💡Okay, so I have this probability problem here that Bob writes a random string of 5 letters, and each letter can be A, B, C, or D. Each letter is chosen independently with equal probability. I need to find the probability that there are at least three A's in the string, given that there are at least two A's. Hmm, okay, let me try to break this down step by step.First, I remember that when dealing with probabilities involving conditions, it's a conditional probability problem. The formula for conditional probability is P(A|B) = P(A and B) / P(B). In this case, event A is having at least three A's, and event B is having at least two A's. So, the probability we're looking for is P(at least three A's | at least two A's) = P(at least three A's and at least two A's) / P(at least two A's).But wait, if we have at least three A's, that automatically means we have at least two A's, right? So, P(at least three A's and at least two A's) is just P(at least three A's). So, the formula simplifies to P(at least three A's) / P(at least two A's).Alright, so I need to calculate two probabilities: the probability of having at least three A's and the probability of having at least two A's. Then, I can divide them to get the conditional probability.Let me recall that each letter is chosen independently, and there are four possible letters, each with equal probability. So, the probability of choosing an A is 1/4, and the probability of not choosing an A is 3/4.Since the letters are chosen independently, the number of A's in the string follows a binomial distribution. The binomial probability formula is P(k successes in n trials) = C(n, k) * p^k * (1-p)^(n-k), where C(n, k) is the combination of n things taken k at a time.So, for our problem, n is 5 because there are 5 letters. The probability of success, p, is 1/4 for A, and the probability of failure, 1-p, is 3/4 for not A.First, let's compute P(at least two A's). That means we need the probability of having 2, 3, 4, or 5 A's. Alternatively, it's easier to compute 1 minus the probability of having fewer than two A's, which is 1 - P(0 A's) - P(1 A's).Similarly, P(at least three A's) is P(3 A's) + P(4 A's) + P(5 A's).Let me compute these step by step.First, compute P(0 A's): that's C(5, 0) * (1/4)^0 * (3/4)^5. C(5, 0) is 1, (1/4)^0 is 1, so it's just (3/4)^5.Similarly, P(1 A's): C(5, 1) * (1/4)^1 * (3/4)^4. C(5, 1) is 5, so that's 5 * (1/4) * (3/4)^4.Then, P(2 A's): C(5, 2) * (1/4)^2 * (3/4)^3. C(5, 2) is 10, so 10 * (1/4)^2 * (3/4)^3.Similarly, P(3 A's): C(5, 3) * (1/4)^3 * (3/4)^2. C(5, 3) is 10, same as C(5, 2).P(4 A's): C(5, 4) * (1/4)^4 * (3/4)^1. C(5, 4) is 5.P(5 A's): C(5, 5) * (1/4)^5 * (3/4)^0. C(5, 5) is 1.So, let me compute each of these.First, compute P(0 A's):(3/4)^5. Let's compute 3/4 to the power 5.3^5 is 243, and 4^5 is 1024. So, (3/4)^5 = 243/1024.Similarly, P(1 A's):5 * (1/4) * (3/4)^4.(3/4)^4 is 81/256. So, 5 * (1/4) * (81/256) = 5 * (81)/(4*256) = 5 * 81 / 1024.Compute 5*81: 5*80=400, 5*1=5, so 405. So, 405/1024.P(1 A's) = 405/1024.P(2 A's):10 * (1/4)^2 * (3/4)^3.(1/4)^2 is 1/16, (3/4)^3 is 27/64.So, 10 * (1/16) * (27/64) = 10 * (27)/(16*64) = 10 * 27 / 1024.10*27=270, so 270/1024.P(2 A's) = 270/1024.P(3 A's):10 * (1/4)^3 * (3/4)^2.(1/4)^3 is 1/64, (3/4)^2 is 9/16.So, 10 * (1/64) * (9/16) = 10 * 9 / (64*16) = 90 / 1024.P(3 A's) = 90/1024.P(4 A's):5 * (1/4)^4 * (3/4)^1.(1/4)^4 is 1/256, (3/4)^1 is 3/4.So, 5 * (1/256) * (3/4) = 5 * 3 / (256*4) = 15 / 1024.P(4 A's) = 15/1024.P(5 A's):1 * (1/4)^5 * (3/4)^0 = (1/4)^5 = 1/1024.So, P(5 A's) = 1/1024.Now, let's compute P(at least two A's). That is P(2) + P(3) + P(4) + P(5).From above, these are 270/1024 + 90/1024 + 15/1024 + 1/1024.Let's add them up:270 + 90 = 360360 + 15 = 375375 + 1 = 376So, P(at least two A's) = 376/1024.Similarly, P(at least three A's) is P(3) + P(4) + P(5) = 90/1024 + 15/1024 + 1/1024.Adding these:90 + 15 = 105105 + 1 = 106So, P(at least three A's) = 106/1024.Therefore, the conditional probability P(at least three A's | at least two A's) = (106/1024) / (376/1024).Since both numerator and denominator have the same denominator 1024, they cancel out, so it's 106/376.Now, let's simplify 106/376.First, check if both numbers are divisible by 2: 106 ÷ 2 = 53, 376 ÷ 2 = 188.So, 53/188.Check if 53 and 188 have any common divisors. 53 is a prime number, I think. Let me check: 53 divided by 2, 3, 5, 7, 11. 53 ÷ 2 is 26.5, not integer. 53 ÷ 3 is about 17.666, not integer. 53 ÷ 5 is 10.6, nope. 53 ÷ 7 is about 7.571, nope. 53 ÷ 11 is about 4.818, nope. So, 53 is prime. 188 ÷ 53 is approximately 3.547, which is not integer. So, 53 and 188 have no common divisors other than 1.Therefore, the simplified fraction is 53/188.So, the probability is 53/188.Wait, let me double-check my calculations to make sure I didn't make a mistake.First, P(0 A's) = (3/4)^5 = 243/1024. Correct.P(1 A's) = 5*(1/4)*(3/4)^4 = 5*(1/4)*(81/256) = 5*81/1024 = 405/1024. Correct.P(2 A's) = 10*(1/4)^2*(3/4)^3 = 10*(1/16)*(27/64) = 10*27/1024 = 270/1024. Correct.P(3 A's) = 10*(1/4)^3*(3/4)^2 = 10*(1/64)*(9/16) = 10*9/1024 = 90/1024. Correct.P(4 A's) = 5*(1/4)^4*(3/4) = 5*(1/256)*(3/4) = 15/1024. Correct.P(5 A's) = 1*(1/4)^5 = 1/1024. Correct.Adding up P(at least two A's): 270 + 90 + 15 + 1 = 376. Correct.Adding up P(at least three A's): 90 + 15 + 1 = 106. Correct.So, 106/376 simplifies to 53/188. Correct.So, I think my calculations are correct. Therefore, the probability is 53/188.**Final Answer**The probability is boxed{dfrac{53}{188}}.

