🧠:Okay, so I need to prove that these two functions, f(x) = sin|x| and y = sin(cube root of x), are not periodic. Hmm, let's start with the first one: sin|x|. First, I remember that a periodic function repeats its values in regular intervals or periods. So, for a function f to be periodic, there must exist some positive number T such that f(x + T) = f(x) for all x in the domain. If I can show that no such T exists, then the function isn't periodic.Starting with f(x) = sin|x|. Let's think about the absolute value inside the sine. The absolute value affects the function by reflecting the negative x values to positive. So, for x ≥ 0, |x| = x, so f(x) = sinx, which is periodic with period 2π. But for x < 0, |x| = -x, so f(x) = sin(-x) = -sinx. Wait, no, hold on: sin(-x) is -sinx, but here we have sin|x|, so when x is negative, |x| is positive, so sin|x| is sin(-x) when x is negative? Wait, no. Wait, no, if x is negative, |x| is -x (since x is negative, -x is positive). So sin|x| = sin(-x) when x is negative? Wait, no, |x| is always non-negative. So, for x < 0, |x| = -x, so f(x) = sin(-x) = -sinx. Wait, that's not right. Wait, sin(-x) is -sinx, but |x| is just the magnitude. So if x is negative, |x| is -x (which is positive), so sin|x| = sin(-x) = -sinx? Wait, no. Wait, no. Let's take an example. If x = -π/2, then |x| = π/2, so sin|x| = sin(π/2) = 1. But sinx would be sin(-π/2) = -1. So sin|x| is equal to sin(|x|), which is always non-negative argument. So actually, for x < 0, sin|x| = sin(-x) = sin|x|, but since |x| is positive, sin|x| is just the sine of a positive number. Wait, hold on, maybe I'm confusing here. Let's clarify:For any real x, |x| is non-negative. So, f(x) = sin|x| is equal to sin(x) when x ≥ 0, and sin(-x) when x < 0. But sin(-x) = -sinx. Wait, that's conflicting. Wait, no. Wait, if x is negative, say x = -a where a > 0, then |x| = a, so sin|x| = sin(a). So, sin|x| is equal to sin(a) where a = |x|. So, regardless of x being positive or negative, sin|x| is sin(|x|). Therefore, sin|x| is actually equal to sin(x) for x ≥ 0, and sin(-x) = sinx because |x| = -x when x < 0. Wait, no. Wait, sin|x| is sin(|x|), which is just sin of a non-negative number. So, whether x is positive or negative, |x| is positive, so sin|x| is sin of a positive number. Therefore, f(x) = sin|x| is actually equal to sin(x) for x ≥ 0, and sin(-x) = sin(x) because |-x| = x. Wait, no, that's not right. Let's take x = -π/2. Then |x| = π/2, so sin|x| = sin(π/2) = 1. Whereas sin(x) would be sin(-π/2) = -1. So sin|x| is not equal to sin(x) for x < 0. Instead, for x < 0, sin|x| = sin(-x) = -sinx. Wait, no. Wait, sin|x| when x is negative is sin(-x). But sin(-x) is equal to -sinx, but here x is negative, so -x is positive. For example, if x = -a, then sin|x| = sin(a). So sin|x| is sin(a) = sin|x|. Therefore, actually, sin|x| is equal to sin(x) when x ≥ 0 and sin(-x) = sin(x) when x < 0. Wait, no. Wait, sin(-x) is -sinx, but here |x| is just the magnitude, so sin|x| is always sin of a non-negative value. So, if x is positive, sin|x| = sinx; if x is negative, sin|x| = sin(-x) = sin|x|. Wait, perhaps I'm overcomplicating this.Let me graph the function mentally. The standard sinx function is odd, with period 2π, symmetric about the origin. But sin|x| would be the sine of the absolute value, which for x ≥0 is the same as sinx, but for x <0, it's sin(-x) = -sinx. Wait, no, because |x| is -x when x <0, but the sine of |x| is sin(-x) when x <0. Wait, sin(-x) is -sinx. Wait, so for x <0, sin|x| = sin(-x) = -sinx. But for x >=0, sin|x| = sinx. So, the graph of sin|x| would be sinx for x >=0 and -sinx for x <0. Wait, but that would make the function even, because sin|x| is equal to sin|x|, which is even. Because |x| is even, and sine is odd, so sin| x | would be an odd function composed with an even function. Wait, let's see: if f(x) = sin|x|, then f(-x) = sin| -x | = sin|x| = f(x). So f is even. Therefore, the function is symmetric about the y-axis. So, the left side is a mirror image of the right side. However, the standard sinx is odd, so sin(-x) = -sinx. But here, since we take the absolute value first, the function becomes even.So, sin|x| is an even function. For x ≥0, it's sinx, which has period 2π. But for x <0, it's sin(-x) = -sinx, but since |x| is positive, it's sin(|x|) which is sinx for x positive. Wait, maybe I need to rephrase this. Let me think again: when x is positive, sin|x| = sinx; when x is negative, sin|x| = sin(-x) = sinx as well because |x| = -x when x is negative, but sin(-x) = -sinx. Wait, wait. This is confusing. Let's take an example. Let x = π/2. Then sin|x| = sin(π/2) = 1. If x = -π/2, then sin|x| = sin(π/2) = 1. Wait, so sin|x| is always sin(|x|), which is sin of a non-negative number, so when x is negative, it's the same as sin(|x|). Therefore, sin|x| is equal to sin(|x|), which is an even function because |x| is even. Therefore, the graph of sin|x| is the same as the graph of sinx for x ≥0, and mirrored across the y-axis for x <0. So, instead of the usual sine wave that oscillates symmetrically about the origin, sin|x| is a combination of the right half of sinx and the left half as a mirror image. So, if we consider the function sin|x|, for x ≥0, it's sinx, which is periodic with period 2π. But for x <0, it's also sinx mirrored. Now, to check if the entire function is periodic, we need to see if there exists a T >0 such that shifting the graph by T results in the same graph. But since the function is even, it's symmetric about the y-axis. Let's suppose, for contradiction, that there exists a period T. Then, for all x, sin|x + T| = sin|x|. Let's check x =0: sin|T| = sin0 =0, which implies sinT =0. So T must be a multiple of π. Let's check x = π/2: sin|π/2 + T| = sin(π/2) =1. So |π/2 + T| must be π/2 + 2π n or 3π/2 + 2π n, where n is integer. But since T is positive, |π/2 + T| = π/2 + T. So π/2 + T must be equal to π/2 + 2π n or 3π/2 + 2π n. Therefore, T = 2π n or T = π + 2π n. But T must satisfy the equation for all x. Let's test T = π. Then, check x = π/2: sin|π/2 + π| = sin(3π/2) = -1, but sin|π/2| =1. So -1 ≠1, so T=π doesn't work. If T=2π, then for x=π/2, sin|π/2 +2π|=sin(5π/2)=1, which matches. But let's check x=-π/2. sin| -π/2 +2π | = sin(3π/2)= -1, but sin| -π/2 | = sin(π/2)=1. So again, -1 ≠1. Therefore, even T=2π doesn't work. Because when you shift x by T, the right side (x ≥0) shifts and maintains the sine wave, but the left side (x <0) would have shifted into different regions. Because the function is even, shifting it to the right by T would cause the left side to not match the original function. Therefore, there's a contradiction, implying no such T exists.Alternatively, another approach: suppose that f(x) = sin|x| is periodic with period T. Then, for x = -T/2, we have f(-T/2 + T) = f(-T/2). But f(-T/2 + T) = f(T/2) = sin(T/2), and f(-T/2) = sin(T/2). So that's okay. But let's consider another point, say x = T. Then f(T + T) = f(T) => sin|2T| = sinT. Similarly, for x = 2T, f(3T) = f(2T), so sin3T = sin2T. This seems like we might get a system of equations. If sin2T = sinT and sin3T = sin2T. Let's solve sin2T = sinT. This equation implies that 2T = T + 2πn or 2T = π - T + 2πn. Solving the first: 2T = T + 2πn → T = 2πn. The second: 2T = π - T + 2πn → 3T = π + 2πn → T = (π + 2πn)/3. Then, check sin3T = sin2T. If T=2πn, then sin3T = sin6πn=0, and sin2T=sin4πn=0, so that holds. But if T=(π + 2πn)/3, then sin3T = sin(π + 2πn) = sinπ=0, and sin2T = sin(2(π + 2πn)/3) = sin(2π/3 + 4πn/3). For n=0: sin(2π/3)=√3/2 ≠0. So that's invalid. For n=1: sin(2π/3 +4π/3)=sin(2π)=0. So, when n=1, T=(π +2π)/3=π. Then, sin2T=sin2π=0, and sin3T=sin3π=0. So that works. But T=π. Let's check if T=π is a period. Take x=0: f(0 + π)=f(π)=sinπ=0, f(0)=0, so that's okay. x=π/2: f(π/2 +π)=f(3π/2)=sin|3π/2|=sin(3π/2)=-1, but f(π/2)=1. So -1≠1, so T=π is invalid. Therefore, even though T=π satisfies sin2T=sinT and sin3T=sin2T, it fails as a period because it doesn't hold for all x. Hence, the only possible T is multiple of 2π, but as before, shifting by 2π doesn't work on the negative side. Therefore, there's no period T that works for all x. Hence, sin|x| is not periodic.Alternatively, another approach: the function sin|x| has a derivative that is different on the left and right sides of x=0. For x>0, f'(x)=cosx. For x<0, f'(x)= -cosx (since |x|=-x, so derivative is cos(-x)*(-1)= -cosx). At x=0, the derivative from the right is cos0=1, from the left is -cos0=-1, so there's a corner at x=0. However, periodic functions are smooth or have consistent periodicity. The presence of this corner at x=0 might disrupt the periodicity. If the function were periodic with period T, then all points separated by T would have the same behavior. However, the corner at x=0 would imply corners at x=T, x=2T, etc., but the function sin|x| only has a corner at x=0. For example, at x=T, the derivative from the right would be cos(T), and from the left would be -cos(T). But unless cos(T) = -cos(T), which would imply cos(T)=0, but that would mean T=π/2 +kπ. However, if T is such that cos(T)=0, then the derivative from both sides would be 0 and -0, which is 0. But even so, this would require the function to have corners at all x=nT, which it doesn't, because sin|x| only has a corner at x=0. Therefore, this inconsistency suggests that the function cannot be periodic.For the second function, y = sin∛x. We need to show this isn't periodic. Let's consider the properties of periodic functions. Suppose there exists T>0 such that sin∛(x + T) = sin∛x for all x. Let’s let u = ∛x, so x = u³. Then the equation becomes sin(∛(u³ + T)) = sinu for all u. Hmm, maybe not helpful. Alternatively, consider the zeros of the function. If the function is periodic with period T, then the spacing between consecutive zeros should be consistent or have a repeating pattern. Let’s find the zeros of sin∛x. sin∛x =0 when ∛x = nπ, so x = (nπ)³, where n is integer. The zeros are at x = ..., (-π)³, 0, (π)³, (2π)³, etc. The spacing between consecutive zeros is ( (n+1)π )³ - (nπ)³ = π³ [ (n+1)³ - n³ ] = π³ (3n² + 3n +1 ). This spacing depends on n, which increases as n increases. Therefore, the distance between zeros is not constant, nor does it repeat periodically. In a periodic function, the spacing between zeros should be periodic or have a fixed pattern, but here the gaps increase as n increases, implying the function cannot be periodic.Alternatively, suppose for contradiction that T is a period. Then sin∛(x + T) = sin∛x for all x. Let’s consider x=0: sin∛T = sin0 =0. Therefore, ∛T = kπ, so T=(kπ)³ for some integer k. Similarly, take x=T: sin∛(2T) = sin∛T =0. Therefore, ∛(2T) = mπ, so 2T=(mπ)³. But T=(kπ)³, so 2(kπ)³=(mπ)³ → 2k³ =m³. But 2k³=m³ implies that m= k * ∛2, which is not an integer unless k=0, which gives T=0, not valid. Therefore, there's no integer k,m satisfying this, so such T does not exist. Therefore, the function cannot be periodic.Another approach: the derivative of y=sin∛x is y’= (1/(3x^(2/3))) cos∛x. For the function to be periodic, its derivative must also be periodic with the same period. However, as x increases, the amplitude of the derivative decreases because of the 1/(3x^(2/3)) term. In a periodic function, the derivative's amplitude should remain constant, since the function repeats its behavior. Since the amplitude here diminishes as x grows, the function cannot be periodic.Alternatively, consider the function ∛x is a strictly increasing function with a decreasing slope. The composition with sine would inherit the non-linear spacing of the zeros and the varying amplitude in the derivative, which conflicts with the requirements of periodicity.Therefore, both functions are not periodic because they fail to satisfy the necessary conditions for periodicity, such as consistent zero spacing, derivative behavior, or invariance under translation by any T>0 for all x.Final Answer(1) boxed{f(x) = sin |x|} is not periodic.(2) boxed{y = sin sqrt[3]{x}} is not periodic.

