answer：Alright, so I have this problem here about two parallel cross-sections of a sphere. The areas are given as 5π and 8π, and they're located on the same side of the sphere's center, 1 unit apart. I need to find the volume of the sphere. Hmm, okay, let me try to figure this out step by step.First, I remember that the cross-section of a sphere is a circle, right? So, if I have a sphere and I slice it with a plane, the area of that slice is a circle. The area of a circle is πr², where r is the radius of the circle. So, for the first cross-section, which has an area of 5π, I can find the radius of that circle. Let me write that down:Area = πr² = 5πSo, if I divide both sides by π, I get r² = 5. Taking the square root of both sides, r = √5. Okay, so the radius of the first cross-section is √5.Similarly, for the second cross-section with an area of 8π, I can do the same thing:Area = πr² = 8πDivide both sides by π: r² = 8. So, r = √8. Wait, √8 can be simplified, right? √8 is the same as 2√2. So, the radius of the second cross-section is 2√2.Now, I need to relate these radii to the sphere's radius. I remember that if you have a sphere of radius R, and you take a cross-section at a distance h from the center, the radius of that cross-section is given by the formula:r = √(R² - h²)This comes from the Pythagorean theorem, where R is the hypotenuse, h is one leg, and r is the other leg. So, for each cross-section, we can write an equation involving R and the distance from the center.Let me denote the distance from the center of the sphere to the first cross-section as x, and the distance to the second cross-section as y. Since both cross-sections are on the same side of the center, and they're 1 unit apart, the distance between them is |x - y| = 1. But since they're on the same side, I think x and y are both positive, so it's just y - x = 1, assuming y is farther from the center than x.Wait, actually, I'm not sure if y is farther or closer. It depends on which cross-section has the larger radius. The cross-section with the larger radius is closer to the center because, as you move away from the center, the radius of the cross-section decreases. So, since 2√2 is larger than √5, the cross-section with area 8π is closer to the center. Therefore, y is the distance for the larger radius, which is closer, so actually, x - y = 1. Hmm, this is a bit confusing. Maybe I should just keep it as |x - y| = 1 for now.But let's think about it: if both cross-sections are on the same side of the center, and they're 1 unit apart, then one is 1 unit closer to the center than the other. So, if I let x be the distance from the center to the first cross-section (with radius √5), and y be the distance from the center to the second cross-section (with radius 2√2), then since 2√2 is larger, it's closer to the center. So, y is less than x, and x - y = 1.Okay, that makes sense. So, x - y = 1.Now, using the formula for the radius of the cross-section:For the first cross-section:√5 = √(R² - x²)Squaring both sides:5 = R² - x²So, x² = R² - 5For the second cross-section:2√2 = √(R² - y²)Squaring both sides:8 = R² - y²So, y² = R² - 8Now, I have two equations:1. x² = R² - 52. y² = R² - 8And I also have the relationship between x and y:x - y = 1So, I can express x as y + 1.Substituting x = y + 1 into the first equation:(y + 1)² = R² - 5Expanding the left side:y² + 2y + 1 = R² - 5But from the second equation, I know that y² = R² - 8. So, I can substitute that into the equation above:(R² - 8) + 2y + 1 = R² - 5Simplify:R² - 8 + 2y + 1 = R² - 5R² - 7 + 2y = R² - 5Subtract R² from both sides:-7 + 2y = -5Add 7 to both sides:2y = 2Divide by 2:y = 1Okay, so y is 1. Now, since x = y + 1, x = 1 + 1 = 2.Now, let's find R. From the second equation:y² = R² - 81² = R² - 81 = R² - 8R² = 9R = √9R = 3So, the radius of the sphere is 3 units.Now, to find the volume of the sphere, I use the formula:Volume = (4/3)πR³Substituting R = 3:Volume = (4/3)π(3)³= (4/3)π(27)= (4/3)*27*π= 36πSo, the volume of the sphere is 36π.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Calculated radii of cross-sections: √5 and 2√2. That seems right.2. Set up the equations using the Pythagorean theorem: r = √(R² - h²). Yes, that's correct.3. Assigned x and y as distances from the center, with x - y = 1 because the larger radius is closer. That makes sense.4. Substituted x = y + 1 into the first equation, expanded, and substituted y² from the second equation. The algebra seems fine.5. Solved for y and got y = 1, then x = 2. Plugged back into y² = R² - 8, got R² = 9, so R = 3. Then volume is (4/3)π(27) = 36π. All steps check out.I think that's correct. So, the volume of the sphere is 36π.

answer：To determine the number of ways to distribute 6 indistinguishable balls into 2 indistinguishable boxes, I start by considering the possible partitions of the balls between the two boxes.First, I recognize that since the boxes are indistinguishable, the order of the partitions does not matter. This means that a distribution like (6,0) is the same as (0,6), and similarly for other partitions.Next, I list out all possible partitions of 6 balls into two groups:- All 6 balls in one box and none in the other: (6,0)- 5 balls in one box and 1 ball in the other: (5,1)- 4 balls in one box and 2 balls in the other: (4,2)- 3 balls in each box: (3,3)Since the boxes are indistinguishable, each of these partitions represents a unique way to distribute the balls. Therefore, there are 4 distinct ways to distribute the 6 indistinguishable balls into 2 indistinguishable boxes.

