a^2 + 2)x = 4a - frac{4}{x} ). The question is asking for the range of the real number ( a ) such that this equation has four distinct positive roots. The options are given as A: ( a > 3sqrt{2} ), B: ( a > 2sqrt{2} ), C: ( 2sqrt{2} < a < 3sqrt{2} ), and D: ( -3sqrt{2} < a < 3sqrt{2} ).First, I need to understand the equation better. It's a cubic equation on the left side and a term with ( frac{4}{x} ) on the right. Since we're dealing with positive roots, ( x ) must be positive, so I don't have to worry about division by zero or negative values complicating things.Maybe I can start by moving all terms to one side to set the equation to zero. Let me try that:( x^3 - 2a x^2 + (a^2 + 2)x - 4a + frac{4}{x} = 0 ).Hmm, that looks a bit messy because of the ( frac{4}{x} ) term. Maybe I can multiply both sides by ( x ) to eliminate the denominator. Let's try that:Multiplying each term by ( x ) gives:( x^4 - 2a x^3 + (a^2 + 2)x^2 - 4a x + 4 = 0 ).Okay, now it's a quartic equation: ( x^4 - 2a x^3 + (a^2 + 2)x^2 - 4a x + 4 = 0 ). Quartic equations can be tricky, but perhaps this one can be factored or simplified somehow.Looking at the equation, I notice that it's a fourth-degree polynomial, and we're told it has four distinct positive roots. So, all roots are positive and distinct. That gives me some information about the possible factors or the behavior of the polynomial.Maybe I can factor this quartic equation. Let me try to see if it can be factored into quadratics or something else.Let me attempt to factor it as a product of two quadratics:( (x^2 + p x + q)(x^2 + r x + s) = x^4 - 2a x^3 + (a^2 + 2)x^2 - 4a x + 4 ).Expanding the left side:( x^4 + (p + r)x^3 + (q + s + pr)x^2 + (ps + qr)x + q s ).Now, equate the coefficients:1. Coefficient of ( x^4 ): 1 = 1, which is fine.2. Coefficient of ( x^3 ): ( p + r = -2a ).3. Coefficient of ( x^2 ): ( q + s + p r = a^2 + 2 ).4. Coefficient of ( x ): ( p s + q r = -4a ).5. Constant term: ( q s = 4 ).So, we have a system of equations:1. ( p + r = -2a )2. ( q + s + p r = a^2 + 2 )3. ( p s + q r = -4a )4. ( q s = 4 )This seems complicated, but maybe I can make some assumptions to simplify it. Let me assume that ( q = s ). Then, from equation 4, ( q^2 = 4 ), so ( q = 2 ) or ( q = -2 ). Since we're dealing with positive roots, maybe ( q ) and ( s ) should be positive? Let me check.If ( q = 2 ), then ( s = 2 ). Let's see if that works.So, ( q = s = 2 ).Then, equation 3 becomes: ( p * 2 + 2 * r = -4a ) => ( 2p + 2r = -4a ) => ( p + r = -2a ). But from equation 1, ( p + r = -2a ). So that's consistent.Equation 2 becomes: ( 2 + 2 + p r = a^2 + 2 ) => ( 4 + p r = a^2 + 2 ) => ( p r = a^2 - 2 ).So, now, we have:1. ( p + r = -2a )2. ( p r = a^2 - 2 )This is a system of equations in ( p ) and ( r ). It looks like the sum and product of two numbers, so we can think of ( p ) and ( r ) as roots of the quadratic equation ( t^2 - (p + r) t + p r = 0 ), which becomes ( t^2 + 2a t + (a^2 - 2) = 0 ).Let me compute the discriminant of this quadratic:Discriminant ( D = (2a)^2 - 4 * 1 * (a^2 - 2) = 4a^2 - 4a^2 + 8 = 8 ).Since the discriminant is positive, there are two distinct real roots for ( p ) and ( r ). So, ( p ) and ( r ) are real and distinct.Therefore, the quartic can be factored as ( (x^2 + p x + 2)(x^2 + r x + 