✅:To determine whether the given functions are periodic, we need to check if there exists a positive number ( T ) such that ( f(x + T) = f(x) ) for all ( x ) in the domain. (1) Proving ( f(x) = sin |x| ) is not periodic1. Understanding the Function: - For ( x geq 0 ), ( f(x) = sin x ). - For ( x < 0 ), ( f(x) = sin(-x) = sin |x| ), which is symmetric about the y-axis, making ( f(x) ) an even function.2. Periodicity Analysis: - If ( f(x) ) were periodic with period ( T ), then ( sin |x + T| = sin |x| ) must hold for all ( x ). - Considering ( x = 0 ), we get ( sin T = 0 ), implying ( T = kpi ) for some integer ( k ). - Testing ( T = pi ): For ( x = pi/2 ), ( sin |pi/2 + pi| = sin(3pi/2) = -1 ), but ( sin |pi/2| = 1 ), which is a contradiction. - Testing ( T = 2pi ): For ( x = -pi/2 ), ( sin |-pi/2 + 2pi| = sin(3pi/2) = -1 ), but ( sin |-pi/2| = 1 ), another contradiction.3. Conclusion: - The function ( sin |x| ) has a corner at ( x = 0 ) and cannot maintain periodicity due to inconsistent behavior on both sides of the y-axis. Thus, ( f(x) = sin |x| ) is not periodic. (2) Proving ( y = sin sqrt[3]{x} ) is not periodic1. Understanding the Function: - The zeros of ( sin sqrt[3]{x} ) occur at ( sqrt[3]{x} = npi ), i.e., ( x = (npi)^3 ) for integers ( n ).2. Periodicity Analysis: - If ( sin sqrt[3]{x} ) were periodic with period ( T ), then ( sin sqrt[3]{x + T} = sin sqrt[3]{x} ) must hold for all ( x ). - Considering ( x = 0 ), we get ( sin sqrt[3]{T} = 0 ), implying ( T = (kpi)^3 ). - Testing ( T = (kpi)^3 ): For ( x = T ), ( sin sqrt[3]{2T} = sin sqrt[3]{T} ). This leads to ( sqrt[3]{2T} = mpi ), which requires ( 2(kpi)^3 = (mpi)^3 ), or ( 2k^3 = m^3 ). This is impossible for integers ( k ) and ( m ).3. Conclusion: - The spacing between consecutive zeros increases as ( n ) increases, which is inconsistent with periodicity. Additionally, the derivative's amplitude decreases, further indicating non-periodicity. Thus, ( y = sin sqrt[3]{x} ) is not periodic. Final Answers(1) boxed{f(x) = sin |x|} is not periodic.(2) boxed{y = sin sqrt[3]{x}} is not periodic.

🔑:Let's tackle both parts of the problem step by step.# Part 1: Proving (f(x) = sin |x|) is not a periodic function1. Assume (f(x) = sin |x|) is a periodic function with period (T). This implies: [ f(x + T) = f(x) quad text{for all} quad x in mathbb{R} ]2. Consider (x = frac{pi}{2}): [ fleft(frac{pi}{2} + Tright) = sin left|frac{pi}{2} + Tright| = 1 ] Since (fleft(frac{pi}{2} + Tright) = fleft(frac{pi}{2}right)) implies: [ sin left|frac{pi}{2} + Tright| = sin left|frac{pi}{2}right| ] This means (T) must satisfy the equation: [ left|frac{pi}{2} + Tright| = 2npi + frac{pi}{2} implies T = 2npi quad (n in mathbb{N}^*) ]3. Next, consider (x = -frac{pi}{2}): [ fleft(-frac{pi}{2} + Tright) = sin left|-frac{pi}{2} + 2npiright| ] If (f(-frac{pi}{2} + T) = f(-frac{pi}{2})), then we have: [ sin left|-frac{pi}{2} + 2npiright| = 1 ]4. Evaluate: [ sin left(-frac{pi}{2} + 2npiright) = sin left(-frac{pi}{2}right) = -1 quad text{for} quad n in mathbb{N} ] Clearly: [ fleft(-frac{pi}{2} + Tright) neq fleft(-frac{pi}{2}right) ]5. Therefore, there does not exist (T neq 0) such that (f(x+T) = f(x)) for all (x in mathbb{R}). Thus, (f(x) = sin |x|) is not a periodic function.Conclusion:[boxed{f(x) = sin |x| text{ is not a periodic function.}}]# Part 2: Proving (y = sin sqrt[3]{x}) is not a periodic function1. Assume (y = sin sqrt[3]{x}) is a periodic function with period (T neq 0). Thus: [ sin sqrt[3]{x+T} = sin sqrt[3]{x} quad text{for all} quad x in mathbb{R} ] 2. Set (x = 0): [ sin sqrt[3]{T} = 0 implies sqrt[3]{T} = k pi quad (k in mathbb{Z}, k neq 0) ]3. Then: [ T = k^3 pi^3 quad (k in mathbb{Z}, k neq 0) ]4. For (x = T): [ sin sqrt[3]{2T} = 0 implies sqrt[3]{2T} = n pi quad (n in mathbb{Z}, n neq 0) ] This gives: [ 2T = n^3 pi^3 implies 2k^3 pi^3 = n^3 pi^3 implies 2k^3 = n^3 ]5. Simplify: [ sqrt[3]{2} = frac{n}{k} implies sqrt[3]{2} text{ is a rational number} ] 6. However, (sqrt[3]{2}) is an irrational number, creating a contradiction.Conclusion:[boxed{y = sin sqrt[3]{x} text{ is not a periodic function.}}]