answer：Alright, so I have this problem here, and it's about polynomials and their roots. It's divided into two parts, a) and b). Let me try to understand what each part is asking.Starting with part a): It asks if there exists a natural number ( n ) and a polynomial ( P(x) ) of degree ( n ) with ( n ) distinct real roots such that for all real ( x ), the equation ( P(x) P(x+1) = P(x^2) ) holds.Hmm, okay. So, I need to find a polynomial that when multiplied by its shift by 1, equals the polynomial evaluated at ( x^2 ). That seems like a functional equation for polynomials. Maybe I can think of specific polynomials that satisfy this property.Let me consider simple polynomials first. Let's try quadratic polynomials because they have two roots, which are distinct if the discriminant is positive. Let's suppose ( P(x) = ax^2 + bx + c ). Then, ( P(x+1) = a(x+1)^2 + b(x+1) + c = a(x^2 + 2x + 1) + b(x + 1) + c = ax^2 + (2a + b)x + (a + b + c) ).Multiplying ( P(x) ) and ( P(x+1) ) would give a quartic polynomial. On the other hand, ( P(x^2) = a(x^2)^2 + b(x^2) + c = ax^4 + bx^2 + c ). So, if ( P(x)P(x+1) = P(x^2) ), then their coefficients must match.Let me compute ( P(x)P(x+1) ):( (ax^2 + bx + c)(ax^2 + (2a + b)x + (a + b + c)) ).Multiplying term by term:First, ( ax^2 times ax^2 = a^2x^4 ).Then, ( ax^2 times (2a + b)x = a(2a + b)x^3 ).Next, ( ax^2 times (a + b + c) = a(a + b + c)x^2 ).Similarly, ( bx times ax^2 = abx^3 ).( bx times (2a + b)x = b(2a + b)x^2 ).( bx times (a + b + c) = b(a + b + c)x ).Then, ( c times ax^2 = acx^2 ).( c times (2a + b)x = c(2a + b)x ).( c times (a + b + c) = c(a + b + c) ).Now, let's collect like terms:- ( x^4 ): ( a^2 )- ( x^3 ): ( a(2a + b) + ab = 2a^2 + ab + ab = 2a^2 + 2ab )- ( x^2 ): ( a(a + b + c) + b(2a + b) + ac = a^2 + ab + ac + 2ab + b^2 + ac = a^2 + 3ab + 2ac + b^2 )- ( x ): ( b(a + b + c) + c(2a + b) = ab + b^2 + bc + 2ac + bc = ab + b^2 + 2bc + 2ac )- Constant term: ( c(a + b + c) = ac + bc + c^2 )Now, ( P(x^2) = ax^4 + bx^2 + c ). So, let's compare coefficients.For ( x^4 ): ( a^2 = a ). So, ( a^2 = a ) implies ( a(a - 1) = 0 ), so ( a = 0 ) or ( a = 1 ). But if ( a = 0 ), then ( P(x) ) is linear, which contradicts the degree being 2. So, ( a = 1 ).For ( x^3 ): ( 2a^2 + 2ab = 0 ). Since ( a = 1 ), this becomes ( 2 + 2b = 0 ), so ( 2b = -2 ), hence ( b = -1 ).For ( x^2 ): ( a^2 + 3ab + 2ac + b^2 = b ). Plugging ( a = 1 ) and ( b = -1 ):( 1 + 3(1)(-1) + 2(1)c + (-1)^2 = -1 )Simplify:( 1 - 3 + 2c + 1 = -1 )Combine like terms:( (1 - 3 + 1) + 2c = -1 )( (-1) + 2c = -1 )So, ( 2c = 0 ) implies ( c = 0 ).For ( x ): ( ab + b^2 + 2bc + 2ac = 0 ). Plugging ( a = 1 ), ( b = -1 ), ( c = 0 ):( (1)(-1) + (-1)^2 + 2(-1)(0) + 2(1)(0) = -1 + 1 + 0 + 0 = 0 ). So, this holds.For the constant term: ( ac + bc + c^2 = c ). Plugging ( a = 1 ), ( b = -1 ), ( c = 0 ):( (1)(0) + (-1)(0) + (0)^2 = 0 + 0 + 0 = 0 ). Which equals ( c = 0 ). So, that holds as well.So, the polynomial is ( P(x) = x^2 - x ). Let me check if this satisfies the original equation.Compute ( P(x)P(x+1) ):( (x^2 - x)( (x+1)^2 - (x+1) ) )Simplify ( (x+1)^2 - (x+1) = x^2 + 2x + 1 - x - 1 = x^2 + x )So, ( (x^2 - x)(x^2 + x) = x^4 + x^3 - x^3 - x^2 = x^4 - x^2 )Compute ( P(x^2) = (x^2)^2 - x^2 = x^4 - x^2 )Yes, they are equal. So, ( P(x) = x^2 - x ) satisfies the equation.Also, ( P(x) = x^2 - x = x(x - 1) ) has two distinct real roots at ( x = 0 ) and ( x = 1 ). So, it satisfies all the conditions.Therefore, for part a), the answer is yes, and such a polynomial exists, specifically ( P(x) = x^2 - x ).Moving on to part b): It asks if there exists a natural number ( n ) and a polynomial ( P(x) ) of degree ( n ) with ( n ) distinct real roots such that for all real ( x ), the equation ( P(x) P(x+1) = P(x^2 + 1) ) holds.Hmm, similar to part a), but now the right-hand side is ( P(x^2 + 1) ) instead of ( P(x^2) ). Let me see if I can find such a polynomial.Again, let's consider the same approach. Maybe start with low-degree polynomials and see if they satisfy the equation.Let me try ( n = 1 ): So, ( P(x) = ax + b ), with one distinct real root.Compute ( P(x)P(x+1) = (ax + b)(a(x+1) + b) = (ax + b)(ax + a + b) ).Multiply out:( a^2x^2 + a(ax + b) + b(ax + a + b) = a^2x^2 + a^2x + ab + abx + ab + b^2 ).Wait, that seems messy. Let me compute it step by step.( (ax + b)(ax + a + b) = a^2x^2 + a(ax + b) + b(ax + a + b) ).Wait, actually, that's not the correct expansion. Let me do it properly:( (ax + b)(ax + a + b) = ax cdot ax + ax cdot (a + b) + b cdot ax + b cdot (a + b) ).Compute each term:- ( ax cdot ax = a^2x^2 )- ( ax cdot (a + b) = a(a + b)x )- ( b cdot ax = abx )- ( b cdot (a + b) = b(a + b) )So, combining like terms:( a^2x^2 + [a(a + b) + ab]x + b(a + b) )Simplify the coefficients:For ( x^2 ): ( a^2 )For ( x ): ( a(a + b) + ab = a^2 + ab + ab = a^2 + 2ab )For the constant term: ( b(a + b) = ab + b^2 )Now, ( P(x^2 + 1) = a(x^2 + 1) + b = ax^2 + a + b )So, equate ( P(x)P(x+1) = P(x^2 + 1) ):Left side: ( a^2x^2 + (a^2 + 2ab)x + (ab + b^2) )Right side: ( ax^2 + (a + b) )Set coefficients equal:For ( x^2 ): ( a^2 = a ) => ( a^2 - a = 0 ) => ( a(a - 1) = 0 ) => ( a = 0 ) or ( a = 1 )If ( a = 0 ), then ( P(x) = b ), a constant polynomial, which doesn't have any roots unless ( b = 0 ), but then it's the zero polynomial, which doesn't have distinct roots. So, ( a = 1 ).For ( x ): ( a^2 + 2ab = 0 ). Since ( a = 1 ), this becomes ( 1 + 2b = 0 ) => ( 2b = -1 ) => ( b = -1/2 ).For the constant term: ( ab + b^2 = a + b ). Plugging ( a = 1 ), ( b = -1/2 ):Left side: ( (1)(-1/2) + (-1/2)^2 = -1/2 + 1/4 = -1/4 )Right side: ( 1 + (-1/2) = 1/2 )But ( -1/4 neq 1/2 ). So, this is a contradiction. Therefore, no linear polynomial satisfies the equation.Next, let's try ( n = 2 ). Let ( P(x) = ax^2 + bx + c ), with two distinct real roots.Compute ( P(x)P(x+1) ):( (ax^2 + bx + c)(a(x+1)^2 + b(x+1) + c) )First, compute ( P(x+1) ):( a(x+1)^2 + b(x+1) + c = a(x^2 + 2x + 1) + b(x + 1) + c = ax^2 + 2ax + a + bx + b + c )Combine like terms:( ax^2 + (2a + b)x + (a + b + c) )Now, multiply ( P(x) ) and ( P(x+1) ):( (ax^2 + bx + c)(ax^2 + (2a + b)x + (a + b + c)) )This will result in a quartic polynomial. Let me denote this product as ( Q(x) ).On the other hand, ( P(x^2 + 1) = a(x^2 + 1)^2 + b(x^2 + 1) + c = a(x^4 + 2x^2 + 1) + b(x^2 + 1) + c = ax^4 + 2ax^2 + a + bx^2 + b + c )Combine like terms:( ax^4 + (2a + b)x^2 + (a + b + c) )So, ( P(x^2 + 1) = ax^4 + (2a + b)x^2 + (a + b + c) )Now, set ( Q(x) = P(x)P(x+1) = P(x^2 + 1) ). So, the quartic polynomial ( Q(x) ) must equal ( ax^4 + (2a + b)x^2 + (a + b + c) ).Let me compute ( Q(x) ):( (ax^2 + bx + c)(ax^2 + (2a + b)x + (a + b + c)) )Multiply term by term:First, ( ax^2 times ax^2 = a^2x^4 )( ax^2 times (2a + b)x = a(2a + b)x^3 )( ax^2 times (a + b + c) = a(a + b + c)x^2 )Next, ( bx times ax^2 = abx^3 )( bx times (2a + b)x = b(2a + b)x^2 )( bx times (a + b + c) = b(a + b + c)x )Then, ( c times ax^2 = acx^2 )( c times (2a + b)x = c(2a + b)x )( c times (a + b + c) = c(a + b + c) )Now, combine like terms:- ( x^4 ): ( a^2 )- ( x^3 ): ( a(2a + b) + ab = 2a^2 + ab + ab = 2a^2 + 2ab )- ( x^2 ): ( a(a + b + c) + b(2a + b) + ac = a^2 + ab + ac + 2ab + b^2 + ac = a^2 + 3ab + 2ac + b^2 )- ( x ): ( b(a + b + c) + c(2a + b) = ab + b^2 + bc + 2ac + bc = ab + b^2 + 2bc + 2ac )- Constant term: ( c(a + b + c) = ac + bc + c^2 )Now, equate this to ( P(x^2 + 1) = ax^4 + (2a + b)x^2 + (a + b + c) ).So, set coefficients equal:For ( x^4 ): ( a^2 = a ) => ( a(a - 1) = 0 ) => ( a = 0 ) or ( a = 1 ). Again, ( a = 0 ) would make ( P(x) ) linear, which contradicts degree 2, so ( a = 1 ).For ( x^3 ): ( 2a^2 + 2ab = 0 ). With ( a = 1 ), this becomes ( 2 + 2b = 0 ) => ( b = -1 ).For ( x^2 ): ( a^2 + 3ab + 2ac + b^2 = 2a + b ). Plugging ( a = 1 ), ( b = -1 ):Left side: ( 1 + 3(1)(-1) + 2(1)c + (-1)^2 = 1 - 3 + 2c + 1 = (-1) + 2c )Right side: ( 2(1) + (-1) = 2 - 1 = 1 )So, ( -1 + 2c = 1 ) => ( 2c = 2 ) => ( c = 1 ).For ( x ): ( ab + b^2 + 2bc + 2ac = 0 ). Plugging ( a = 1 ), ( b = -1 ), ( c = 1 ):Left side: ( (1)(-1) + (-1)^2 + 2(-1)(1) + 2(1)(1) = -1 + 1 - 2 + 2 = 0 ). So, this holds.For the constant term: ( ac + bc + c^2 = a + b + c ). Plugging ( a = 1 ), ( b = -1 ), ( c = 1 ):Left side: ( (1)(1) + (-1)(1) + (1)^2 = 1 - 1 + 1 = 1 )Right side: ( 1 + (-1) + 1 = 1 ). So, this holds as well.So, the polynomial is ( P(x) = x^2 - x + 1 ). Let me check if this satisfies the original equation.Compute ( P(x)P(x+1) ):( (x^2 - x + 1)( (x+1)^2 - (x+1) + 1 ) )Simplify ( (x+1)^2 - (x+1) + 1 = x^2 + 2x + 1 - x - 1 + 1 = x^2 + x + 1 )So, ( (x^2 - x + 1)(x^2 + x + 1) )Multiply:( x^2(x^2 + x + 1) - x(x^2 + x + 1) + 1(x^2 + x + 1) )= ( x^4 + x^3 + x^2 - x^3 - x^2 - x + x^2 + x + 1 )Combine like terms:- ( x^4 )- ( x^3 - x^3 = 0 )- ( x^2 - x^2 + x^2 = x^2 )- ( -x + x = 0 )- Constant term: 1So, ( x^4 + x^2 + 1 )Compute ( P(x^2 + 1) = (x^2 + 1)^2 - (x^2 + 1) + 1 = x^4 + 2x^2 + 1 - x^2 - 1 + 1 = x^4 + x^2 + 1 )Yes, they are equal. So, ( P(x) = x^2 - x + 1 ) satisfies the equation.However, wait a minute. The problem states that ( P(x) ) must have ( n ) distinct real roots. Let's check the roots of ( P(x) = x^2 - x + 1 ).The discriminant is ( b^2 - 4ac = (-1)^2 - 4(1)(1) = 1 - 4 = -3 ), which is negative. So, this polynomial has no real roots, only complex ones. But the problem requires ( n ) distinct real roots. Therefore, this polynomial doesn't satisfy the condition.Hmm, so even though ( P(x) = x^2 - x + 1 ) satisfies the functional equation, it doesn't have real roots. So, maybe there's no such polynomial with real roots?Wait, but let me think again. Maybe I made a mistake in assuming ( n = 2 ). Maybe I need to try higher degrees.Alternatively, perhaps such a polynomial doesn't exist. Let me consider the roots.Suppose ( P(x) ) has ( n ) distinct real roots ( r_1, r_2, ldots, r_n ). Then, ( P(x) = a(x - r_1)(x - r_2)cdots(x - r_n) ).Given that ( P(x)P(x+1) = P(x^2 + 1) ), let's analyze the roots.If ( P(x)P(x+1) = P(x^2 + 1) ), then the roots of ( P(x^2 + 1) ) must be the union of the roots of ( P(x) ) and ( P(x+1) ).But ( P(x^2 + 1) = 0 ) implies ( x^2 + 1 = r_i ) for some ( i ), so ( x = pm sqrt{r_i - 1} ). But for real roots, ( r_i - 1 geq 0 ), so ( r_i geq 1 ).On the other hand, the roots of ( P(x) ) are ( r_1, r_2, ldots, r_n ), and the roots of ( P(x+1) ) are ( r_1 - 1, r_2 - 1, ldots, r_n - 1 ).So, the roots of ( P(x)P(x+1) ) are ( r_1, r_2, ldots, r_n, r_1 - 1, r_2 - 1, ldots, r_n - 1 ).But the roots of ( P(x^2 + 1) ) are ( pm sqrt{r_i - 1} ) for each ( r_i geq 1 ).So, for these sets of roots to match, every root of ( P(x)P(x+1) ) must be a root of ( P(x^2 + 1) ), and vice versa.But let's consider the largest root of ( P(x) ). Suppose ( r_n ) is the largest root. Then, ( r_n - 1 ) is a root of ( P(x+1) ), so it must be a root of ( P(x^2 + 1) ). Therefore, there exists some ( r_i ) such that ( r_n - 1 = sqrt{r_i - 1} ) or ( r_n - 1 = -sqrt{r_i - 1} ).But ( r_n - 1 ) is a real number, and ( sqrt{r_i - 1} ) is non-negative. So, ( r_n - 1 = sqrt{r_i - 1} ) implies ( (r_n - 1)^2 = r_i - 1 ). So, ( r_i = (r_n - 1)^2 + 1 ).But ( r_i ) is a root of ( P(x) ), which is less than or equal to ( r_n ). So, ( (r_n - 1)^2 + 1 leq r_n ).Let me solve this inequality:( (r_n - 1)^2 + 1 leq r_n )Expand:( r_n^2 - 2r_n + 1 + 1 leq r_n )Simplify:( r_n^2 - 2r_n + 2 leq r_n )Bring all terms to one side:( r_n^2 - 3r_n + 2 leq 0 )Factor:( (r_n - 1)(r_n - 2) leq 0 )So, this inequality holds when ( 1 leq r_n leq 2 ).But ( r_n ) is the largest root, so it must be at least as large as all other roots. If ( r_n leq 2 ), then all roots are less than or equal to 2.But let's consider the root ( r_i = (r_n - 1)^2 + 1 ). Since ( r_n leq 2 ), ( r_n - 1 leq 1 ), so ( (r_n - 1)^2 leq 1 ), hence ( r_i leq 2 ).But also, ( r_i geq 1 ) because ( r_i = (r_n - 1)^2 + 1 geq 1 ).So, all roots are between 1 and 2.But let's think about the roots of ( P(x+1) ), which are ( r_1 - 1, r_2 - 1, ldots, r_n - 1 ). Since ( r_i geq 1 ), these roots are ( geq 0 ).But ( P(x^2 + 1) ) has roots ( pm sqrt{r_i - 1} ). So, for each root ( r_i geq 1 ), we have two roots ( sqrt{r_i - 1} ) and ( -sqrt{r_i - 1} ).But ( P(x)P(x+1) ) has roots ( r_1, r_2, ldots, r_n, r_1 - 1, r_2 - 1, ldots, r_n - 1 ).So, for the roots to match, every ( sqrt{r_i - 1} ) and ( -sqrt{r_i - 1} ) must be among the roots of ( P(x)P(x+1) ).But ( P(x)P(x+1) ) has roots only at ( r_j ) and ( r_j - 1 ). So, ( sqrt{r_i - 1} ) must be equal to some ( r_k ) or ( r_k - 1 ).Similarly, ( -sqrt{r_i - 1} ) must be equal to some ( r_k ) or ( r_k - 1 ).But ( r_k geq 1 ) and ( r_k - 1 geq 0 ). So, ( -sqrt{r_i - 1} ) would have to be equal to some ( r_k - 1 ), but ( r_k - 1 geq 0 ), while ( -sqrt{r_i - 1} leq 0 ). The only way this can happen is if ( sqrt{r_i - 1} = 0 ), which implies ( r_i = 1 ).But if ( r_i = 1 ), then ( -sqrt{r_i - 1} = 0 ), which would have to be a root of ( P(x)P(x+1) ). So, either ( 0 ) is a root of ( P(x) ) or ( P(x+1) ).If ( 0 ) is a root of ( P(x) ), then ( P(0) = 0 ), so ( 0 ) is one of the ( r_j ). Similarly, if ( 0 ) is a root of ( P(x+1) ), then ( P(1) = 0 ), so ( 1 ) is a root of ( P(x) ).But earlier, we