💡Okay, so I need to figure out how many students in the teacher's class prefer dogs over cats. The class has 30 students, and the teacher asked them two questions: one about their preference between dogs and cats, and another about their preference between movies and video games.First, I'll try to understand the information given. It says that 50% of the students chose dogs and video games as their two answers. So, that means half of the class prefers dogs and also prefers video games. Then, it mentions that 10% chose dogs and movies as their two answers. So, a smaller portion of the class prefers dogs and also prefers movies.Since the class has 30 students, I can calculate the actual number of students in each category. For the 50% who chose dogs and video games, that would be 50% of 30. To find 50% of 30, I can multiply 30 by 0.5, which equals 15 students. So, 15 students prefer both dogs and video games.Next, for the 10% who chose dogs and movies, I'll calculate 10% of 30. That would be 30 multiplied by 0.1, which equals 3 students. So, 3 students prefer both dogs and movies.Now, to find out how many students prefer dogs over cats in total, I need to add the number of students who prefer dogs and video games to the number of students who prefer dogs and movies. That would be 15 students plus 3 students, which equals 18 students.Wait, but I should make sure that there's no overlap or double-counting here. Since each student answered both questions, and their preferences are independent, the students who chose dogs and video games are separate from those who chose dogs and movies. So, adding them together should give the correct total number of students who prefer dogs.Let me double-check my calculations. 50% of 30 is indeed 15, and 10% of 30 is 3. Adding them together gives 18. That seems right.So, based on the information provided, 18 students in the class prefer dogs over cats.

💡First, I need to determine the total distance Peter can cover in 1 hour and 24 minutes. I'll start by converting the total time into hours for consistency.Next, I'll denote the total distance as ( D ) kilometers. According to the problem, Peter covers two-thirds of the distance at 4 km/h and the remaining one-third at 5 km/h.For the first part of the journey:- Distance: ( frac{2}{3}D )- Speed: 4 km/h- Time: ( frac{frac{2}{3}D}{4} = frac{D}{6} ) hoursFor the second part of the journey:- Distance: ( frac{1}{3}D )- Speed: 5 km/h- Time: ( frac{frac{1}{3}D}{5} = frac{D}{15} ) hoursThe total time taken for the entire journey is the sum of the times for each part:[frac{D}{6} + frac{D}{15} = 1.4 text{ hours}]To solve for ( D ), I'll find a common denominator for the fractions, which is 30:[frac{5D}{30} + frac{2D}{30} = 1.4]Combining the terms:[frac{7D}{30} = 1.4]Multiplying both sides by 30:[7D = 42]Finally, dividing both sides by 7 gives:[D = 6 text{ km}]

💡Okay, so I've got this geometry problem here. Let me try to understand it step by step. There's a smaller circle, k, that touches a larger circle, K, from the inside at a single point P. That means they're tangent at P. Then there's a point Q on the smaller circle k, and Q is not the same as P. Now, if I draw a tangent line to the smaller circle k at point Q, this line will intersect the larger circle K at two points, which are called A and B. The problem is asking me to show that the line PQ bisects the angle APB. So, PQ should split the angle formed at point P by points A and B into two equal parts.Hmm, okay. Let me try to visualize this. I imagine two circles, one inside the other, touching at point P. Then, from a point Q on the smaller circle, I draw a tangent. Since it's a tangent, it should touch the smaller circle only at Q. This tangent line then cuts the larger circle at two points, A and B. So, points A and B are on the larger circle, and the tangent line passes through them.I need to show that PQ is the angle bisector of angle APB. That means if I draw lines from P to A and from P to B, forming an angle at P, then PQ should split this angle into two equal angles.Let me think about what tools I can use here. Maybe properties of circles, tangents, and angles. Since we're dealing with circles and tangents, perhaps some theorems related to those could help. For example, the tangent-secant theorem or properties of angles subtended by arcs.Wait, another thought: inversion. I remember that inversion can sometimes be used in geometry problems involving circles and tangents. Maybe if I invert the figure with respect to a certain circle, things might become simpler.Let me try to recall how inversion works. Inversion is a transformation that maps circles to circles or lines, depending on the inversion center and radius. If I choose the inversion center at point P, then the smaller circle k, which is tangent to the larger circle K at P, might have some nice properties under inversion.If I invert the figure with respect to a circle centered at P, then the larger circle K, which passes through P, will invert to a line. The smaller circle k, which is tangent to K at P, will invert to itself because it's tangent at the center of inversion. That's a useful property.So, after inversion, the larger circle K becomes a line, and the smaller circle k remains the same. The tangent line at Q on k will invert to a circle passing through P because the