🧠:Okay, so I need to find the minimum value of the function ( y = (sin x + 1)(cos x + 1) ) in the interval ( -frac{pi}{6} leqslant x leqslant frac{pi}{2} ). Hmm, let's start by understanding the problem. First, the function is a product of two trigonometric expressions, each shifted by 1. Since both sine and cosine functions oscillate between -1 and 1, adding 1 shifts their ranges to between 0 and 2. So, ( sin x + 1 ) and ( cos x + 1 ) are both non-negative in the given interval. Therefore, their product ( y ) will also be non-negative. But I need the minimum value here, not the maximum. To find the minimum of a continuous function on a closed interval, I can use calculus. The standard approach is to find the critical points by taking the derivative and setting it equal to zero, then evaluate the function at those critical points and at the endpoints of the interval. The smallest of those values will be the minimum. So, first, let's find the derivative of ( y ) with respect to ( x ). The function is ( y = (sin x + 1)(cos x + 1) ). To differentiate this, I can use the product rule. The product rule states that the derivative of ( u cdot v ) is ( u'v + uv' ). Let me set ( u = sin x + 1 ) and ( v = cos x + 1 ). Then, ( u' = cos x ) and ( v' = -sin x ). Applying the product rule, the derivative ( y' ) is:( y' = cos x (cos x + 1) + (sin x + 1)(-sin x) ).Let me simplify this expression step by step. First term: ( cos x (cos x + 1) = cos^2 x + cos x ).Second term: ( (sin x + 1)(-sin x) = -sin^2 x - sin x ).Combine both terms:( y' = cos^2 x + cos x - sin^2 x - sin x ).Hmm, maybe I can simplify this further. Let me recall some trigonometric identities. I know that ( cos^2 x - sin^2 x = cos 2x ). So, that part simplifies to ( cos 2x ). Then the remaining terms are ( cos x - sin x ). So, ( y' = cos 2x + cos x - sin x ).Alternatively, maybe expanding ( cos^2 x ) and ( sin^2 x ) in terms of double angles isn't the best way. Let me check if there's another identity or if I can factor terms differently.Wait, perhaps grouping terms differently. Let me write:( y' = (cos^2 x - sin^2 x) + (cos x - sin x) ).So, that's ( cos 2x + (cos x - sin x) ). Hmm, not sure if that helps. Alternatively, maybe factor ( cos x - sin x ).Let me factor out ( cos x - sin x ) from the expression. Wait, looking at the original derivative:( y' = cos^2 x + cos x - sin^2 x - sin x ).Let me rearrange the terms:( (cos^2 x - sin^2 x) + (cos x - sin x) ).Yes, that's the same as before. Then, as I mentioned, ( cos^2 x - sin^2 x = cos 2x ), so:( y' = cos 2x + (cos x - sin x) ).Alternatively, perhaps factor ( cos x - sin x ). Let me see:( cos x - sin x = sqrt{2} cos left( x + frac{pi}{4} right) ). But not sure if that's helpful here. Alternatively, perhaps write the entire derivative in terms of sine or cosine. Alternatively, set ( t = x + frac{pi}{4} ) or something. Maybe that's overcomplicating.Alternatively, maybe square the terms? Hmm. Let me consider another approach.Alternatively, since the original function is ( y = (sin x + 1)(cos x + 1) ), maybe expanding it first before differentiating could make things simpler. Let's try that.Expanding ( y ):( y = sin x cos x + sin x + cos x + 1 ).So, ( y = sin x cos x + sin x + cos x + 1 ).Now, maybe differentiate term by term.The derivative of ( sin x cos x ) is ( cos^2 x - sin^2 x ) (using product rule: derivative of sin is cos, derivative of cos is -sin, so cos*cos + sin*(-sin) = cos² - sin²).Then derivative of ( sin x ) is ( cos x ).Derivative of ( cos x ) is ( -sin x ).Derivative of 1 is 0.So, altogether:( y' = (cos^2 x - sin^2 x) + cos x - sin x ).Which is the same expression as before: ( y' = cos 2x + cos x - sin x ). So, whether I expand first or use product rule initially, I get the same derivative. So, maybe this is as simplified as it gets. So, the critical points are where ( cos 2x + cos x - sin x = 0 ).Hmm, solving ( cos 2x + cos x - sin x = 0 ). That seems a bit tricky. Let me see if I can manipulate this equation.Perhaps express ( cos 2x ) in terms of ( sin x ) or ( cos x ). Recall that ( cos 2x = 1 - 2 sin^2 x ), or ( cos 2x = 2 cos^2 x - 1 ). Let me try substituting that.Using ( cos 2x = 2 cos^2 x - 1 ):So, equation becomes:( 2 cos^2 x - 1 + cos x - sin x = 0 ).Hmm, not sure if that helps. Alternatively, using ( cos 2x = 1 - 2 sin^2 x ):Equation becomes:( 1 - 2 sin^2 x + cos x - sin x = 0 ).Rearranged:( -2 sin^2 x - sin x + cos x + 1 = 0 ).Still complicated. Alternatively, let's consider substituting ( t = sin x + cos x ). Because in such equations, sometimes this substitution helps. Let's see.We know that ( t = sin x + cos x ), and ( t^2 = sin^2 x + 2 sin x cos x + cos^2 x = 1 + sin 2x ). So, ( sin 2x = t^2 - 1 ). But our equation is ( cos 2x + cos x - sin x = 0 ). Let's see:( cos 2x = cos^2 x - sin^2 x = (cos x - sin x)(cos x + sin x) = (cos x - sin x) t ).So, substituting into the equation:( (cos x - sin x) t + (cos x - sin x) = 0 ).Factor out ( cos x - sin x ):( (cos x - sin x)(t + 1) = 0 ).Therefore, either ( cos x - sin x = 0 ) or ( t + 1 = 0 ).But ( t = sin x + cos x ), so ( t + 1 = 0 ) implies ( sin x + cos x = -1 ).Let's consider each case.First case: ( cos x - sin x = 0 ).This implies ( cos x = sin x ), so ( tan x = 1 ). Therefore, ( x = frac{pi}{4} + kpi ), where ( k ) is an integer. In our interval ( -frac{pi}{6} leq x leq frac{pi}{2} ), the solutions are ( x = frac{pi}{4} ), since ( frac{pi}{4} ) is within the interval, and the next solution would be ( frac{5pi}{4} ), which is outside the interval. So, only ( x = frac{pi}{4} ) in the interval.Second case: ( sin x + cos x = -1 ).Let me solve this equation. Let's square both sides to see possible solutions:( (sin x + cos x)^2 = 1 ).Which gives ( sin^2 x + 2 sin x cos x + cos^2 x = 1 ).Simplify: ( 1 + sin 2x = 1 ), so ( sin 2x = 0 ).Thus, ( 2x = kpi ), so ( x = frac{kpi}{2} ).But original equation is ( sin x + cos x = -1 ). Let's check these solutions.For ( x = 0 ): ( sin 0 + cos 0 = 0 + 1 = 1 neq -1 ).For ( x = frac{pi}{2} ): ( sin frac{pi}{2} + cos frac{pi}{2} = 1 + 0 = 1 neq -1 ).For ( x = pi ): Outside the interval.For ( x = frac{3pi}{2} ): Outside the interval.Wait, so squaring introduced extraneous solutions. Therefore, the equation ( sin x + cos x = -1 ) might have solutions only at certain points.Alternatively, express ( sin x + cos x ) as ( sqrt{2} sin left( x + frac{pi}{4} right) ). So, ( sqrt{2} sin left( x + frac{pi}{4} right) = -1 ).Thus, ( sin left( x + frac{pi}{4} right) = -frac{1}{sqrt{2}} ).Solutions are ( x + frac{pi}{4} = frac{5pi}{4} + 2kpi ) or ( x + frac{pi}{4} = frac{7pi}{4} + 2kpi ).Therefore, ( x = frac{5pi}{4} - frac{pi}{4} + 2kpi = pi + 2kpi ) or ( x = frac{7pi}{4} - frac{pi}{4} + 2kpi = frac{3pi}{2} + 2kpi ).But these solutions are at ( x = pi ) or ( x = frac{3pi}{2} ), which are outside our interval ( -frac{pi}{6} leq x leq frac{pi}{2} ). Therefore, there are no solutions in this case within the given interval. Therefore, the only critical point is at ( x = frac{pi}{4} ).Therefore, the critical points in the interval are at ( x = frac{pi}{4} ), and the endpoints at ( x = -frac{pi}{6} ) and ( x = frac{pi}{2} ). Now, we need to evaluate the function ( y = (sin x + 1)(cos x + 1) ) at these three points and determine which is the minimum.Let's compute each one:1. At ( x = -frac{pi}{6} ):First, compute ( sin(-frac{pi}{6}) = -frac{1}{2} ), so ( sin x + 1 = -frac{1}{2} + 1 = frac{1}{2} ).( cos(-frac{pi}{6}) = cos(frac{pi}{6}) = frac{sqrt{3}}{2} ), so ( cos x + 1 = frac{sqrt{3}}{2} + 1 = frac{sqrt{3} + 2}{2} ).Therefore, ( y = frac{1}{2} times frac{sqrt{3} + 2}{2} = frac{sqrt{3} + 2}{4} approx frac{1.732 + 2}{4} = frac{3.732}{4} approx 0.933 ).2. At ( x = frac{pi}{4} ):( sin frac{pi}{4} = frac{sqrt{2}}{2} ), so ( sin x + 1 = frac{sqrt{2}}{2} + 1 approx 0.707 + 1 = 1.707 ).( cos frac{pi}{4} = frac{sqrt{2}}{2} ), so ( cos x + 1 = frac{sqrt{2}}{2} + 1 approx 1.707 ).Therefore, ( y = 1.707 times 1.707 approx (1 + frac{sqrt{2}}{2})^2 ). Let me compute it exactly:( y = left(1 + frac{sqrt{2}}{2}right)^2 = 1 + sqrt{2} + frac{1}{2} = frac{3}{2} + sqrt{2} approx 1.5 + 1.414 = 2.914 ).3. At ( x = frac{pi}{2} ):( sin frac{pi}{2} = 1 ), so ( sin x + 1 = 1 + 1 = 2 ).( cos frac{pi}{2} = 0 ), so ( cos x + 1 = 0 + 1 = 1 ).Therefore, ( y = 2 times 1 = 2 ).So, summarizing the values:- ( x = -pi/6 ): ( approx 0.933 )- ( x = pi/4 ): ( approx 2.914 )- ( x = pi/2 ): ( 2 )Therefore, the minimum value seems to be at ( x = -pi/6 ) with ( y approx 0.933 ). But let me confirm the exact value at ( x = -pi/6 ):( y = frac{sqrt{3} + 2}{4} ). To express this in a more precise form, perhaps rationalize or see if this can be simplified.Wait, ( frac{sqrt{3} + 2}{4} ) is already simplified. Alternatively, we can write it as ( frac{2 + sqrt{3}}{4} ). However, maybe there's a way to check if this is indeed the minimum. But we only have one critical point in the interval, which is a maximum. Wait, but when we checked the critical points, we only found one critical point at ( x = pi/4 ), which gave a higher value. So, the minimum occurs at the left endpoint ( x = -pi/6 ). But wait, before finalizing this conclusion, I need to make sure that there are no other critical points in the interval. Earlier, when solving for ( y' = 0 ), we found only ( x = pi/4 ) as critical point. So, the function's derivative is zero only at ( x = pi/4 ). Therefore, the extrema can only be at endpoints or at ( x = pi/4 ). Therefore, since at ( x = pi/4 ) the function reaches a local maximum (as the value is higher than at both endpoints), then the minimum must be at one of the endpoints. Comparing the two endpoints, at ( x = -pi/6 ) we have ( y approx 0.933 ), and at ( x = pi/2 ), ( y = 2 ). Therefore, the minimum is at ( x = -pi/6 ).However, just to be thorough, let me check if there could be any other critical points that we might have missed. We considered the case when ( cos x - sin x = 0 ), which gave ( x = pi/4 ), and the case ( sin x + cos x = -1 ), which had no solutions in the interval. Therefore, the only critical point is indeed ( x = pi/4 ). Therefore, the conclusion is that the minimum value of the function on the interval is ( frac{2 + sqrt{3}}{4} ), which is approximately 0.933, achieved at ( x = -pi/6 ).But let me verify this calculation once again to avoid any arithmetic errors.At ( x = -pi/6 ):( sin(-pi/6) = -sin(pi/6) = -1/2 ), so ( sin x + 1 = 1 - 1/2 = 1/2 ).( cos(-pi/6) = cos(pi/6) = sqrt{3}/2 ), so ( cos x + 1 = sqrt{3}/2 + 1 ).Multiplying these together:( (1/2)(sqrt{3}/2 + 1) = (1/2)(1 + sqrt{3}/2) = 1/2 + sqrt{3}/4 = (2 + sqrt{3})/4 ).Yes, that's correct.Therefore, the exact value is ( frac{2 + sqrt{3}}{4} ).Alternatively, we can rationalize this expression or convert it to another form, but this seems as simplified as possible.Therefore, the minimum value of the function ( y = (sin x + 1)(cos x + 1) ) in the interval ( -frac{pi}{6} leqslant x leqslant frac{pi}{2} ) is ( frac{2 + sqrt{3}}{4} ).However, just to ensure that there are no mistakes in the derivative calculation, let me double-check that.Original function: ( y = (sin x + 1)(cos x + 1) ).Product rule: derivative is ( cos x (cos x + 1) + (sin x + 1)(- sin x ) ).Calculating each part:First term: ( cos x (cos x + 1) = cos^2 x + cos x ).Second term: ( (sin x + 1)(- sin x ) = - sin^2 x - sin x ).So, total derivative: ( cos^2 x + cos x - sin^2 x - sin x ).Which is equal to ( (cos^2 x - sin^2 x) + (cos x - sin x) = cos 2x + (cos x - sin x) ). Correct.Then, setting derivative equal to zero: ( cos 2x + cos x - sin x = 0 ).Then, as previously, factoring:( (cos x - sin x)(sin x + cos x + 1) = 0 ).Wait, how did we get that? Let me check:Earlier, when substituting ( t = sin x + cos x ), we had:Original equation: ( cos 2x + cos x - sin x = 0 ).Expressed as ( (cos x - sin x)(sin x + cos x) + (cos x - sin x) = 0 ).Factor out ( (cos x - sin x) ):( (cos x - sin x)(sin x + cos x + 1) = 0 ).Ah, yes! That's a better way to factor it. So, the equation factors into ( (cos x - sin x)(sin x + cos x + 1) = 0 ).Therefore, either ( cos x - sin x = 0 ) or ( sin x + cos x + 1 = 0 ).So, the first factor gives ( cos x = sin x ), which leads to ( x = pi/4 + kpi ), and in the interval, only ( x = pi/4 ).The second factor gives ( sin x + cos x = -1 ), which we already saw has no solutions in the interval.Therefore, this confirms that the only critical point is at ( x = pi/4 ).Thus, the calculations are correct. Therefore, the minimum is indeed at ( x = -pi/6 ), and the value is ( frac{2 + sqrt{3}}{4} ).Therefore, the final answer is ( boxed{dfrac{2 + sqrt{3}}{4}} ).Final AnswerThe minimum value of the function is boxed{dfrac{2 + sqrt{3}}{4}}.

✅:To find the minimum value of the function ( y = (sin x + 1)(cos x + 1) ) in the interval ( -frac{pi}{6} leqslant x leqslant frac{pi}{2} ), we follow these steps:1. Expand the Function: [ y = (sin x + 1)(cos x + 1) = sin x cos x + sin x + cos x + 1 ]2. Find the Derivative: Using the product rule and simplifying, we get: [ y' = cos 2x + cos x - sin x ]3. Solve for Critical Points: Set the derivative equal to zero: [ cos 2x + cos x - sin x = 0 ] Factoring, we find: [ (cos x - sin x)(sin x + cos x + 1) = 0 ] This gives two cases: - ( cos x - sin x = 0 ) leading to ( x = frac{pi}{4} ) - ( sin x + cos x = -1 ) which has no solutions in the interval4. Evaluate the Function at Critical Points and Endpoints: - At ( x = -frac{pi}{6} ): [ sinleft(-frac{pi}{6}right) = -frac{1}{2}, quad cosleft(-frac{pi}{6}right) = frac{sqrt{3}}{2} ] [ y = left(frac{1}{2}right)left(frac{sqrt{3}}{2} + 1right) = frac{sqrt{3} + 2}{4} ] - At ( x = frac{pi}{4} ): [ sinleft(frac{pi}{4}right) = cosleft(frac{pi}{4}right) = frac{sqrt{2}}{2} ] [ y = left(1 + frac{sqrt{2}}{2}right)^2 = frac{3}{2} + sqrt{2} ] - At ( x = frac{pi}{2} ): [ y = 2 times 1 = 2 ]5. Determine the Minimum Value: The values at the critical points and endpoints are: - ( x = -frac{pi}{6} ): ( frac{sqrt{3} + 2}{4} approx 0.933 ) - ( x = frac{pi}{4} ): ( approx 2.914 ) - ( x = frac{pi}{2} ): ( 2 )The minimum value is at ( x = -frac{pi}{6} ), giving the exact value:[boxed{dfrac{2 + sqrt{3}}{4}}]

