💡First, I need to understand the set ( A ) and the values of ( a ) and ( b ).The set ( A ) is defined as all real numbers ( x ) such that ( x leq 2sqrt{3} ).Given:- ( a = sqrt{14} )- ( b = 2sqrt{2} )Next, I'll compare ( a ) and ( b ) with ( 2sqrt{3} ) to determine their membership in set ( A ).**Step 1: Compare ( a = sqrt{14} ) with ( 2sqrt{3} )**- Calculate ( sqrt{14} approx 3.7417 )- Calculate ( 2sqrt{3} approx 3.4641 )- Since ( 3.7417 > 3.4641 ), ( a ) is not an element of ( A ).**Step 2: Compare ( b = 2sqrt{2} ) with ( 2sqrt{3} )**- Calculate ( 2sqrt{2} approx 2.8284 )- Calculate ( 2sqrt{3} approx 3.4641 )- Since ( 2.8284 < 3.4641 ), ( b ) is an element of ( A ).**Conclusion:**- ( a notin A )- ( b in A )Therefore, the correct choice is **B**.

💡To find the possible values of ( a ) such that (-3) is an element of the set ( M = {a - 3, 2a - 1, a^2 + 4} ), we consider each element of the set and set it equal to (-3).1. **Case 1:** ( a - 3 = -3 ) [ a - 3 = -3 implies a = 0 ] Substituting ( a = 0 ) into the set: [ M = {0 - 3, 2(0) - 1, 0^2 + 4} = {-3, -1, 4} ] This satisfies the condition.2. **Case 2:** ( 2a - 1 = -3 ) [ 2a - 1 = -3 implies 2a = -2 implies a = -1 ] Substituting ( a = -1 ) into the set: [ M = {-1 - 3, 2(-1) - 1, (-1)^2 + 4} = {-4, -3, 5} ] This also satisfies the condition.3. **Case 3:** ( a^2 + 4 = -3 ) [ a^2 + 4 = -3 implies a^2 = -7 ] Since ( a^2 ) cannot be negative for real numbers, this case has no real solutions.Therefore, the possible real values of ( a ) are ( 0 ) and ( -1 ).

💡Alright, so I have to compute arccos(1) and express the answer in radians. Hmm, okay, let's start by recalling what arccos is. Arccos is the inverse function of cosine, right? So, if I have arccos(x), it gives me the angle whose cosine is x. In this case, x is 1, so I'm looking for the angle whose cosine is 1.Now, I remember that the cosine of an angle in a right triangle is adjacent over hypotenuse. But since we're dealing with the unit circle here, the cosine of an angle is the x-coordinate of the point on the unit circle corresponding to that angle. So, when is the x-coordinate equal to 1 on the unit circle?Thinking about the unit circle, I know that at 0 radians, the point is (1,0). So, the cosine of 0 radians is 1. That seems straightforward. But wait, are there other angles where the cosine is also 1? Let me think. Cosine has a period of 2π, so adding any multiple of 2π to 0 radians would also give a cosine of 1. For example, 2π, 4π, 6π, and so on. Similarly, negative angles like -2π, -4π, etc., would also have a cosine of 1.But when we talk about the principal value of arccos, which is the value we get when we compute arccos(x) without any additional context, it's defined to be in the range [0, π]. So, even though there are infinitely many angles where cosine is 1, the principal value is just 0 radians.Let me double-check this. If I take the cosine of 0 radians, cos(0) is indeed 1. And since 0 is within the principal range of [0, π], that must be the answer. I don't think I'm missing anything here. It seems pretty straightforward.Wait, but just to be thorough, let me consider if there's any other angle within [0, π] where cosine is 1. Well, at π/2 radians, cosine is 0, and at π radians, cosine is -1. So, between 0 and π, the cosine function starts at 1, goes down to 0 at π/2, and then to -1 at π. Therefore, the only angle in that interval where cosine is 1 is at 0 radians.Okay, so I think I'm confident now that arccos(1) is 0 radians.