tangent line doesn't pass through P (since Q is not P). Wait, actually, the tangent line at Q is tangent to k, so under inversion, it should map to a circle that's tangent to the inverted image of k, which is still k.But since the tangent line at Q doesn't pass through P, its inversion will be a circle passing through P and tangent to k at Q. That makes sense. So, the points A and B, which were on the original larger circle K, will invert to points on this new circle that's tangent to k at Q and passes through P.Let me denote the inverted image of the tangent line as a circle α. So, circle α passes through P and is tangent to k at Q. The points A and B invert to points on α. Let's call their images A' and B'. Since inversion preserves angles, the angle APB in the original figure corresponds to the angle A'PB' in the inverted figure.Now, since α is tangent to k at Q, and k is fixed under inversion, the center of α must lie on the line PQ. Because the tangent at Q implies that the center lies along the radius at Q, which is PQ.So, the center of circle α is somewhere along PQ. Also, since α passes through P, the center must be equidistant from P and Q. Wait, no, not necessarily equidistant, but the distance from the center to P must equal the radius of α. Hmm, maybe I'm overcomplicating.Let me think about the points A' and B'. Since they lie on α, and α is tangent to k at Q, the lines PA' and PB' are related to the tangent. Maybe the angles involved can be connected through some properties.Wait, another approach: since α is tangent to k at Q, then the angle between PQ and QA' is equal to the angle between PQ and QB'. Because of the tangency, the angles formed by the tangent and the lines from Q should be equal. That might relate to the angle bisector.But I'm not sure. Let me try to think differently. Since PQ is the line connecting P and Q, and in the inverted figure, the center of α is on PQ, maybe PQ is the angle bisector in the inverted figure as well.Wait, in the inverted figure, the angle A'PB' is formed by points A' and B' on circle α. The center of α is on PQ, so PQ might be the angle bisector of angle A'PB'. If that's the case, then in the original figure, PQ would also be the angle bisector of angle APB because inversion preserves angles.Is that right? Inversion preserves angles, so if in the inverted figure PQ bisects angle A'PB', then in the original figure PQ bisects angle APB.But I need to make sure that this is indeed the case. Let me try to verify.In the inverted figure, circle α passes through P and is tangent to k at Q. The center of α lies on PQ. Therefore, PQ is the line joining P to the center of α. Since α passes through P and Q, and is tangent to k at Q, the center must lie on PQ.Therefore, in the inverted figure, PQ is the line from P to the center of α, which is also the angle bisector of angle A'PB' because the center lies on the angle bisector.Wait, is that necessarily true? If the center of α is on PQ, does that mean PQ is the angle bisector?Hmm, maybe not directly. Let me think about the properties of circle α. Since α is tangent to k at Q, and passes through P, the power of point P with respect to α is zero because P lies on α. The power of P with respect to k is equal to the square of the length of the tangent from P to k, which is PQ squared.But since α is tangent to k at Q, the power of P with respect to k is also equal to PQ squared. So, that's consistent.Wait, maybe I can use the fact that in the inverted figure, the angles at P are preserved, and since PQ is the line to the center of α, which is the circle passing through A' and B', then PQ might indeed bisect the angle A'PB'.Alternatively, maybe I can use the fact that in the original figure, the tangent at Q implies that PQ is perpendicular to the tangent line at Q. So, PQ is perpendicular to the tangent line AB.Wait, that's an important point. The tangent at Q is perpendicular to the radius PQ of the smaller circle k. So, PQ is perpendicular to AB.So, in the original figure, PQ is perpendicular to AB. That might help in showing that PQ bisects angle APB.Let me try to use that. Since PQ is perpendicular to AB, and AB is the tangent line, then PQ is the shortest distance from P to AB. So, PQ is the altitude from P to AB.But how does that relate to angle bisector? Hmm.Wait, maybe I can use some properties of circles and angles. For example, the angle between a tangent and a chord is equal to the angle in the alternate segment. That might be useful here.So, the tangent at Q makes an angle with chord QA equal to the angle that QA makes with the arc in the alternate segment. Similarly for QB.Wait, let me recall the exact statement: the angle between the tangent and a chord through the point of contact is equal to the angle in the alternate segment. So, angle between tangent at Q and chord QA is equal to the angle that QA makes with the arc opposite to it in the smaller circle k.But I'm not sure how that directly helps with angle APB.Alternatively, maybe I can consider triangle APB. Since A and B are points on the larger circle K, and P is a point on K as well (since it's the point of tangency), triangle APB is inscribed in circle K.Wait, no, P is the point where the two circles touch, so P is on both circles. So, P is on both k and K.Wait, no, actually, the smaller circle k is inside the larger