saw that ( r_i = 1 ) leads to ( -sqrt{0} = 0 ), which would require ( 0 ) to be a root of ( P(x) ) or ( P(x+1) ). If ( 0 ) is a root of ( P(x) ), then ( r_j = 0 ), but earlier we concluded all roots ( r_i geq 1 ). Contradiction.Alternatively, if ( 0 ) is a root of ( P(x+1) ), then ( P(1) = 0 ), so ( 1 ) is a root of ( P(x) ). But then, ( r_i = 1 ), which leads to ( sqrt{1 - 1} = 0 ), which again requires ( 0 ) to be a root of ( P(x) ) or ( P(x+1) ). If ( 0 ) is a root of ( P(x) ), then ( r_j = 0 ), but ( r_j geq 1 ), contradiction. If ( 0 ) is a root of ( P(x+1) ), then ( P(1) = 0 ), which we already have.This seems to create a loop where ( 0 ) must be a root, but it can't be because all roots are ( geq 1 ). Therefore, this suggests that such a polynomial cannot exist.Alternatively, maybe I can approach this by considering the degrees. Let me think about the degrees.Suppose ( P(x) ) is of degree ( n ). Then, ( P(x)P(x+1) ) is of degree ( 2n ). On the other hand, ( P(x^2 + 1) ) is of degree ( 2n ) as well, since ( x^2 + 1 ) is quadratic.So, the degrees match. But the problem is with the roots. As we saw earlier, the roots of ( P(x^2 + 1) ) are symmetric around 0, while the roots of ( P(x)P(x+1) ) are not necessarily symmetric.Moreover, for each root ( r ) of ( P(x) ), ( r - 1 ) is a root of ( P(x+1) ). But ( P(x^2 + 1) ) has roots ( pm sqrt{r_i - 1} ). So, unless ( r_i - 1 ) is a perfect square, which is not necessarily the case, the roots won't match.But even if ( r_i - 1 ) is a perfect square, say ( s^2 ), then ( r_i = s^2 + 1 ), and ( sqrt{r_i - 1} = s ). So, ( s ) must be a root of ( P(x) ) or ( P(x+1) ).But ( s ) could be positive or negative. If ( s ) is positive, it could be a root of ( P(x) ). If ( s ) is negative, it could be a root of ( P(x+1) ), meaning ( s + 1 ) is a root of ( P(x) ).But this creates a chain of roots:If ( r ) is a root, then ( sqrt{r - 1} ) must be a root or ( sqrt{r - 1} - 1 ) must be a root, and so on.This seems to lead to an infinite sequence of roots unless it terminates, which would require ( sqrt{r - 1} ) to be an integer or something, but polynomials have finite roots.Therefore, unless the polynomial is of a specific form, this chain cannot terminate, implying that such a polynomial cannot exist.Alternatively, let's consider the case where ( P(x) ) has roots that are all of the form ( k + 1 ) where ( k ) is a root. But this would require the roots to be in an arithmetic progression, which might not necessarily satisfy the functional equation.Wait, but in part a), we had ( P(x) = x(x - 1) ), which worked. But in part b), the functional equation is different, involving ( x^2 + 1 ) instead of ( x^2 ). So, maybe the approach is different.Alternatively, perhaps such a polynomial doesn't exist because the roots cannot satisfy the necessary conditions. As we saw earlier, trying to satisfy the root conditions leads to contradictions, especially regarding the existence of roots at 0 and 1, which conflict with the requirement that all roots are ( geq 1 ).Therefore, it seems that no such polynomial exists for part b).So, summarizing:For part a), yes, such a polynomial exists, specifically ( P(x) = x^2 - x ).For part b), no such polynomial exists because the root conditions lead to contradictions.

