answer：Okay, so I have this equation ( x|x| + px + q = 0 ) and I need to figure out which of the given conclusions is incorrect. The options are (A) through (D). Let me try to break this down step by step.First, I remember that the absolute value function ( |x| ) behaves differently depending on whether ( x ) is positive or negative. So, maybe I should split the equation into two cases: one where ( x geq 0 ) and another where ( x < 0 ).For ( x geq 0 ), ( |x| = x ), so the equation becomes:[ x^2 + px + q = 0 ]Let me call this equation (1).For ( x < 0 ), ( |x| = -x ), so the equation becomes:[ x^2 - px - q = 0 ]Let me call this equation (2).Now, I can analyze each case separately.Starting with equation (1): ( x^2 + px + q = 0 ). This is a quadratic equation, and the number of real roots it has depends on its discriminant. The discriminant ( D_1 ) is:[ D_1 = p^2 - 4q ]If ( D_1 geq 0 ), equation (1) has real roots. If ( D_1 < 0 ), it doesn't.Similarly, for equation (2): ( x^2 - px - q = 0 ). The discriminant ( D_2 ) is:[ D_2 = p^2 + 4q ]Since ( p^2 ) is always non-negative and ( 4q ) could be positive or negative, ( D_2 ) is always greater than or equal to ( p^2 - 4|q| ). But since ( D_2 = p^2 + 4q ), if ( q ) is positive, ( D_2 ) is definitely positive. If ( q ) is negative, ( D_2 ) could still be positive or negative depending on the value of ( p ).Wait, actually, ( D_2 = p^2 + 4q ). So, if ( q ) is positive, ( D_2 ) is definitely positive because both ( p^2 ) and ( 4q ) are non-negative. If ( q ) is negative, ( D_2 ) could be positive or negative. For example, if ( p = 0 ) and ( q = -1 ), then ( D_2 = 0 + 4(-1) = -4 ), which is negative. But if ( p = 3 ) and ( q = -1 ), then ( D_2 = 9 - 4 = 5 ), which is positive.So, equation (2) can have real roots depending on the values of ( p ) and ( q ).Now, let's go back to the original problem. The equation ( x|x| + px + q = 0 ) can have roots from both equation (1) and equation (2). So, the total number of real roots is the sum of the real roots from both equations, considering the domain for each case.Let me analyze each conclusion one by one.**Conclusion (A): There are at most three real roots.**Hmm, let's see. Equation (1) can have at most two real roots (if ( D_1 geq 0 )), and equation (2) can also have at most two real roots (if ( D_2 geq 0 )). However, for equation (1), the roots must satisfy ( x geq 0 ), and for equation (2), the roots must satisfy ( x < 0 ). So, in the best case, equation (1) can have two positive roots, and equation (2) can have two negative roots, but wait, equation (2) is ( x^2 - px - q = 0 ). The roots of equation (2) could be both positive or both negative or one positive and one negative.Wait, actually, for equation (2): ( x^2 - px - q = 0 ). The product of the roots is ( -q ). So, if ( q > 0 ), the product is negative, meaning one root is positive and one is negative. But since we're considering ( x < 0 ) for equation (2), only the negative root would be valid. So, equation (2) can contribute at most one real root when ( q > 0 ).Similarly, if ( q < 0 ), the product of the roots is positive, so both roots have the same sign. Since equation (2) is for ( x < 0 ), both roots would be negative, so equation (2) can contribute two real roots.Similarly, for equation (1): if ( q > 0 ), the product of the roots is ( q ), which is positive, so both roots are either positive or both negative. But since equation (1) is for ( x geq 0 ), only the positive roots are valid. So, if ( D_1 geq 0 ), equation (1) can contribute two real roots.Wait, let me think again. For equation (1): ( x^2 + px + q = 0 ). The product of the roots is ( q ). So, if ( q > 0 ), both roots have the same sign. Since equation (1) is for ( x geq 0 ), both roots must be non-negative. So, if ( D_1 geq 0 ), equation (1) can have two real roots, both non-negative.If ( q < 0 ), the product of the roots is negative, so one root is positive and one is negative. But since equation (1) is for ( x geq 0 ), only the positive root is valid, so equation (1) can contribute one real root.Similarly, for equation (2): ( x^2 - px - q = 0 ). The product of the roots is ( -q ). So, if ( q > 0 ), the product is negative, meaning one root is positive and one is negative. Since equation (2) is for ( x < 0 ), only the negative root is valid, so equation (2) contributes one real root.If ( q < 0 ), the product is positive, so both roots have the same sign. Since equation (2) is for ( x < 0 ), both roots must be negative, so equation (2) can contribute two real roots.So, putting it all together:- If ( q > 0 ): - Equation (1): Two real roots (both non-negative) if ( D_1 geq 0 ). - Equation (2): One real root (negative). - Total: Up to three real roots.