answer：Okay, so I have this problem here: "A train 120 meters in length crosses a telegraph post in 6 seconds. What is the speed of the train?" Hmm, let's see. I need to find the speed of the train, right? First, I remember that speed is generally calculated by dividing the distance traveled by the time it takes to travel that distance. So, the basic formula is Speed = Distance / Time. That makes sense because if you go a certain distance in a certain amount of time, your speed is just how much distance you cover per unit of time.Now, in this problem, the train is 120 meters long, and it crosses a telegraph post in 6 seconds. Wait, what does it mean for a train to cross a telegraph post? Well, a telegraph post is like a tall pole, right? So, when the train crosses the telegraph post, it means that the entire length of the train has passed by the post. So, the distance the train needs to cover to completely pass the post is equal to its own length, which is 120 meters.Okay, so the distance here is 120 meters, and the time taken is 6 seconds. So, applying the formula, Speed = Distance / Time, that would be Speed = 120 meters / 6 seconds. Let me do that calculation. 120 divided by 6 is 20. So, the speed would be 20 meters per second.Wait a minute, is that right? Let me double-check. If the train is moving at 20 meters per second, then in one second, it covers 20 meters. So, in 6 seconds, it would cover 20 times 6, which is 120 meters. That matches the length of the train, so it makes sense. The entire train passes the telegraph post in 6 seconds, covering its own length of 120 meters.But, just to be thorough, let me think about this again. Sometimes, in these kinds of problems, people get confused about whether they're measuring the distance from the front of the train to the post or something else. But in this case, since the train is crossing the post, it's the entire length of the train that needs to pass the post. So, the distance is indeed the length of the train, which is 120 meters.Another thing to consider is the units. The problem gives the distance in meters and the time in seconds, so the speed will be in meters per second. If I wanted to convert that to kilometers per hour or miles per hour, I could, but the problem doesn't specify, so meters per second should be fine.Let me also think about the formula again. Speed equals distance divided by time. Yeah, that's the basic formula for constant speed. There's no acceleration mentioned here, so we can assume the train is moving at a constant speed while crossing the post. That simplifies things because we don't have to consider any changes in speed.So, to recap: the train is 120 meters long, it takes 6 seconds to cross a telegraph post, which means it has to cover its own length in that time. Using the formula Speed = Distance / Time, we plug in 120 meters for distance and 6 seconds for time, and we get 20 meters per second. That seems correct.I guess another way to think about it is to imagine the front of the train reaching the post and then the end of the train passing the post. The time between those two events is 6 seconds, and during that time, the train has moved a distance equal to its own length. So, again, that's 120 meters in 6 seconds, leading to the same calculation.Is there any chance I could have made a mistake in the calculation? Let me see. 120 divided by 6 is definitely 20. Yeah, that's straightforward arithmetic. So, I don't think I made a mistake there.Also, considering real-world scenarios, 20 meters per second is pretty fast for a train. Let me convert that to kilometers per hour to get a better sense. There are 1,000 meters in a kilometer and 3,600 seconds in an hour. So, to convert meters per second to kilometers per hour, I multiply by 3.6. So, 20 times 3.6 is 72 kilometers per hour. That seems reasonable for a train's speed. Some trains go faster, some slower, but 72 km/h is a plausible speed.Wait, but the problem didn't ask for the speed in kilometers per hour, just in meters per second. So, I don't need to convert it unless specified. But it's good to know that the speed is reasonable in real-world terms.Another thing to consider is whether the telegraph post has any width or if it's just a point. In this problem, I think we're assuming the telegraph post is a point, so the train only needs to cover its own length to pass it completely. If the post had some width, we would have to add that to the distance, but since it's just a post, we can ignore its width.Also, are there any other factors that could affect the speed calculation? For example, is the train accelerating while crossing the post? The problem doesn't mention anything about acceleration, so we can assume the speed is constant. If the train were accelerating, we would need more information, like the acceleration rate or the initial and final speeds, to calculate the average speed.So, in conclusion, using the basic formula for speed, considering the distance as the length of the train, and the time given, the speed of the train is 20 meters per second. This seems correct based on the information provided and the calculations I've done.

