answer：First, identify the signs of the coordinates for a point in the third quadrant. In the third quadrant, both the x-coordinate and y-coordinate are negative.Next, use the given distances to determine the specific coordinates. The distance to the x-axis is 3, which corresponds to the absolute value of the y-coordinate. Therefore, the y-coordinate is -3. Similarly, the distance to the y-axis is 5, which corresponds to the absolute value of the x-coordinate. Thus, the x-coordinate is -5.Combining these, the coordinates of point P are (-5, -3).

answer：Alright, so I have this problem where I need to find the range of the function ( g(x) = (arccos x)^4 + (arcsin x)^4 ) for ( x in [-1, 1] ). Both arccos and arcsin are measured in radians. Hmm, okay, let me think about how to approach this.First, I remember that ( arccos x ) and ( arcsin x ) are related. Specifically, ( arccos x + arcsin x = frac{pi}{2} ). That's a useful identity because it means I can express one in terms of the other. Let me set ( alpha = arccos x ) and ( beta = arcsin x ). So, ( alpha + beta = frac{pi}{2} ). That simplifies things a bit because now I can express ( g(x) ) in terms of just one variable.So, ( g(x) = alpha^4 + beta^4 ). Since ( beta = frac{pi}{2} - alpha ), I can substitute that into the equation:( g(x) = alpha^4 + left( frac{pi}{2} - alpha right)^4 ).Now, I need to find the range of this function as ( alpha ) varies. But what are the possible values of ( alpha )? Since ( x in [-1, 1] ), ( arccos x ) ranges from 0 to ( pi ). However, because ( arcsin x ) must also be real, ( x ) is restricted to ( [-1, 1] ), which we already have. So, ( alpha ) goes from 0 to ( pi ), but since ( beta = frac{pi}{2} - alpha ), and ( beta ) must also be between 0 and ( frac{pi}{2} ), actually ( alpha ) ranges from 0 to ( frac{pi}{2} ).Wait, let me double-check that. If ( x = 1 ), then ( arccos 1 = 0 ) and ( arcsin 1 = frac{pi}{2} ). If ( x = -1 ), ( arccos(-1) = pi ) and ( arcsin(-1) = -frac{pi}{2} ). Hmm, so actually, ( beta ) can range from ( -frac{pi}{2} ) to ( frac{pi}{2} ), but since ( arcsin x ) is defined as the inverse function, it's actually between ( -frac{pi}{2} ) and ( frac{pi}{2} ). However, in our case, since ( x in [-1, 1] ), ( arcsin x ) is between ( -frac{pi}{2} ) and ( frac{pi}{2} ), but ( arccos x ) is between 0 and ( pi ).But since ( alpha + beta = frac{pi}{2} ), if ( alpha ) is between 0 and ( pi ), then ( beta ) must be between ( -frac{pi}{2} ) and ( frac{pi}{2} ). Wait, that doesn't seem right because ( alpha ) can't be more than ( pi ), but ( beta ) can't be less than ( -frac{pi}{2} ). So, actually, ( alpha ) is between 0 and ( pi ), but ( beta ) is between ( -frac{pi}{2} ) and ( frac{pi}{2} ). However, since ( alpha + beta = frac{pi}{2} ), if ( alpha ) is 0, ( beta = frac{pi}{2} ); if ( alpha = frac{pi}{2} ), ( beta = 0 ); and if ( alpha = pi ), ( beta = -frac{pi}{2} ). So, ( alpha ) ranges from 0 to ( pi ), but ( beta ) ranges from ( -frac{pi}{2} ) to ( frac{pi}{2} ).But in our case, since ( x in [-1, 1] ), and ( arccos x ) is always between 0 and ( pi ), and ( arcsin x ) is between ( -frac{pi}{2} ) and ( frac{pi}{2} ). So, ( alpha ) is between 0 and ( pi ), and ( beta ) is between ( -frac{pi}{2} ) and ( frac{pi}{2} ). But since ( alpha + beta = frac{pi}{2} ), the values of ( alpha ) and ( beta ) are dependent on each other.So, to find the range of ( g(x) = alpha^4 + beta^4 ), I can consider ( alpha ) as a variable ranging from 0 to ( pi ), and ( beta = frac{pi}{2} - alpha ). Therefore, ( g(x) ) becomes a function of ( alpha ):( g(alpha) = alpha^4 + left( frac{pi}{2} - alpha right)^4 ).Now, I need to find the minimum and maximum values of this function as ( alpha ) varies from 0 to ( pi ). To do this, I can take the derivative of ( g(alpha) ) with respect to ( alpha ), set it equal to zero, and solve for ( alpha ). The critical points will give me potential minima or maxima.Let's compute the derivative:( g'(alpha) = 4alpha^3 - 4left( frac{pi}{2} - alpha right)^3 ).Set ( g'(alpha) = 0 ):( 4alpha^3 - 4left( frac{pi}{2} - alpha right)^3 = 0 ).Divide both sides by 4:( alpha^3 - left( frac{pi}{2} - alpha right)^3 = 0 ).So,( alpha^3 = left( frac{pi}{2} - alpha right)^3 ).Taking cube roots on both sides:( alpha = frac{pi}{2} - alpha ).Solving for ( alpha ):( 2alpha = frac{pi}{2} ).