💡Okay, so I have this problem here where I need to find the probability that a randomly selected number x₀ from the interval (0,16) makes the function f(x) = 2^x - sqrt(x) - 14 greater than zero. The options are A: 1/4, B: 1/3, C: 2/3, and D: 3/4. First, I think I need to understand the function f(x) better. It's composed of three parts: 2^x, which is an exponential function, sqrt(x), which is a square root function, and the constant -14. The function f(x) is the difference between 2^x and the sum of sqrt(x) and 14. So, to find when f(x) is greater than zero, I need to find the values of x where 2^x is greater than sqrt(x) + 14. That means I need to solve the inequality 2^x - sqrt(x) - 14 > 0. I remember that for continuous functions, the sign of the function can change at its roots. So, if I can find the point where f(x) = 0, that will help me determine the intervals where f(x) is positive or negative. Let me try to solve f(x) = 0:2^x - sqrt(x) - 14 = 0This seems a bit tricky because it's a transcendental equation, meaning it's not straightforward to solve algebraically. Maybe I can try plugging in some integer values for x to see where the function crosses zero.Let's try x = 4:2^4 = 16sqrt(4) = 2So, f(4) = 16 - 2 - 14 = 0Oh, so x = 4 is a root of the function. That means when x is 4, f(x) is zero. Now, I need to check the behavior of f(x) around x = 4 to see where it's positive and where it's negative. Let me pick a value less than 4, say x = 3:2^3 = 8sqrt(3) ≈ 1.732f(3) = 8 - 1.732 - 14 ≈ -7.732So, f(3) is negative. Now, let's pick a value greater than 4, say x = 5:2^5 = 32sqrt(5) ≈ 2.236f(5) = 32 - 2.236 - 14 ≈ 15.764So, f(5) is positive. This suggests that f(x) is negative when x < 4 and positive when x > 4. To confirm, let's check another point, maybe x = 2:2^2 = 4sqrt(2) ≈ 1.414f(2) = 4 - 1.414 - 14 ≈ -11.414Negative again. And x = 6:2^6 = 64sqrt(6) ≈ 2.449f(6) = 64 - 2.449 - 14 ≈ 47.551Positive as well. So, it seems that x = 4 is the point where f(x) transitions from negative to positive. Therefore, in the interval (0,4), f(x) is negative, and in the interval (4,16), f(x) is positive. Now, the problem asks for the probability that f(x₀) > 0 when x₀ is randomly selected from (0,16). Probability, in this case, would be the length of the interval where f(x) is positive divided by the total length of the interval. The total interval is from 0 to 16, which is 16 units. The interval where f(x) is positive is from 4 to 16, which is 12 units. So, the probability is 12/16, which simplifies to 3/4. Looking back at the options, D is 3/4, so that should be the answer. Wait, let me double-check my reasoning. I found that f(x) is positive from x=4 to x=16, which is 12 units. The total interval is 16 units, so 12/16 is indeed 3/4. Is there a possibility that f(x) could cross zero again beyond x=4? Let me test x=16:2^16 = 65536sqrt(16) = 4f(16) = 65536 - 4 - 14 = 65518That's definitely positive. What about x=0? f(0) = 2^0 - sqrt(0) -14 = 1 - 0 -14 = -13, which is negative. So, the function only crosses zero once at x=4, and it's negative before that and positive after that. Therefore, my conclusion is correct. The probability is 3/4, which is option D.