- If ( q < 0 ): - Equation (1): One real root (positive) if ( D_1 geq 0 ). - Equation (2): Two real roots (both negative). - Total: Up to three real roots.- If ( q = 0 ): - Equation (1): ( x^2 + px = 0 ) → roots at ( x = 0 ) and ( x = -p ). But since ( x geq 0 ), only ( x = 0 ) is valid if ( p neq 0 ). - Equation (2): ( x^2 - px = 0 ) → roots at ( x = 0 ) and ( x = p ). But since ( x < 0 ), only ( x = p ) is valid if ( p < 0 ). - So, depending on ( p ), we can have up to two real roots.Wait, but if ( q = 0 ), equation (1) can have ( x = 0 ) and ( x = -p ). But ( x = -p ) must be non-negative, so ( -p geq 0 ) → ( p leq 0 ). Similarly, equation (2) can have ( x = 0 ) and ( x = p ). But ( x = p ) must be negative, so ( p < 0 ).So, if ( p < 0 ), equation (1) has ( x = 0 ) and ( x = -p ) (which is positive), and equation (2) has ( x = p ) (negative). So, total three real roots: ( 0, -p, p ). But ( p < 0 ), so ( -p > 0 ). So, three real roots.If ( p = 0 ), equation (1): ( x^2 = 0 ) → ( x = 0 ). Equation (2): ( x^2 = 0 ) → ( x = 0 ). But since equation (2) is for ( x < 0 ), ( x = 0 ) is not valid. So, only one real root at ( x = 0 ).So, in the case ( q = 0 ), depending on ( p ), we can have one or three real roots.Therefore, in all cases, the equation can have at most three real roots. So, conclusion (A) is correct.**Conclusion (B): There is at least one real root.**Hmm, is this always true? Let's see.If ( q > 0 ):- Equation (2) always has one real root because ( D_2 = p^2 + 4q > 0 ), so equation (2) has two real roots, but only one is negative.- So, at least one real root.If ( q < 0 ):- Equation (2) can have two real roots if ( D_2 geq 0 ), which is ( p^2 + 4q geq 0 ). But since ( q < 0 ), ( p^2 + 4q ) could be positive or negative.- If ( p^2 + 4q geq 0 ), equation (2) has two real roots, both negative, so two real roots.- If ( p^2 + 4q < 0 ), equation (2) has no real roots.- But equation (1): ( x^2 + px + q = 0 ). The discriminant is ( D_1 = p^2 - 4q ). Since ( q < 0 ), ( -4q > 0 ), so ( D_1 = p^2 - 4q > p^2 geq 0 ). So, equation (1) always has two real roots, both non-negative because the product is ( q < 0 ), so one positive and one negative. But since equation (1) is for ( x geq 0 ), only the positive root is valid. So, equation (1) contributes one real root.Therefore, even if equation (2) has no real roots, equation (1) still has one real root. So, overall, there is always at least one real root.If ( q = 0 ):- As discussed earlier, depending on ( p ), we can have one or three real roots, but at least one.Therefore, conclusion (B) is correct.**Conclusion (C): There are real roots only if ( p^2 - 4q geq 0 ).**Hmm, this is saying that for the equation to have real roots, the discriminant of equation (1) must be non-negative. But wait, equation (2) also contributes real roots. So, even if ( p^2 - 4q < 0 ), equation (2) might still have real roots.Let me test with an example. Let me choose ( p = 1 ) and ( q = 1 ). Then, equation (1): ( x^2 + x + 1 = 0 ). Discriminant ( D_1 = 1 - 4 = -3 < 0 ), so no real roots. Equation (2): ( x^2 - x - 1 = 0 ). Discriminant ( D_2 = 1 + 4 = 5 > 0 ), so two real roots. But since equation (2) is for ( x < 0 ), only the negative root is valid. So, the equation ( x|x| + x + 1 = 0 ) has one real root. So, even though ( p^2 - 4q < 0 ), the equation still has a real root. Therefore, conclusion (C) is incorrect because it's not necessary for ( p^2 - 4q geq 0 ) to have real roots.Wait, but conclusion (C) says "only if", which means that having real roots implies ( p^2 - 4q geq 0 ). But in my example, we have real roots even when ( p^2 - 4q < 0 ). So, conclusion (C) is incorrect.But let me double-check. Maybe I misinterpreted the conclusion. It says "There are real roots only if ( p^2 - 4q geq 0 )", which is logically equivalent to "If there are real roots, then ( p^2 - 4q geq 0 )". But in my example, there are real roots, but ( p^2 - 4q < 0 ). So, the conclusion is incorrect because the implication is false.Therefore, conclusion (C) is incorrect.