answer：Okay, so I have this problem where I'm given that the tangent of theta is 3, and I need to find the tangent of three times theta. Hmm, I think I remember there's a formula for tan of multiple angles, maybe triple angle? Let me try to recall. I think it's something like tan(3θ) equals (3 tan θ minus tan cubed θ) over (1 minus 3 tan squared θ). Yeah, that sounds right. Let me write that down:tan(3θ) = (3 tan θ - tan³ θ) / (1 - 3 tan² θ)Alright, so I have tan θ = 3. Let me substitute that into the formula. So, replacing tan θ with 3, I get:tan(3θ) = (3*3 - 3³) / (1 - 3*(3)²)Let me compute the numerator first. 3 times 3 is 9, and 3 cubed is 27. So, 9 minus 27 is... hmm, that's -18. Okay, so the numerator is -18.Now, the denominator. 3 squared is 9, so 3 times 9 is 27. Then, 1 minus 27 is... that's -26. So, the denominator is -26.So now, tan(3θ) is (-18)/(-26). Two negatives make a positive, so that simplifies to 18/26. Wait, can I reduce that fraction? Both 18 and 26 are divisible by 2. So, 18 divided by 2 is 9, and 26 divided by 2 is 13. So, it simplifies to 9/13.Let me double-check my steps to make sure I didn't make any mistakes. First, I used the triple angle formula correctly. I substituted tan θ = 3 into the formula. Calculated the numerator: 3*3 is 9, 3³ is 27, 9-27 is -18. Denominator: 3*(3²) is 27, 1-27 is -26. So, -18/-26 simplifies to 9/13. Yeah, that seems right.Wait, just to be thorough, maybe I should consider if there's another way to approach this problem, just to confirm the answer. Maybe using the addition formula for tangent? Because tan(3θ) can also be written as tan(θ + 2θ). Let me try that.The addition formula is tan(A + B) = (tan A + tan B) / (1 - tan A tan B). So, if I let A be θ and B be 2θ, then tan(3θ) = tan(θ + 2θ) = (tan θ + tan 2θ) / (1 - tan θ tan 2θ).But wait, I don't know tan 2θ yet. I need to find tan 2θ first. There's a double angle formula for tangent as well: tan(2θ) = (2 tan θ) / (1 - tan² θ). Let me use that.Given tan θ = 3, tan 2θ = (2*3)/(1 - 3²) = 6/(1 - 9) = 6/(-8) = -3/4.Okay, so tan 2θ is -3/4. Now, plug that back into the addition formula:tan(3θ) = (tan θ + tan 2θ) / (1 - tan θ tan 2θ) = (3 + (-3/4)) / (1 - 3*(-3/4)).Let me compute the numerator: 3 is 12/4, so 12/4 - 3/4 is 9/4.Denominator: 1 is 4/4, and 3*(-3/4) is -9/4, so 1 - (-9/4) is 4/4 + 9/4 = 13/4.So, tan(3θ) = (9/4) / (13/4) = (9/4)*(4/13) = 9/13.Hmm, same result as before. So, that confirms it. Using both the triple angle formula directly and the addition formula with the double angle formula, I get tan(3θ) = 9/13.Just to make sure, maybe I can think about the unit circle or some other method, but I think these two methods are solid. The triple angle formula is straightforward, and the addition formula with double angle also checks out. Both give me the same answer, so I feel confident that 9/13 is correct.I guess another way to think about it is to consider the angle whose tangent is 3. Let me find θ such that tan θ = 3. θ is arctan(3), which is approximately 71.565 degrees. Then, 3θ would be approximately 214.695 degrees. Let me check what tan(214.695 degrees) is.But wait, 214.695 degrees is in the third quadrant, where tangent is positive because both sine and cosine are negative, and their ratio is positive. So, tan(214.695) should be positive, which matches our result of 9/13, a positive number.Alternatively, if I compute tan(3θ) numerically, using θ ≈ 71.565 degrees:3θ ≈ 214.695 degrees.Calculating tan(214.695 degrees):Since 214.695 - 180 = 34.695 degrees, so tan(214.695) = tan(180 + 34.695) = tan(34.695). And tan(34.695) is approximately 0.666, which is roughly 2/3, but wait, that doesn't match 9/13. Hmm, maybe my approximation is off.Wait, 9/13 is approximately 0.6923, which is close to 0.666, but not exact. Maybe my calculator is using a different approximation.Alternatively, perhaps I should compute it more accurately. Let me use a calculator for tan(3θ):Given tan θ = 3, θ ≈ 71.56505118 degrees.3θ ≈ 214.6951535 degrees.Now, tan(214.6951535 degrees):Since 214.6951535 - 180 = 34.6951535 degrees.tan(34.6951535 degrees) ≈ tan(34.6951535) ≈ 0.6832.Wait, 9/13 is approximately 0.6923, which is slightly higher. Hmm, maybe my calculator's precision is limited. Alternatively, perhaps I made an error in the angle calculation.Wait, let me check the exact value. If tan(3θ) = 9/13, then 9/13 is approximately 0.6923, which is close to the calculator's 0.6832, but not exact. Maybe the discrepancy is due to rounding errors in the angle approximation.Alternatively, perhaps I can compute it more precisely. Let me use more decimal places for θ.Given tan θ = 3, θ = arctan(3). Let me compute arctan(3) more accurately. Using a calculator, arctan(3) is approximately 1.2490 radians, which is about 71.56505118 degrees.Then, 3θ is 3 * 1.2490 ≈ 3.7470 radians. Converting that to degrees: 3.7470 * (180/π) ≈ 214.695 degrees.Now, tan(3.7470 radians). Let me compute that.Using a calculator, tan(3.7470) ≈ tan(π + 0.6054) ≈ tan(0.6054) ≈ 0.6923, which is exactly 9/13. So, that matches.Earlier, when I converted 3.7470 radians to degrees, I got approximately 214.695 degrees, and tan(214.695 degrees) is tan(180 + 34.695 degrees) = tan(34.695 degrees) ≈ 0.6923, which is 9/13. So, that checks out.Therefore, all methods confirm that tan(3θ) is 9/13 when tan θ = 3.I think I've thoroughly checked this problem from multiple angles, both algebraically and numerically, and all paths lead to the same answer. So, I'm confident that tan(3θ) is indeed 9/13.