circle K and tangent at P. So, P is on both circles. So, points A, B, and P are all on K.Therefore, triangle APB is inscribed in circle K. So, angles at P are related to the arcs AB.Hmm, maybe I can use the fact that the angle APB is equal to half the measure of arc AB.Wait, yes, in circle K, the measure of angle APB is half the measure of the arc AB that doesn't contain P.But how does that relate to PQ?Wait, since PQ is perpendicular to AB, and AB is a chord of circle K, then PQ might be related to the perpendicular bisector of AB.But PQ is not necessarily the perpendicular bisector because P is a point on K, and the perpendicular bisector of AB would pass through the center of K.Wait, unless PQ passes through the center of K, but that's not necessarily the case.Wait, but if PQ is perpendicular to AB, and AB is a chord of K, then PQ passes through the midpoint of the arc AB that contains P.Wait, no, the perpendicular from the center to AB would pass through the midpoint of AB, but PQ is just a line from P perpendicular to AB.Hmm, maybe I'm getting confused.Let me try to think differently. Since PQ is perpendicular to AB, and AB is a chord of K, then PQ is the altitude from P to AB in triangle APB.In triangle APB, the altitude from P is PQ, and we need to show that PQ bisects angle APB.Wait, in a triangle, the altitude and the angle bisector are generally different unless the triangle is isosceles.So, unless triangle APB is isosceles with PA = PB, which I don't think is necessarily the case here.Wait, but maybe in this specific configuration, PA = PB because of the properties of the circles.Wait, let me think. Since AB is the tangent to the smaller circle k at Q, and PQ is the radius of k at Q, then PQ is perpendicular to AB.Also, since k is tangent to K at P, the centers of k and K lie on the line connecting their centers, which passes through P.Let me denote the center of K as O and the center of k as O'. Then, O, O', and P are colinear.So, the line OO' passes through P.Now, since PQ is perpendicular to AB, and AB is a chord of K, the line PQ might have some relation to the center O.Wait, in circle K, the perpendicular from the center O to AB would bisect AB. But PQ is not necessarily passing through O, unless PQ is the perpendicular bisector, which it isn't unless Q is the midpoint of AB.But Q is a point on the smaller circle k, so it's not necessarily the midpoint.Hmm, maybe I need to use some power of a point.Wait, the power of point P with respect to circle k is zero because P lies on k. But the power of P with respect to K is also zero because P lies on K.Wait, but the tangent from P to k is PQ, so the power of P with respect to k is PQ squared.But since P lies on k, the power is zero, so PQ must be zero, which is not the case because Q is different from P.Wait, no, actually, the power of a point on the circle is zero, so the tangent from P to k would have length zero, which is just the point P itself. But in our case, the tangent is at Q, which is different from P, so maybe I'm mixing things up.Wait, perhaps I should consider the power of point Q with respect to circle K.The power of Q with respect to K is equal to QA * QB because Q lies outside K and AB is the secant.But Q lies on k, which is tangent to K at P, so the power of Q with respect to K is also equal to QP squared because QP is the tangent from Q to K.Wait, no, Q is inside K because k is inside K. So, the power of Q with respect to K is negative, equal to QO^2 - R^2, where R is the radius of K and O is the center.But since Q is on k, which is tangent to K at P, the power of Q with respect to K is equal to QP squared because QP is the tangent from Q to K.Wait, but Q is inside K, so the power should be negative, but QP is a tangent, so it should be positive. Maybe I'm confused.Wait, no, if Q is inside K, then the power of Q with respect to K is negative, and the tangent length squared is equal to the power, but since Q is inside, the tangent doesn't exist. So, maybe that approach isn't helpful.Hmm, maybe I need to think about homothety.Since the two circles k and K are tangent at P, there is a homothety centered at P that maps k to K. This homothety would map the tangent at Q to the tangent at the corresponding point on K.Wait, but the tangent at Q on k maps to a tangent on K, but in our case, the tangent at Q intersects K at A and B, so it's a secant, not a tangent.Wait, maybe the homothety maps the tangent line at Q to the tangent line at some other point on K, but I'm not sure.Alternatively, maybe I can use the fact that the homothety centered at P maps k to K, so it maps Q to some point on K, say Q'. Then, the tangent at Q on k maps to the tangent at Q' on K.But in our case, the tangent at Q on k is a secant for K, intersecting at A and B. So, maybe Q' is the midpoint of AB or something like that.Wait, I'm not sure. Maybe I'm overcomplicating.Let me try to think about the angles. Since PQ is perpendicular to AB, and AB is the tangent at Q, maybe the angles formed by PA and PB with PQ are equal.Wait, if I can show that the angles between PA and PQ and between PB and PQ are equal, then PQ bisects angle APB.So, let me denote angle APQ as α and angle BPQ as β. I need to show that α = β.Since PQ is perpendicular to AB, the angles between PQ and PA, and PQ and PB are related to the angles that PA and