answer：Alright, so I have this problem here. It says that given ( a > 0 ) and ( a neq 1 ), the function ( f(x) = a^{log(x^2 - 2x + 3)} ) has a maximum value. Then, I need to find the solution set of the inequality ( log_a(x^2 - 5x + 7) > 0 ).Hmm, okay. Let me try to break this down step by step. First, I need to understand what's going on with the function ( f(x) ). It's an exponential function with base ( a ) and the exponent is a logarithm of a quadratic expression. So, ( f(x) = a^{log(x^2 - 2x + 3)} ). I know that ( x^2 - 2x + 3 ) is a quadratic, so maybe I should complete the square to see its minimum or maximum value. Let me do that.Completing the square for ( x^2 - 2x + 3 ):- Take the coefficient of ( x ), which is -2, divide by 2 to get -1, square it to get 1.- So, ( x^2 - 2x + 1 + 2 = (x - 1)^2 + 2 ).Okay, so ( x^2 - 2x + 3 = (x - 1)^2 + 2 ). That means the quadratic expression is always greater than or equal to 2 because ( (x - 1)^2 ) is always non-negative. So, ( x^2 - 2x + 3 geq 2 ).Therefore, the argument inside the logarithm, ( log(x^2 - 2x + 3) ), is always at least ( log(2) ). Since the logarithm function is increasing, the minimum value of the exponent is ( log(2) ).Now, the function ( f(x) = a^{log(x^2 - 2x + 3)} ) is an exponential function. The behavior of this function depends on the base ( a ). If ( a > 1 ), the exponential function is increasing, and if ( 0 < a < 1 ), it's decreasing.The problem states that ( f(x) ) has a maximum value. So, for ( f(x) ) to have a maximum, the exponent ( log(x^2 - 2x + 3) ) must have a minimum because the exponential function will attain its maximum when the exponent is at its minimum if ( a < 1 ), or at its maximum if ( a > 1 ).Wait, let me think again. If ( a > 1 ), then as the exponent increases, ( f(x) ) increases. But since the exponent ( log(x^2 - 2x + 3) ) can go to infinity as ( x ) becomes large, ( f(x) ) would also go to infinity, meaning there's no maximum. On the other hand, if ( 0 < a < 1 ), then as the exponent increases, ( f(x) ) decreases, and as the exponent decreases, ( f(x) ) increases. Since the exponent has a minimum value of ( log(2) ), ( f(x) ) will attain its maximum at this minimum exponent. Therefore, ( f(x) ) has a maximum value only if ( 0 < a < 1 ).Okay, so that tells me that ( a ) must be between 0 and 1.Now, moving on to the inequality ( log_a(x^2 - 5x + 7) > 0 ). Since ( a ) is between 0 and 1, the logarithmic function ( log_a ) is decreasing. That means the inequality ( log_a(y) > 0 ) is equivalent to ( y < 1 ) because as the argument ( y ) decreases, the logarithm increases.So, translating the inequality ( log_a(x^2 - 5x + 7) > 0 ) into something without the logarithm, we get:( x^2 - 5x + 7 < 1 ).Simplifying that:( x^2 - 5x + 7 - 1 < 0 )( x^2 - 5x + 6 < 0 ).Now, I need to solve the quadratic inequality ( x^2 - 5x + 6 < 0 ). Let me factor the quadratic:( x^2 - 5x + 6 = (x - 2)(x - 3) ).So, the inequality becomes:( (x - 2)(x - 3) < 0 ).To solve this, I can use a sign chart. The roots are at ( x = 2 ) and ( x = 3 ). These divide the number line into three intervals:1. ( x < 2 )2. ( 2 < x < 3 )3. ( x > 3 )Testing each interval:- For ( x < 2 ), say ( x = 1 ): ( (1 - 2)(1 - 3) = (-1)(-2) = 2 > 0 ).- For ( 2 < x < 3 ), say ( x = 2.5 ): ( (2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25 < 0 ).- For ( x > 3 ), say ( x = 4 ): ( (4 - 2)(4 - 3) = (2)(1) = 2 > 0 ).So, the inequality ( (x - 2)(x - 3) < 0 ) holds true when ( 2 < x < 3 ).Therefore, the solution set of the inequality ( log_a(x^2 - 5x + 7) > 0 ) is ( 2 < x < 3 ).But wait, I should also make sure that the argument of the logarithm is positive because the logarithm is only defined for positive numbers. So, ( x^2 - 5x + 7 > 0 ).Let me check if ( x^2 - 5x + 7 ) is always positive. The discriminant is ( (-5)^2 - 4(1)(7) = 25 - 28 = -3 ). Since the discriminant is negative, the quadratic doesn't cross the x-axis and since the coefficient of ( x^2 ) is positive, it's always positive. So, ( x^2 - 5x + 7 > 0 ) for all real ( x ). Therefore, the domain of the logarithm is all real numbers, and the solution ( 2 < x < 3 ) is valid without any restrictions.So, putting it all together, the solution set is ( 2 < x < 3 ).**Final Answer**The solution set of the inequality is boxed{ {x mid 2 < x < 3 } }.