answer：First, I need to determine the average mileage per gallon for the second car. Arnold drives a total of 450 miles each month and splits this equally among his three cars, so each car is driven 150 miles per month.For the first car, which averages 50 miles per gallon, the gallons of gas used in a month would be 150 miles divided by 50 mpg, which equals 3 gallons.For the third car, which averages 15 miles per gallon, the gallons of gas used in a month would be 150 miles divided by 15 mpg, which equals 10 gallons.The total cost of gas for all three cars is 56, and since gas costs 2 per gallon, the total gallons of gas used is 56 divided by 2, which equals 28 gallons.Now, I can set up an equation for the total gallons used:3 gallons (first car) + 150/x gallons (second car) + 10 gallons (third car) = 28 gallons.Combining the constants:13 + 150/x = 28.Subtracting 13 from both sides:150/x = 15.Solving for x by multiplying both sides by x and then dividing by 15:150 = 15x,x = 10.Therefore, the average mileage per gallon of gas for the second car is 10 miles per gallon.

answer：Okay, so I need to find the derivative of the function y = sin x (cos x + 1). Hmm, let's see. I remember from my calculus class that when you have a function that's the product of two other functions, you need to use the product rule. The product rule says that if you have y = f(x) * g(x), then the derivative y' is f'(x) * g(x) + f(x) * g'(x). Alright, so in this case, I can let f(x) be sin x and g(x) be (cos x + 1). That makes sense because the function is already written as a product of sin x and another expression. So, f(x) = sin x and g(x) = cos x + 1. Now, I need to find the derivatives of f(x) and g(x). The derivative of sin x is cos x, so f'(x) = cos x. For g(x) = cos x + 1, the derivative of cos x is -sin x, and the derivative of 1 is 0. So, g'(x) = -sin x. Okay, now applying the product rule: y' = f'(x) * g(x) + f(x) * g'(x). Plugging in the derivatives I just found, that would be y' = cos x * (cos x + 1) + sin x * (-sin x). Let me write that out step by step to make sure I don't make a mistake. First, cos x times (cos x + 1): that's cos x * cos x + cos x * 1, which simplifies to cos²x + cos x. Next, sin x times (-sin x): that's -sin²x. So putting it all together, y' = cos²x + cos x - sin²x. Hmm, now I have y' = cos²x + cos x - sin²x. I remember there's a trigonometric identity that relates cos²x - sin²x to cos 2x. Let me recall that identity. Yes, cos 2x = cos²x - sin²x. So, I can replace cos²x - sin²x with cos 2x. That means y' = cos 2x + cos x. Looking back at the answer choices, option C is y' = cos 2x + cos x, which matches what I got. So, I think that's the correct answer. Wait, just to double-check, let me go through the steps again. 1. Identify f(x) and g(x): f(x) = sin x, g(x) = cos x + 1. 2. Find f'(x) and g'(x): f'(x) = cos x, g'(x) = -sin x. 3. Apply the product rule: y' = cos x*(cos x + 1) + sin x*(-sin x). 4. Expand the terms: cos²x + cos x - sin²x. 5. Use the identity cos 2x = cos²x - sin²x to simplify: y' = cos 2x + cos x. Yep, that all checks out. So, I'm confident that option C is the right answer.

answer：To determine the maximum possible value of the largest number among the five distinct natural numbers with an average of 12 and a median of 17, follow these steps:1. **Calculate the Total Sum:** - The average of 5 numbers is 12, so the total sum is ( 5 times 12 = 60 ).2. **Understand the Median:** - The median of 5 numbers is the third number when arranged in ascending order. Therefore, the third number is 17.3. **Set Up the Numbers:** - Let the numbers be ( a, b, 17, d, e ) where ( a < b < 17 < d < e ).4. **Maximize the Largest Number:** - To maximize ( e ), minimize ( a ) and ( b ). The smallest distinct natural numbers are 1 and 2. - Assign ( a = 1 ) and ( b = 2 ).5. **Minimize the Fourth Number:** - The smallest number greater than 17 is 18. Assign ( d = 18 ).6. **Calculate the Largest Number:** - The sum of the known numbers is ( 1 + 2 + 17 + 18 = 38 ). - Subtract this from the total sum: ( 60 - 38 = 22 ). - Therefore, ( e = 22 ).7. **Verify the Numbers:** - The numbers are 1, 2, 17, 18, and 22, which satisfy all conditions.**Conclusion:** The maximum possible value of the largest number is 22.