( alpha = frac{pi}{4} ).So, the critical point is at ( alpha = frac{pi}{4} ). Now, I need to check the value of ( g(alpha) ) at this critical point as well as at the endpoints of the interval ( alpha = 0 ) and ( alpha = pi ).First, let's compute ( g(0) ):( g(0) = 0^4 + left( frac{pi}{2} - 0 right)^4 = 0 + left( frac{pi}{2} right)^4 = frac{pi^4}{16} ).Next, ( g(pi) ):( g(pi) = pi^4 + left( frac{pi}{2} - pi right)^4 = pi^4 + left( -frac{pi}{2} right)^4 = pi^4 + frac{pi^4}{16} = frac{17pi^4}{16} ).Wait, that seems a bit high. Let me double-check. ( left( -frac{pi}{2} right)^4 ) is indeed ( frac{pi^4}{16} ), so ( g(pi) = pi^4 + frac{pi^4}{16} = frac{16pi^4 + pi^4}{16} = frac{17pi^4}{16} ). Okay, that's correct.Now, let's compute ( gleft( frac{pi}{4} right) ):( gleft( frac{pi}{4} right) = left( frac{pi}{4} right)^4 + left( frac{pi}{2} - frac{pi}{4} right)^4 = left( frac{pi}{4} right)^4 + left( frac{pi}{4} right)^4 = 2 times left( frac{pi}{4} right)^4 = 2 times frac{pi^4}{256} = frac{pi^4}{128} ).Wait, that seems too low. Let me check the calculations again. ( left( frac{pi}{4} right)^4 = frac{pi^4}{256} ), so adding two of them gives ( frac{pi^4}{128} ). Hmm, but earlier, I had ( g(0) = frac{pi^4}{16} ) and ( g(pi) = frac{17pi^4}{16} ). So, ( frac{pi^4}{128} ) is actually smaller than ( frac{pi^4}{16} ), which is about 6.1685 (since ( pi^4 approx 97.4091 )), so ( frac{pi^4}{16} approx 6.1685 ), and ( frac{pi^4}{128} approx 0.761 ). That seems inconsistent because ( g(pi) ) is much larger.Wait a minute, I think I made a mistake in calculating ( gleft( frac{pi}{4} right) ). Let me recalculate:( left( frac{pi}{4} right)^4 = left( frac{pi}{4} right)^4 = frac{pi^4}{256} ).Similarly, ( left( frac{pi}{4} right)^4 = frac{pi^4}{256} ).So, adding them together: ( frac{pi^4}{256} + frac{pi^4}{256} = frac{pi^4}{128} ).Wait, that's correct. So, ( gleft( frac{pi}{4} right) = frac{pi^4}{128} approx 0.761 ), which is indeed less than ( g(0) = frac{pi^4}{16} approx 6.1685 ). So, the function reaches a minimum at ( alpha = frac{pi}{4} ) and maximums at the endpoints ( alpha = 0 ) and ( alpha = pi ).But wait, when ( alpha = pi ), ( beta = -frac{pi}{2} ), but ( arcsin x ) can be negative, so that's okay. However, when ( alpha = pi ), ( x = -1 ), and ( arcsin(-1) = -frac{pi}{2} ). So, that's valid.But let me think again. Is ( g(pi) = frac{17pi^4}{16} ) the maximum? Let me check the behavior of ( g(alpha) ) as ( alpha ) approaches ( pi ). Since ( alpha ) is approaching ( pi ), ( beta ) approaches ( -frac{pi}{2} ), but since we're raising to the fourth power, the negative sign doesn't matter. So, ( g(alpha) ) approaches ( pi^4 + left( -frac{pi}{2} right)^4 = pi^4 + frac{pi^4}{16} = frac{17pi^4}{16} ).Similarly, when ( alpha = 0 ), ( beta = frac{pi}{2} ), so ( g(0) = 0 + left( frac{pi}{2} right)^4 = frac{pi^4}{16} ).Wait, but when ( alpha = frac{pi}{4} ), ( g(alpha) = frac{pi^4}{128} ), which is much smaller. So, the function has a minimum at ( alpha = frac{pi}{4} ) and maximums at the endpoints.But wait, is that the case? Let me plot the function or think about its behavior. Since ( g(alpha) = alpha^4 + left( frac{pi}{2} - alpha right)^4 ), it's a function that is symmetric around ( alpha = frac{pi}{4} ) because ( alpha ) and ( frac{pi}{2} - alpha ) are symmetric around ( frac{pi}{4} ). So, the function should have a minimum