answer：Okay, so I have this problem where I'm given two functions, f(x) and g(x). The function f(x) is defined as x squared plus e to the power of x minus one half, but only for x less than zero. The function g(x) is x squared plus the natural logarithm of (x plus a). The problem states that the graphs of these functions have points that are symmetric about the y-axis, and I need to determine the range of values for a.Hmm, symmetric about the y-axis. I remember that if two points are symmetric about the y-axis, then if one point is (x, y), the other point should be (-x, y). So, for the functions f and g, if they have points symmetric about the y-axis, then there must exist some x such that f(x) equals g(-x). That makes sense because if (x, f(x)) is a point on f, then (-x, f(x)) should be a point on g, right?So, mathematically, that means f(x) = g(-x) for some x. Let me write that down:f(x) = g(-x)Given that f(x) is defined for x < 0, and g(x) is defined for x + a > 0, which implies x > -a. So, when we plug in -x into g, we have g(-x) = (-x)^2 + ln(-x + a). Simplifying that, it's x squared plus ln(a - x). So, substituting into the equation f(x) = g(-x), we get:x² + e^x - 1/2 = x² + ln(a - x)Wait, the x² terms on both sides cancel out, so we're left with:e^x - 1/2 = ln(a - x)That's an equation we need to solve for x, where x is less than 0 because f(x) is only defined for x < 0. So, we have:e^x - 1/2 - ln(a - x) = 0Let me define a new function m(x) = e^x - 1/2 - ln(a - x). We need to find the values of a such that m(x) = 0 has a solution for x < 0.Alright, so m(x) is defined for x < 0 and a - x > 0, which means a > x. Since x is negative, a must be greater than a negative number. But we need to find the range of a such that m(x) = 0 has a solution.Let me analyze the behavior of m(x). As x approaches negative infinity, e^x approaches 0, so the first term goes to 0. The term ln(a - x) as x approaches negative infinity becomes ln(a - x) which is ln(a - (-∞)) = ln(∞) which goes to infinity. So, m(x) approaches 0 - 1/2 - ∞, which is negative infinity. So, as x approaches negative infinity, m(x) tends to negative infinity.Now, let's look at the behavior as x approaches 0 from the left. As x approaches 0-, e^x approaches 1, so the first term is 1. The term ln(a - x) as x approaches 0- becomes ln(a - 0) = ln(a). So, m(x) approaches 1 - 1/2 - ln(a) = 1/2 - ln(a).So, m(x) approaches 1/2 - ln(a) as x approaches 0 from the left.Now, since m(x) is continuous on its domain (x < 0 and a - x > 0), and it goes from negative infinity to 1/2 - ln(a). For m(x) to have a zero, the function must cross the x-axis somewhere. So, if the limit as x approaches 0 from the left is positive, then by the Intermediate Value Theorem, since m(x) goes from negative infinity to a positive value, it must cross zero somewhere in between.So, for m(x) to have a solution, we need that 1/2 - ln(a) > 0. Let's solve that inequality:1/2 - ln(a) > 0=> ln(a) < 1/2=> a < e^(1/2)=> a < sqrt(e)So, that gives us an upper bound for a: a must be less than sqrt(e).But wait, what about the lower bound? The problem didn't specify any lower bound, but we need to make sure that the function m(x) is defined for x < 0. The term ln(a - x) requires that a - x > 0, which is equivalent to a > x. Since x is less than 0, a just needs to be greater than some negative number. But since a can be any real number as long as a > x for x < 0, but we need to ensure that the domain is consistent.Wait, actually, for m(x) to be defined for all x < 0, a must be greater than x for all x < 0. But x can be any negative