💡First, I recognize that the point (4, -5) being on the graph of ( y = g(x) ) means that ( g(4) = -5 ).Next, I need to determine the corresponding point on the graph of ( y = h(x) ), where ( h(x) = |g(x)|^3 ).To find ( h(4) ), I substitute ( x = 4 ) into the function:[h(4) = |g(4)|^3 = |-5|^3 = 5^3 = 125]Therefore, the point (4, 125) must be on the graph of ( y = h(x) ).Finally, I calculate the sum of the coordinates of this point:[4 + 125 = 129]

💡Okay, so I have this equation here: ( x^2 - y^2 = 0 ). Hmm, I need to figure out what kind of graph this represents. The options are two intersecting straight lines, two parallel straight lines, an ellipse, or a point. Alright, let me think step by step.First, I remember that equations involving ( x^2 ) and ( y^2 ) often represent conic sections like circles, ellipses, hyperbolas, or parabolas. But this equation is ( x^2 - y^2 = 0 ), which looks a bit different. It doesn't have a constant term on the other side, so maybe it's a degenerate conic?Wait, degenerate conics are cases where the conic doesn't have its usual shape. For example, a circle could degenerate into a single point or two intersecting lines. In this case, since it's ( x^2 - y^2 = 0 ), maybe it's similar to a hyperbola, but without the constant term. So, perhaps it's two lines?Let me try to factor the equation. ( x^2 - y^2 ) is a difference of squares, right? So, it can be factored as ( (x - y)(x + y) = 0 ). That means either ( x - y = 0 ) or ( x + y = 0 ). So, these are two linear equations.Solving ( x - y = 0 ) gives ( y = x ), and solving ( x + y = 0 ) gives ( y = -x ). So, these are two straight lines with slopes 1 and -1, respectively. They both pass through the origin, so they intersect at the origin.Therefore, the graph of ( x^2 - y^2 = 0 ) consists of two straight lines that intersect at the origin. That would be option A: Two intersecting straight lines.Just to make sure I didn't make a mistake, let me think again. If I plug in some points, say (1,1), does it satisfy the equation? ( 1^2 - 1^2 = 0 ), which is true. Similarly, (1,-1): ( 1^2 - (-1)^2 = 0 ), also true. So, points on both lines satisfy the equation. If I pick a point not on these lines, like (1,0), then ( 1^2 - 0^2 = 1 ), which doesn't equal 0, so it doesn't lie on the graph. That seems consistent.I don't think it's two parallel lines because parallel lines would have the same slope, but here we have slopes 1 and -1, which are different. It's definitely not an ellipse because an ellipse has both ( x^2 ) and ( y^2 ) terms with the same sign, and it would be set equal to a positive constant. And it's not a single point because there are infinitely many solutions along those two lines.So, yeah, I'm pretty confident that the answer is A: Two intersecting straight lines.