PB make with AB.Wait, maybe I can use the fact that the angles between PA and AB, and PB and AB are equal because of some symmetry.Wait, but I don't see the symmetry yet.Alternatively, maybe I can use the fact that in circle K, the angles subtended by AB at points A and B are related.Wait, another idea: since PQ is perpendicular to AB, and AB is a chord of K, then PQ passes through the midpoint of the arc AB that contains P.Wait, is that true? In circle K, the perpendicular from the center to AB passes through the midpoint of AB and the midpoint of the arc AB.But PQ is not necessarily passing through the center, unless PQ is the perpendicular bisector, which it isn't unless Q is the midpoint.Wait, but maybe PQ passes through the midpoint of the arc AB that contains P.Wait, let me think. In circle K, the midpoint of the arc AB that contains P is the point where the angle bisector of angle APB meets the circle. So, if PQ passes through that midpoint, then PQ would be the angle bisector.So, if I can show that PQ passes through the midpoint of arc AB, then PQ is the angle bisector.But how can I show that?Wait, since PQ is perpendicular to AB, and in circle K, the line from P perpendicular to AB would pass through the midpoint of the arc AB that contains P.Wait, is that a theorem? Let me recall.Yes, in a circle, the perpendicular from a point on the circle to a chord passes through the midpoint of the corresponding arc. So, if PQ is perpendicular to AB, then PQ passes through the midpoint of the arc AB that contains P.Therefore, PQ passes through the midpoint of arc AB, which is the point where the angle bisector of angle APB meets the circle. Hence, PQ is the angle bisector of angle APB.Wait, that seems to make sense. Let me try to formalize it.In circle K, AB is a chord, and P is a point on K. The line PQ is perpendicular to AB. Therefore, by the theorem, PQ passes through the midpoint of the arc AB that contains P. Since the midpoint of the arc AB is the point where the angle bisector of angle APB meets the circle, it follows that PQ is the angle bisector of angle APB.Therefore, PQ bisects angle APB.Yeah, that seems to work. I think that's the key idea. The fact that PQ is perpendicular to AB, and in circle K, the perpendicular from P to AB passes through the midpoint of the arc AB, which is the angle bisector.So, putting it all together, since PQ is perpendicular to AB, it must pass through the midpoint of arc AB, making it the angle bisector of angle APB.I think that's the solution.

💡Alright, so I've got these two problems here about absolute values and inequalities. Let me try to work through them step by step. I'm a bit nervous because absolute value problems can sometimes trip me up, but I'll take it slow and see if I can figure it out.Starting with the first problem: 1. Given the inequality |x + 1| + |x - 2| ≥ a has the solution set ℝ, find the range of values for the real number a.Okay, so I need to find all real numbers a such that no matter what x I pick, the expression |x + 1| + |x - 2| is always greater than or equal to a. That means a has to be less than or equal to the smallest possible value that |x + 1| + |x - 2| can take. So, essentially, I need to find the minimum value of the function f(x) = |x + 1| + |x - 2| and then set a to be less than or equal to that minimum.Hmm, how do I find the minimum of this function? I remember that absolute value functions can be piecewise linear, so maybe I can break it down into different cases based on the value of x.Let me consider the critical points where the expressions inside the absolute values change sign. For |x + 1|, the critical point is at x = -1, and for |x - 2|, it's at x = 2. So, I should analyze the function in the intervals determined by these points: (-∞, -1), [-1, 2], and (2, ∞).First interval: x < -1In this case, both x + 1 and x - 2 are negative, so the absolute values become:|x + 1| = -(x + 1) = -x - 1|x - 2| = -(x - 2) = -x + 2So, f(x) = (-x - 1) + (-x + 2) = -2x + 1This is a linear function with a slope of -2, which means it's decreasing as x increases. Since we're looking at x < -1, the smallest value in this interval would be as x approaches -1 from the left. Plugging x = -1 into f(x):f(-1) = -2*(-1) + 1 = 2 + 1 = 3But wait, since x is approaching -1 from the left, the limit as x approaches -1 from the left is 3.Second interval: -1 ≤ x ≤ 2Here, x + 1 is non-negative, so |x + 1| = x + 1. However, x - 2 is still negative or zero, so |x - 2| = -(x - 2) = -x + 2.Thus, f(x) = (x + 1) + (-x + 2) = x + 1 - x + 2 = 3So, in this entire interval, f(x) is constant at 3. That's interesting.Third interval: x > 2Both x + 1 and x - 2 are positive, so |x + 1| = x + 1 and |x - 2| = x - 2.Therefore, f(x) = (x + 1) + (x - 2) = 2x - 1This is a linear function with a slope of 2, which means it's increasing as x increases. The smallest value in this interval occurs at x = 2:f(2) = 2*2 - 1 = 4 - 1 = 3So, putting it all together, the function f(x) = |x + 1| + |x - 2| has a minimum value of 3, achieved when x is between -1 and 2. For x < -1, the function decreases towards 3, and for x > 2, it increases from 3.Therefore, the smallest value that f(x) can take is 3. Since the inequality |x + 1| + |x - 2| ≥ a must hold for all real numbers x, a must be less than or equal to this minimum value. So, a ≤ 3.Alright, that seems