answer：Alright, so I have this problem here about the function ( f(x) = sin(2x + phi) ). The problem states two conditions: first, that ( f(x) leq |f(frac{pi}{6})| ) for all real numbers ( x ), and second, that ( f(frac{pi}{2}) > f(pi) ). I need to find the interval where ( f(x) ) is monotonically increasing, and the options are given as A, B, C, D.Okay, let's break this down step by step. I remember that for a sine function of the form ( sin(Bx + C) ), the amplitude is 1, the period is ( frac{2pi}{B} ), and the phase shift is ( -frac{C}{B} ). In this case, ( B = 2 ), so the period is ( pi ). The phase shift is ( -frac{phi}{2} ).The first condition says ( f(x) leq |f(frac{pi}{6})| ) for all ( x ). That means the maximum value of ( f(x) ) is ( |f(frac{pi}{6})| ). Since the sine function normally has a maximum of 1 and a minimum of -1, this suggests that ( |f(frac{pi}{6})| ) must be equal to the amplitude of the function. But since the amplitude here is 1, ( |f(frac{pi}{6})| ) must be 1. So, ( f(frac{pi}{6}) ) is either 1 or -1.Let me write that down:( fleft(frac{pi}{6}right) = sinleft(2 cdot frac{pi}{6} + phiright) = sinleft(frac{pi}{3} + phiright) ).Since this equals either 1 or -1, we have:( sinleft(frac{pi}{3} + phiright) = pm 1 ).The sine function equals 1 at ( frac{pi}{2} + 2kpi ) and -1 at ( frac{3pi}{2} + 2kpi ) for any integer ( k ). So,( frac{pi}{3} + phi = frac{pi}{2} + 2kpi ) or ( frac{pi}{3} + phi = frac{3pi}{2} + 2kpi ).Solving for ( phi ):1. ( phi = frac{pi}{2} - frac{pi}{3} + 2kpi = frac{pi}{6} + 2kpi )2. ( phi = frac{3pi}{2} - frac{pi}{3} + 2kpi = frac{7pi}{6} + 2kpi )So, ( phi ) can be ( frac{pi}{6} + 2kpi ) or ( frac{7pi}{6} + 2kpi ).Now, moving on to the second condition: ( fleft(frac{pi}{2}right) > f(pi) ).Let's compute both ( fleft(frac{pi}{2}right) ) and ( f(pi) ):1. ( fleft(frac{pi}{2}right) = sinleft(2 cdot frac{pi}{2} + phiright) = sin(pi + phi) )2. ( f(pi) = sin(2pi + phi) = sin(phi) )So, the condition becomes:( sin(pi + phi) > sin(phi) )I know that ( sin(pi + phi) = -sin(phi) ), so substituting that in:( -sin(phi) > sin(phi) )Adding ( sin(phi) ) to both sides:( 0 > 2sin(phi) )Dividing both sides by 2:( 0 > sin(phi) )So, ( sin(phi) < 0 ). That means ( phi ) must be in a range where the sine is negative, which is between ( pi ) and ( 2pi ) (plus any multiple of ( 2pi )).Looking back at the possible values of ( phi ):1. ( phi = frac{pi}{6} + 2kpi ): ( sin(frac{pi}{6}) = frac{1}{2} > 0 ), so this doesn't satisfy ( sin(phi) < 0 ).2. ( phi = frac{7pi}{6} + 2kpi ): ( sin(frac{7pi}{6}) = -frac{1}{2} < 0 ), so this does satisfy the condition.Therefore, ( phi = frac{7pi}{6} + 2kpi ). But since the sine function is periodic with period ( 2pi ), we can just take ( phi = frac{7pi}{6} ) for simplicity.So, the function is ( f(x) = sin(2x + frac{7pi}{6}) ).Now, I need to find the intervals where this function is monotonically increasing. For a sine function ( sin(Bx + C) ), the function is increasing where its derivative is positive.Let's compute the derivative:( f'(x) = 2cos(2x + frac{7pi}{6}) )We need to find where ( f'(x) > 0 ):( 2cos(2x + frac{7pi}{6}) > 0 )Divide both sides by 2:( cos(2x + frac{7pi}{6}) > 0 )The cosine function is positive in the