💡Okay, so I have this problem where I'm given a function ( f(x) ) that's differentiable everywhere on the real line, and it's always true that ( f(x) < f'(x) ) for all ( x ). I need to figure out the relationship between ( f(2) ) and ( e^2 f(0) ), and similarly between ( f(2017) ) and ( e^{2017} f(0) ). The options are A, B, C, D, each giving different inequalities.Hmm, let's see. The condition ( f(x) < f'(x) ) seems important. Maybe I can rearrange this inequality to get something more manageable. If I subtract ( f(x) ) from both sides, I get ( f'(x) - f(x) > 0 ). That looks like the derivative of some function might be positive, which would imply that function is increasing.Wait, how can I relate ( f'(x) - f(x) ) to another function? Maybe I can think of a function that when differentiated gives ( f'(x) - f(x) ). Let me try multiplying both sides by an integrating factor. In differential equations, when you have something like ( f'(x) + P(x)f(x) = Q(x) ), you multiply by an integrating factor ( e^{int P(x) dx} ). Here, my equation is ( f'(x) - f(x) > 0 ), which is similar to ( f'(x) + (-1)f(x) > 0 ). So, the integrating factor would be ( e^{int -1 dx} = e^{-x} ).Let me define a new function ( g(x) = e^{-x} f(x) ). Then, let's compute its derivative:( g'(x) = frac{d}{dx} [e^{-x} f(x)] = e^{-x} f'(x) - e^{-x} f(x) = e^{-x} (f'(x) - f(x)) ).From the given condition, ( f'(x) - f(x) > 0 ), so ( g'(x) = e^{-x} (f'(x) - f(x)) > 0 ) because ( e^{-x} ) is always positive. Therefore, ( g(x) ) is an increasing function.Since ( g(x) ) is increasing, for any ( a < b ), ( g(a) < g(b) ). Let's apply this to specific points. For ( x = 0 ) and ( x = 2 ):( g(0) < g(2) ).Substituting back into ( g(x) ):( e^{-0} f(0) < e^{-2} f(2) ).Simplifying:( f(0) < e^{-2} f(2) ).Multiplying both sides by ( e^{2} ):( e^{2} f(0) < f(2) ).So, ( f(2) > e^{2} f(0) ).Similarly, let's apply this to ( x = 0 ) and ( x = 2017 ):( g(0) < g(2017) ).Which gives:( e^{-0} f(0) < e^{-2017} f(2017) ).Simplifying:( f(0) < e^{-2017} f(2017) ).Multiplying both sides by ( e^{2017} ):( e^{2017} f(0) < f(2017) ).So, ( f(2017) > e^{2017} f(0) ).Looking back at the options:A: ( f(2) > e^{2}f(0) ), ( f(2017) > e^{2017}f(0) )B: ( f(2) < e^{2}f(0) ), ( f(2017) > e^{2017}f(0) )C: ( f(2) > e^{2}f(0) ), ( f(2017) < e^{2017}f(0) )D: ( f(2) < e^{2}f(0) ), ( f(2017) < e^{2017}f(0) )From my deductions, both ( f(2) > e^{2}f(0) ) and ( f(2017) > e^{2017}f(0) ) are true. So, the correct answer should be option A.Wait, let me double-check my steps to make sure I didn't make a mistake. I defined ( g(x) = e^{-x} f(x) ), found its derivative, and since ( g'(x) > 0 ), ( g(x) ) is increasing. Therefore, at higher ( x ), ( g(x) ) is larger, meaning ( f(x) ) grows faster than ( e^{x} ). So, yes, multiplying back, ( f(x) ) should be greater than ( e^{x} f(0) ) for positive ( x ). That makes sense.I think my reasoning is solid. So, I'm confident that option A is correct.**Final Answer**boxed{A}

💡Okay, so I've got this problem here: In the decimal notation of a number, there's a zero. When you remove this zero, the number decreases by 9 times. I need to figure out where this zero was located in the original number.Alright, let's break this down. First, I need to understand what it means for a number to decrease by 9 times when a zero is removed. That suggests that the original number is nine times larger than the number you get after removing the zero. So, if I have a number N, and after removing a zero it becomes M, then N = 9 * M.Now, I need to think about how removing a zero affects the number. Let's consider the structure of the number. Suppose the original number has digits like A0B, where A and B are sequences of digits, and the zero is somewhere in the middle. When we remove the zero, it becomes AB.For example, if the number is 102, removing the zero gives 12. But 102 is not nine times 12 because 12 * 9 = 108, which is not 102. So, that doesn't work. Maybe I need a different example.Let me try a number like 108. If I remove the zero, it becomes 18. Now, 18 * 9 = 162, which is not 108. Hmm, that's not right either. Maybe I need a larger number.Let's think about a three-digit number. Suppose the number is 100. Removing the zero gives 10. But 10 * 9 = 90, which is not 100. Nope. Maybe a four-digit number.How about 1008? Removing the zero gives 108. Now, 108 * 9 = 972, which is not 1008. Still not matching. Maybe I'm approaching this wrong.Wait, maybe the zero isn't in the middle. What if it's at the end? For example, 10. Removing the zero gives 1. But 1 * 9 = 9, which is not 10. Doesn't work. How about 20? Removing the zero gives 2. 2 * 9 = 18, not 20.Hmm, maybe the zero is in the hundreds place. Let's try 1000. Removing the zero gives 100. 100 * 9 = 900, not 1000. Not matching.Wait, maybe the number has more digits. Let's try a five-digit number: 10000. Removing the zero gives 1000. 