🔑:Given the function:[ y = (sin x + 1)(cos x + 1) ]where ( -frac{pi}{6} leq x leq frac{pi}{2} ).We want to find the minimum value of this function in the given interval.Step 1: Consider the range of (sin x) and (cos x). - (sin x) ranges from (sin(-frac{pi}{6}) = -frac{1}{2}) to (sin(frac{pi}{2}) = 1).- (cos x) ranges from (cos(frac{pi}{2}) = 0) to (cos(-frac{pi}{6}) = frac{sqrt{3}}{2}).Step 2: Transform these trigonometric functions. Define:[u = sin x + 1 ][v = cos x + 1 ]We want to minimize the product:[ y = uv ]Step 3: Analyze critical points of the function by examining the range of (u) and (v).- The range of (u) will be from ( sin(-frac{pi}{6}) + 1 = frac{1}{2} ) to ( sin(frac{pi}{2}) + 1 = 2 ).- The range of (v) will be from ( cos(frac{pi}{2}) + 1 = 1 ) to ( cos(-frac{pi}{6}) + 1 = 1 + frac{sqrt{3}}{2} ).Step 4: Using calculus to find minimum by taking derivative.- The function to minimize is ( y = (sin x + 1)(cos x + 1) ).- Taking the derivative using the product rule:[frac{d}{dx} left[ (sin x + 1)(cos x + 1) right] = (cos x + 1)(cos x) + (sin x + 1)(-sin x)][= cos^2 x + cos x + sin x - sin^2 x]Setting (frac{d}{dx}(y) = 0) to find critical points.[cos^2 x + cos x + sin x - sin^2 x = 0]Step 5: Solve for (x) that satisfies the above condition:There are no easy algebraic solutions, so consider boundary and notable points:- At (x = -frac{pi}{6}), (sin(-frac{pi}{6}) = -frac{1}{2}), (cos(-frac{pi}{6}) = frac{sqrt{3}}{2}),[ y = left( -frac{1}{2} + 1 right) left( frac{sqrt{3}}{2} + 1 right) = frac{1}{2} left( frac{sqrt{3}}{2} + 1 right) = frac{1}{2} left( frac{sqrt{3} + 2}{2} right) = frac{sqrt{3} + 2}{4} ]- At (x = frac{pi}{2}), (sin(frac{pi}{2}) = 1), (cos(frac{pi}{2}) = 0),[ y = (1 + 1)(0 + 1) = 2 ]Conclusion: The minimum value of ((sin x + 1)(cos x + 1)) in the given interval is:[boxed{frac{2 + sqrt{3}}{4}}]

🧠:Okay, so I need to figure out the value of this expression: (√6 + √2)(√3 - 2)√(√3 + 2). The options are from (a) to (e), and I have to choose the correct one. Let me start by breaking down the problem step by step.First, let's look at the expression structure. There are three main parts multiplied together: (√6 + √2), (√3 - 2), and √(√3 + 2). Maybe if I simplify each part or find some relationships between them, it will help.Starting with the first part, (√6 + √2). I wonder if this can be simplified. Let me see: √6 is √2*√3, so maybe I can factor out √2? That would give √2(√3 + 1). Let me check:√2(√3 + 1) = √2*√3 + √2*1 = √6 + √2. Yes, that's correct. So, (√6 + √2) = √2(√3 + 1). That might be helpful later.Next, the second part is (√3 - 2). That seems straightforward, but maybe there's a way to relate it to the third part, which is √(√3 + 2). Hmm, the third part is a square root of (√3 + 2). Let me note that (√3 + 2) and (√3 - 2) might be conjugates. If I multiply them together, (√3 + 2)(√3 - 2) = (√3)^2 - (2)^2 = 3 - 4 = -1. Oh, interesting! So their product is -1. That could be useful.So, (√3 - 2) = -1/(√3 + 2). Therefore, maybe the product of the second and third parts can be simplified. Let's see:(√3 - 2) * √(√3 + 2) = (-1)/(√3 + 2) * √(√3 + 2) = -√(√3 + 2)/(√3 + 2). Hmm, not sure if that helps yet. Maybe another approach.Alternatively, let's denote x = √3. Then, the expression becomes (√(2x) + √2)(x - 2)√(x + 2). Maybe substituting x = √3 complicates things, but perhaps it can be simplified. Let's see:First, √(2x) = √(2√3), which is not very helpful. Alternatively, maybe squaring the entire expression to eliminate the square roots. But that might get messy. Let's think again.Looking at the third term, √(√3 + 2). Let me see if √3 + 2 can be expressed as a square. Suppose √3 + 2 = (√a + b)^2. Then expanding the right side gives a + 2b√a + b². Comparing to √3 + 2, we need:a + b² = 2 and 2b√a = √3.Let me solve these equations. Let me assume a and b are rational numbers.From the second equation: 2b√a = √3. Let me square both sides: 4b²a = 3.From the first equation: a + b² = 2. Let me let b² = 2 - a. Substitute into the second equation:4(2 - a)a = 3 → 4(2a - a²) = 3 → 8a - 4a² = 3 → 4a² -8a +3=0.Quadratic equation: 4a² -8a +3=0. Discriminant D = 64 - 48 = 16. Solutions: a = [8 ± √16]/8 = [8 ±4]/8. So a = (8+4)/8=12/8=3/2, or a=(8-4)/8=4/8=1/2.Check for a=3/2: Then b² = 2 - 3/2=1/2. Then from 2b√a=√3: 2b*(√(3/2))=√3. Let's compute left side: 2b*(√6/2)=b√6. Set equal to √3: b√6=√3 → b=√3/√6=1/√2. But b²=(1/√2)^2=1/2, which matches. So yes, a=3/2 and b=1/√2. Therefore:√(√3 + 2) = √( (3/2) + (1/√2)^2 + 2*(1/√2)*√(3/2) )? Wait, no, the original assumption was that √3 + 2 = (√a + b)^2. So with a=3/2 and b=1/√2, then:(√(3/2) + 1/√2)^2 = (3/2) + 2*(√(3/2)*(1/√2)) + (1/2) = 3/2 + 1/2 + 2*(√3/(√2*√2)) = 2 + 2*(√3/2) = 2 + √3. Yes! Perfect. Therefore, √(√3 + 2) = √(3/2) + 1/√2. Let me compute that:√(3/2) = √3 / √2, and 1/√2 is already simplified. So, √(√3 + 2) = (√3 + 1)/√2. Because:√3 / √2 + 1/√2 = (√3 + 1)/√2. Yes, that's right. Therefore, √(√3 + 2) = (√3 + 1)/√2.Okay, so that simplifies the third term. Let's substitute that back into the original expression:Original expression: (√6 + √2)(√3 - 2) * (√3 + 1)/√2.Earlier, we found that (√6 + √2) = √2(√3 + 1). Let me substitute that:So, √2(√3 + 1) * (√3 - 2) * (√3 + 1)/√2.Notice that √2 and 1/√2 will multiply to 1. So those cancel out. Then, we have (√3 + 1)*(√3 - 2)*(√3 + 1).Wait, actually, the order is (√3 + 1) * (√3 - 2) * (√3 + 1). Which can be written as [(√3 + 1)^2]*(√3 - 2).Let me compute (√3 + 1)^2 first. That is (√3)^2 + 2*√3*1 + 1^2 = 3 + 2√3 + 1 = 4 + 2√3. Then multiply by (√3 - 2):So, (4 + 2√3)(√3 - 2). Let's expand this:4*√3 - 8 + 2√3*√3 - 4√3.Compute each term:4√3 -8 + 2*(3) -4√3 = 4√3 -8 +6 -4√3. The 4√3 and -4√3 cancel out. Then -8 +6 = -2. So the result is -2.Wait, so after all that, the expression simplifies to -2? Which is option (c). Let me check again to make sure I didn't make a mistake.First step: (√6 + √2) becomes √2(√3 +1). Then multiplied by (√3 -2) and √(√3 +2) which is (√3 +1)/√2. Then √2 * 1/√2 cancels to 1. Then we have (√3 +1)*(√3 -2)*(√3 +1). Which is (√3 +1)^2*(√3 -2). Then (√3 +1)^2 is 4 +2√3, multiplied by (√3 -2) gives:(4 +2√3)(√3 -2) = 4√3 -8 +2*3 -4√3 = (4√3 -4√3) + (-8 +6) = 0 -2 = -2. Yep, that's correct. So the answer should be (c) -2.But let me confirm again using another method to ensure I didn't make a mistake.Alternative approach: Let me compute the entire expression numerically.First, compute each component approximately:√3 ≈ 1.732, √6 ≈ 2.449, √2 ≈ 1.414.Compute (√6 + √2) ≈ 2.449 + 1.414 ≈ 3.863.Compute (√3 -2) ≈ 1.732 - 2 ≈ -0.268.Compute √(√3 +2). First, √3 +2 ≈ 1.732 +2 ≈ 3.732. Then square root of that is √3.732 ≈ 1.931.Multiply all together: 3.863 * (-0.268) * 1.931 ≈ First, 3.863 * (-0.268) ≈ -1.036. Then multiply by 1.931 ≈ -1.036 * 1.931 ≈ -2.000. So approximately -2. Therefore, the answer is (c) -2. That matches the algebraic result. So that seems solid.Another way: Maybe rationalizing or manipulating the expression differently.Original expression: (√6 + √2)(√3 - 2)√(√3 + 2).Let me denote A = (√6 + √2), B = (√3 - 2), C = √(√3 + 2). So the expression is A*B*C.Earlier, we found that C = (√3 +1)/√2. So substituting, A*B*C = (√6 +√2)*(√3 -2)*(√3 +1)/√2.But note that (√6 +√2)/√2 = (√6/√2 + √2/√2) = √3 +1. Therefore, A/√2 = √3 +1. Therefore, the entire expression becomes (√3 +1)*(√3 -2)*(√3 +1) as before, which equals -2.So multiple methods confirm that the answer is -2, which is option (c). I think that's solid. Let me check if there's any other approach.Alternatively, perhaps multiply all terms together first:Original expression: (√6 + √2)(√3 - 2)√(√3 + 2).Let me first compute (√6 + √2)(√3 - 2). Let's expand this:√6*√3 + √6*(-2) + √2*√3 + √2*(-2).Compute each term:√6*√3 = √(6*3) = √18 = 3√2.√6*(-2) = -2√6.√2*√3 = √6.√2*(-2) = -2√2.So combining all terms: 3√2 -2√6 + √6 -2√2.Combine like terms:3√2 -2√2 = √2.-2√6 + √6 = -√6.So overall, (√6 + √2)(√3 -2) = √2 - √6.Now multiply this by √(√3 +2):(√2 - √6) * √(√3 +2).Hmm, let's see. Let me substitute √(√3 +2) as (√3 +1)/√2, as before. Then:(√2 - √6) * (√3 +1)/√2.Factor out √2 in the numerator:√2*(1 - √3) * (√3 +1)/√2.The √2 cancels with 1/√2, leaving:(1 - √3)(√3 +1).Multiply this out: (1)(√3) +1(1) -√3(√3) -√3(1).Simplify:√3 +1 -3 -√3 = (√3 -√3) + (1 -3) = 0 -2 = -2.Again, the same result. So regardless of the method, it's -2. Therefore, the answer is (c) -2.I think this is thorough. All approaches lead to the same conclusion. So I'm confident the answer is (c).Final Answerboxed{C}

✅:To determine the value of the expression ((sqrt{6} + sqrt{2})(sqrt{3} - 2) sqrt{sqrt{3} + 2}), we can break it down into manageable parts and simplify step by step.1. Simplify ((sqrt{6} + sqrt{2})): [ sqrt{6} + sqrt{2} = sqrt{2}(sqrt{3} + 1) ]2. Simplify (sqrt{sqrt{3} + 2}): Recognizing that (sqrt{3} + 2) can be expressed as a square: [ sqrt{sqrt{3} + 2} = frac{sqrt{3} + 1}{sqrt{2}} ]3. Substitute the simplified forms back into the original expression: [ (sqrt{2}(sqrt{3} + 1))(sqrt{3} - 2)left(frac{sqrt{3} + 1}{sqrt{2}}right) ]4. Cancel out (sqrt{2}) and (frac{1}{sqrt{2}}): [ (sqrt{3} + 1)(sqrt{3} - 2)(sqrt{3} + 1) ]5. Simplify the product: [ (sqrt{3} + 1)^2 (sqrt{3} - 2) ]6. Expand ((sqrt{3} + 1)^2): [ (sqrt{3} + 1)^2 = 3 + 2sqrt{3} + 1 = 4 + 2sqrt{3} ]7. Multiply by ((sqrt{3} - 2)): [ (4 + 2sqrt{3})(sqrt{3} - 2) ]8. Expand and simplify: [ 4sqrt{3} - 8 + 2 cdot 3 - 4sqrt{3} = 4sqrt{3} - 8 + 6 - 4sqrt{3} = -2 ]Thus, the value of the expression is (-2), which corresponds to option (c).[boxed{C}]