at ( alpha = frac{pi}{4} ) and maximums at the endpoints.But wait, when ( alpha = frac{pi}{2} ), ( beta = 0 ), so ( gleft( frac{pi}{2} right) = left( frac{pi}{2} right)^4 + 0 = frac{pi^4}{16} ), which is the same as ( g(0) ). So, actually, the function has the same value at ( alpha = 0 ) and ( alpha = frac{pi}{2} ), and a minimum at ( alpha = frac{pi}{4} ). But when ( alpha = pi ), ( g(pi) = frac{17pi^4}{16} ), which is larger than ( g(0) ) and ( gleft( frac{pi}{2} right) ).Wait, so actually, the function ( g(alpha) ) is not symmetric around ( frac{pi}{4} ) when considering the entire interval from 0 to ( pi ). Because when ( alpha ) exceeds ( frac{pi}{2} ), ( beta ) becomes negative, but since we're raising to the fourth power, it's still positive. So, the function continues to increase as ( alpha ) increases beyond ( frac{pi}{2} ) up to ( pi ).Therefore, the maximum value of ( g(alpha) ) occurs at ( alpha = pi ), giving ( g(pi) = frac{17pi^4}{16} ), and the minimum occurs at ( alpha = frac{pi}{4} ), giving ( gleft( frac{pi}{4} right) = frac{pi^4}{128} ).But wait, let me verify this by considering the second derivative to confirm if ( alpha = frac{pi}{4} ) is indeed a minimum.Compute the second derivative ( g''(alpha) ):First, ( g'(alpha) = 4alpha^3 - 4left( frac{pi}{2} - alpha right)^3 ).So, ( g''(alpha) = 12alpha^2 + 12left( frac{pi}{2} - alpha right)^2 ).At ( alpha = frac{pi}{4} ):( g''left( frac{pi}{4} right) = 12left( frac{pi}{4} right)^2 + 12left( frac{pi}{2} - frac{pi}{4} right)^2 = 12left( frac{pi^2}{16} right) + 12left( frac{pi}{4} right)^2 = 12 times frac{pi^2}{16} + 12 times frac{pi^2}{16} = 24 times frac{pi^2}{16} = frac{3pi^2}{2} ).Since ( g''left( frac{pi}{4} right) > 0 ), this critical point is indeed a local minimum.Now, to ensure that there are no other critical points, we can observe that ( g'(alpha) = 0 ) only at ( alpha = frac{pi}{4} ) within the interval ( [0, pi] ). Therefore, the function ( g(alpha) ) has only one critical point, which is a minimum, and the maximum occurs at the endpoint ( alpha = pi ).Thus, the range of ( g(x) ) is from ( frac{pi^4}{128} ) to ( frac{17pi^4}{16} ).Wait, but earlier I thought the maximum was at ( alpha = pi ), giving ( frac{17pi^4}{16} ), and the minimum at ( alpha = frac{pi}{4} ), giving ( frac{pi^4}{128} ). However, when I think about the function ( g(alpha) = alpha^4 + left( frac{pi}{2} - alpha right)^4 ), it's symmetric around ( alpha = frac{pi}{4} ) only if we consider ( alpha ) in ( [0, frac{pi}{2}] ). Beyond ( frac{pi}{2} ), the function continues to increase because ( alpha^4 ) grows faster than the decrease in ( left( frac{pi}{2} - alpha right)^4 ).But let me confirm by evaluating ( g(alpha) ) at ( alpha = frac{pi}{2} ):( gleft( frac{pi}{2} right) = left( frac{pi}{2} right)^4 + 0 = frac{pi^4}{16} ).So, at ( alpha = frac{pi}{2} ), ( g(alpha) = frac{pi^4}{16} ), which is the same as at ( alpha = 0 ). Therefore, the function decreases from ( alpha = 0 ) to ( alpha = frac{pi}{4} ), reaching a minimum, and then increases again from ( alpha = frac{pi}{4} ) to ( alpha = pi ), reaching a maximum at ( alpha = pi ).Therefore, the range of ( g(x) ) is from ( frac{pi^4}{128} ) to ( frac{17pi^4}{16} ).But wait, let me compute these values numerically to get a sense:( pi approx 3.1416 ), so ( pi^4 approx (3.1416)^4 approx 97.4091 ).So, ( frac{pi^4}{128} approx frac{97.4091}{128} approx 0.761 ).And ( frac{17pi^4}{16} approx frac{17 times 97.4091}{16} approx frac{1655.9547}{16} approx 103.497 ).So, the function ranges from approximately 0.761 to 103.497.But wait, when ( alpha = frac{pi}{4} ), ( g(alpha) = frac{pi^4}{128} approx 0.761 ), and when ( alpha = pi ), ( g(alpha) approx 103.497 ). That seems like a huge range. Is that correct?Let me think again. When ( x = 1 ), ( arccos 1 = 0 ), ( arcsin 1 = frac{pi}{2} ), so ( g(1) = 0 + left( frac{pi}{2} right)^4 = frac{pi^4}{16} approx 6.1685 ).When ( x = -1 ), ( arccos(-1) = pi ), ( arcsin(-1) = -frac{pi}{2} ), so ( g(-1) = pi^4 + left( -frac{pi}{2} right)^4 = pi^4 + frac{pi^4}{16} = frac{17pi^4}{16} approx 103.497 ).Wait, so at ( x = 1 ), ( g(x) approx 6.1685 ), and at ( x = -1 ), ( g(x) approx 103.497 ). But earlier, I found that the minimum is at ( x ) corresponding to ( alpha = frac{pi}{4} ), which is ( x = cosleft( frac{pi}{4} right) = frac{sqrt{2}}{2} approx 0.7071 ). So, at ( x = frac{sqrt{2}}{2} ), ( g(x) approx 0.761 ).But wait, that seems inconsistent because when ( x = frac{sqrt{2}}{2} ), both ( arccos x ) and ( arcsin x ) are equal to ( frac{pi}{4} ), so ( g(x) = 2 times left( frac{pi}{4} right)^4 = 2 times frac{pi^4}{256} = frac{pi^4}{128} approx 0.761 ). That's correct.But then, when ( x = 1 ), ( g(x) approx 6.1685 ), which is higher than the minimum but lower than the maximum at ( x = -1 ). So, the function reaches its minimum at ( x = frac{sqrt{2}}{2} ) and its maximum at ( x = -1 ).Therefore, the range of ( g(x) ) is from ( frac{pi^4}{128} ) to ( frac{17pi^4}{16} ).But wait, let me check if there are any other critical points beyond ( alpha = frac{pi}{4} ). Since we only found one critical point in the interval ( [0, pi] ), which is a minimum, and the function increases beyond that point towards ( alpha = pi ), there are no other extrema.Therefore, the range of ( g(x) ) is ( left[ frac{pi^4}{128}, frac{17pi^4}{16} right] ).But wait, in the initial problem statement, the user mentioned that the range is ( left[ frac{pi^4}{32}, frac{2pi^4}{5} right] ). Hmm, that doesn't match my calculations. Did I make a mistake somewhere?Let me go back and check.I set ( alpha = arccos x ) and ( beta = arcsin x ), with ( alpha + beta = frac{pi}{2} ). Then, ( g(x) = alpha^4 + beta^4 ). Substituting ( beta = frac{pi}{2} - alpha ), we get ( g(alpha) = alpha^4 + left( frac{pi}{2} - alpha right)^4 ).Taking the derivative:( g'(alpha) = 4alpha^3 - 4left( frac{pi}{2} - alpha right)^3 ).Setting to zero:( 4alpha^3 = 4left( frac{pi}{2} - alpha right)^3 ).Dividing both sides by 4:( alpha^3 = left( frac{pi}{2} - alpha right)^3 ).Taking cube roots:( alpha = frac{pi}{2} - alpha ).So,( 2alpha = frac{pi}{2} ).( alpha = frac{pi}{4} ).That seems correct. Then, evaluating ( g(alpha) ) at ( alpha = frac{pi}{4} ):( gleft( frac{pi}{4} right) = 2 times left( frac{pi}{4} right)^4 = 2 times frac{pi^4}{256} = frac{pi^4}{128} ).At ( alpha = 0 ):( g(0) = left( frac{pi}{2} right)^4 = frac{pi^4}{16} ).At ( alpha = pi ):( g(pi) = pi^4 + left( -frac{pi}{2} right)^4 = pi^4 + frac{pi^4}{16} = frac{17pi^4}{16} ).So, my calculations seem correct. But the initial problem's solution mentioned a range of ( left[ frac{pi^4}{32}, frac{2pi^4}{5} right] ). That suggests that perhaps I made a mistake in interpreting the problem or in my calculations.Wait, let me check the initial problem again. It says:"Let ( g(x) = (arccos x)^4 + (arcsin x)^4 ). Find the range of ( g(x) ) where ( x in [-1,1] ) and both ( arccos x ) and ( arcsin x ) are measured in radians."So, that's exactly what I did. Maybe the initial solution was incorrect, or perhaps I misapplied the identity.Wait, another thought: when ( x = -1 ), ( arccos(-1) = pi ), and ( arcsin(-1) = -frac{pi}{2} ). So, ( g(-1) = pi^4 + left( -frac{pi}{2} right)^4 = pi^4 + frac{pi^4}{16} = frac{17pi^4}{16} ).But perhaps the problem expects the range to be between ( frac{pi^4}{32} ) and ( frac{2pi^4}{5} ). Let me compute these values:( frac{pi^4}{32} approx frac{97.4091}{32} approx 3.044 ).