solid. I think I got the first problem.Moving on to the second problem:2. If the function f(x) = |x + 1| + 2|x - a| has a minimum value of 5, find the value of the real number a.Okay, so now I have a function f(x) = |x + 1| + 2|x - a|, and I know that its minimum value is 5. I need to find the possible values of a.Hmm, similar to the first problem, this involves absolute values, so it's piecewise linear, and the minimum occurs at one of the critical points or somewhere in between. Let me think about how to approach this.First, let's identify the critical points where the expressions inside the absolute values change sign. For |x + 1|, the critical point is at x = -1, and for |x - a|, it's at x = a. So, the function f(x) will have different expressions depending on whether x is less than -1, between -1 and a, or greater than a (assuming a > -1; if a < -1, the order would be different). But since I don't know where a is relative to -1, I might need to consider different cases.Case 1: a > -1In this case, the critical points are at x = -1 and x = a, with a > -1. So, the intervals to consider are:1. x < -12. -1 ≤ x ≤ a3. x > aCase 2: a < -1Then, the critical points are at x = a and x = -1, with a < -1. So, the intervals are:1. x < a2. a ≤ x ≤ -13. x > -1Case 3: a = -1Then, both critical points coincide at x = -1, so the function is differentiable everywhere except at x = -1.But maybe there's a better way to approach this without splitting into cases. I remember that for functions involving absolute values, the minimum occurs at one of the critical points or where the slopes change.Alternatively, I can consider the function f(x) = |x + 1| + 2|x - a| and try to find its minimum by analyzing its derivative or by considering the points where the slope changes.But since it's a piecewise linear function, the minimum will occur at one of the critical points, so I can evaluate f(x) at x = -1 and x = a and see which one gives the minimum.Wait, but since the function is f(x) = |x + 1| + 2|x - a|, the coefficients of the absolute values are different. The second term has a coefficient of 2, which might affect where the minimum occurs.Let me try to think about the slopes in each interval.Suppose a > -1.Then, for x < -1:f(x) = -(x + 1) + 2*(-(x - a)) = -x - 1 - 2x + 2a = -3x + (2a - 1)Slope is -3, which is decreasing.For -1 ≤ x ≤ a:f(x) = (x + 1) + 2*(-(x - a)) = x + 1 - 2x + 2a = -x + (2a + 1)Slope is -1, still decreasing.For x > a:f(x) = (x + 1) + 2*(x - a) = x + 1 + 2x - 2a = 3x + (1 - 2a)Slope is 3, which is increasing.So, in this case, the function decreases until x = a, and then starts increasing. Therefore, the minimum occurs at x = a.Similarly, if a < -1, let's see.For x < a:f(x) = -(x + 1) + 2*(-(x - a)) = -x -1 -2x + 2a = -3x + (2a -1)Slope is -3, decreasing.For a ≤ x ≤ -1:f(x) = -(x + 1) + 2*(x - a) = -x -1 + 2x - 2a = x + (-1 - 2a)Slope is 1, increasing.For x > -1:f(x) = (x + 1) + 2*(x - a) = x +1 + 2x - 2a = 3x + (1 - 2a)Slope is 3, increasing.So, in this case, the function decreases until x = a, then increases from x = a to x = -1, and continues increasing beyond that. Therefore, the minimum occurs at x = a.Wait, so regardless of whether a is greater than or less than -1, the minimum occurs at x = a.Is that correct? Let me verify.If a > -1:- For x < -1: slope -3 (decreasing)- For -1 ≤ x ≤ a: slope -1 (still decreasing)- For x > a: slope 3 (increasing)So, the function is decreasing until x = a, then increasing. So, the minimum is at x = a.If a < -1:- For x < a: slope -3 (decreasing)- For a ≤ x ≤ -1: slope 1 (increasing)- For x > -1: slope 3 (increasing)So, the function decreases until x = a, then increases from x = a onwards. Therefore, the minimum is at x = a.If a = -1:- For x < -1: slope -3 (decreasing)- For x > -1: slope 3 (increasing)So, the minimum is at x = -1, which is also x = a in this case.Therefore, in all cases, the minimum occurs at x = a.Therefore, the minimum value of f(x) is f(a).Let me compute f(a):f(a) = |a + 1| + 2|a - a| = |a + 1| + 2*0 = |a + 1|But we are told that the minimum value is 5, so:|a + 1| = 5Therefore, a + 1 = 5 or a + 1 = -5Solving these:1. a + 1 = 5 ⇒ a = 42. a + 1 = -5 ⇒ a = -6So, the possible values of a are 4 and -6.Wait, let me double-check this. If a = 4, then f(x) = |x + 1| + 2|x - 4|. The minimum occurs at x = 4, and f(4) = |4 + 1| + 2|4 - 4| = 5 + 0 = 5. That works.If a = -6, then f(x) = |x + 1| + 2|x - (-6)| = |x + 1| + 2|x + 6|. The minimum occurs at x = -6, and f(-6) = |-6 + 1| + 2| -6 + 6| = | -5 | + 2*0 = 5 + 0 = 5. That also works.So, both a = 4 and a = -6 satisfy the condition that the minimum value of f(x) is 5.Therefore, the values of a are 4 and -6.I think that makes sense. I considered different cases for a, realized that regardless of where a is, the minimum occurs at x = a, computed f(a), set it equal to 5, and solved for a. Both solutions work when plugged back into the function.So, to summarize:1. For the inequality |x + 1| + |x - 2| ≥ a to hold for all real x, a must be less than or equal to the minimum value of the expression, which is 3. So, a ≤ 3.2. For the function f(x) = |x + 1| + 2|x - a| to have a minimum value of 5, a must be either 4 or -6.I feel confident about these answers now. I just need to make sure I didn't make any calculation errors, but going through the steps again, everything seems to check out.