🔑:1. Denote the given expression by ( A ): [ A = (sqrt{6} + sqrt{2})(sqrt{3} - 2) sqrt{sqrt{3} + 2} ]2. Compute ( A^2 ): [ A^2 = left[ (sqrt{6} + sqrt{2})(sqrt{3} - 2) sqrt{sqrt{3} + 2} right]^2 ] 3. Simplify ( A^2 ): [ A^2 = (sqrt{6} + sqrt{2})^2 (sqrt{3} - 2)^2 (sqrt{sqrt{3} + 2})^2 ] [ = (sqrt{6} + sqrt{2})^2 (sqrt{3} - 2)^2 (sqrt{3} + 2) ]4. Express ((sqrt{3} - 2)^2(sqrt{3} + 2)): [ = (sqrt{3} - 2)[(sqrt{3} - 2)(sqrt{3} + 2)] ] Notice that ((sqrt{3} - 2)(sqrt{3} + 2)) can be expanded using the difference of squares formula, ((a-b)(a+b)=a^2-b^2): [ = (sqrt{3} - 2)((sqrt{3})^2 - 2^2) ] [ = (sqrt{3} - 2)(3 - 4) = (sqrt{3} - 2)(-1) ] [ = -(sqrt{3} - 2) ]5. Compute ((sqrt{6} + sqrt{2})^2): [ (sqrt{6} + sqrt{2})^2 = (sqrt{6})^2 + 2(sqrt{6})(sqrt{2}) + (sqrt{2})^2 ] [ = 6 + 2sqrt{12} + 2 = 6 + 2sqrt{12} + 2 = 8 + 4sqrt{3} ]6. Combine all parts: [ A^2 = (8 + 4sqrt{3})(-sqrt{3} + 2) ] [ = -(8 + 4sqrt{3})(sqrt{3} - 2) ]7. Simplify the expression: [ = -(8sqrt{3} - 16 + 4(sqrt{3})^2 - 8sqrt{3}) ] [ = -(8sqrt{3} - 16 + 12 - 8sqrt{3}) ] [ = -(-4) ] [ = 4 ]8. Hence, [ A^2 = 4 implies A = pm 2 ] Since ((sqrt{3} - 2)) is negative while ((sqrt{6} + sqrt{2})(sqrt{sqrt{3} + 2})) are positive, thus making the product negative. Thus, 9. Conclusion: [ A = -2 ]Therefore, the correct option is ( boxed{text{c}} ).

🧠:Okay, so I need to find the square of the distance between two points A and B on the parabola y = x², and both points are at a distance of 8√2 units from the line y = -x - 4. Hmm, let's start by recalling the formula for the distance from a point to a line. The distance d from a point (x₀, y₀) to the line ax + by + c = 0 is |ax₀ + by₀ + c| / sqrt(a² + b²). Right?First, let's write the given line in standard form. The line is y = -x - 4. If I rearrange that, it becomes x + y + 4 = 0. So, a = 1, b = 1, c = 4. The distance from a point (x, y) to this line would be |x + y + 4| / sqrt(1² + 1²) = |x + y + 4| / sqrt(2). Since both points A and B lie on the parabola y = x², their coordinates can be written as (a, a²) and (b, b²), right? Wait, but maybe A and B could be the same point? But the problem says "points" plural, so they must be distinct. So we need to find all points (x, x²) on the parabola where the distance to the line y = -x -4 is 8√2. Then, once we have those points, compute the square of the distance between them.So, let's set up the equation for the distance. For a point (x, x²) on the parabola, the distance to the line is |x + x² + 4| / sqrt(2) = 8√2. Multiply both sides by sqrt(2): |x + x² + 4| = 8√2 * sqrt(2) = 8*2 = 16.Therefore, |x² + x + 4| = 16. So this gives two equations:1. x² + x + 4 = 162. x² + x + 4 = -16Let's solve the first equation: x² + x + 4 = 16 => x² + x - 12 = 0. Let's factor this quadratic. Looking for two numbers that multiply to -12 and add to 1. 4 and -3: 4 * (-3) = -12, 4 + (-3) = 1. So factors are (x + 4)(x - 3) = 0. So solutions x = -4 and x = 3.Second equation: x² + x + 4 = -16 => x² + x + 20 = 0. Let's compute the discriminant: D = 1² - 4*1*20 = 1 - 80 = -79. Since discriminant is negative, no real solutions here. So the only real solutions are x = -4 and x = 3.Therefore, the points A and B are (-4, (-4)²) = (-4, 16) and (3, 3²) = (3, 9). Now, compute the distance squared between these two points. The distance squared is (x2 - x1)^2 + (y2 - y1)^2.So, (3 - (-4))² + (9 - 16)^2 = (7)² + (-7)^2 = 49 + 49 = 98. Therefore, the square of the distance between A and B is 98.Wait a second, but hold on. Let me check if I did everything right. Let me verify the distance from each point to the line. For point (-4, 16):Using the distance formula |x + y + 4| / sqrt(2) = |-4 + 16 + 4| / sqrt(2) = |16| / sqrt(2) = 16 / sqrt(2) = 8*sqrt(2). Correct. For point (3,9):|3 + 9 + 4| / sqrt(2) = |16| / sqrt(2) = 16/sqrt(2) = 8*sqrt(2). Also correct. So both points are indeed at the required distance. Then the distance squared between (-4,16) and (3,9):Difference in x: 3 - (-4) = 7, squared is 49.Difference in y: 9 - 16 = -7, squared is 49.Total distance squared: 49 + 49 = 98. Yes, that seems correct. So the answer should be 98.But wait, the problem says "Find the square of the distance between A and B." So 98 is the answer. But let me check again if there are more than two points. Wait, the quadratic equation gave us two real solutions, so only two points. So A and B are those two points. Therefore, their distance squared is 98. But just to make sure, maybe I made a mistake in solving the quadratic? Let's check x² + x -12 = 0. The roots are x = [-1 ± sqrt(1 + 48)] / 2 = [-1 ± 7]/2. So (-1 +7)/2 = 6/2=3, (-1 -7)/2=-8/2=-4. Correct. So the roots are indeed 3 and -4. And the other quadratic had discriminant negative, so no real roots. So only two points. So the answer is 98. Hmm, seems correct.Final Answerboxed{98}

✅:To find the square of the distance between points ( A ) and ( B ) on the parabola ( y = x^2 ) that are both at a distance of ( 8sqrt{2} ) units from the line ( y = -x - 4 ):1. Convert the line equation to standard form: The line ( y = -x - 4 ) becomes ( x + y + 4 = 0 ).2. Distance formula: The distance from a point ((x, x^2)) to the line ( x + y + 4 = 0 ) is given by: [ frac{|x + x^2 + 4|}{sqrt{2}} = 8sqrt{2} ]3. Solve for ( x ): [ |x^2 + x + 4| = 16 ] This results in two equations: [ x^2 + x + 4 = 16 quad text{and} quad x^2 + x + 4 = -16 ]4. Solve the quadratic equations: - For ( x^2 + x - 12 = 0 ): [ (x + 4)(x - 3) = 0 implies x = -4 text{ or } x = 3 ] - For ( x^2 + x + 20 = 0 ), the discriminant is negative, so no real solutions.5. Identify points ( A ) and ( B ): - ( A = (-4, 16) ) - ( B = (3, 9) )6. Calculate the square of the distance between ( A ) and ( B ): [ text{Distance squared} = (3 - (-4))^2 + (9 - 16)^2 = 7^2 + (-7)^2 = 49 + 49 = 98 ]The square of the distance between ( A ) and ( B ) is (boxed{98}).

🔑:Let's solve the problem step by step, ensuring we explain every detail.1. Identify and Set Up Coordinates: - Given that A and B lie on the parabola y = x^2. - These points are at a distance of 8sqrt{2} units from the line y = -x - 4.2. Distance Calculation: - To find points on the parabola that are equidistant from the given line y = -x - 4, consider a general point on the parabola (x, x^2). - The distance (d) from a point (h, k) to a line (Ax + By + C = 0) is given by: [ d = frac{|Ax + By + C|}{sqrt{A^2 + B^2}} ] - For the line (y = -x - 4) or equivalently (x + y + 4 = 0), (A = 1), (B = 1), and (C = 4).3. Using the Distance Formula: - Substitute the point (x, x^2) and the distance 8sqrt{2}: [ 8sqrt{2} = frac{|x + x^2 + 4|}{sqrt{1^2 + 1^2}} = frac{|x + x^2 + 4|}{sqrt{2}} ] - Simplify: [ 8sqrt{2} cdot sqrt{2} = |x + x^2 + 4| ] [ 16 = |x + x^2 + 4| ] - This splits into two cases: [ x + x^2 + 4 = 16 quad text{or} quad x + x^2 + 4 = -16 ]4. Solve Each Equation: - For (x + x^2 + 4 = 16): [ x^2 + x + 4 = 16 ] [ x^2 + x - 12 = 0 ] - Solving the quadratic: [ x = frac{-1 pm sqrt{1^2 - 4 cdot 1 cdot (-12)}}{2 cdot 1} ] [ x = frac{-1 pm sqrt{1 + 48}}{2} = frac{-1 pm 7}{2} ] - Therefore, (x = 3) or (x = -4). - For (x + x^2 + 4 = -16): [ x^2 + x + 4 = -16 ] [ x^2 + x + 20 = 0 ] - The discriminant ((b^2 - 4ac)) is negative: [ 1^2 - 4 cdot 1 cdot 20 < 0 ] - This equation has no real solutions.5. Find the Coordinates of Points (A) and (B): - The solutions from the first case are (x = 3) and (x = -4). - Corresponding coordinates are: - For (x = 3): (3, 9) - For (x = -4): (-4, 16)6. Calculate the Distance Between Points (A) and (B): - (A = (3, 9)) and (B = (-4, 16)) - Use the distance formula: [ |AB|^2 = (3 - (-4))^2 + (9 - 16)^2 ] [ |AB|^2 = (3 + 4)^2 + (9 - 16)^2 ] [ |AB|^2 = 7^2 + (-7)^2 ] [ |AB|^2 = 49 + 49 ] [ |AB|^2 = 98 ] Conclusion:The square of the distance between points (A) and (B) is (boxed{98}).