number, so the smallest x is negative infinity. However, a is a constant, so a must be greater than the supremum of x, which is 0. Wait, no, that's not correct.Wait, let me think again. For m(x) to be defined for some x < 0, we need a - x > 0. Since x < 0, a - x = a + |x|, which is always positive as long as a is a real number because |x| is positive. So, actually, m(x) is defined for all x < 0 as long as a is a real number because a - x will always be positive since x is negative. So, a can be any real number, but we have the condition from earlier that a < sqrt(e).But wait, if a is very large, say a approaches infinity, then ln(a - x) approaches ln(a), which is large. So, m(x) = e^x - 1/2 - ln(a - x) would be dominated by the negative term ln(a - x), so m(x) would be negative for large a. But we need m(x) to cross zero, so we need that as x approaches 0, m(x) is positive, which requires a < sqrt(e). But what if a is negative? Let's see. If a is negative, say a = -k where k > 0, then ln(a - x) becomes ln(-k - x). But since x < 0, -k - x = -k + |x|. If |x| > k, then -k + |x| is positive, but if |x| < k, then it's negative. So, ln(a - x) is only defined when -k - x > 0, which is equivalent to x < -k. So, for a negative a, the domain of m(x) is x < -k, which is a subset of x < 0.So, in that case, as x approaches -k from the left, ln(a - x) approaches ln(0+) which tends to negative infinity. So, m(x) approaches e^x - 1/2 - (-infty), which is positive infinity. As x approaches negative infinity, m(x) approaches negative infinity as before. So, in this case, m(x) goes from negative infinity to positive infinity as x approaches -k from the left, so by the Intermediate Value Theorem, m(x) must cross zero somewhere in between. Therefore, for a negative a, m(x) = 0 has a solution.So, combining both cases, when a is negative, m(x) = 0 has a solution, and when a is positive but less than sqrt(e), m(x) = 0 also has a solution. Therefore, the range of a is all real numbers less than sqrt(e).Wait, but what about when a is exactly sqrt(e)? Then, 1/2 - ln(a) = 1/2 - ln(sqrt(e)) = 1/2 - (1/2) = 0. So, m(x) approaches 0 as x approaches 0 from the left. But does m(x) actually reach zero at some point?Since m(x) is continuous and approaches 0 from above (because as x approaches 0-, m(x) approaches 0 from the positive side), and as x approaches negative infinity, m(x) approaches negative infinity. So, by the Intermediate Value Theorem, there must be some x where m(x) = 0. So, even when a = sqrt(e), m(x) = 0 has a solution. Therefore, a can be equal to sqrt(e) as well.Wait, but earlier I thought a must be less than sqrt(e). Let me double-check. If a = sqrt(e), then as x approaches 0-, m(x) approaches 0. So, does m(x) actually reach zero at x = 0? But x = 0 is not in the domain of f(x). So, does m(x) = 0 have a solution at some x < 0?Yes, because m(x) is continuous, and as x approaches 0 from the left, m(x) approaches 0, and as x approaches negative infinity, m(x) approaches negative infinity. So, there must be some x < 0 where m(x) = 0. Therefore, a can be equal to sqrt(e).So, putting it all together, the range of a is all real numbers less than or equal to sqrt(e). But wait, earlier I considered a negative a and concluded that m(x) = 0 has a solution. So, the range of a is (-infty, sqrt(e)].But let me check again. If a is negative, say a = -1, then ln(a - x) = ln(-1 - x). For x < 0, -1 - x = -1 + |x|. So, if |x| > 1, then -1 + |x| > 0, so ln is defined. So, for a = -1, m(x) is defined for x < -1. As x approaches -1 from the left, ln(a - x) approaches ln(0+) which is