( frac{2pi^4}{5} approx frac{2 times 97.4091}{5} approx frac{194.8182}{5} approx 38.9636 ).But according to my calculations, the minimum is approximately 0.761 and the maximum is approximately 103.497. So, the initial solution's range is much smaller. Therefore, I must have made a mistake.Wait, perhaps I misapplied the identity. Let me double-check the relationship between ( arccos x ) and ( arcsin x ). It's true that ( arccos x + arcsin x = frac{pi}{2} ) for all ( x in [-1, 1] ). So, that part is correct.But when I set ( alpha = arccos x ) and ( beta = arcsin x ), then ( alpha + beta = frac{pi}{2} ). So, ( beta = frac{pi}{2} - alpha ). Therefore, ( g(x) = alpha^4 + left( frac{pi}{2} - alpha right)^4 ).Wait, but when ( alpha = pi ), ( beta = -frac{pi}{2} ), which is correct because ( arcsin(-1) = -frac{pi}{2} ). So, that's fine.But perhaps the function ( g(x) ) is not defined for ( alpha > frac{pi}{2} ) because ( arccos x ) is always between 0 and ( pi ), but ( arcsin x ) is between ( -frac{pi}{2} ) and ( frac{pi}{2} ). However, since ( arccos x ) can be up to ( pi ), ( beta = frac{pi}{2} - alpha ) can be as low as ( -frac{pi}{2} ).Wait, but in the initial problem, the user's solution concluded the range as ( left[ frac{pi^4}{32}, frac{2pi^4}{5} right] ). Let me compute these values:( frac{pi^4}{32} approx 3.044 ).( frac{2pi^4}{5} approx 38.9636 ).But according to my calculations, the minimum is ( frac{pi^4}{128} approx 0.761 ) and the maximum is ( frac{17pi^4}{16} approx 103.497 ). So, there's a discrepancy.Wait, perhaps the initial solution considered only ( x in [0, 1] ), which would make ( arccos x ) between 0 and ( frac{pi}{2} ), and ( arcsin x ) between 0 and ( frac{pi}{2} ). In that case, ( alpha ) would range from 0 to ( frac{pi}{2} ), and ( beta ) would range from ( frac{pi}{2} ) to 0. Then, the function ( g(alpha) = alpha^4 + left( frac{pi}{2} - alpha right)^4 ) would have a minimum at ( alpha = frac{pi}{4} ) and maximums at ( alpha = 0 ) and ( alpha = frac{pi}{2} ), both giving ( g(alpha) = frac{pi^4}{16} approx 6.1685 ).But that doesn't align with the initial solution's range either. Alternatively, perhaps the initial solution considered a different approach, such as expressing ( g(x) ) in terms of ( theta ) where ( theta = arccos x ), and then using trigonometric identities to simplify.Let me try that approach.Let ( theta = arccos x ), so ( x = cos theta ), and ( arcsin x = arcsin (cos theta) ). But ( arcsin (cos theta) ) can be expressed in terms of ( theta ). Since ( cos theta = sinleft( frac{pi}{2} - theta right) ), so ( arcsin (cos theta) = arcsin left( sinleft( frac{pi}{2} - theta right) right) ).However, ( arcsin (sin phi) = phi ) only if ( phi in left[ -frac{pi}{2}, frac{pi}{2} right] ). Since ( theta in [0, pi] ), ( frac{pi}{2} - theta ) ranges from ( -frac{pi}{2} ) to ( frac{pi}{2} ). Therefore, ( arcsin (cos theta) = frac{pi}{2} - theta ) when ( theta in [0, pi] ).Wait, that's consistent with our earlier identity ( arccos x + arcsin x = frac{pi}{2} ). So, ( arcsin x = frac{pi}{2} - theta ).Therefore, ( g(x) = theta^4 + left( frac{pi}{2} - theta right)^4 ), which is the same as before.So, my initial approach was correct. Therefore, the range should indeed be from ( frac{pi^4}{128} ) to ( frac{17pi^4}{16} ).But the initial solution mentioned a different range. Perhaps there was a miscalculation there. Alternatively, maybe the problem was intended to have a different function, such as ( (arccos x)^2 + (arcsin x)^2 ), which would have a different range.Alternatively, perhaps the initial solution considered only the principal values where ( arcsin x ) is non-negative, i.e., ( x in [0, 1] ), which would limit ( alpha ) to ( [0, frac{pi}{2}] ), and thus ( g(alpha) ) would range from ( frac{pi^4}{128} ) to ( frac{pi^4}{16} ). But even then, ( frac{pi^4}{16} approx 6.1685 ), which is still not matching the initial solution's upper bound of ( frac{2pi^4}{5} approx 38.9636 ).Wait, perhaps the initial solution made a mistake in the derivative. Let me re-examine the derivative step.Given ( g(alpha) = alpha^4 + left( frac{pi}{2} - alpha right)^4 ).Then, ( g'(alpha) = 4alpha^3 - 4left( frac{pi}{2} - alpha right)^3 ).Setting to zero:( 4alpha^3 = 4left( frac{pi}{2} - alpha right)^3 ).Divide both sides by 4:( alpha^3 = left( frac{pi}{2} - alpha right)^3 ).Taking cube roots:( alpha = frac{pi}{2} - alpha ).So,( 2alpha = frac{pi}{2} ).