🧠:Okay, so I need to find the sum of the x-coordinates of the distinct points where these two plane curves intersect. The equations given are (x^2 = x + y + 4) and (y^2 = y - 15x + 36). Hmm, let me start by understanding what each equation represents and then figure out how to solve them simultaneously.First, the first equation is (x^2 = x + y + 4). That looks like a quadratic equation in x, but since it has both x and y, it's probably a parabola. Maybe it's a parabola that opens either upwards or downwards depending on the coefficient. Since the coefficient of (x^2) is positive (1), it opens upwards. Let me rearrange the equation to express y in terms of x. If I subtract x and 4 from both sides, I get (y = x^2 - x - 4). So this is a parabola opening upwards, with vertex somewhere. Not sure if the vertex is important here, but maybe useful for visualization.The second equation is (y^2 = y - 15x + 36). This seems like a quadratic in y, but again with both x and y. Let me try to rearrange this as well. Subtract y and add 15x, subtract 36 from both sides to get (y^2 - y + 15x - 36 = 0). Hmm, maybe it's easier to express x in terms of y here. Let me try that. Starting from (y^2 = y -15x +36), subtract y and 36 from both sides: (y^2 - y - 36 = -15x), then divide both sides by -15: (x = frac{-y^2 + y + 36}{15}). So x is expressed in terms of y here. That might be helpful.Now, since I need to find the points of intersection, I need to solve these two equations simultaneously. That means finding (x, y) that satisfy both equations. Since I have y expressed in terms of x from the first equation, maybe I can substitute that into the second equation. Alternatively, since I have x in terms of y from the second equation, I could substitute that into the first. Let me try both approaches and see which is easier.Starting with the first approach: substitute (y = x^2 - x - 4) into the second equation. The second equation is (y^2 = y -15x +36). So replacing y with (x^2 - x -4), we get:[(x^2 - x -4)^2 = (x^2 - x -4) -15x +36]Let me expand the left side first. ((x^2 - x -4)^2). Let me compute that step by step. First, square the first term: (x^4). Then cross terms: 2*(x^2)*(-x) = -2x^3, 2*(x^2)*(-4) = -8x^2, then cross terms for (-x)*(-4): 2*(-x)*(-4) = 8x. Then square the second term: (-x)^2 = x^2, cross term with -4: 2*(-x)*(-4) = 8x, then square the last term: (-4)^2 = 16. Wait, maybe it's better to just multiply it out properly.Let me write it as (x^2 - x -4)(x^2 - x -4):Multiply x^2 by each term in the second polynomial: x^2*x^2 = x^4, x^2*(-x) = -x^3, x^2*(-4) = -4x^2.Multiply -x by each term: -x*x^2 = -x^3, -x*(-x) = x^2, -x*(-4) = 4x.Multiply -4 by each term: -4*x^2 = -4x^2, -4*(-x) = 4x, -4*(-4) = 16.Now combine all terms:x^4 -x^3 -4x^2 -x^3 +x^2 +4x -4x^2 +4x +16Combine like terms:x^4 + (-x^3 -x^3) + (-4x^2 +x^2 -4x^2) + (4x +4x) +16Simplify:x^4 -2x^3 -7x^2 +8x +16So the left side is (x^4 -2x^3 -7x^2 +8x +16).Now the right side is ((x^2 -x -4) -15x +36). Let's compute that:First, combine like terms: x^2 -x -4 -15x +36.Combine the x terms: -x -15x = -16xConstants: -4 +36 = 32So the right side simplifies to (x^2 -16x +32).Now set the left side equal to the right side:(x^4 -2x^3 -7x^2 +8x +16 = x^2 -16x +32)Bring all terms to the left side:(x^4 -2x^3 -7x^2 +8x +16 -x^2 +16x -32 = 0)Combine like terms:x^4 -2x^3 + (-7x^2 -x^2) + (8x +16x) + (16 -32) = 0Simplify:x^4 -2x^3 -8x^2 +24x -16 = 0So now we have a quartic equation: (x^4 -2x^3 -8x^2 +24x -16 = 0). Hmm, quartic equations can be tricky, but maybe this factors nicely. Let's try to factor this polynomial.First, let's check if there are any rational roots using the Rational Root Theorem. The possible rational roots are the factors of the constant term (-16) divided by the factors of the leading coefficient (1). So possible roots are ±1, ±2, ±4, ±8, ±16.Let's test x=1: Plugging into the polynomial: 1 -2 -8 +24 -16 = (1 -2) + (-8 +24) + (-16) = (-1) +16 -16 = -1. Not zero.x=2: 16 -16 -32 +48 -16 = (16-16) + (-32+48) -16 = 0 +16 -16 = 0. So x=2 is a root. Great. So (x - 2) is a factor.Let's perform polynomial division or use synthetic division to factor out (x - 2).Using synthetic division with root 2:Coefficients: 1 (x^4), -2 (x^3), -8 (x^2), 24 (x), -16 (constant)Bring down the 1.Multiply 1 by 2: 2. Add to -2: 0.Multiply 0 by 2: 0. Add to -8: -8.Multiply -8 by 2: -16. Add to 24: 8.Multiply 8 by 2: 16. Add to -16: 0.So the result is x^3 +0x^2 -8x +8. So the polynomial factors as (x - 2)(x^3 -8x +8).Now let's factor x^3 -8x +8. Let's check for rational roots here. Possible roots are ±1, ±2, ±4, ±8.Testing x=1: 1 -8 +8 =1. Not zero.x=2: 8 -16 +8=0. So x=2 is another root. So (x -2) is a factor again.Performing synthetic division on x^3 -8x +8 with root 2:Coefficients: 1 (x^3), 0 (x^2), -8 (x), 8 (constant)Bring down 1.Multiply 1 by 2: 2. Add to 0: 2.Multiply 2 by 2: 4. Add to -8: -4.Multiply -4 by 2: -8. Add to 8: 0.So the quotient polynomial is x^2 +2x -4. Therefore, the original polynomial factors as (x -2)^2 (x^2 +2x -4).Now, we need to factor or solve x^2 +2x -4. Let's find its roots using quadratic formula.x = [-2 ± sqrt(4 +16)]/2 = [-2 ± sqrt(20)]/2 = [-2 ± 2*sqrt(5)]/2 = -1 ± sqrt(5)So the roots are x = 2 (double root), and x = -1 + sqrt(5), x = -1 - sqrt(5)Therefore, the quartic equation factors as (x -2)^2 (x +1 - sqrt(5))(x +1 + sqrt(5)) =0.Hence, the solutions for x are x=2 (with multiplicity 2), x= -1 + sqrt(5), and x= -1 - sqrt(5).But wait, since we are looking for distinct points of intersection, even though x=2 is a double root, we need to check if each x corresponds to a distinct y or if they might result in the same point.So, first, let's find the corresponding y for each x.Starting with x=2. From the first equation, (y = x^2 -x -4). Plugging in x=2: y = 4 -2 -4 = -2. So the point is (2, -2). Now check if this satisfies the second equation (y^2 = y -15x +36). Plugging in y=-2 and x=2: Left side: (-2)^2 =4. Right side: -2 -30 +36=4. So yes, it works. So (2, -2) is a valid intersection point.Now check if x=2 with multiplicity 2 gives the same point. Since we have a double root, but since the equations are algebraic, maybe there's a tangency here? But the y is uniquely determined once x is given, so even though the root is double, the point is the same. So only one distinct point here.Next, x= -1 + sqrt(5). Let's compute y for this x. Again using (y = x^2 -x -4). Let's compute x^2:x = -1 + sqrt(5). Then x +1 = sqrt(5). Squaring both sides: (x +1)^2 = 5. So x^2 +2x +1 =5. Therefore, x^2 =5 -2x -1 =4 -2x. Therefore, x^2 =4 -2x.So plugging back into y:y = x^2 -x -4 = (4 -2x) -x -4 =4 -2x -x -4 = -3x. Therefore, y= -3x for x = -1 + sqrt(5). Wait, is that correct? Let me verify.Alternatively, compute x^2 directly.x = -1 + sqrt(5). So x^2 = (-1 + sqrt(5))^2 = 1 -2 sqrt(5) +5 =6 -2 sqrt(5). Then y = x^2 -x -4 = (6 -2 sqrt(5)) - (-1 + sqrt(5)) -4 =6 -2 sqrt(5) +1 - sqrt(5) -4 = (6 +1 -4) + (-2 sqrt(5) - sqrt(5)) =3 -3 sqrt(5). So y=3 -3 sqrt(5). Alternatively, if we use y= -3x: y= -3*(-1 + sqrt(5))=3 -3 sqrt(5). Yes, same result. So the point is (-1 + sqrt(5), 3 -3 sqrt(5)). Let's check this in the second equation (y^2 = y -15x +36).Compute left side: (3 -3 sqrt(5))^2 =9 -18 sqrt(5) +45=54 -18 sqrt(5).Right side: y -15x +36 = (3 -3 sqrt(5)) -15*(-1 + sqrt(5)) +36 =3 -3 sqrt(5) +15 -15 sqrt(5) +36 = (3 +15 +36) + (-3 sqrt(5) -15 sqrt(5)) =54 -18 sqrt(5). So it matches. Good.Similarly, for x= -1 - sqrt(5). Compute y:Again using y = x^2 -x -4. Let's compute x^2:x = -1 - sqrt(5). So x +1 = -sqrt(5). Squaring both sides: (x +1)^2 =5. So x^2 +2x +1=5. Therefore, x^2=5 -2x -1=4 -2x. Therefore, y= x^2 -x -4=4 -2x -x -4= -3x. So y= -3*(-1 - sqrt(5))=3 +3 sqrt(5). Alternatively, compute directly:x^2 = (-1 - sqrt(5))^2=1 +2 sqrt(5) +5=6 +2 sqrt(5). Then y=6 +2 sqrt(5) - (-1 - sqrt(5)) -4=6 +2 sqrt(5) +1 + sqrt(5) -4= (6 +1 -4)+(2 sqrt(5) + sqrt(5))=3 +3 sqrt(5). So the point is (-1 - sqrt(5), 3 +3 sqrt(5)). Check in the second equation:Left side: (3 +3 sqrt(5))^2=9 +18 sqrt(5) +45=54 +18 sqrt(5).Right side: y -15x +36= (3 +3 sqrt(5)) -15*(-1 - sqrt(5)) +36=3 +3 sqrt(5) +15 +15 sqrt(5) +36=(3+15+36)+(3 sqrt(5)+15 sqrt(5))=54 +18 sqrt(5). Correct.So all three x-values (2, -1 + sqrt(5), -1 - sqrt(5)) correspond to valid points. Even though x=2 was a double root, the point (2, -2) is a single distinct point. So the distinct intersection points are:1. (2, -2)2. (-1 + sqrt(5), 3 -3 sqrt(5))3. (-1 - sqrt(5), 3 +3 sqrt(5))So there are three distinct points. Therefore, the x-coordinates are 2, -1 + sqrt(5), and -1 - sqrt(5). The sum of these x-coordinates is 2 + (-1 + sqrt(5)) + (-1 - sqrt(5)) = 2 -1 -1 + sqrt(5) - sqrt(5) =0.Wait, that's interesting. The sqrt(5) terms cancel out, and the constants add up to 0. So the sum is 0.But let me verify that, because it's possible that when dealing with roots of equations, the sum of roots relates to coefficients. Wait, the quartic equation was x^4 -2x^3 -8x^2 +24x -16=0. The sum of all roots (counted with multiplicity) is equal to the coefficient of x^3 term with opposite sign. Since the quartic is degree 4, the sum of roots (with multiplicity) is 2. But since we have roots at x=2 (twice), x=-1 + sqrt(5), and x=-1 - sqrt(5)), sum is 2 +2 + (-1 + sqrt(5)) + (-1 - sqrt(5)) = 2 +2 -1 -1 +0=2. Which matches the coefficient. However, when considering distinct roots, the sum is 2 + (-1 + sqrt(5)) + (-1 - sqrt(5)) =0. So that's correct.Therefore, even though the quartic had a double root at x=2, the sum of distinct x-coordinates is 0.But let me cross-check. Maybe I made a mistake when assuming the sum is 0. Let me compute the sum step by step.Sum = 2 + (-1 + sqrt(5)) + (-1 - sqrt(5)).Compute constants: 2 -1 -1 =0.Compute radicals: sqrt(5) - sqrt(5)=0.So total sum=0. Yes, that's correct.Therefore, the sum of the x-coordinates of the distinct intersection points is 0.But let me just verify once again to make sure there are no mistakes in the process.First, solving the system:1. (y = x^2 -x -4)2. (y^2 = y -15x +36)Substituted equation 1 into equation 2, expanded, and got a quartic equation. Factored it, found roots, and calculated the sum. The key steps seem correct.Another way to approach: Since we have two equations, perhaps using substitution the other way. Express x from equation 2 in terms of y and substitute into equation 1. Let me try this method quickly to confirm.From equation 2: (y^2 = y -15x +36). Solving for x:15x = y - y^2 +36x = (y - y^2 +36)/15Now substitute into equation 1: (x^2 =x + y +4). So:[(y - y^2 +36)/15]^2 = (y - y^2 +36)/15 + y +4This seems more complicated, but let's compute left and right sides.Left side: [( -y^2 + y +36)/15]^2 = [(-y^2 + y +36)^2]/225Right side: [(-y^2 + y +36)/15] + y +4 = (-y^2 + y +36)/15 + (15y)/15 +60/15 = (-y^2 + y +36 +15y +60)/15 = (-y^2 +16y +96)/15So set left side equal to right side:[(-y^2 + y +36)^2]/225 = (-y^2 +16y +96)/15Multiply both sides by 225:(-y^2 + y +36)^2 = 15*(-y^2 +16y +96)Expand left side: (-y^2 + y +36)^2. Let's compute that.Let me write it as ( -y^2 + y +36 )^2. Let me denote A = -y^2, B = y, C=36.Then (A + B + C)^2 = A^2 + B^2 + C^2 +2AB +2AC +2BCCompute each term:A^2 = (-y^2)^2 = y^4B^2 = y^2C^2 = 36^2 = 12962AB = 2*(-y^2)(y) = -2y^32AC = 2*(-y^2)(36) = -72y^22BC = 2*(y)(36) =72ySo combining all terms:y^4 + y^2 +1296 -2y^3 -72y^2 +72ySimplify:y^4 -2y^3 + (y^2 -72y^2) +72y +1296Which is:y^4 -2y^3 -71y^2 +72y +1296So left side is y^4 -2y^3 -71y^2 +72y +1296Right side is 15*(-y^2 +16y +96)= -15y^2 +240y +1440Set left minus right equal to zero:y^4 -2y^3 -71y^2 +72y +1296 +15y^2 -240y -1440 =0Combine like terms:y^4 -2y^3 + (-71y^2 +15y^2) + (72y -240y) + (1296 -1440) =0Simplify:y^4 -2y^3 -56y^2 -168y -144 =0So we have a quartic in y: y^4 -2y^3 -56y^2 -168y -144 =0. Hmm, this seems more complicated than the previous quartic. Let's check if we can factor this.Again, using Rational Root Theorem: possible roots are factors of 144 over 1, so ±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±16, ±18, ±24, ±36, ±48, ±72, ±144.Let's test y= -2: Plug into equation: (-2)^4 -2*(-2)^3 -56*(-2)^2 -168*(-2) -144 =16 +16 -224 +336 -144= (16+16) + (-224 +336) + (-144)=32 +112 -144=0. So y=-2 is a root. So (y +2) is a factor.Performing synthetic division on the quartic with root y=-2.Coefficients: 1 (y^4), -2 (y^3), -56 (y^2), -168 (y), -144 (constant)Bring down 1.Multiply 1 by -2: -2. Add to -2: -4.Multiply -4 by -2:8. Add to -56: -48.Multiply -48 by -2:96. Add to -168: -72.Multiply -72 by -2:144. Add to -144:0.So the quotient is y^3 -4y^2 -48y -72. Now factor this cubic.Testing possible roots. Let's try y=6: 216 - 144 -288 -72= (216-144) + (-288-72)=72 -360= -288≠0.y= -3: (-27) -4*(9) -48*(-3) -72= -27 -36 +144 -72= (-27-36) + (144-72)= -63 +72=9≠0.y= -4: (-64) -4*(16) -48*(-4) -72= -64 -64 +192 -72= (-64-64)+(192-72)= -128 +120=-8≠0.y= -6: (-216) -4*(36) -48*(-6) -72= -216 -144 +288 -72= (-216-144)+(288-72)= -360 +216= -144≠0.y=12: 1728 -4*144 -48*12 -72=1728 -576 -576 -72=1728 -1224=504≠0.Wait, maybe y= -2 again? Let's try y= -2 again in the cubic:(-8) -4*(4) -48*(-2) -72= -8 -16 +96 -72= (-8-16)+(96-72)= -24 +24=0. So y=-2 is another root.So factor out (y +2) again.Using synthetic division on y^3 -4y^2 -48y -72 with y=-2:Coefficients:1, -4, -48, -72Bring down 1.Multiply 1 by -2: -2. Add to -4: -6.Multiply -6 by -2:12. Add to -48: -36.Multiply -36 by -2:72. Add to -72:0.So quotient polynomial is y^2 -6y -36. So the quartic factors as (y +2)^2(y^2 -6y -36)=0.Now solve y^2 -6y -36=0. Using quadratic formula:y = [6 ± sqrt(36 +144)]/2 = [6 ± sqrt(180)]/2 = [6 ± 6*sqrt(5)]/2 = 3 ± 3 sqrt(5).Therefore, the roots are y=-2 (double root), y=3 +3 sqrt(5), and y=3 -3 sqrt(5).Now, corresponding x for each y:For y=-2: From equation x=(y - y^2 +36)/15. Plugging in y=-2:x=( -2 -4 +36)/15=(30)/15=2. So x=2. So the point (2, -2), as before.For y=3 +3 sqrt(5): x=( (3 +3 sqrt(5)) - (3 +3 sqrt(5))^2 +36)/15.First compute (3 +3 sqrt(5))^2=9 +18 sqrt(5) +45=54 +18 sqrt(5).Then compute numerator: 3 +3 sqrt(5) -54 -18 sqrt(5) +36= (3 -54 +36) + (3 sqrt(5) -18 sqrt(5))= (-15) + (-15 sqrt(5))= -15(1 + sqrt(5)).Thus x= -15(1 + sqrt(5))/15= - (1 + sqrt(5))= -1 - sqrt(5). So x= -1 - sqrt(5), which matches the previous result.Similarly for y=3 -3 sqrt(5): x=(3 -3 sqrt(5) - (3 -3 sqrt(5))^2 +36)/15.Compute (3 -3 sqrt(5))^2=9 -18 sqrt(5) +45=54 -18 sqrt(5).Numerator:3 -3 sqrt(5) -54 +18 sqrt(5) +36= (3 -54 +36) + (-3 sqrt(5) +18 sqrt(5))= (-15) +15 sqrt(5).Thus x= (-15 +15 sqrt(5))/15= (-1 + sqrt(5)). So x= -1 + sqrt(5). Correct.Therefore, the points are:(2, -2), (-1 - sqrt(5), 3 +3 sqrt(5)), (-1 + sqrt(5), 3 -3 sqrt(5)).Same as before. So sum of x-coordinates is 2 + (-1 - sqrt(5)) + (-1 + sqrt(5))= 0. Therefore, this method also gives the same result. Hence, the answer is 0.I think this confirms the previous conclusion. Therefore, the sum of the x-coordinates is indeed 0.Final AnswerThe sum of the (x)-coordinates is boxed{0}.