negative infinity, so m(x) approaches e^x - 1/2 - (-infty) which is positive infinity. As x approaches negative infinity, m(x) approaches negative infinity. So, m(x) goes from negative infinity to positive infinity, so by IVT, there is a solution. Therefore, a can be any real number less than or equal to sqrt(e).Wait, but earlier I thought a can be any real number, but actually, when a is greater than sqrt(e), does m(x) = 0 have a solution?If a > sqrt(e), then 1/2 - ln(a) < 0. So, as x approaches 0 from the left, m(x) approaches 1/2 - ln(a) < 0. And as x approaches negative infinity, m(x) approaches negative infinity. So, m(x) is always negative in this case. Therefore, m(x) = 0 has no solution when a > sqrt(e).Therefore, the range of a is all real numbers less than or equal to sqrt(e). So, a ∈ (-infty, sqrt(e)].Wait, but in the initial analysis, when a is negative, m(x) is defined for x < a, but a is negative, so x < a is a subset of x < 0. So, as long as a is less than or equal to sqrt(e), m(x) = 0 has a solution.Therefore, the range of a is (-infty, sqrt(e)].But let me confirm with a specific example. Let's take a = sqrt(e). Then, m(x) = e^x - 1/2 - ln(sqrt(e) - x). As x approaches 0-, m(x) approaches 1 - 1/2 - ln(sqrt(e)) = 1/2 - 1/2 = 0. So, m(x) approaches 0 from above. Since m(x) is continuous and approaches negative infinity as x approaches negative infinity, there must be some x < 0 where m(x) = 0.Another example: a = 0. Then, m(x) = e^x - 1/2 - ln(-x). As x approaches 0-, ln(-x) approaches negative infinity, so m(x) approaches positive infinity. As x approaches negative infinity, m(x) approaches negative infinity. So, m(x) must cross zero somewhere.Another example: a = 1, which is less than sqrt(e) (~1.648). Then, m(x) = e^x - 1/2 - ln(1 - x). As x approaches 0-, m(x) approaches 1 - 1/2 - ln(1) = 1/2 - 0 = 1/2 > 0. As x approaches negative infinity, m(x) approaches negative infinity. So, m(x) must cross zero.If a = 2, which is greater than sqrt(e), then m(x) = e^x - 1/2 - ln(2 - x). As x approaches 0-, m(x) approaches 1 - 1/2 - ln(2) ≈ 1/2 - 0.693 ≈ -0.193 < 0. As x approaches negative infinity, m(x) approaches negative infinity. So, m(x) is always negative, so no solution.Therefore, the range of a is all real numbers less than or equal to sqrt(e). So, a ∈ (-infty, sqrt(e)].But wait, in the initial problem statement, the user wrote "the graphs of these functions have points symmetric about the y-axis." So, does that mean that there exists at least one pair of symmetric points, or that all points are symmetric? I think it just means that there exists at least one pair of points symmetric about the y-axis. So, as long as there is at least one x < 0 such that f(x) = g(-x), which translates to m(x) = 0 having at least one solution.Therefore, the range of a is (-infty, sqrt(e)].But wait, in the initial analysis, when a is negative, m(x) is defined for x < a, but a is negative, so x < a is a subset of x < 0. So, as long as a is less than or equal to sqrt(e), m(x) = 0 has a solution.Therefore, the final answer is a ∈ (-infty, sqrt(e)].But in the initial problem, the user wrote "the graphs of these functions have points symmetric about the y-axis." So, does that mean that for every point on f, there is a corresponding point on g symmetric about the y-axis? Or just that there exists at least one such pair?I think it's the latter, that there exists at least one pair of points symmetric about the y-axis. So, as long as m(x) = 0 has at least one solution for x < 0, which happens when a ≤ sqrt(e).Therefore, the range of a is (-infty, sqrt(e)].