**Conclusion (D): When ( p < 0 ) and ( q > 0 ), there are three real roots.**Let me test this with an example. Let me choose ( p = -1 ) and ( q = 1 ). Then, equation (1): ( x^2 - x + 1 = 0 ). Discriminant ( D_1 = 1 - 4 = -3 < 0 ), so no real roots. Equation (2): ( x^2 + x - 1 = 0 ). Discriminant ( D_2 = 1 + 4 = 5 > 0 ), so two real roots. Since equation (2) is for ( x < 0 ), only the negative root is valid. So, the equation ( x|x| - x + 1 = 0 ) has one real root. Therefore, when ( p < 0 ) and ( q > 0 ), there is only one real root, not three. So, conclusion (D) is incorrect.Wait, but maybe I should try another example where ( p < 0 ) and ( q > 0 ) and see if it can have three real roots.Let me choose ( p = -3 ) and ( q = 2 ). Then, equation (1): ( x^2 - 3x + 2 = 0 ). Factoring: ( (x - 1)(x - 2) = 0 ), so roots at ( x = 1 ) and ( x = 2 ). Both are non-negative, so equation (1) contributes two real roots.Equation (2): ( x^2 + 3x - 2 = 0 ). Discriminant ( D_2 = 9 + 8 = 17 > 0 ), so two real roots. The roots are ( x = [-3 pm sqrt{17}]/2 ). Since ( sqrt{17} approx 4.123 ), the roots are approximately ( (-3 + 4.123)/2 approx 0.5615 ) and ( (-3 - 4.123)/2 approx -3.5615 ). Since equation (2) is for ( x < 0 ), only the negative root is valid. So, equation (2) contributes one real root.Therefore, total real roots: two from equation (1) and one from equation (2), totaling three real roots.Wait, so in this case, when ( p < 0 ) and ( q > 0 ), there are three real roots. But in my previous example, when ( p = -1 ) and ( q = 1 ), there was only one real root. So, conclusion (D) says "When ( p < 0 ) and ( q > 0 ), there are three real roots." But in my first example, it was only one real root, so conclusion (D) is not always true. Therefore, conclusion (D) is incorrect.Wait, but in the second example, with ( p = -3 ) and ( q = 2 ), there were three real roots. So, does it depend on the specific values of ( p ) and ( q )?Let me analyze this more carefully. When ( p < 0 ) and ( q > 0 ):- Equation (1): ( x^2 + px + q = 0 ). Since ( p < 0 ), let me denote ( p = -k ) where ( k > 0 ). So, equation (1) becomes ( x^2 - kx + q = 0 ). The discriminant is ( D_1 = k^2 - 4q ). So, if ( k^2 - 4q geq 0 ), equation (1) has two real roots, both non-negative because the product is ( q > 0 ) and the sum is ( k > 0 ). So, both roots are positive.- Equation (2): ( x^2 - px - q = 0 ). Substituting ( p = -k ), it becomes ( x^2 + kx - q = 0 ). The discriminant is ( D_2 = k^2 + 4q > 0 ). So, equation (2) has two real roots. The product of the roots is ( -q < 0 ), so one root is positive and one is negative. Since equation (2) is for ( x < 0 ), only the negative root is valid.Therefore, if ( k^2 - 4q geq 0 ), equation (1) contributes two real roots, and equation (2) contributes one real root, totaling three real roots.If ( k^2 - 4q < 0 ), equation (1) has no real roots, but equation (2) still has one real root (the negative one). So, in this case, only one real root.Therefore, conclusion (D) is not always true. It depends on whether ( p^2 - 4q geq 0 ). If ( p^2 - 4q geq 0 ), then there are three real roots; otherwise, only one. So, conclusion (D) is incorrect because it claims that when ( p < 0 ) and ( q > 0 ), there are three real roots, but this is only true if ( p^2 - 4q geq 0 ). If ( p^2 - 4q < 0 ), there is only one real root.Therefore, conclusion (D) is incorrect.Wait, but in my first example with ( p = -1 ) and ( q = 1 ), ( p^2 - 4q = 1 - 4 = -3 < 0 ), so equation (1) has no real roots, and equation (2) has one real root. So, only one real root.In the second example, ( p = -3 ) and ( q = 2 ), ( p^2 - 4q = 9 - 8 = 1 geq 0 ), so equation (1) has two real roots, and equation (2) has one real root, totaling three real roots.Therefore, conclusion (D) is incorrect because it doesn't hold when ( p^2 - 4q < 0 ).So, to summarize:- Conclusion (A): Correct.- Conclusion (B): Correct.- Conclusion (C): Incorrect.- Conclusion (D): Incorrect.Therefore, the incorrect conclusions are (C) and (D).