💡Alright, so I've got this problem here about a sequence defined by some least common multiples and stuff. Let me try to wrap my head around it step by step.Okay, the sequence is given as x₁ = 19, x₂ = 95, and then for n > 1, x_{n+2} = lcm(x_{n+1}, x_n) + x_n. I need to find the greatest common divisor (GCD) of x_{1995} and x_{1996}. That seems pretty far out in the sequence, so calculating each term up to that point isn't feasible. There must be a pattern or a property I can use.First, let me recall what lcm(a, b) means. It's the least common multiple of a and b, which is the smallest number that both a and b divide into. Also, I remember that lcm(a, b) can be expressed in terms of the GCD: lcm(a, b) = (a * b) / gcd(a, b). Maybe that will come in handy.So, the recursive formula is x_{n+2} = lcm(x_{n+1}, x_n) + x_n. Let me try to compute the first few terms to see if I can spot a pattern.We have x₁ = 19 and x₂ = 95. Let's compute x₃:x₃ = lcm(x₂, x₁) + x₁= lcm(95, 19) + 19Since 95 is a multiple of 19 (because 19 * 5 = 95), the lcm of 95 and 19 is just 95. So,x₃ = 95 + 19 = 114Okay, x₃ is 114. Now, let's compute x₄:x₄ = lcm(x₃, x₂) + x₂= lcm(114, 95) + 95Hmm, what's the lcm of 114 and 95? Let's factor both:114 = 2 * 3 * 1995 = 5 * 19So, the lcm is the product of the highest powers of all primes present: 2 * 3 * 5 * 19 = 570Therefore, x₄ = 570 + 95 = 665Alright, x₄ is 665. Now, x₅:x₅ = lcm(x₄, x₃) + x₃= lcm(665, 114) + 114Let me factor these:665 = 5 * 7 * 19114 = 2 * 3 * 19So, lcm is 2 * 3 * 5 * 7 * 19 = 3990Thus, x₅ = 3990 + 114 = 4104Hmm, okay, x₅ is 4104. Let's do one more term, x₆:x₆ = lcm(x₅, x₄) + x₄= lcm(4104, 665) + 665Factoring 4104:4104 ÷ 2 = 20522052 ÷ 2 = 10261026 ÷ 2 = 513513 ÷ 3 = 171171 ÷ 3 = 5757 ÷ 3 = 19So, 4104 = 2³ * 3³ * 19And 665 is 5 * 7 * 19So, lcm is 2³ * 3³ * 5 * 7 * 19 = 8 * 27 * 5 * 7 * 19Calculating that:8 * 27 = 216216 * 5 = 10801080 * 7 = 75607560 * 19 = Let's compute 7560 * 20 = 151200, subtract 7560: 151200 - 7560 = 143640So, lcm(4104, 665) = 143640Therefore, x₆ = 143640 + 665 = 144305Wow, that's getting big quickly. Maybe computing more terms isn't helpful. Let me think about the GCD instead.I need to find gcd(x_{1995}, x_{1996}). Maybe there's a pattern in the GCDs of consecutive terms.Let me compute the GCDs of the terms I have so far:gcd(x₁, x₂) = gcd(19, 95) = 19gcd(x₂, x₃) = gcd(95, 114). Let's compute that:95 = 5 * 19114 = 2 * 3 * 19So, gcd is 19.gcd(x₃, x₄) = gcd(114, 665)114 = 2 * 3 * 19665 = 5 * 7 * 19gcd is 19.gcd(x₄, x₅) = gcd(665, 4104)665 = 5 * 7 * 194104 = 2³ * 3³ * 19gcd is 19.gcd(x₅, x₆) = gcd(4104, 144305)4104 = 2³ * 3³ * 19144305: Let's see if 19 divides it. 144305 ÷ 19. 19*7000=133000, 144305-133000=11305. 19*500=9500, 11305-9500=1805. 19*95=1805. So total is 7000+500+95=7595. So, 144305 = 19 * 7595Thus, gcd(4104, 144305) = 19Hmm, so it seems like the GCD of any two consecutive terms is 19. If this pattern continues, then gcd(x_{1995}, x_{1996}) = 19.But why is that? Let me try to see if I can prove it in general.Suppose that for some n, gcd(x_n, x_{n+1}) = d. Then, let's see what gcd(x_{n+1}, x_{n+2}) would be.x_{n+2} = lcm(x_{n+1}, x_n) + x_nWe know that lcm(x_{n+1}, x_n) = (x_{n+1} * x_n) / gcd(x_{n+1}, x_n) = (x_{n+1} * x_n) / dSo, x_{n+2} = (x_{n+1} * x_n) / d + x_n = x_n * (x_{n+1} / d + 1)Now, let's compute gcd(x_{n+1}, x_{n+2}):gcd(x_{n+1}, x_n * (x_{n+1}/d + 1))Since d divides x_{n+1} and x_n, let's write x_{n+1} = d * a and x_n = d * b, where gcd(a, b) = 1 (since d is the GCD).Then, x_{n+2} = d * b * ( (d * a) / d + 1 ) = d * b * (a + 1)So, x_{n+2} = d * b * (a + 1)Now, compute gcd(d * a, d * b * (a + 1)) = d * gcd(a, b * (a + 1))But since gcd(a, b) = 1, and a and a + 1 are consecutive integers, so gcd(a, a + 1) = 1. Therefore, gcd(a, b * (a + 1)) = 1.Thus, gcd(x_{n+1}, x_{n+2}) = d * 1 = d.So, this shows that if gcd(x_n, x_{n+1}) = d, then gcd(x_{n+1}, x_{n+2}) = d as well. Therefore, the GCD remains constant throughout the sequence.Since we saw that gcd(x₁, x₂) = 19, this GCD will be 19 for all consecutive terms in the sequence. Therefore, gcd(x_{1995}, x_{1996}) = 19.I think that makes sense. The key was realizing that the GCD remains invariant through the recursive definition, which I proved using the properties of LCM and GCD. So, regardless of how far out in the sequence we go, the GCD of consecutive terms stays the same.