✅:To find the sum of the (x)-coordinates of the distinct points of intersection of the plane curves given by (x^2 = x + y + 4) and (y^2 = y - 15x + 36), we start by solving these equations simultaneously.1. Express (y) from the first equation: [ y = x^2 - x - 4 ]2. Substitute (y) into the second equation: [ (x^2 - x - 4)^2 = (x^2 - x - 4) - 15x + 36 ]3. Expand and simplify the equation: [ (x^2 - x - 4)^2 = x^2 - 16x + 32 ] Expanding the left side: [ x^4 - 2x^3 - 7x^2 + 8x + 16 = x^2 - 16x + 32 ] Bringing all terms to one side: [ x^4 - 2x^3 - 8x^2 + 24x - 16 = 0 ]4. Factor the quartic equation: Using the Rational Root Theorem and synthetic division, we factor the polynomial: [ (x - 2)^2 (x^2 + 2x - 4) = 0 ] Solving (x^2 + 2x - 4 = 0) using the quadratic formula: [ x = -1 pm sqrt{5} ]5. Determine the distinct roots: The roots are (x = 2) (with multiplicity 2), (x = -1 + sqrt{5}), and (x = -1 - sqrt{5}). Since (x = 2) is a double root but corresponds to a single distinct point, the distinct (x)-coordinates are (2), (-1 + sqrt{5}), and (-1 - sqrt{5}).6. Sum the distinct (x)-coordinates: [ 2 + (-1 + sqrt{5}) + (-1 - sqrt{5}) = 2 - 1 - 1 + sqrt{5} - sqrt{5} = 0 ]Thus, the sum of the (x)-coordinates of the distinct points of intersection is (boxed{0}).

🔑:To find the sum of the x-coordinates of the distinct points of intersection of the curves given by the equations (x^{2} = x + y + 4) and (y^{2} = y - 15x + 36), we will follow these steps:1. Solve for y from the first equation: [ x^2 = x + y + 4 ] Rearrange to isolate y: [ y = x^2 - x - 4 ]2. Substitute y = x^2 - x - 4 into the second equation y^2 = y - 15x + 36: [ (x^2 - x - 4)^2 = (x^2 - x - 4) - 15x + 36 ] Expand the left-hand side: [ (x^2 - x - 4)^2 = (x^2 - x - 4)(x^2 - x - 4) = x^4 - 2x^3 + x^2 - 4x^2 + 8x + 16 (Expand) = x^4 - 2x^3 - 7x^2 + 8x + 16 ]3. Simplify and rearrange the equation: [ x^4 - 2x^3 - 7x^2 + 8x + 16 = x^2 - x - 4 + 15x - 36 ] Combine like terms: [ x^4 - 2x^3 - 7x^2 + 8x + 16 = -x^2 + x + 4 + 15x - 36 ] [ x^4 - 2x^3 - 7x^2 + 8x + 16 = -x^2 + 16x - 32 ]4. Combine all terms into a polynomial: [ x^4 - 2x^3 - 8x^2 + 24x - 16 = 0 ]5. Factor the polynomial: Factor out ( (x-2) ): [ x^4 - 2x^3 - 8x^2 + 24x - 16 = (x-2)(x^3 - 8x + 8) ]6. Further factorization: [ x^3 - 8x + 8 = (x-2)(x^2 + 2x - 4) ] [ x^4 - 2x^3 - 8x^2 + 24x - 16 = (x-2)^2 (x^2 + 2x - 4) ]7. Find the roots: [ (x-2)^2 (x^2 + 2x - 4) = 0 ] [ (x-2)(x-2)(x^2 + 2x - 4) = 0 ] Solve for x: [ x-2 = 0 text{ or } x^2 + 2x - 4 = 0 ] [ x = 2 quad text{(with multiplicity 2)} ] [ x^2 + 2x - 4 = 0 rightarrow text{Use quadratic formula } ] [ x = frac{-b pm sqrt{b^2 - 4ac}} {2a} ] [ x = frac{-2 pm sqrt{4 + 16}}{2} implies x = frac{-2 pm 4.472} {2} ] [ x = 1 + sqrt{5} ] [ x = 1 - sqrt{5} ]8. Sum of the x coordinates: [ x_1 = 2, x_2 = -1 + sqrt{5}, x_3 = -1 - sqrt{5} ] [ text{Sum} = 2 + (-1+sqrt{5}) + (-1-sqrt{5}) = 2 - 1 - 1 = 0 ] Conclusion:[boxed{0}]