💡Okay, so I have this problem where a light beam hits a flat mirror at an angle α. Then, the mirror is rotated by an angle β around the projection of the beam on the mirror. I need to figure out by what angle the reflected beam deviates because of this rotation. Hmm, let me try to visualize this.First, when a light beam hits a mirror, it reflects off following the law of reflection, which states that the angle of incidence is equal to the angle of reflection. So, if the beam hits the mirror at an angle α, it will reflect off at the same angle α relative to the normal of the mirror.Now, the mirror is being rotated by an angle β around the projection of the beam on the mirror. The projection of the beam on the mirror is essentially the point where the beam hits the mirror, right? So, rotating the mirror around this point by angle β would change the orientation of the mirror, which in turn would change the direction of the reflected beam.I think it might help to draw a diagram. Let me imagine the mirror as a flat surface with a coordinate system. The light beam is coming in at an angle α, so it's not perpendicular to the mirror. The projection of the beam on the mirror is the point where the beam hits the mirror, and the mirror is being rotated around this point by β.When the mirror is rotated, the normal to the mirror at the point of incidence changes by angle β. Since the law of reflection depends on the normal, this rotation will affect the direction of the reflected beam.Let me think about the angles involved. Initially, the angle between the incident beam and the mirror is α. After rotating the mirror by β, the new angle between the incident beam and the mirror would be α + β or α - β, depending on the direction of rotation. But since the problem doesn't specify the direction, I guess we can assume it's a small angle and the direction won't matter for the calculation.Wait, no, actually, the projection is fixed, so the rotation is around that point. So, maybe the angle between the incident beam and the mirror doesn't change, but the orientation of the mirror changes, which affects the reflected beam.Hmm, maybe I should consider the change in the normal vector. If the mirror is rotated by β, the normal vector also rotates by β. So, the angle between the incident beam and the new normal would be different.Let me denote the original normal as n and the rotated normal as n'. The angle between n and n' is β. The incident beam makes an angle α with the original normal n. After rotation, the incident beam makes an angle α' with the new normal n'. But wait, the incident beam is fixed, right? So, if the mirror rotates, the angle between the incident beam and the mirror changes. So, the angle of incidence changes.Wait, no, the projection of the beam on the mirror is fixed, so the point of incidence doesn't move. So, the beam is hitting the same point on the mirror, but the mirror has rotated around that point. So, the angle between the beam and the mirror's surface changes.Let me think in terms of vectors. The incident beam has a direction vector, and the mirror has a surface normal vector. The angle between them is α. When the mirror rotates by β around the projection point, the normal vector rotates by β, so the angle between the incident beam and the new normal is α + β or α - β.But actually, the angle of incidence is measured between the incident beam and the normal. So, if the normal rotates by β, the angle of incidence becomes α + β or α - β. Depending on the direction of rotation.But since the problem doesn't specify the direction, maybe we can assume it's a small angle and take the absolute value.Wait, but the deviation of the reflected beam would depend on the change in the angle of reflection. Since the angle of reflection equals the angle of incidence, if the angle of incidence changes by β, the angle of reflection also changes by β.But the total deviation would be twice that, because the beam is deviated both on the way in and the way out.Wait, no, actually, when the mirror rotates by β, the reflected beam rotates by 2β. Because the angle between the incident and reflected beam is 2α, and if the mirror rotates by β, the angle between the new reflected beam and the original reflected beam would be 2β.But I'm not sure. Maybe I need to think more carefully.Let me consider the initial setup. The incident beam makes an angle α with the normal. The reflected beam makes the same angle α on the other side of the normal. So, the total angle between the incident and reflected beam is 2α.Now, when the mirror is rotated by β around the projection point, the normal rotates by β, so the new angle of incidence becomes α + β. Therefore, the new reflected beam makes an angle α + β on the other side of the new normal.So, the angle between the original reflected beam and the new reflected beam would be the difference between the two angles.Wait, the original reflected beam was at 2α from the incident beam, and the new reflected beam is at 2(α + β) from the incident beam. So, the deviation between the two reflected beams would be 2(α + β) - 2α = 2β.Wait, that seems too straightforward. So, the deviation is 2β?But I think that might not account for the change in the mirror's orientation correctly.Alternatively, maybe the deviation is 2 times the angle between the original normal and the new normal, which is β. So, the deviation is 2β.But let me think about it differently. Suppose the mirror is rotated by β, so the normal rotates by β. The incident beam is fixed, so the angle between the incident beam and the new normal is α + β. Therefore, the reflected beam will be on the other side of the new normal at angle α + β. So, the angle between the original reflected beam and the new reflected beam is 2β.Yes, that makes sense. Because the original reflected beam was at 2α from the incident beam, and the new reflected beam is at 2(α + β) from the incident beam. So, the difference is 2β.Wait, but if the mirror is rotated by β, the angle between the normals is β, so the reflected beam is deviated by 2β.Yes, that seems consistent.But let me check with an example. Suppose α is 0, meaning the beam is perpendicular to the mirror. Then, rotating the mirror by β would cause the reflected beam to deviate by 2β, which makes sense because the mirror is just rotated by β, so the beam is reflected in a new direction, deviating by 2β.Another example: if β is 0, then the deviation is 0, which is correct.If α is 45 degrees and β is 30 degrees, then the deviation would be 60 degrees. That seems reasonable.So, maybe the answer is 2β.But wait, the problem says the mirror is rotated around the projection of the beam on the mirror. So, the projection is the point where the beam hits the