answer：Alright, so I have this problem about a geometric sequence. Let me try to understand what it's asking.First, it says the common ratio of the geometric sequence {a_n} is 2. Okay, so that means each term is twice the previous one. So, if the first term is a_1, then a_2 = 2*a_1, a_3 = 2*a_2 = 4*a_1, and so on. Got it.Next, it says that a_2, a_3 + 2, a_4 form an arithmetic sequence. Hmm, arithmetic sequence means that the difference between consecutive terms is constant. So, if I have three terms in an arithmetic sequence, the middle term is the average of the first and the third.So, for a_2, a_3 + 2, a_4 to be an arithmetic sequence, the following should hold:2*(a_3 + 2) = a_2 + a_4Let me write that down:2*(a_3 + 2) = a_2 + a_4Since it's a geometric sequence with ratio 2, I can express a_2, a_3, and a_4 in terms of a_1.a_2 = 2*a_1a_3 = 4*a_1a_4 = 8*a_1Substituting these into the equation:2*(4*a_1 + 2) = 2*a_1 + 8*a_1Let me compute the left side:2*(4*a_1 + 2) = 8*a_1 + 4And the right side:2*a_1 + 8*a_1 = 10*a_1So now we have:8*a_1 + 4 = 10*a_1Subtracting 8*a_1 from both sides:4 = 2*a_1Dividing both sides by 2:a_1 = 2Okay, so the first term is 2. Therefore, the general formula for the geometric sequence is:a_n = a_1 * 2^{n-1} = 2 * 2^{n-1} = 2^nSo, part (1) is solved, and the general formula is a_n = 2^n.Now, moving on to part (2). It says that b_n = log_2(a_n) + a_n, and we need to find the sum of the first n terms of {b_n}, denoted as T_n.First, let's express b_n in terms of n. Since a_n = 2^n, then log_2(a_n) = log_2(2^n) = n.So, b_n = n + 2^nTherefore, the sequence {b_n} is just the sum of two sequences: one is the sequence of natural numbers (1, 2, 3, ..., n) and the other is the geometric sequence (2^1, 2^2, 2^3, ..., 2^n).To find T_n, the sum of the first n terms of {b_n}, we can split it into two separate sums:T_n = sum_{k=1}^n b_k = sum_{k=1}^n (k + 2^k) = sum_{k=1}^n k + sum_{k=1}^n 2^kSo, we have two sums to compute: the sum of the first n natural numbers and the sum of the first n terms of a geometric series with ratio 2.Starting with the sum of the first n natural numbers:sum_{k=1}^n k = n(n + 1)/2That's a well-known formula.Now, for the geometric series sum_{k=1}^n 2^k. The formula for the sum of a geometric series is:sum_{k=0}^{n} ar^k = a*(r^{n+1} - 1)/(r - 1)But in our case, the series starts at k=1, not k=0, and the first term is 2^1 = 2. So, let's adjust the formula accordingly.sum_{k=1}^n 2^k = 2*(2^n - 1)/(2 - 1) = 2*(2^n - 1)/1 = 2^{n+1} - 2Wait, let me verify that:sum_{k=1}^n 2^k = 2 + 4 + 8 + ... + 2^nThis is a geometric series with first term a = 2, common ratio r = 2, and number of terms n.The sum is a*(r^n - 1)/(r - 1) = 2*(2^n - 1)/(2 - 1) = 2*(2^n - 1) = 2^{n+1} - 2Yes, that's correct.So, putting it all together:T_n = sum_{k=1}^n k + sum_{k=1}^n 2^k = n(n + 1)/2 + (2^{n+1} - 2)Therefore, the sum of the first n terms of {b_n} is:T_n = (n(n + 1))/2 + 2^{n+1} - 2I think that's the final answer for part (2).Let me just recap:1. For part (1), we used the property of arithmetic sequences to set up an equation involving terms of the geometric sequence, solved for the first term, and found the general formula.2. For part (2), we expressed b_n in terms of n, recognized it as the sum of two sequences, computed each sum separately using known formulas, and combined them to get T_n.Everything seems to check out. I don't see any mistakes in the calculations.**Final Answer**(1) The general formula is boxed{a_n = 2^n}.(2) The sum of the first n terms is boxed{T_n = frac{n(n + 1)}{2} + 2^{n+1} - 2}.