answer：Okay, so I have this problem about curves in the Cartesian coordinate system, and I need to convert them into polar coordinates and then find the maximum distance between two points on these curves along a ray. Let me try to break this down step by step.First, part (1) asks for the polar equations of curves (C₁) and (C₂). Starting with (C₁), which is given parametrically:[begin{cases}x = cos t y = 1 + sin tend{cases}]I remember that parametric equations can often be converted to Cartesian form by eliminating the parameter. Here, the parameter is ( t ). So, if I square both ( x ) and ( y - 1 ), I should get something useful.Calculating ( x^2 ) gives ( cos^2 t ), and ( (y - 1)^2 ) gives ( sin^2 t ). Adding these together:[x^2 + (y - 1)^2 = cos^2 t + sin^2 t = 1]So, the Cartesian equation for (C₁) is ( x^2 + (y - 1)^2 = 1 ). That looks like a circle with radius 1 centered at (0, 1).Now, to convert this into polar coordinates. I know that ( x = rho cos theta ) and ( y = rho sin theta ). Substituting these into the equation:[(rho cos theta)^2 + (rho sin theta - 1)^2 = 1]Expanding this:[rho^2 cos^2 theta + rho^2 sin^2 theta - 2 rho sin theta + 1 = 1]Simplifying:[rho^2 (cos^2 theta + sin^2 theta) - 2 rho sin theta + 1 = 1]Since ( cos^2 theta + sin^2 theta = 1 ), this reduces to:[rho^2 - 2 rho sin theta + 1 = 1]Subtracting 1 from both sides:[rho^2 - 2 rho sin theta = 0]Factoring out ( rho ):[rho (rho - 2 sin theta) = 0]So, ( rho = 0 ) or ( rho = 2 sin theta ). Since ( rho = 0 ) is just the origin, the polar equation for (C₁) is ( rho = 2 sin theta ).Moving on to (C₂), which is given by the Cartesian equation:[x^2 + (y - 2)^2 = 4]This is a circle with radius 2 centered at (0, 2). Converting this to polar coordinates, substitute ( x = rho cos theta ) and ( y = rho sin theta ):[(rho cos theta)^2 + (rho sin theta - 2)^2 = 4]Expanding:[rho^2 cos^2 theta + rho^2 sin^2 theta - 4 rho sin theta + 4 = 4]Simplifying:[rho^2 (cos^2 theta + sin^2 theta) - 4 rho sin theta + 4 = 4]Again, ( cos^2 theta + sin^2 theta = 1 ), so:[rho^2 - 4 rho sin theta + 4 = 4]Subtracting 4 from both sides:[rho^2 - 4 rho sin theta = 0]Factoring out ( rho ):[rho (rho - 4 sin theta) = 0]So, ( rho = 0 ) or ( rho = 4 sin theta ). Excluding the origin, the polar equation for (C₂) is ( rho = 4 sin theta ).Alright, part (1) seems done. Now, part (2) is about finding the maximum value of |AB|, where A and B are the intersection points of the ray ( theta = a ) with curves (C₁) and (C₂), respectively, excluding the origin.So, the ray is given by ( theta = a ), which is a straight line from the origin at an angle ( a ) from the polar axis. Since ( 0 < a < pi ), this ray is in the upper half-plane.To find points A and B, I need to find the points where this ray intersects (C₁) and (C₂). From part (1), we have the polar equations for both curves.For (C₁): ( rho = 2 sin theta ). So, when ( theta = a ), ( rho = 2 sin a ). Therefore, point A has polar coordinates ( (2 sin a, a) ).Similarly, for (C₂): ( rho = 4 sin theta ). So, when ( theta = a ), ( rho = 4 sin a ). Therefore, point B has polar coordinates ( (4 sin a, a) ).Now, to find |AB|, the distance between points A and B. Since both points lie on the same ray ( theta = a ), the distance between them is simply the difference in their radii.So, |AB| = |OB| - |OA| = ( 4 sin a - 2 sin a = 2 sin a ).Wait, that seems straightforward. So, |AB| = 2 sin a. Now, we need to find the maximum value of this expression for ( 0 < a < pi ).The sine function reaches its maximum value of 1 at ( a = pi/2 ). Therefore, the maximum value of |AB| is ( 2 * 1 = 2 ).But let me double-check. Is |AB| always equal to 2 sin a? Since both points are along the same ray, and (C₂) is just a scaled-up version of (C₁), their radii are proportional. So, the distance between them is indeed the difference in radii, which is 2 sin a.And since sin a is maximized at a = pi/2, the maximum distance is 2.Hmm, seems correct. I don't think I made any mistakes here. The key was recognizing that both points lie on the same ray, so their distance is just the difference in their radial distances.**Final Answer**The maximum value of ( |AB| ) is boxed{2}.

answer：First, I need to determine how many shares the man purchased. Since each share is priced at a 20% premium over Rs. 100, the cost per share is Rs. 120.Next, I'll calculate the number of shares he bought by dividing his total investment of Rs. 14,400 by the cost per share:Number of shares = Rs. 14,400 / Rs. 120 = 120 shares.At the end of the year, he receives a total dividend of Rs. 600. To find the dividend per share, I'll divide the total dividend by the number of shares:Dividend per share = Rs. 600 / 120 = Rs. 5.Finally, to determine the dividend percentage, I'll calculate the dividend per share as a percentage of the face value of the share:Dividend percentage = (Rs. 5 / Rs. 100) * 100 = 5%.Therefore, the dividend percentage declared by the company is 5%.