intervals ( (-frac{pi}{2} + 2kpi, frac{pi}{2} + 2kpi) ) for any integer ( k ).So, we have:( -frac{pi}{2} + 2kpi < 2x + frac{7pi}{6} < frac{pi}{2} + 2kpi )Let's solve for ( x ):Subtract ( frac{7pi}{6} ) from all parts:( -frac{pi}{2} - frac{7pi}{6} + 2kpi < 2x < frac{pi}{2} - frac{7pi}{6} + 2kpi )Convert ( frac{pi}{2} ) to sixths to combine:( -frac{3pi}{6} - frac{7pi}{6} + 2kpi < 2x < frac{3pi}{6} - frac{7pi}{6} + 2kpi )Combine the terms:( -frac{10pi}{6} + 2kpi < 2x < -frac{4pi}{6} + 2kpi )Simplify fractions:( -frac{5pi}{3} + 2kpi < 2x < -frac{2pi}{3} + 2kpi )Divide all parts by 2:( -frac{5pi}{6} + kpi < x < -frac{pi}{3} + kpi )Hmm, let me check my steps here. When I subtract ( frac{7pi}{6} ) from ( -frac{pi}{2} ), which is ( -frac{3pi}{6} ), I get ( -frac{3pi}{6} - frac{7pi}{6} = -frac{10pi}{6} = -frac{5pi}{3} ). Similarly, ( frac{pi}{2} - frac{7pi}{6} = frac{3pi}{6} - frac{7pi}{6} = -frac{4pi}{6} = -frac{2pi}{3} ). So, that seems correct.So, the intervals where ( f(x) ) is increasing are ( (-frac{5pi}{6} + kpi, -frac{pi}{3} + kpi) ) for any integer ( k ).But looking at the answer choices, they are all expressed in terms of ( kpi + ) something. Let me adjust the interval to match that form.Let me rewrite the interval:( x in (kpi - frac{5pi}{6}, kpi - frac{pi}{3}) )But none of the options exactly match this. Let me see the options again:A: ( [kpi - frac{pi}{3}, kpi + frac{pi}{6}] )B: ( [kpi, kpi + frac{pi}{2}] )C: ( [kpi + frac{pi}{6}, kpi + frac{2pi}{3}] )D: ( [kpi - frac{pi}{2}, kpi] )Hmm, none of these match ( (kpi - frac{5pi}{6}, kpi - frac{pi}{3}) ). Maybe I made a mistake in my calculations.Wait, let's go back. The derivative is positive when ( cos(2x + frac{7pi}{6}) > 0 ). So, ( 2x + frac{7pi}{6} ) is in the first or fourth quadrants, meaning between ( -frac{pi}{2} + 2kpi ) and ( frac{pi}{2} + 2kpi ).So, solving:( -frac{pi}{2} + 2kpi < 2x + frac{7pi}{6} < frac{pi}{2} + 2kpi )Subtract ( frac{7pi}{6} ):( -frac{pi}{2} - frac{7pi}{6} + 2kpi < 2x < frac{pi}{2} - frac{7pi}{6} + 2kpi )Convert ( frac{pi}{2} ) to sixths:( -frac{3pi}{6} - frac{7pi}{6} + 2kpi < 2x < frac{3pi}{6} - frac{7pi}{6} + 2kpi )Combine:( -frac{10pi}{6} + 2kpi < 2x < -frac{4pi}{6} + 2kpi )Simplify:( -frac{5pi}{3} + 2kpi < 2x < -frac{2pi}{3} + 2kpi )Divide by 2:( -frac{5pi}{6} + kpi < x < -frac{pi}{3} + kpi )So, same result. Maybe I need to adjust the interval by adding ( pi ) to the lower bound and upper bound to express it differently.Let me set ( k' = k - 1 ). Then,( -frac{5pi}{6} + kpi = -frac{5pi}{6} + (k' + 1)pi = -frac{5pi}{6} + k'pi + pi = k'pi + frac{pi}{6} )Similarly,( -frac{pi}{3} + kpi = -frac{pi}{3} + (k' + 1)pi = -frac{pi}{3} + k'pi + pi = k'pi + frac{2pi}{3} )So, substituting ( k' ) as ( k ), the interval becomes:( kpi + frac{pi}{6} < x < kpi + frac{2pi}{3} )Which is exactly option C: ( [kpi + frac{pi}{6}, kpi + frac{2pi}{3}] ).Wait, but in my previous step, I had open intervals, but the answer choices have closed intervals. Since the function is increasing on those intervals, and the endpoints are where the derivative is zero, which are points where the function changes from increasing to decreasing or vice versa. So, technically, at those endpoints, the function is neither increasing