answer：Okay, so I have this problem here: "The sum of the interior angles of a polygon is 4 times the sum of its exterior angles. Find the number of sides of this polygon." Hmm, let's see. I remember that polygons have certain properties when it comes to their angles, but I need to recall the exact formulas.First, I think the sum of the interior angles of a polygon depends on the number of sides. I believe the formula is something like (n-2) multiplied by 180 degrees, where n is the number of sides. Yeah, that sounds right. So, for a triangle, which has 3 sides, the sum of interior angles is (3-2)*180 = 180 degrees, which makes sense. For a quadrilateral, it's (4-2)*180 = 360 degrees, which also checks out. Okay, so I'm confident about that formula.Now, what about the sum of the exterior angles? I think it's a constant value, regardless of the number of sides. I recall that the sum of the exterior angles of any polygon is 360 degrees. Wait, is that right? Let me think. If I imagine walking around the polygon, each exterior angle is the angle I turn at each vertex. Since I make a full circle when I go all the way around, that should be 360 degrees. Yeah, that makes sense. So, the sum of exterior angles is always 360 degrees, no matter how many sides the polygon has.Alright, so the problem says that the sum of the interior angles is 4 times the sum of the exterior angles. Let me write that down as an equation. Let n be the number of sides. Then, the sum of interior angles is (n-2)*180, and the sum of exterior angles is 360. According to the problem, (n-2)*180 = 4*360.Let me compute 4 times 360 first. 4*360 is 1440. So, the equation becomes (n-2)*180 = 1440. Now, I need to solve for n. Let's divide both sides by 180 to simplify. So, (n-2) = 1440 / 180. Let's calculate that. 1440 divided by 180. Hmm, 180 times 8 is 1440, right? Because 180*8 = 1440. So, (n-2) = 8.Now, to find n, I just add 2 to both sides. So, n = 8 + 2 = 10. Therefore, the polygon has 10 sides. Let me just double-check my calculations to make sure I didn't make a mistake. Starting with the formula: (n-2)*180 = 4*360. 4*360 is indeed 1440. Dividing 1440 by 180 gives 8, so n-2=8, which means n=10. Yeah, that seems correct. I also remember that a decagon has 10 sides, so this must be a decagon. Just to be thorough, let me verify the sums. The sum of interior angles would be (10-2)*180 = 8*180 = 1440 degrees. The sum of exterior angles is 360 degrees. Is 1440 equal to 4 times 360? Yes, because 4*360 is 1440. So, the condition given in the problem is satisfied.I think that's solid. I don't see any mistakes in my reasoning. The key was remembering the formulas for the sums of interior and exterior angles and setting up the equation correctly. It's always good to double-check, especially since sometimes it's easy to mix up formulas or make arithmetic errors. But in this case, everything adds up.Another thing I could consider is whether there are any special cases or exceptions. For example, does this apply to convex polygons only? I think the formulas for the sum of interior and exterior angles generally apply to convex polygons. If the polygon is concave, some exterior angles might be negative or reflex angles, but I think the sum still holds as 360 degrees because it's based on the turning angle when traversing the polygon. So, even for concave polygons, the sum of exterior angles is 360 degrees. Therefore, the solution should still be valid regardless of whether the polygon is convex or concave.Also, just to make sure, let's think about a regular polygon, where all sides and angles are equal. In a regular decagon, each interior angle would be 1440/10 = 144 degrees, and each exterior angle would be 360/10 = 36 degrees. Checking if 144 is indeed 4 times 36: 4*36 is 144, so that works out. So, even in the regular case, the relationship holds. This gives me more confidence that n=10 is the correct answer. It's always nice when different approaches or checks confirm the same result. It reinforces that the solution is likely correct.In summary, by using the formulas for the sum of interior and exterior angles, setting up the equation based on the given condition, solving for n, and then verifying the result, I've arrived at the conclusion that the polygon has 10 sides. I don't see any issues with the logic or calculations, so I feel confident about this answer.

answer：To determine the percentage increase on the price the trader bought the car for, we'll follow these steps:1. **Define the Original Price:** Let the original price of the car be ( P ).2. **Calculate the Purchase Price:** The trader bought the car at a 20% discount, so he paid: [ P - 0.20P = 0.80P ]3. **Determine the Selling Price:** The trader made a profit of approximately 16% on the original price, so he sold it for: [ P + 0.16P = 1.16P ]4. **Find the Increase in Price:** The difference between the selling price and the purchase price is: [ 1.16P - 0.80P = 0.36P ]5. **Calculate the Percentage Increase:** To find the percentage increase relative to the purchase price: [ frac{0.36P}{0.80P} times 100 = frac{0.36}{0.80} times 100 = 0.45 times 100 = 45% ]Therefore, the trader sold the car at a 45% increase on the price he bought it for.

answer：First, I recognize that in triangle DEF, the measure of angle F is triple the measure of angle E, and angle E is given as 15 degrees.I calculate angle F by multiplying angle E by 3:3 × 15° = 45°.Next, I use the fact that the sum of all angles in a triangle is 180 degrees. I set up the equation:Angle D + Angle E + Angle F = 180°.Substituting the known values:Angle D + 15° + 45° = 180°.Combining the known angles:Angle D + 60° = 180°.Finally, I solve for Angle D:Angle D = 180° - 60° = 120°.