🧠:Okay, so I need to figure out for which natural numbers n (where n is at least 2), the least common multiple of all the proper divisors of n is not equal to n itself. The example given is n=18, where the proper divisors are 1, 2, 3, 6, 9, and their lcm is 18, which is equal to n. So, I need to find all n where this lcm is different from n.First, let me make sure I understand the problem correctly. The proper divisors of n are all the divisors except n itself. So for any n ≥ 2, we list all divisors d such that 1 ≤ d < n and d divides n. Then compute the lcm of all those d's and see if it's not equal to n. The task is to find all n where this happens.Hmm. So, when would the lcm of the proper divisors not equal n? Let's think of some examples. Let's take n=6. The proper divisors are 1, 2, 3. The lcm of 1,2,3 is 6, which is equal to n. So, 6 is not one of the numbers we want. Similarly, n=4: divisors are 1, 2. lcm(1,2)=2, which is not equal to 4. Wait, so 4 would be one of the numbers we want. Wait, but hold on. Wait, n=4: divisors less than 4 are 1 and 2. lcm(1,2) is 2, which is not 4. So, n=4 is a solution.Wait, so maybe primes? Let's check a prime number. Take n=5. Proper divisors are just 1. The lcm of 1 is 1, which is not equal to 5. So, primes would also be in the solution set. So, primes and composite numbers where the lcm of their proper divisors is not equal to themselves. Wait, but n=4 is composite, and it works. So, need to find all primes and composite numbers where this condition holds.Wait, but wait. Let me check another composite number. Take n=9. Proper divisors are 1, 3. lcm(1,3)=3≠9. So, 9 would also be a solution. Similarly, n=8: divisors 1,2,4. lcm(1,2,4)=4≠8. So, 8 is a solution. Hmm. So, primes are solutions, and composite numbers might also be solutions depending on their divisors.Wait, let's take n=12. Divisors are 1,2,3,4,6. lcm(1,2,3,4,6)=12. So, n=12 is not a solution. Similarly, n=6: lcm(1,2,3)=6. So, n=6 is not a solution. So, some composite numbers are not solutions. So, what differentiates composite numbers where the lcm of proper divisors equals n vs. those where it doesn't?Looking at the examples:n=4: divisors 1,2. lcm=2≠4. So, 4 is a solution.n=6: divisors 1,2,3. lcm=6. Not a solution.n=8: divisors 1,2,4. lcm=4≠8. Solution.n=9: divisors 1,3. lcm=3≠9. Solution.n=12: divisors 1,2,3,4,6. lcm=12. Not a solution.n=10: divisors 1,2,5. lcm=10. Wait, 1,2,5: lcm(1,2,5)=10. So, n=10 is not a solution. Wait, but n=10 is composite. So why is the lcm equal to n here? Let me check. 1,2,5. The lcm of 1,2,5 is indeed 10. So, n=10 is not a solution. Hmm.Wait, so what's the difference between n=4,8,9 (which are solutions) and n=6,10,12 (which are not)?n=4: prime power, specifically 2^2. Divisors are 1,2. lcm=2.n=8: 2^3. Divisors are 1,2,4. lcm=4.n=9: 3^2. Divisors 1,3. lcm=3.n=6: 2*3. Divisors 1,2,3. lcm=6.n=10: 2*5. Divisors 1,2,5. lcm=10.n=12: 2^2*3. Divisors 1,2,3,4,6. lcm=12.So, maybe if n is a prime power (p^k where k ≥ 2), then the lcm of its proper divisors is p^{k-1}, which is less than n. Therefore, prime powers would be solutions. On the other hand, if n has at least two distinct prime factors, then maybe the lcm of the proper divisors is n itself. Let me check another number. Take n=15=3*5. Divisors are 1,3,5. The lcm of 1,3,5 is 15. So, n=15 is not a solution. So, this seems to hold. Similarly, n=21=3*7. Divisors 1,3,7. lcm=21. Not a solution. So, maybe the key is whether n is a prime power or not.But wait, let's test n=16. Divisors are 1,2,4,8. lcm=8≠16. So, n=16 is a prime power (2^4) and it's a solution. Similarly, n=25=5^2. Divisors 1,5. lcm=5≠25. So, yes.But then, n=12=2^2*3. Its divisors are 1,2,3,4,6. The lcm of these is 12. So, if n is composite but not a prime power, then the lcm is n. So, seems like the answer is that n must be a prime power, i.e., n = p^k where p is a prime and k ≥1. But wait, for primes, n=p^1. Let me check. For n=2 (prime). Proper divisors are 1. lcm(1)=1≠2. So, primes are also solutions. So, primes are p^1, which are prime powers. So, the answer would be all prime powers. So, n is a prime power. So, if n is a prime power (including primes themselves, which are prime powers with exponent 1), then the lcm of the proper divisors is less than n, otherwise, if n has more than one distinct prime factor, the lcm is n.Wait, let me check n=1. Wait, but n≥2. So, primes (p) are included as prime powers. Let me check n=2. Divisors less than 2: just 1. lcm(1)=1≠2. So, yes. n=3: divisors less than 3: 1. lcm=1≠3. Similarly, n=5: lcm=1≠5. So primes are included.Now, let's check n=12 again. It has divisors 1,2,3,4,6. The lcm is 12. So, when n is composite with multiple prime factors, the lcm of the proper divisors is n. But why? Because in such cases, the proper divisors include all the prime factors and possibly some composites. For example, in n=6=2*3: divisors 1,2,3. The lcm of 2 and 3 is 6. So, even though they are primes, their lcm is n. Similarly, in n=10=2*5: divisors 1,2,5. The lcm of 2 and 5 is 10. So, the presence of two different primes in the factors causes their lcm to be n. However, if n is a prime power, like 4=2^2, the divisors are 1 and 2. The lcm is 2, which is less than 4. Similarly, 8=2^3: divisors 1,2,4. lcm=4. So, in prime powers, the highest proper divisor is p^{k-1}, and since all divisors are lower powers of p, their lcm is p^{k-1}, which is less than n. However, in composite numbers with multiple primes, the lcm of the divisors is n, because the primes and their combinations give the necessary factors to reach n.Therefore, the conclusion is that n must be a prime power. So, all natural numbers n ≥2 which are prime powers (i.e., n = p^k where p is prime and k ≥1) satisfy lcm(d₁,...,dᵣ) ≠n. For composite numbers with at least two distinct prime factors, the lcm of their proper divisors is equal to n.But wait, let me check n=12 again. Divisors:1,2,3,4,6. lcm(1,2,3,4,6). The lcm of 2,3,4,6. Let's compute step by step. lcm(2,3)=6. lcm(6,4)=12. lcm(12,6)=12. So, yes, lcm is 12. So, n=12 is composite with multiple primes, and the lcm is n.Another example: n=18=2*3^2. Divisors:1,2,3,6,9. lcm(1,2,3,6,9)=18. The example in the problem statement. So, even though 9 is a higher power, the lcm still ends up being n because of the presence of 2 and 3. So, in that case, even if there's a higher power of one prime, but another prime is present, the lcm of all the divisors will be n.Therefore, the answer seems to be that all prime powers (n = p^k for prime p and integer k ≥1) satisfy lcm(d₁,...,dᵣ) ≠n, and all composite numbers with at least two distinct prime factors have lcm equal to n. Therefore, the numbers n where the lcm is not equal to n are exactly the prime powers.Wait, but let me check n=1. Wait, n≥2, so primes start at 2, which is okay. So, primes and prime powers. So, to confirm, the numbers where the lcm of the proper divisors is not equal to n are exactly the prime powers. Therefore, the answer is all prime powers.But let's check n=16. Divisors:1,2,4,8. lcm(1,2,4,8)=8≠16. So, yes, prime power. Similarly, n=25=5^2: divisors 1,5. lcm=5≠25. So, prime power.Wait, but wait a second. Let me check n=49=7^2. Divisors:1,7. lcm=7≠49. Yes. So, that's consistent.Another check: n=27=3^3. Divisors:1,3,9. lcm=9≠27. So, yes.But then, n=12=2^2*3. Divisors include both 2 and 3. So, their lcm is 12. So, the presence of multiple primes makes the lcm equal to n.Therefore, the conclusion is that the numbers n for which the lcm of their proper divisors is not equal to n are exactly the prime powers. So, all n that are primes or prime powers.But let me think again. Wait, primes are included in prime powers (k=1). For a prime p, the proper divisors are just 1. The lcm of 1 is 1, which is not equal to p. So, primes are included. For prime squares, cubes, etc., the proper divisors are 1, p, p²,…,p^{k-1}, and their lcm is p^{k-1}, which is less than p^k. Therefore, yes, all prime powers.Hence, the solution is that all n≥2 which are prime powers (n=p^k with prime p and k≥1) satisfy the condition that the lcm of their proper divisors is not equal to n. All composite numbers with at least two distinct prime factors do not satisfy the condition (i.e., their lcm is equal to n).Therefore, the answer is: All natural numbers n ≥2 that are powers of primes. In other words, n must be of the form p^k, where p is a prime number and k is a positive integer.But let me check if there is any composite number that is not a prime power but still has lcm of proper divisors not equal to n. Suppose n is a product of two primes, like n=6=2*3. Divisors:1,2,3. lcm(1,2,3)=6. So, equal. If n is a product of three primes, like n=30=2*3*5. Divisors include 1,2,3,5,6,10,15. The lcm of all those is 30. So, equal. If n is a prime power times another prime, like n=12=2^2*3. Divisors include 1,2,3,4,6. The lcm is 12. So, equal. So, seems like as long as there are two distinct primes, the lcm of the proper divisors is n. So, only when there is only one prime involved (i.e., prime powers) does the lcm of proper divisors not equal n. Therefore, the answer is all prime powers.Thus, the final answer is: All prime powers, that is, all numbers of the form ( p^k ), where ( p ) is a prime and ( k geq 1 ).Final AnswerThe natural numbers ( n ) are all prime powers. Thus, the solution is (boxed{n = p^k}) where ( p ) is a prime and ( k geq 1 ).

✅:To determine all natural numbers ( n geq 2 ) for which the least common multiple (LCM) of all proper divisors of ( n ) is not equal to ( n ), we analyze the structure of ( n ) and its divisors.1. Prime Numbers: For a prime number ( p ), the only proper divisor is 1. The LCM of 1 is 1, which is not equal to ( p ). Hence, prime numbers satisfy the condition.2. Prime Powers: Consider ( n = p^k ) where ( p ) is a prime and ( k geq 2 ). The proper divisors are ( 1, p, p^2, ldots, p^{k-1} ). The LCM of these divisors is ( p^{k-1} ), which is less than ( p^k ). Therefore, prime powers satisfy the condition.3. Composite Numbers with Multiple Prime Factors: For a composite number ( n ) with at least two distinct prime factors, the proper divisors include all prime factors and their combinations. The LCM of these divisors includes all prime factors and their highest powers present in ( n ), resulting in the LCM being equal to ( n ). Hence, such numbers do not satisfy the condition.Examples:- ( n = 6 ) (proper divisors: 1, 2, 3; LCM = 6)- ( n = 12 ) (proper divisors: 1, 2, 3, 4, 6; LCM = 12)- ( n = 15 ) (proper divisors: 1, 3, 5; LCM = 15)Thus, the numbers ( n ) for which the LCM of the proper divisors is not equal to ( n ) are exactly the prime powers (including primes themselves). Final AnswerThe natural numbers ( n ) are all prime powers. Thus, the solution is (boxed{n = p^k}) where ( p ) is a prime and ( k geq 1 ).

🔑:```markdownLet's solve the problem step-by-step by examining the conditions under which (operatorname{kgV}(d_1, d_2, ldots, d_r) neq n).# Step 1: Analyze the case when ( n ) is a prime power.1. Suppose ( n ) is a prime power, say ( n = p^alpha ) where ( p ) is a prime and ( alpha geq 1 ).2. The divisors of ( n ) less than ( n ) are ( d_1 = 1, d_2 = p, ldots, d_r = p^{alpha-1} ).3. The least common multiple (LCM) of these divisors is given by: [ operatorname{kgV}(1, p, p^2, ldots, p^{alpha-1}) = p^{alpha-1} ] because including the highest power ( p^{alpha-1} ) covers all the others.4. Therefore, in this case: [ operatorname{kgV}(d_1, d_2, ldots, d_r) = p^{alpha-1} neq n = p^alpha ]This proves that for any ( n ) that is a prime power, ( operatorname{kgV}(d_1, d_2, ldots, d_r) neq n ).# Step 2: Analyze the case when ( n ) is not a prime power.1. If ( n ) is not a prime power, then ( n ) can be expressed as a product of distinct primes raised to positive integer powers. Let: [ n = p_1^{alpha_1} p_2^{alpha_2} ldots p_k^{alpha_k} ] where ( k geq 2 ).2. Identify two significant divisors of ( n ), specifically ( p_1^{alpha_1} ) and ((p_2^{alpha_2} cdots p_k^{alpha_k})), which are significantly less than ( n ).3. The least common multiple (LCM) of all divisors less than ( n ) should be at least the LCM of these two divisors: [ operatorname{kgV}left(p_1^{alpha_1}, p_2^{alpha_2} ldots p_k^{alpha_k}right) = n ] because the LCM includes all prime power components of ( n ).4. Since every divisor ( d_i ) of ( n ) is a product of powers of the primes ( p_1, p_2, ldots, p_k ): [ operatorname{kgV}(d_1, d_2, ldots, d_r) mid n ] Thus, we conclude that: [ operatorname{kgV}(d_1, d_2, ldots, d_r) = n ]This concludes that for any ( n ) which is not a prime power, ( operatorname{kgV}(d_1, d_2, ldots, d_r) = n ). ConclusionCombining both cases:- ( operatorname{kgV}(d_1, d_2, ldots, d_r) neq n ) for ( n ) being a prime power.- ( operatorname{kgV}(d_1, d_2, ldots, d_r) = n ) for ( n ) not being a prime power.Thus, the ( n ) we are looking for (i.e., for which ( operatorname{kgV}(d_1, d_2, ldots, d_r) neq n )) are exactly the prime powers. (boxed{text{Prime powers}})```