2) ), where ( p ) and ( r ) satisfy ( p + r = -2a ) and ( p r = a^2 - 2 ).So, now, the quartic equation factors into two quadratics:( (x^2 + p x + 2)(x^2 + r x + 2) = 0 ).Therefore, the roots of the quartic are the roots of each quadratic. So, each quadratic ( x^2 + p x + 2 = 0 ) and ( x^2 + r x + 2 = 0 ) must have two distinct positive roots each.Wait, but each quadratic is ( x^2 + p x + 2 ). For a quadratic ( x^2 + b x + c ), the roots are real and distinct if the discriminant is positive, and both roots are positive if the sum of roots is positive and the product is positive.Given that ( x^2 + p x + 2 = 0 ), the discriminant is ( p^2 - 8 ). For real and distinct roots, we need ( p^2 - 8 > 0 ) => ( |p| > 2sqrt{2} ).Similarly, for the quadratic ( x^2 + r x + 2 = 0 ), discriminant ( r^2 - 8 > 0 ) => ( |r| > 2sqrt{2} ).But since ( p + r = -2a ), and ( p r = a^2 - 2 ), we can analyze the conditions.Let me denote ( p ) and ( r ) as roots of the quadratic ( t^2 + 2a t + (a^2 - 2) = 0 ). So, ( p ) and ( r ) are:( t = frac{ -2a pm sqrt{(2a)^2 - 4(a^2 - 2)} }{2} = frac{ -2a pm sqrt{8} }{2} = -a pm sqrt{2} ).So, ( p = -a + sqrt{2} ) and ( r = -a - sqrt{2} ), or vice versa.Therefore, ( p = -a + sqrt{2} ) and ( r = -a - sqrt{2} ).Now, we need both quadratics ( x^2 + p x + 2 ) and ( x^2 + r x + 2 ) to have two distinct positive roots each.So, for each quadratic ( x^2 + p x + 2 = 0 ), the roots are:( x = frac{ -p pm sqrt{p^2 - 8} }{2} ).Similarly, for ( x^2 + r x + 2 = 0 ), the roots are:( x = frac{ -r pm sqrt{r^2 - 8} }{2} ).Since we need the roots to be positive, the expressions ( frac{ -p pm sqrt{p^2 - 8} }{2} ) and ( frac{ -r pm sqrt{r^2 - 8} }{2} ) must be positive.Let me analyze the first quadratic with ( p = -a + sqrt{2} ):The roots are:( x = frac{ -(-a + sqrt{2}) pm sqrt{ (-a + sqrt{2})^2 - 8 } }{2} = frac{ a - sqrt{2} pm sqrt{ a^2 - 2asqrt{2} + 2 - 8 } }{2} ).Simplify the discriminant:( a^2 - 2asqrt{2} + 2 - 8 = a^2 - 2asqrt{2} - 6 ).So, the roots are:( x = frac{ a - sqrt{2} pm sqrt{ a^2 - 2asqrt{2} - 6 } }{2} ).Similarly, for the quadratic with ( r = -a - sqrt{2} ):The roots are:( x = frac{ -(-a - sqrt{2}) pm sqrt{ (-a - sqrt{2})^2 - 8 } }{2} = frac{ a + sqrt{2} pm sqrt{ a^2 + 2asqrt{2} + 2 - 8 } }{2} ).Simplify the discriminant:( a^2 + 2asqrt{2} + 2 - 8 = a^2 + 2asqrt{2} - 6 ).So, the roots are:( x = frac{ a + sqrt{2} pm sqrt{ a^2 + 2asqrt{2} - 6 } }{2} ).Now, for both quadratics to have two distinct positive roots, several conditions must be satisfied:1. The discriminants must be positive for both quadratics.2. The expressions for the roots must be positive.Let's start with the discriminants.For the first quadratic, discriminant ( D_1 = a^2 - 2asqrt{2} - 6 > 0 ).For the second quadratic, discriminant ( D_2 = a^2 + 2asqrt{2} - 6 > 0 ).So, both ( D_1 ) and ( D_2 ) must be positive.Let me solve ( D_1 > 0 ):( a^2 - 2asqrt{2} - 6 > 0 ).This is a quadratic inequality. Let's find the roots of the equation ( a^2 - 2asqrt{2} - 6 = 0 ).Using quadratic formula:( a = frac{ 2sqrt{2} pm sqrt{ (2sqrt{2})^2 + 24 } }{2} = frac{ 2sqrt{2} pm sqrt{ 8 + 24 } }{2} = frac{ 2sqrt{2} pm sqrt{32} }{2} = frac{ 2sqrt{2} pm 4sqrt{2} }{2} ).So, ( a = frac{ 2sqrt{2} + 4sqrt{2} }{2} = frac{6sqrt{2}}{2} = 3sqrt{2} ), or ( a = frac{ 2sqrt{2} - 4sqrt{2} }{2} = frac{ -2sqrt{2} }{2} = -sqrt{2} ).Since the coefficient of ( a^2 ) is positive, the inequality ( a^2 - 2asqrt{2} - 6 > 0 ) holds when ( a > 3sqrt{2} ) or ( a < -sqrt{2} ).Similarly, for ( D_2 > 0 ):( a^2 + 2asqrt{2} - 6 > 0 ).Again, solving ( a^2 + 2asqrt{2} - 6 = 0 ):Using quadratic formula:( a = frac{ -2sqrt{2} pm sqrt{ (2sqrt{2})^2 + 24 } }{2} = frac{ -2sqrt{2} pm sqrt{8 + 24} }{2} = frac{ -2sqrt{2} pm sqrt{32} }{2} = frac{ -2sqrt{2} pm 4sqrt{2} }{2} ).So, ( a = frac{ -2sqrt{2} + 4sqrt{2} }{2} = frac{2sqrt{2}}{2} = sqrt{2} ), or ( a = frac{ -2sqrt{2} - 4sqrt{2} }{2} = frac{ -6sqrt{2} }{2} = -3sqrt{2} ).Again, since the coefficient of ( a^2 ) is positive, the inequality ( a^2 + 2asqrt{2} - 6 > 0 ) holds when ( a > sqrt{2} ) or ( a < -3sqrt{2} ).Now, combining the conditions from both discriminants:From ( D_1 > 0 ): ( a > 3sqrt{2} ) or ( a < -sqrt{2} ).From ( D_2 > 0 ): ( a > sqrt{2} ) or ( a < -3sqrt{2} ).So, the intersection of these conditions is:For positive ( a ): ( a > 3sqrt{2} ) (since ( a > 3sqrt{2} ) is more restrictive than ( a > sqrt{2} )).For negative ( a ): ( a < -3sqrt{2} ) (since ( a < -3sqrt{2} ) is more restrictive than ( a < -sqrt{2} )).However, the original problem specifies that the equation has four distinct positive roots. So, all roots must be positive, which implies that ( a ) must be positive because if ( a ) were negative, the coefficients in the original equation might lead to negative roots or complicate the positivity.Therefore, we can focus on the positive ( a ) condition: ( a > 3sqrt{2} ).But let's verify if this is sufficient for all roots to be positive.Looking back at the roots of the quadratics:For the first quadratic with ( p = -a + sqrt{2} ):The roots are ( frac{ a - sqrt{2} pm sqrt{ a^2 - 2asqrt{2} - 6 } }{2} ).Given that ( a > 3sqrt{2} ), let's check if these roots are positive.First, the numerator for the positive root is ( a - sqrt{2} + sqrt{ a^2 - 2asqrt{2} - 6 } ). Since ( a > 3sqrt{2} ), ( a - sqrt{2} ) is positive, and the square root term is also positive, so the entire numerator is positive, making the root positive.For the negative root, it's ( a - sqrt{2} - sqrt{ a^2 - 2asqrt{2} - 6 } ). We need to ensure this is positive as well.Let me compute:( a - sqrt{2} - sqrt{ a^2 - 2asqrt{2} - 6 } > 0 ).Let me denote ( S = a - sqrt{2} ) and ( T = sqrt{ a^2 - 2asqrt{2} - 6 } ).So, the inequality is ( S - T > 0 ) => ( S > T ).Square both sides (since both are positive):( S^2 > T^2 ).Compute ( S^2 = (a - sqrt{2})^2 = a^2 - 2asqrt{2} + 2 ).Compute ( T^2 = a^2 - 2asqrt{2} - 6 ).So, the inequality becomes:( a^2 - 2asqrt{2} + 2 > a^2 - 2asqrt{2} - 6 ).Simplify:( 2 > -6 ), which is always true.Therefore, ( S > T ), so ( a - sqrt{2} - T > 0 ). Thus, both roots from the first quadratic are positive.Similarly, for the second quadratic with ( r = -a - sqrt{2} ):The roots are ( frac{ a + sqrt{2} pm sqrt{ a^2 + 2asqrt{2} - 6 } }{2} ).Again, since ( a > 3sqrt{2} ), ( a + sqrt{2} ) is positive, and the square root term is positive, so the positive root is clearly positive.For the negative root:( frac{ a + sqrt{2} - sqrt{ a^2 + 2asqrt{2} - 6 } }{2} ).Again, let me denote ( S = a + sqrt{2} ) and ( T = sqrt{ a^2 + 2asqrt{2} - 6 } ).We need ( S - T > 0 ).Square both sides:( S^2 > T^2 ).Compute ( S^2 = (a + sqrt{2})^2 = a^2 + 2asqrt{2} + 2 ).Compute ( T^2 = a^2 + 2asqrt{2} - 6 ).So, the inequality becomes:( a^2 + 2asqrt{2} + 2 > a^2 + 2asqrt{2} - 6 ).Simplify:( 2 > -6 ), which is always true.Therefore, ( S > T ), so ( a + sqrt{2} - T > 0 ). Thus, both roots from the second quadratic are positive.Therefore, when ( a > 3sqrt{2} ), both quadratics have two distinct positive roots, giving the quartic equation four distinct positive roots.Now, let's check if ( a = 3sqrt{2} ) is allowed. If ( a = 3sqrt{2} ), then the discriminant ( D_1 = (3sqrt{2})^2 - 2*(3sqrt{2})*sqrt{2} - 6 = 18 - 12 - 6 = 0 ). So, the quadratic would have a repeated root, which means the quartic would have a repeated root, violating the condition of four distinct roots. Therefore, ( a ) must be strictly greater than ( 3sqrt{2} ).Similarly, for ( a < -3sqrt{2} ), although the discriminants are positive, the roots of the quadratics may not all be positive. Let me check:If ( a < -3sqrt{2} ), then ( p = -a + sqrt{2} ) becomes positive because ( -a ) is positive and greater than ( 3sqrt{2} ), so ( p = -a + sqrt{2} > 3sqrt{2} + sqrt{2} = 4sqrt{2} ). Similarly, ( r = -a - sqrt{2} ) is also positive and greater than ( 3sqrt{2} - sqrt{2} = 2sqrt{2} ).But then, looking at the roots of the quadratics:For ( x^2 + p x + 2 = 0 ), with ( p > 4sqrt{2} ), the roots would be:( x = frac{ -p pm sqrt{p^2 - 8} }{2} ).But since ( p ) is positive, ( -p ) is negative, and the square root term is positive. So, the roots would be:( frac{ -p + sqrt{p^2 - 8} }{2} ) and ( frac{ -p - sqrt{p^2 - 8} }{2} ).Both numerators are negative, so both roots would be negative. Similarly, for the other quadratic, ( x^2 + r x + 2 = 0 ), with ( r > 2sqrt{2} ), the roots would also be negative.Therefore, if ( a < -3sqrt{2} ), all roots of the quartic would be negative, which contradicts the requirement of four distinct positive roots. Hence, ( a ) cannot be less than ( -3sqrt{2} ).Therefore, the only valid range for ( a ) is ( a > 3sqrt{2} ).So, the correct answer is option A: ( a > 3sqrt{2} ).**Final Answer**boxed{A}

💡Okay, so I need to figure out how many three-digit numbers there are where the middle digit is twice the average of the first and last digits. Hmm, let's break this down step by step.First, let's represent a three-digit number as ABC, where A is the first digit, B is the middle digit, and C is the last digit. Since it's a three-digit number, A can't be zero; it has to be from 1 to 9. B and C, on the other hand, can be from 0 to 9.The problem states that the middle digit B is twice the average of the first and last digits. So, mathematically, that would be:B = 2 * (A + C) / 2Wait, simplifying that, the 2 in the numerator and denominator cancel out, so it becomes:B = A + COkay, so the middle digit is equal to the sum of the first and last digits. But here's the catch: since B is a single digit, the sum A + C must also be a single digit. That means A + C has to be between 0 and 9 inclusive. However, since A is at least 1, the minimum value of A + C is 1 (when A=1 and C=0), and the maximum is 9 (when A=9 and C=0, or A=8 and C=1, etc.).So, for each possible value of A (from 1 to 9), I need to find how many possible values of C will satisfy the condition that A + C is a single digit (i.e., less than or equal to 9). Let's list this out:- If A = 1, then C can be 0, 1, 2, 3, 4, 5, 6, 7, 8 (since 1 + 8 = 9). That's 9 possibilities.