answer：Alright, so I have this function ( f(x) = frac{1}{x^2} + frac{1}{x^4} ), and I need to find the range of values for ( a ) such that ( f(a-2) < f(2a+1) ). Hmm, okay. Let me try to break this down step by step.First, I should probably understand what the function ( f(x) ) looks like. It's composed of two terms: ( frac{1}{x^2} ) and ( frac{1}{x^4} ). Both of these are even functions because replacing ( x ) with ( -x ) doesn't change their values. So, ( f(x) ) is also an even function. That means ( f(x) = f(-x) ).Now, since both ( frac{1}{x^2} ) and ( frac{1}{x^4} ) are positive for all ( x neq 0 ), the function ( f(x) ) is always positive in its domain. Also, as ( |x| ) increases, both ( frac{1}{x^2} ) and ( frac{1}{x^4} ) decrease. So, ( f(x) ) is a decreasing function for ( x > 0 ) and increasing for ( x < 0 ) because of the even nature.Wait, no, actually, since ( f(x) ) is even, its behavior on the positive side mirrors the negative side. So, for ( x > 0 ), as ( x ) increases, ( f(x) ) decreases, and for ( x < 0 ), as ( x ) becomes more negative (i.e., ( |x| ) increases), ( f(x) ) also decreases. So, ( f(x) ) is decreasing on ( (0, infty) ) and increasing on ( (-infty, 0) ).But when dealing with inequalities involving ( f(a-2) ) and ( f(2a+1) ), I need to consider the absolute values because ( f(x) ) depends on ( |x| ). So, ( f(a-2) < f(2a+1) ) implies that ( |a - 2| > |2a + 1| ) because ( f ) is decreasing for positive arguments. Wait, is that right?Let me think. If ( f ) is decreasing on ( (0, infty) ), then a larger input gives a smaller output. So, if ( f(a-2) < f(2a+1) ), then ( |a - 2| > |2a + 1| ). Yeah, that seems correct because if ( |a - 2| ) is larger, then ( f(|a - 2|) ) would be smaller than ( f(|2a + 1|) ) since ( f ) is decreasing.So, the inequality reduces to ( |a - 2| > |2a + 1| ). Now, I need to solve this absolute value inequality. Absolute value inequalities can be tricky, but I remember that ( |A| > |B| ) implies that either ( A > B ) or ( A < -B ). So, applying that here, I get two cases:1. ( a - 2 > 2a + 1 )2. ( a - 2 < -(2a + 1) )Let me solve each case separately.**Case 1:** ( a - 2 > 2a + 1 )Subtract ( a ) from both sides:( -2 > a + 1 )Subtract 1 from both sides:( -3 > a )Which can be rewritten as:( a < -3 )**Case 2:** ( a - 2 < -(2a + 1) )First, distribute the negative sign on the right side:( a - 2 < -2a - 1 )Add ( 2a ) to both sides:( 3a - 2 < -1 )Add 2 to both sides:( 3a < 1 )Divide both sides by 3:( a < frac{1}{3} )So, from the two cases, I have ( a < -3 ) or ( a < frac{1}{3} ). Hmm, but wait, that seems a bit confusing because if ( a < -3 ) is already included in ( a < frac{1}{3} ), right? Because ( -3 ) is less than ( frac{1}{3} ). So, does that mean the solution is just ( a < frac{1}{3} )?But hold on, I think I might have made a mistake here. Let me double-check. The original inequality is ( |a - 2| > |2a + 1| ). When solving absolute value inequalities, it's not just about solving each case separately but also considering the domains where each case applies.Alternatively, another approach is to square both sides to eliminate the absolute values. Since both sides are non-negative, squaring should preserve the inequality.So, let's try that:( |a - 2|^2 > |2a + 1|^2 )Which simplifies to:( (a - 2)^2 > (2a + 1)^2 )Expanding both sides:Left side: ( a^2 - 4a + 4 )Right side: ( 4a^2 + 4a + 1 )So, the inequality becomes:( a^2 - 4a + 4 > 4a^2 + 4a + 1 )Bring all terms to the left side:( a^2 - 4a + 4 - 4a^2 - 4a - 1 > 0 )Combine like terms:( -3a^2 - 8a + 3 > 0 )Multiply both sides by -1 (remember to reverse the