answer：Okay, so I have this problem here: Given that f(x) = ln(1 + x) has a domain of set M, and g(x) = 2^x + 1 has a range of set N, then M ∩ N = __________. Hmm, let me try to figure this out step by step.First, I need to understand what the domain of f(x) is. The function f(x) is a natural logarithm function, ln(1 + x). I remember that the natural logarithm function, ln(y), is only defined when y is greater than 0. So, for ln(1 + x) to be defined, the expression inside the logarithm, which is (1 + x), must be greater than 0.So, let me write that down: 1 + x > 0. Solving for x, I subtract 1 from both sides, which gives me x > -1. Therefore, the domain of f(x) is all real numbers x such that x is greater than -1. In interval notation, that would be (-1, ∞). So, set M is (-1, ∞).Next, I need to find the range of g(x) = 2^x + 1. The function g(x) is an exponential function. I recall that for exponential functions of the form a^x, where a > 0 and a ≠ 1, the range is always (0, ∞). In this case, a is 2, which is greater than 0 and not equal to 1, so the range of 2^x is (0, ∞).But wait, our function is 2^x + 1. So, we're adding 1 to the exponential function. That means the entire graph of 2^x is shifted upward by 1 unit. Therefore, the range of 2^x + 1 should be (0 + 1, ∞ + 1), which simplifies to (1, ∞). So, set N is (1, ∞).Now, the problem asks for the intersection of sets M and N, which is M ∩ N. Set M is (-1, ∞) and set N is (1, ∞). To find the intersection, I need to determine the overlap between these two intervals.Visualizing the intervals on a number line might help. Set M starts just above -1 and goes to infinity, while set N starts just above 1 and goes to infinity. The overlapping part of these two intervals is where both sets agree, which is from just above 1 to infinity. So, the intersection M ∩ N is (1, ∞).Let me double-check my reasoning. For f(x) = ln(1 + x), the domain is indeed x > -1 because the argument of the logarithm must be positive. For g(x) = 2^x + 1, since 2^x is always positive and greater than 0, adding 1 shifts the entire range up by 1, making it start at 1 instead of 0. Therefore, the range is (1, ∞). The intersection of (-1, ∞) and (1, ∞) is (1, ∞) because that's where both intervals overlap.I think that makes sense. I don't see any mistakes in my logic. So, the intersection of M and N is (1, ∞).**Final Answer**The intersection of sets ( M ) and ( N ) is boxed{(1, +infty)}.

answer：Okay, so I have this problem here about a regular polygon. It says that if you start from one vertex, you can draw at most 5 diagonals. Then it asks for one of its interior angles, with options given. Hmm, let me try to figure this out step by step.First, I remember that in a regular polygon, the number of diagonals you can draw from a single vertex is related to the number of sides the polygon has. I think the formula is something like n minus 3, where n is the number of sides. Let me recall why that is. If you have a polygon with n sides, each vertex is connected to n minus 3 other vertices by diagonals because you can't draw a diagonal to itself or to its two adjacent vertices. So, yeah, the number of diagonals from one vertex is n - 3.The problem says that at most 5 diagonals can be drawn from a vertex. So, if I set up the equation n - 3 = 5, that should give me the number of sides. Let me write that down:n - 3 = 5Solving for n, I add 3 to both sides:n = 5 + 3n = 8Okay, so the polygon has 8 sides. That makes it an octagon. Now, I need to find one of its interior angles. I remember there's a formula for the sum of the interior angles of a polygon, which is (n - 2) multiplied by 180 degrees. Let me write that formula:Sum of interior angles = (n - 2) × 180°Since it's a regular polygon, all the interior angles are equal, so to find one interior angle, I can divide the sum by the number of sides, n. So, the formula for one interior angle is:Interior angle = [(n - 2) × 180°] / nPlugging in n = 8:Interior angle = [(8 - 2) × 180°] / 8Interior angle = [6 × 180°] / 8Interior angle = 1080° / 8Interior angle = 135°So, one of the interior angles is 135 degrees. Let me check the options given:A: 1080°B: 720°C: 140°D: 135°135° is option D. That must be the correct answer. Just to make sure I didn't make any mistakes, let me recap:1. Number of diagonals from one vertex: n - 3 = 5 ⇒ n = 8.2. Sum of interior angles: (8 - 2) × 180° = 1080°.3. Each interior angle: 1080° / 8 = 135°.Everything checks out. I think I'm confident that D is the right choice.