- If A = 2, then C can be 0, 1, 2, 3, 4, 5, 6, 7 (since 2 + 7 = 9). That's 8 possibilities.- If A = 3, then C can be 0, 1, 2, 3, 4, 5, 6 (since 3 + 6 = 9). That's 7 possibilities.- If A = 4, then C can be 0, 1, 2, 3, 4, 5 (since 4 + 5 = 9). That's 6 possibilities.- If A = 5, then C can be 0, 1, 2, 3, 4 (since 5 + 4 = 9). That's 5 possibilities.- If A = 6, then C can be 0, 1, 2, 3 (since 6 + 3 = 9). That's 4 possibilities.- If A = 7, then C can be 0, 1, 2 (since 7 + 2 = 9). That's 3 possibilities.- If A = 8, then C can be 0, 1 (since 8 + 1 = 9). That's 2 possibilities.- If A = 9, then C can only be 0 (since 9 + 0 = 9). That's 1 possibility.Now, if I add up all these possibilities: 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1. Let's compute that:9 + 8 = 1717 + 7 = 2424 + 6 = 3030 + 5 = 3535 + 4 = 3939 + 3 = 4242 + 2 = 4444 + 1 = 45So, there are 45 possible three-digit numbers that satisfy the condition where the middle digit is twice the average of the first and last digits.Wait a second, let me double-check to make sure I didn't miss anything. Each A from 1 to 9 has a corresponding number of Cs that make A + C ≤ 9. Adding them up gives 45. That seems right because it's the sum of the first 9 natural numbers, which is (9*10)/2 = 45. Yeah, that checks out.I think I've covered all the cases and the arithmetic seems correct. So, I'm confident that the answer is 45.

💡Okay, so I need to figure out the total percentage increase in John's weekly earnings after ten months. Let's see, John starts making 30 a week. After six months, he gets a raise to 40 a week. Then, four months later, he gets another raise to 50 a week. First, I should understand what the question is asking. It wants the total percentage increase in his weekly earnings after ten months. That means I need to compare his starting salary to his salary after ten months and see what the percentage increase is.So, initially, he makes 30 per week. After six months, he gets a raise to 40. Then, four months after that, he gets another raise to 50. Wait, six months plus four months is ten months, so after ten months, he's making 50 per week.To find the percentage increase, I think I need to calculate how much his earnings have increased in total and then express that increase as a percentage of his original earnings.So, the increase in his earnings is 50 minus 30, which is 20. Now, to find the percentage increase, I can use the formula:Percentage Increase = (Increase / Original Amount) * 100%Plugging in the numbers:Percentage Increase = (20 / 30) * 100%Let me calculate that. 20 divided by 30 is 0.666..., and multiplying that by 100% gives approximately 66.67%.Wait, does that make sense? He started at 30 and ended at 50, so the increase is 20. Yes, 20 is two-thirds of 30, and two-thirds as a percentage is about 66.67%. So, the total percentage increase in his weekly earnings after ten months is 66.67%.I think that's it. I don't think I need to consider the time periods separately because the question is asking for the total percentage increase, not the average or anything like that. It's just comparing the starting point to the ending point after ten months.