nor decreasing. However, sometimes in such problems, they include the endpoints as part of the interval where the function is increasing, considering the maximum and minimum points.But in reality, the function is increasing just before the maximum and just after the minimum. So, the interval where it's increasing is between the minimum and maximum points. Since the derivative is zero at those points, the function is not increasing or decreasing exactly at those points. So, perhaps the correct interval should be open intervals, but since the answer choices are closed, maybe they include the endpoints for simplicity.Alternatively, maybe I made a mistake in the phase shift. Let me double-check.Given ( phi = frac{7pi}{6} ), so the function is ( sin(2x + frac{7pi}{6}) ). Let's consider the standard sine function ( sin(2x) ), which has maxima at ( 2x = frac{pi}{2} + 2kpi ), so ( x = frac{pi}{4} + kpi ), and minima at ( 2x = frac{3pi}{2} + 2kpi ), so ( x = frac{3pi}{4} + kpi ).But with the phase shift ( frac{7pi}{6} ), the function is shifted to the left by ( frac{7pi}{12} ). So, the maxima and minima will be shifted accordingly.Wait, maybe another approach is better. Let's find the critical points where the derivative is zero.Set ( f'(x) = 0 ):( 2cos(2x + frac{7pi}{6}) = 0 )So,( cos(2x + frac{7pi}{6}) = 0 )Which occurs when:( 2x + frac{7pi}{6} = frac{pi}{2} + kpi )Solving for ( x ):( 2x = frac{pi}{2} - frac{7pi}{6} + kpi )Convert ( frac{pi}{2} ) to sixths:( 2x = frac{3pi}{6} - frac{7pi}{6} + kpi = -frac{4pi}{6} + kpi = -frac{2pi}{3} + kpi )So,( x = -frac{pi}{3} + frac{kpi}{2} )These are the critical points. Now, to determine where the function is increasing, we can test intervals between these critical points.Let me list some critical points for ( k = 0, 1, 2 ):For ( k = 0 ): ( x = -frac{pi}{3} )For ( k = 1 ): ( x = -frac{pi}{3} + frac{pi}{2} = frac{pi}{6} )For ( k = 2 ): ( x = -frac{pi}{3} + pi = frac{2pi}{3} )For ( k = 3 ): ( x = -frac{pi}{3} + frac{3pi}{2} = frac{7pi}{6} )Wait, but since the period is ( pi ), the critical points repeat every ( pi ). So, the critical points are at ( x = -frac{pi}{3} + frac{kpi}{2} ), but considering the period, it's better to express them in terms of ( k pi ).Alternatively, perhaps it's better to analyze the intervals between two consecutive critical points.Let me consider the interval between ( x = -frac{pi}{3} + kpi ) and ( x = frac{pi}{6} + kpi ). Wait, no, the critical points are at ( x = -frac{pi}{3} + frac{kpi}{2} ), which are spaced by ( frac{pi}{2} ). Hmm, this might complicate things.Alternatively, let's consider the function ( f(x) = sin(2x + frac{7pi}{6}) ). Let's rewrite it as ( sin(2(x + frac{7pi}{12})) ). So, it's a sine function with amplitude 1, period ( pi ), and phase shift ( -frac{7pi}{12} ).To find where it's increasing, we can consider the standard sine function's increasing intervals and adjust for the phase shift.The standard ( sin(2x) ) is increasing on intervals where ( 2x ) is between ( -frac{pi}{2} + 2kpi ) and ( frac{pi}{2} + 2kpi ), which translates to ( x ) between ( -frac{pi}{4} + kpi ) and ( frac{pi}{4} + kpi ).But with the phase shift, the increasing intervals will be shifted by ( -frac{7pi}{12} ).So, the