answer：First, I need to find the smallest possible perimeter of a triangle with consecutive integer side lengths, where the smallest side is greater than 2.Let's denote the smallest side as ( n ) (where ( n > 2 )). The other two sides will then be ( n+1 ) and ( n+2 ).To ensure these sides form a valid triangle, they must satisfy the triangle inequality theorem, which states that the sum of any two sides must be greater than the third side.1. **Check the Triangle Inequality:** - ( n + (n+1) > n+2 ) simplifies to ( 2n + 1 > n + 2 ), which leads to ( n > 1 ). Since ( n > 2 ), this condition is satisfied. - ( n + (n+2) > n+1 ) simplifies to ( 2n + 2 > n + 1 ), which leads to ( n + 1 > 1 ). Again, this is true as ( n > 2 ). - ( (n+1) + (n+2) > n ) simplifies to ( 2n + 3 > n ), which leads to ( n > -3 ). This is always true since ( n > 2 ).2. **Determine the Smallest Valid ( n ):** The smallest integer greater than 2 is 3. Let's verify if ( n = 3 ) satisfies the triangle inequality: - ( 3 + 4 > 5 ) → ( 7 > 5 ) (True) - ( 3 + 5 > 4 ) → ( 8 > 4 ) (True) - ( 4 + 5 > 3 ) → ( 9 > 3 ) (True)3. **Calculate the Perimeter:** If ( n = 3 ), the sides of the triangle are 3, 4, and 5 units. The perimeter is ( 3 + 4 + 5 = 12 ) units.Therefore, the smallest possible perimeter of such a triangle is 12 units.

answer：Okay, so I have this problem where a ball is rolling down an inclined plane, and its motion is described by the equation s = t². I need to find the instantaneous velocity of the ball at t = 5. Hmm, let's see. I remember that velocity is related to how fast something is moving at a particular moment, which is different from average velocity over a period of time. So, the equation given is s = t², where s is the displacement and t is time. I think to find the instantaneous velocity, I need to find the derivative of the displacement with respect to time. That's because velocity is the rate of change of displacement over time. So, if I can find the derivative of s with respect to t, that should give me the velocity function, and then I can plug in t = 5 to find the specific velocity at that moment.Alright, let's recall how to take derivatives. For a function like s = t², the derivative with respect to t is straightforward. The power rule says that the derivative of t^n is n*t^(n-1). So, applying that here, the derivative of t² would be 2*t. So, ds/dt = 2t. That means the velocity at any time t is 2t.Now, to find the instantaneous velocity at t = 5, I just need to substitute t = 5 into this velocity function. So, plugging in 5 for t, we get 2*5 = 10. So, the instantaneous velocity at t = 5 is 10 units per time unit. Wait, but what are the units here? The problem doesn't specify, but usually, in physics, displacement is in meters and time is in seconds, so the velocity would be in meters per second. So, 10 meters per second.Let me double-check my steps to make sure I didn't make a mistake. First, I identified that velocity is the derivative of displacement with respect to time. Then, I applied the power rule correctly to s = t², getting ds/dt = 2t. Finally, I substituted t = 5 into the derivative, resulting in 10. That seems right.I also remember that instantaneous velocity is the limit of the average velocity as the time interval approaches zero. So, if I were to calculate the average velocity between t = 5 and t = 5 + h, where h is a very small number, and then take the limit as h approaches zero, I should get the same result. Let's try that quickly.The average velocity between t = 5 and t = 5 + h is [s(5 + h) - s(5)] / h. Substituting s = t², that becomes [(5 + h)² - 5²] / h. Expanding (5 + h)², I get 25 + 10h + h². Subtracting 5², which is 25, leaves me with 10h + h². So, the average velocity is (10h + h²) / h, which simplifies to 10 + h. Now, taking the limit as h approaches zero, the h term goes away, leaving me with 10. So, that confirms that the instantaneous velocity at t = 5 is indeed 10.I think that's solid. I don't see any mistakes in my reasoning. The key steps were recognizing that instantaneous velocity is the derivative of displacement, applying the power rule correctly, and then verifying by using the limit definition of the derivative. Both methods led me to the same answer, so I'm confident that 10 is the correct instantaneous velocity at t = 5.