( alpha = frac{pi}{4} ).That seems correct. So, the critical point is indeed at ( alpha = frac{pi}{4} ).Therefore, unless there's a different approach, my calculations seem correct, and the range should be ( left[ frac{pi^4}{128}, frac{17pi^4}{16} right] ).But perhaps the initial solution considered a different function, such as ( (arccos x)^2 + (arcsin x)^2 ), which would have a different range. Let me compute that quickly.If ( g(x) = (arccos x)^2 + (arcsin x)^2 ), then using ( alpha + beta = frac{pi}{2} ), we have ( g(alpha) = alpha^2 + left( frac{pi}{2} - alpha right)^2 ).Expanding:( g(alpha) = alpha^2 + frac{pi^2}{4} - pi alpha + alpha^2 = 2alpha^2 - pi alpha + frac{pi^2}{4} ).Taking derivative:( g'(alpha) = 4alpha - pi ).Setting to zero:( 4alpha - pi = 0 ).( alpha = frac{pi}{4} ).So, the minimum is at ( alpha = frac{pi}{4} ):( gleft( frac{pi}{4} right) = 2 times left( frac{pi}{4} right)^2 - pi times frac{pi}{4} + frac{pi^2}{4} = 2 times frac{pi^2}{16} - frac{pi^2}{4} + frac{pi^2}{4} = frac{pi^2}{8} ).And the maximums at ( alpha = 0 ) and ( alpha = frac{pi}{2} ):( g(0) = 0 + left( frac{pi}{2} right)^2 = frac{pi^2}{4} ).So, the range would be ( left[ frac{pi^2}{8}, frac{pi^2}{4} right] approx [1.2337, 2.4674] ).But that's for the squared function, not the fourth power.Alternatively, perhaps the initial solution considered a different exponent or a different combination of functions.Alternatively, maybe the initial solution considered the function ( (arccos x)^4 + (arcsin x)^4 ) but restricted ( x ) to ( [0, 1] ), which would limit ( alpha ) to ( [0, frac{pi}{2}] ), and thus ( g(alpha) ) would range from ( frac{pi^4}{128} ) to ( frac{pi^4}{16} ).But even then, ( frac{pi^4}{16} approx 6.1685 ), which is still not matching the initial solution's upper bound of ( frac{2pi^4}{5} approx 38.9636 ).Wait, perhaps the initial solution made a mistake in the derivative calculation. Let me check.If ( g(alpha) = alpha^4 + left( frac{pi}{2} - alpha right)^4 ), then:( g'(alpha) = 4alpha^3 - 4left( frac{pi}{2} - alpha right)^3 ).Setting to zero:( 4alpha^3 = 4left( frac{pi}{2} - alpha right)^3 ).Divide by 4:( alpha^3 = left( frac{pi}{2} - alpha right)^3 ).Taking cube roots:( alpha = frac{pi}{2} - alpha ).So,( 2alpha = frac{pi}{2} ).( alpha = frac{pi}{4} ).That's correct. So, the critical point is indeed at ( alpha = frac{pi}{4} ).Therefore, unless there's a different interpretation of the problem, my calculations seem correct, and the range should be ( left[ frac{pi^4}{128}, frac{17pi^4}{16} right] ).But the initial solution mentioned a different range, so perhaps there was a misunderstanding or a miscalculation there.Alternatively, perhaps the initial solution considered the function ( (arccos x)^4 + (arcsin x)^4 ) but only for ( x in [0, 1] ), which would limit ( alpha ) to ( [0, frac{pi}{2}] ), and thus the maximum would be at ( alpha = 0 ) or ( alpha = frac{pi}{2} ), both giving ( g(alpha) = frac{pi^4}{16} approx 6.1685 ), and the minimum at ( alpha = frac{pi}{4} ), giving ( frac{pi^4}{128} approx 0.761 ). But that still doesn't align with the initial solution's range.Alternatively, perhaps the initial solution considered a different function, such as ( (arccos x)^2 + (arcsin x)^4 ), which would have a different range. But without more information, it's hard to say.In conclusion, based on my calculations, the range of ( g(x) = (arccos x)^4 + (arcsin x)^4 ) for ( x in [-1, 1] ) is from ( frac{pi^4}{128} ) to ( frac{17pi^4}{16} ). Therefore, the range is ( left[ frac{pi^4}{128}, frac{17pi^4}{16} right] ).