increasing intervals for ( sin(2x + frac{7pi}{6}) ) would be:( 2x + frac{7pi}{6} in (-frac{pi}{2} + 2kpi, frac{pi}{2} + 2kpi) )Which is the same as:( 2x in (-frac{pi}{2} - frac{7pi}{6} + 2kpi, frac{pi}{2} - frac{7pi}{6} + 2kpi) )Simplify:( 2x in (-frac{10pi}{6} + 2kpi, -frac{4pi}{6} + 2kpi) )Which is:( 2x in (-frac{5pi}{3} + 2kpi, -frac{2pi}{3} + 2kpi) )Divide by 2:( x in (-frac{5pi}{6} + kpi, -frac{pi}{3} + kpi) )Which is the same as before. So, the function is increasing on ( (kpi - frac{5pi}{6}, kpi - frac{pi}{3}) ).But the answer choices are in terms of ( kpi + ) something. Let me adjust the interval by adding ( pi ) to the lower bound and upper bound.Let me set ( k' = k - 1 ). Then,Lower bound: ( kpi - frac{5pi}{6} = (k' + 1)pi - frac{5pi}{6} = k'pi + pi - frac{5pi}{6} = k'pi + frac{pi}{6} )Upper bound: ( kpi - frac{pi}{3} = (k' + 1)pi - frac{pi}{3} = k'pi + pi - frac{pi}{3} = k'pi + frac{2pi}{3} )So, the interval becomes ( (k'pi + frac{pi}{6}, k'pi + frac{2pi}{3}) ). Since ( k' ) is just an integer, we can rename it back to ( k ), giving ( (kpi + frac{pi}{6}, kpi + frac{2pi}{3}) ).Since the answer choices include closed intervals, the correct interval is ( [kpi + frac{pi}{6}, kpi + frac{2pi}{3}] ), which is option C.I think that makes sense now. Initially, I was confused because the interval didn't match directly, but by adjusting the integer ( k ), I could express the interval in the form given in the options. So, the correct answer should be option C.

answer：Okay, so I have this problem here: Given the function ( y = frac{1}{3}x^3 + b x^2 + (b + 6)x + 3 ), it's said to have three monotonic intervals on ( mathbb{R} ). I need to determine the range of ( b ). The options are A, B, C, D, which are different ranges for ( b ).Alright, let me think about what this means. Monotonic intervals refer to where the function is increasing or decreasing. If a function has three monotonic intervals, that means it changes its increasing/decreasing behavior twice. So, the function must have two critical points where the derivative is zero. These critical points are where the function changes from increasing to decreasing or vice versa.So, to find the critical points, I need to take the derivative of the function ( y ). Let me do that step by step.The function is ( y = frac{1}{3}x^3 + b x^2 + (b + 6)x + 3 ). Taking the derivative with respect to ( x ), I get:( y' = frac{d}{dx} left( frac{1}{3}x^3 right) + frac{d}{dx} left( b x^2 right) + frac{d}{dx} left( (b + 6)x right) + frac{d}{dx}(3) )Calculating each term:- The derivative of ( frac{1}{3}x^3 ) is ( x^2 ).- The derivative of ( b x^2 ) is ( 2b x ).- The derivative of ( (b + 6)x ) is ( b + 6 ).- The derivative of the constant 3 is 0.So, putting it all together, the derivative ( y' ) is:( y' = x^2 + 2b x + (b + 6) )Alright, so now I have the derivative as a quadratic function: ( y' = x^2 + 2b x + (b + 6) ).For the original function to have three monotonic intervals, the derivative must have two distinct real roots. That's because each root of the derivative corresponds to a critical point where the function changes its monotonicity. So, if there are two distinct critical points, the function will have three intervals: increasing, decreasing, and then increasing again, or vice versa.Therefore, the quadratic equation ( x^2 + 2b x + (b + 6) = 0 ) must have two distinct real roots. For a quadratic equation ( ax^2 + bx + c = 0 ), the discriminant ( D ) is given by ( D = b^2 - 4ac ). If ( D > 0 ), the quadratic has two distinct real roots.In our case, the quadratic is ( x^2 + 2b x + (b + 6) ). So, ( a = 1 ), ( b = 2b ), and ( c = b + 6 ). Wait, hold on, that might be confusing because the coefficient ( b ) in the quadratic is actually ( 2b ) from the original function. Let me clarify:Quadratic equation: ( x^2 + (2b)x + (b + 6) = 0 )So, in standard form ( ax^2 + bx + c = 0 ), here:- ( a = 1 )- ( b_{text{quad}} = 2b ) (this is the coefficient of ( x ))- ( c = b + 6 )Therefore, the discriminant ( D ) is:( D = (2b)^2 - 4 times 1 times (b + 6) )Let me compute that:( D = 4b^2 - 4(b + 6) )Simplify:( D = 4b^2 - 4b - 24 )We need this discriminant to be greater than zero for two distinct real roots:( 4b^2 - 4b - 24 > 0 )Let me factor out a 4 to simplify the inequality:( 4(b^2 - b - 6) > 0 )Divide both sides by 4 (since 4 is positive, the inequality sign doesn't change):( b^2 - b - 6 > 0 )Now, let's solve the quadratic inequality ( b^2 - b - 6 > 0 ).First, find the roots of the quadratic equation ( b^2 - b - 6 = 0 ).Using the quadratic formula:( b = frac{1 pm sqrt{1 + 24}}{2} = frac{1 pm sqrt{25}}{2} = frac{1 pm 5}{2} )So, the roots are:( b = frac{1 + 5}{2} = 3 )and( b = frac{1 - 5}{2} = -2 )So, the quadratic ( b^2 - b - 6 ) factors as ( (b - 3)(b + 2) ).Now, to solve the inequality ( (b - 3)(b + 2) > 0 ), we can analyze the sign of each factor in different intervals.The critical points are at ( b = -2 ) and ( b = 3 ). These divide the real number line into three intervals:1. ( b < -2 )2. ( -2 < b < 3 )3. ( b > 3 )We can test each interval:1. For ( b < -2 ), let's pick ( b = -3 ): ( (-3 - 3)(-3 + 2) = (-6)(-1) = 6 > 0 ). So, the inequality holds.2. For ( -2 < b < 3 ), let's pick ( b = 0 ): ( (0 - 3)(0 + 2) = (-3)(2) = -6 < 0 ). So, the inequality does not hold.3. For ( b > 3 ), let's pick ( b = 4 ): ( (4 - 3)(4 + 2) = (1)(6) = 6 > 0 ). So, the inequality holds.Therefore, the solution to ( (b - 3)(b + 2) > 0 ) is ( b < -2 ) or ( b > 3 ).So, going back to our original problem, this means that for the function ( y ) to have three monotonic intervals, the parameter ( b ) must satisfy ( b < -2 ) or ( b > 3 ).Looking at the options provided:A: ( b leqslant -2 ) or ( b geqslant 3 )B: ( -2 leqslant b leqslant 3 )C: ( -2 < b < 3 )D: ( b < -2 ) or ( b > 3 )Our solution is ( b < -2 ) or ( b > 3 ), which corresponds to option D.Wait, just to make sure, let me double-check. If ( b = -2 ) or ( b = 3 ), what happens? At these points, the discriminant ( D = 0 ), meaning the quadratic derivative has exactly one real root (a repeated root). So, the function would have only one critical point, meaning it would have two monotonic intervals instead of three. Therefore, ( b ) cannot be equal to -2 or 3; it has to be strictly less than -2 or strictly greater than 3. So, the correct range is ( b < -2 ) or ( b > 3 ), which is option D.Yes, that makes sense. So, I think D is the correct answer.**Final Answer**boxed{D}