inequality sign):( 3a^2 + 8a - 3 < 0 )Now, we have a quadratic inequality: ( 3a^2 + 8a - 3 < 0 )To solve this, first find the roots of the quadratic equation ( 3a^2 + 8a - 3 = 0 ).Using the quadratic formula:( a = frac{-b pm sqrt{b^2 - 4ac}}{2a} )Here, ( a = 3 ), ( b = 8 ), ( c = -3 ):( a = frac{-8 pm sqrt{64 - 4*3*(-3)}}{6} )Calculate the discriminant:( 64 - 4*3*(-3) = 64 + 36 = 100 )So,( a = frac{-8 pm sqrt{100}}{6} )( a = frac{-8 pm 10}{6} )So, two solutions:1. ( a = frac{-8 + 10}{6} = frac{2}{6} = frac{1}{3} )2. ( a = frac{-8 - 10}{6} = frac{-18}{6} = -3 )So, the quadratic ( 3a^2 + 8a - 3 ) factors as ( 3(a + 3)(a - frac{1}{3}) ). Wait, actually, let me check:Wait, if the roots are at ( a = -3 ) and ( a = frac{1}{3} ), then the quadratic can be written as ( 3(a + 3)(a - frac{1}{3}) ). Let's verify:( 3(a + 3)(a - frac{1}{3}) = 3[(a)(a) + (a)(-frac{1}{3}) + 3(a) + 3*(-frac{1}{3})] )= ( 3[a^2 - frac{1}{3}a + 3a - 1] )= ( 3[a^2 + frac{8}{3}a - 1] )= ( 3a^2 + 8a - 3 )Yes, that's correct.Now, the quadratic ( 3a^2 + 8a - 3 ) opens upwards (since the coefficient of ( a^2 ) is positive). Therefore, the quadratic is less than zero between its roots. So, the solution to ( 3a^2 + 8a - 3 < 0 ) is ( -3 < a < frac{1}{3} ).So, combining this with the previous result, we have ( -3 < a < frac{1}{3} ).But wait, earlier when I solved the absolute value inequality by splitting into cases, I got ( a < -3 ) or ( a < frac{1}{3} ). But now, using the squaring method, I get ( -3 < a < frac{1}{3} ). There seems to be a discrepancy here. Which one is correct?I think the issue arises because when we square both sides of an inequality, we have to be careful about the direction of the inequality and whether both sides are non-negative. In this case, both ( |a - 2| ) and ( |2a + 1| ) are non-negative, so squaring is valid. Therefore, the solution ( -3 < a < frac{1}{3} ) should be the correct one.But why did the case method give a different result? Let me revisit that.In the case method, I considered two cases:1. ( a - 2 > 2a + 1 ) leading to ( a < -3 )2. ( a - 2 < -(2a + 1) ) leading to ( a < frac{1}{3} )But perhaps I didn't consider the domains where each case is valid. For example, in the first case, ( a - 2 > 2a + 1 ) implies ( a < -3 ). But in this case, we need to ensure that both ( a - 2 ) and ( 2a + 1 ) are positive or negative appropriately.Wait, actually, when dealing with absolute value inequalities, the cases are based on the signs of the expressions inside the absolute values. So, perhaps I oversimplified by just splitting into two cases without considering the signs.Let me try a different approach by considering the four possible cases based on the signs of ( a - 2 ) and ( 2a + 1 ).Case 1: Both ( a - 2 ) and ( 2a + 1 ) are positive.So, ( a - 2 > 0 ) implies ( a > 2 )And ( 2a + 1 > 0 ) implies ( a > -frac{1}{2} )So, in this case, ( a > 2 ). Then, the inequality ( |a - 2| > |2a + 1| ) becomes ( a - 2 > 2a + 1 ), which simplifies to ( -3 > a ). But in this case, ( a > 2 ), so there's no solution here because ( a ) can't be both greater than 2 and less than -3.Case 2: ( a - 2 > 0 ) and ( 2a + 1 < 0 )So, ( a > 2 ) and ( a < -frac{1}{2} ). But these two conditions can't be true simultaneously, so no solution here.Case 3: ( a - 2 < 0 ) and ( 2a + 1 > 0 )So, ( a < 2 ) and ( a > -frac{1}{2} ). In this case, the inequality ( |a - 2| > |2a + 1| ) becomes ( -(a - 2) > 2a + 1 ), which simplifies to ( -a + 2 > 2a + 1 ), leading to ( -3a > -1 ), so ( a < frac{1}{3} ).But in this case, ( a ) is between ( -frac{1}{2} ) and 2, so combining with ( a < frac{1}{3} ), the solution here is ( -frac{1}{2} < a < frac{1}{3} ).Case 4: Both ( a - 2 ) and ( 2a + 1 ) are negative.So, ( a < 2 ) and ( a < -frac{1}{2} ). Thus, ( a < -frac{1}{2} ). The inequality becomes ( -(a - 2) > -(2a + 1) ), which simplifies to ( -a + 2 > -2a - 1 ), leading to ( a > -3 ).So, in this case, ( a < -frac{1}{2} ) and ( a > -3 ), so the solution is ( -3 < a < -frac{1}{2} ).Combining the solutions from Case 3 and Case 4, we get ( -3 < a < frac{1}{3} ), excluding ( a = -frac{1}{2} ) because at ( a = -frac{1}{2} ), both ( a - 2 ) and ( 2a + 1 ) are negative, but plugging into the original inequality:( f(a - 2) = f(-frac{1}{2} - 2) = f(-frac{5}{2}) = f(frac{5}{2}) )( f(2a + 1) = f(2*(-frac{1}{2}) + 1) = f(-1 + 1) = f(0) )But ( f(0) ) is undefined because of division by zero. So, ( a = -frac{1}{2} ) is not allowed.Therefore, the solution is ( -3 < a < frac{1}{3} ) with ( a neq -frac{1}{2} ).Wait, but earlier when I squared both sides, I got ( -3 < a < frac{1}{3} ) without considering ( a = -frac{1}{2} ). So, I need to exclude ( a = -frac{1}{2} ) from the solution set.Therefore, the range of values for ( a ) is ( (-3, -frac{1}{2}) cup (-frac{1}{2}, frac{1}{3}) ).Let me verify this with some test values.1. Let's pick ( a = -2 ), which is in ( (-3, -frac{1}{2}) ).Compute ( f(a - 2) = f(-2 - 2) = f(-4) = f(4) = frac{1}{16} + frac{1}{256} approx 0.0625 + 0.0039 = 0.0664 )Compute ( f(2a + 1) = f(2*(-2) + 1) = f(-4 + 1) = f(-3) = f(3) = frac{1}{9} + frac{1}{81} approx 0.1111 + 0.0123 = 0.1234 )So, ( f(a - 2) approx 0.0664 < 0.1234 = f(2a + 1) ). It satisfies the inequality.2. Let's pick ( a = 0 ), which is in ( (-frac{1}{2}, frac{1}{3}) ).Compute ( f(a - 2) = f(0 - 2) = f(-2) = f(2) = frac{1}{4} + frac{1}{16} = 0.25 + 0.0625 = 0.3125 )Compute ( f(2a + 1) = f(0 + 1) = f(1) = 1 + 1 = 2 )So, ( f(a - 2) = 0.3125 < 2 = f(2a + 1) ). It satisfies the inequality.3. Let's pick ( a = -4 ), which is less than -3.Compute ( f(a - 2) = f(-4 - 2) = f(-6) = f(6) = frac{1}{36} + frac{1}{1296} approx 0.0278 + 0.00077 = 0.0286 )Compute ( f(2a + 1) = f(2*(-4) + 1) = f(-8 + 1) = f(-7) = f(7) = frac{1}{49} + frac{1}{2401} approx 0.0204 + 0.000416 = 0.0208 )So, ( f(a - 2) approx 0.0286 > 0.0208 = f(2a + 1) ). It does not satisfy the inequality, which is consistent with our solution that ( a < -3 ) is not part of the solution.4. Let's pick ( a = 1 ), which is greater than ( frac{1}{3} ).Compute ( f(a - 2) = f(1 - 2) = f(-1) = f(1) = 1 + 1 = 2 )Compute ( f(2a + 1) = f(2*1 + 1) = f(3) = frac{1}{9} + frac{1}{81} approx 0.1111 + 0.0123 = 0.1234 )So, ( f(a - 2) = 2 > 0.1234 = f(2a + 1) ). It does not satisfy the inequality, which is consistent with our solution that ( a > frac{1}{3} ) is not part of the solution.5. Let's check ( a = -frac{1}{2} ), which is excluded.Compute ( f(a - 2) = f(-frac{1}{2} - 2) = f(-frac{5}{2}) = f(frac{5}{2}) = frac{1}{(frac{5}{2})^2} + frac{1}{(frac{5}{2})^4} = frac{4}{25} + frac{16}{625} approx 0.16 + 0.0256 = 0.1856 )Compute ( f(2a + 1) = f(2*(-frac{1}{2}) + 1) = f(-1 + 1) = f(0) ). But ( f(0) ) is undefined because we have division by zero. So, ( a = -frac{1}{2} ) is indeed not allowed.Therefore, after verifying with test values, it seems that the solution ( (-3, -frac{1}{2}) cup (-frac{1}{2}, frac{1}{3}) ) is correct.I think I initially confused myself by not considering the domains properly when splitting into cases, but after using the squaring method and then verifying with test values, I'm confident that this is the correct range for ( a ).**Final Answer**The range of values for ( a ) is boxed{left( -3, -frac{1}{2} right) cup left( -frac{1}{2}, frac{1}{3} right)}.