answer：First, I need to determine the number of choices available for each component of the dress design: color, pattern, and material.There are 5 fabric colors to choose from.Next, there are 5 different patterns available.Additionally, there are 2 fabric materials to consider.To find the total number of different dress designs, I will multiply the number of choices for each component together.So, the total number of dress designs is 5 (colors) multiplied by 5 (patterns) multiplied by 2 (materials), which equals 50.

answer：Okay, so I have this problem: Suppose that (2^n + 1) is an odd prime for some positive integer (n). I need to show that (n) must be a power of 2. Hmm, interesting. Let me try to figure this out step by step.First, let me recall what it means for a number to be a power of 2. A number is a power of 2 if it can be expressed as (2^k) for some non-negative integer (k). So, numbers like 1, 2, 4, 8, 16, etc., are powers of 2. Now, the problem is saying that if (2^n + 1) is prime, then (n) has to be one of these numbers.I know that primes are numbers greater than 1 that have no positive divisors other than 1 and themselves. So, (2^n + 1) being prime means it can't be factored into smaller integers except for 1 and itself.Let me think about specific examples. If (n = 1), then (2^1 + 1 = 3), which is prime. If (n = 2), (2^2 + 1 = 5), also prime. For (n = 3), (2^3 + 1 = 9), which is not prime. (n = 4), (2^4 + 1 = 17), prime again. (n = 5), (2^5 + 1 = 33), not prime. Hmm, so it seems like when (n) is a power of 2, (2^n + 1) is prime, but when (n) isn't, it might not be.Wait, but the problem is asking me to show that if (2^n + 1) is prime, then (n) must be a power of 2. So, it's kind of the converse: if the expression is prime, (n) has to be a power of 2. So, I need to prove that (n) can't be anything else.Let me think about how to approach this. Maybe I can use some properties of exponents or modular arithmetic. I remember that if (n) is not a power of 2, then it has some odd divisor greater than 1. Maybe that can help.Suppose (n) is not a power of 2. Then, (n) can be written as (n = k cdot m), where (m) is an odd integer greater than 1, and (k) is some positive integer. So, (2^n + 1 = 2^{k cdot m} + 1).I remember that (a^{ab} + 1) can sometimes be factored if (a) is even or has certain properties. Let me see if I can factor (2^{k cdot m} + 1). Since (m) is odd, maybe I can use the formula for sum of powers.Wait, the formula for (a^m + b^m) when (m) is odd is ((a + b)(a^{m-1} - a^{m-2}b + dots - ab^{m-2} + b^{m-1})). So, in this case, if I set (a = 2^k) and (b = 1), then (2^{k cdot m} + 1 = (2^k)^m + 1^m). Since (m) is odd, this can be factored as ((2^k + 1)(2^{k(m-1)} - 2^{k(m-2)} + dots - 2^k + 1)).So, if (n) is not a power of 2, then (2^n + 1) can be factored into two integers greater than 1, which would mean it's not prime. That seems to make sense. Therefore, for (2^n + 1) to be prime, (n) must be such that it can't be factored in this way, meaning (n) must be a power of 2.Wait, let me test this with an example. Take (n = 6), which is not a power of 2. Then, (2^6 + 1 = 64 + 1 = 65), which factors into 5 and 13, so it's not prime. Another example: (n = 12), (2^{12} + 1 = 4096 + 1 = 4097). I think 4097 is 17 times 241, so again, not prime.But wait, what about (n = 1)? (2^1 + 1 = 3), which is prime, and 1 is (2^0), so it's a power of 2. Similarly, (n = 2), (2^2 + 1 = 5), prime; (n = 4), (2^4 + 1 = 17), prime; (n = 8), (2^8 + 1 = 257), prime; (n = 16), (2^{16} + 1 = 65537), which is also prime. These are known as Fermat primes, right?So, it seems like when (n) is a power of 2, (2^n + 1) is prime, and when it's not, it's composite. Therefore, if (2^n + 1) is prime, (n) must be a power of 2.But let me make sure I haven't missed anything. Is there a case where (n) is not a power of 2, but (2^n + 1) is still prime? From the examples I tried, it doesn't seem so. Also, the factoring argument seems solid: if (n) has an odd factor, then (2^n + 1) can be factored, hence it's composite.Therefore, the conclusion is that (n) must be a power of 2 for (2^n + 1) to be prime.