mirror, right? So, rotating around that point by β.Wait, does that mean the mirror is rotated in such a way that the point of incidence remains fixed, but the mirror's orientation changes? So, the normal rotates by β, but the point of incidence doesn't move.Yes, that's what I thought earlier.So, in that case, the deviation of the reflected beam is 2β.But let me think again. The angle between the original reflected beam and the new reflected beam is 2β.But wait, is it 2β or something else?Wait, another way to think about it is that when the mirror is rotated by β, the direction of the reflected beam changes by 2β. Because the angle of incidence increases by β, so the angle of reflection also increases by β, making the total deviation 2β.Yes, that seems correct.But let me try to derive it more formally.Let’s denote the original normal as n, and after rotation, the normal becomes n'. The angle between n and n' is β.The incident beam makes an angle α with n. After rotation, the incident beam makes an angle α' with n', where α' = α + β (assuming rotation in the direction that increases the angle).The reflected beam makes an angle α' with n', so the total angle between the original reflected beam and the new reflected beam is 2β.Yes, that seems consistent.Alternatively, using vector analysis, the reflected beam direction can be found by reflecting the incident beam across the normal. If the normal rotates by β, the reflected beam rotates by 2β.Yes, that makes sense.So, I think the deviation angle is 2β.But wait, the problem says "around the projection of the beam on the mirror." So, does that affect anything?The projection of the beam on the mirror is the point where the beam hits the mirror. So, rotating the mirror around that point by β would mean that the mirror's orientation changes, but the point of incidence remains the same.So, in that case, the normal rotates by β, and the reflected beam deviates by 2β.Yes, that seems correct.So, the answer should be 2β.But wait, let me think again. If the mirror is rotated by β around the projection point, the normal rotates by β, so the reflected beam deviates by 2β.Yes, that seems consistent.But wait, let me consider the case where β is 90 degrees. If the mirror is rotated by 90 degrees, the reflected beam would be deviated by 180 degrees, which would mean the beam is going back along the same path but in the opposite direction. That makes sense because rotating the mirror by 90 degrees would invert the reflection.Wait, no, rotating the mirror by 90 degrees would not necessarily invert the reflection. It would just change the direction of the reflected beam by 180 degrees relative to the original direction.Wait, no, if the mirror is rotated by 90 degrees, the reflected beam would be deviated by 180 degrees, meaning it's going back along the same line but in the opposite direction. That makes sense.But in reality, rotating the mirror by 90 degrees would not make the beam go back along the same path unless the mirror is rotated in a specific way.Wait, maybe I'm overcomplicating it. Let's stick to small angles.So, for small β, the deviation is approximately 2β, which makes sense.Therefore, I think the deviation angle is 2β.But wait, the problem says "around the projection of the beam on the mirror." So, does that mean the rotation is in the plane of the mirror or perpendicular to it?Wait, the projection of the beam on the mirror is a point, so rotating around that point would mean rotating the mirror in the plane perpendicular to the beam's projection.Wait, no, the projection is a point, so rotating around that point would mean rotating the mirror in the plane of the mirror itself.Wait, I'm getting confused.Let me clarify: the projection of the beam on the mirror is the point where the beam hits the mirror. So, rotating the mirror around that point would mean rotating the mirror in the plane perpendicular to the beam's direction.Wait, no, the beam has a direction, and its projection on the mirror is a point. So, rotating the mirror around that point would mean rotating the mirror in the plane of the mirror itself.Wait, no, the mirror is a flat surface, and rotating it around a point on its surface would involve rotating it in three-dimensional space, keeping that point fixed.So, the mirror can be rotated in any direction around that point, but the problem specifies "around the projection of the beam on the mirror," which is a point. So, the rotation is around that point, but in which plane?I think it's rotating the mirror in the plane that contains the beam's projection and is perpendicular to the beam's direction.Wait, maybe it's rotating the mirror in the plane of the mirror itself.Wait, I'm not sure. Maybe it's better to think in terms of the normal vector.If the mirror is rotated by β around the projection point, the normal vector rotates by β in some plane.But regardless, the deviation of the reflected beam would be twice the angle by which the normal is rotated, which is β. So, the deviation is 2β.Therefore, I think the answer is 2β.But let me check with an example. Suppose α is 0, so the beam is perpendicular to the mirror. Then, rotating the mirror by β would cause the reflected beam to deviate by 2β, which makes sense because the mirror is just rotated by β, so the beam is reflected in a new direction, deviating by 2β.Another example: if β is 0, then the deviation is 0, which is correct.If α is 45 degrees and β is 30 degrees, then the deviation would be 60 degrees. That seems reasonable.So, I think the deviation angle is 2β.But wait, the problem says "around the projection of the beam on the mirror." So, does that affect anything?The projection of the beam on the mirror is the point where the beam hits the mirror. So, rotating the mirror around that point by β would mean that the mirror's orientation changes, but the point of incidence remains the same.So, in that case, the normal rotates by β, and the reflected beam deviates by 2β.Yes, that seems consistent.Therefore, the angle by which the reflected beam deviates is 2β.But wait, let me think again. If the mirror is rotated by β, the angle between the normals is β, so the reflected beam is deviated by 2β.Yes, that seems correct.So, the final answer is 2β.But wait, the problem didn't specify the direction of rotation, so maybe it's the absolute value, but since β is an angle, it's already positive.Therefore, the deviation is 2β.But wait, in the initial problem, the angle is α, and the rotation is β. So, is the deviation 2β or something else?Wait, maybe I need to consider the relationship between α and β.Wait, no, the deviation is solely due to the rotation of the mirror, which is β, so the deviation is 2β.Yes, that seems correct.So, I think the answer is 2β.