answer：Okay, so I have this problem where I'm given two vectors, vector a and vector b. Vector a is (m, -6, 2) and vector b is (4, n, 1). The problem says that vector a is parallel to vector b, and I need to find the product of m and n, which is m multiplied by n.Alright, first, I remember that if two vectors are parallel, they must be scalar multiples of each other. That means each component of one vector is a constant multiple of the corresponding component of the other vector. So, there should be some constant k such that vector a equals k times vector b.Let me write that down:Vector a = k * Vector bSo, breaking it down component-wise:m = k * 4-6 = k * n2 = k * 1Hmm, okay, so from the third component, 2 = k * 1, that should be straightforward. If I solve for k, I can find the value of k.So, 2 = k * 1 implies that k = 2.Now that I know k is 2, I can find m and n.Starting with m:m = k * 4 = 2 * 4 = 8So, m is 8.Next, let's find n using the second component:-6 = k * nWe know k is 2, so:-6 = 2 * nTo solve for n, I'll divide both sides by 2:n = -6 / 2 = -3So, n is -3.Now, the problem asks for m multiplied by n, which is m * n.So, m is 8 and n is -3, so:8 * (-3) = -24Therefore, the value of m multiplied by n is -24.Wait, let me double-check my steps to make sure I didn't make any mistakes.1. I started by recognizing that parallel vectors are scalar multiples of each other, so Vector a = k * Vector b.2. I broke it down component-wise: - m = 4k - -6 = nk - 2 = k3. From the third component, I correctly found k = 2.4. Plugging k back into the first component, m = 4*2 = 8.5. Plugging k into the second component, -6 = 2n, so n = -3.6. Finally, m * n = 8*(-3) = -24.Everything seems to check out. I don't see any errors in my calculations or reasoning. So, I'm confident that the answer is -24.

answer：Okay, so I have this problem here about sets M and N, and I need to find their intersection. Let me try to understand what each set represents.First, set M is defined as {x | |x| < 1}. Hmm, that notation means all real numbers x such that the absolute value of x is less than 1. So, if I think about absolute value, it's the distance from zero on the number line. Therefore, |x| < 1 would include all numbers between -1 and 1, but not including -1 and 1 themselves. So, M is the interval (-1, 1). Got that down.Next, set N is given as {x | x² - x < 0}. Okay, this is a quadratic inequality. I need to solve for x where x squared minus x is less than zero. Let me write that inequality down: x² - x < 0. To solve this, I can factor it. Let's see, x² - x can be factored as x(x - 1). So, the inequality becomes x(x - 1) < 0.Now, to solve x(x - 1) < 0, I need to find the intervals where this product is negative. The critical points are where each factor is zero, which is at x = 0 and x = 1. These points divide the number line into three intervals: (-∞, 0), (0, 1), and (1, ∞). I'll test each interval to see where the product is negative.First interval: Let's pick a number less than 0, say x = -1. Plugging into x(x - 1): (-1)(-1 - 1) = (-1)(-2) = 2, which is positive. So, this interval doesn't satisfy the inequality.Second interval: Let's choose x = 0.5. Plugging in: 0.5(0.5 - 1) = 0.5(-0.5) = -0.25, which is negative. So, this interval satisfies the inequality.Third interval: Let's take x = 2. Plugging in: 2(2 - 1) = 2(1) = 2, which is positive. So, this interval doesn't satisfy the inequality either.Therefore, the solution to x² - x < 0 is the interval (0, 1). So, set N is (0, 1).Now, I need to find the intersection of sets M and N. Set M is (-1, 1) and set N is (0, 1). The intersection of two sets is the set of elements that are common to both. So, overlapping parts of (-1, 1) and (0, 1) would be from 0 to 1, since that's where both intervals overlap.Therefore, M ∩ N is (0, 1). Looking at the options, option B is {x | 0 < x < 1}, which matches my result.Wait, just to double-check, let me visualize both intervals on the number line. M spans from -1 to 1, and N spans from 0 to 1. So, their overlap is indeed from 0 to 1. Yep, that seems right.I think I got it. The intersection is the interval where both conditions are satisfied, which is between 0 and 1.**Final Answer**boxed{B}