1000 * 9 = 9000, not 10000.This isn't working. Maybe I need to think algebraically instead of plugging in numbers.Let me denote the original number as N, and the number after removing the zero as M. So, N = 9 * M.Now, let's think about how removing a zero affects the number. If the zero is in a certain position, removing it changes the place value of the digits around it.Suppose the zero is in the k-th position from the right. Then, removing the zero would effectively divide the number by 10^k and then subtract something. Wait, maybe it's better to think in terms of place value.Let's say the original number is A * 10^{k+1} + 0 * 10^k + B, where A is the part before the zero, and B is the part after the zero. So, N = A * 10^{k+1} + B.When we remove the zero, the number becomes M = A * 10^k + B.Given that N = 9 * M, we have:A * 10^{k+1} + B = 9 * (A * 10^k + B)Let's expand the right side:A * 10^{k+1} + B = 9A * 10^k + 9BNow, let's subtract 9A * 10^k + 9B from both sides:A * 10^{k+1} - 9A * 10^k + B - 9B = 0Factor out A * 10^k:A * 10^k (10 - 9) - 8B = 0Simplify:A * 10^k - 8B = 0So, A * 10^k = 8BNow, A and B are integers, and 10^k is a power of 10. So, 8B must be a multiple of 10^k.Let's think about the smallest k where this can happen. If k=1, then 10^1=10, so 8B must be a multiple of 10. That means B must be a multiple of 10/ gcd(8,10)=10/2=5. So, B must be a multiple of 5.But B is a number, so it can be any digit or sequence of digits. Let's see.If k=1, then A * 10 = 8BSo, A = (8B)/10But A must be an integer, so 8B must be divisible by 10, which means B must end with a 0 or 5.But B is the part after the zero in the original number, so if the zero is in the tens place, B is the units digit.Wait, if k=1, the zero is in the tens place, so B is the units digit.So, B must be a single digit, and A is the part before the zero, which could be multiple digits.But A = (8B)/10Since A must be an integer, 8B must be divisible by 10, so B must be 5 because 8*5=40, which is divisible by 10.So, B=5, then A=40/10=4So, the original number is A0B = 405Removing the zero gives 45Now, 45 * 9 = 405Yes, that works.So, the zero is in the tens place, which is the second position from the right.Wait, but in the problem, it's asking for the position in the decimal notation, so from the left or from the right?In the example, 405, the zero is in the middle, which is the second position from the left.But in terms of place value, it's the tens place, which is the second position from the right.Hmm, the problem says "in which position was this zero located?" without specifying left or right.But in the example, 405, the zero is in the second position from the left.But in the algebraic approach, we found that k=1, meaning the zero is in the tens place, which is the second position from the right.So, there might be some confusion here.Wait, in the original number, A0B, A is the part before the zero, and B is the part after the zero.If k=1, then the zero is in the tens place, so B is the units digit.So, in the number 405, A=4, B=5, and the zero is in the tens place, which is the second position from the right.But in the number, it's written as 405, so the zero is in the middle, which is the second position from the left.So, depending on how we count positions, it could be the second position from the left or the second position from the right.But in the problem statement, it's not specified, so I think it's safer to assume it's from the left.But in the algebraic solution, k=1 corresponds to the tens place, which is the second position from the right.Hmm, this is a bit confusing.Wait, let's clarify.In the number 405:- From the left: 4 (hundreds place), 0 (tens place), 5 (units place)So, the zero is in the second position from the left.From the right: 5 (units place), 0 (tens place), 4 (hundreds place)So, the zero is in the second position from the right.But in the problem, it's asking for the position in the decimal notation, which is typically written from left to right, so the position from the left.But in the algebraic solution, k=1 refers to the place value, which is from the right.So, perhaps the answer is that the zero is in the second position from the left.But let's verify with another example.Suppose k=2, so the zero is in the hundreds place.Then, A * 10^2 = 8BSo, A * 100 = 8BBut B is the part after the zero, which would be the last two digits.But B must be a two-digit number.So, A * 100 = 8BA = (8B)/100But A must be an integer, so 8B must be divisible by 100.So, B must be a multiple of 25 because 8*25=200, which is divisible by 100.So, B=25, then A=200/100=2So, the original number is A0B = 2025Removing the zero gives 225Now, 225 * 9 = 2025Yes, that works.In this case, the zero is in the hundreds place, which is the third position from the right, or the second position from the left in a four-digit number.Wait, 2025: positions from the left are 2 (thousands), 0 (hundreds), 2 (tens), 5 (units). So, the zero is in the second position from the left.Similarly, in the previous example, 405: positions from the left are 4 (hundreds), 0 (tens), 5 (units). So, the zero is in the second position from the left.So, it seems that regardless of k, the zero is in the second position from the left.Wait, but in the first example, k=1, which corresponds to the tens place, which is the second position from the right, but in the number, it's the second position from the left.Similarly, in the second example, k=2, which corresponds to the hundreds place, which is the third position from the right, but in the number, it's the second position from the left.So, it seems that the zero is always in the second position from the left, regardless of the value of k.But wait, let's test with k=3.If k=3, then A * 10^3 = 8BSo, A * 1000 = 8BA = (8B)/1000But B is a three-digit number.So, 8B must be divisible by 1000, meaning B must be a multiple of 125 because 8*125=1000.So, B=125, then A=1000/1000=1So, the original number is A0B = 10125Removing the zero gives 1125Now, 1125 * 9 = 10125Yes, that works.In this case, the zero is in the thousands place, which is the fourth position from the right, but in the number 10125, it's the second position from the left.So, again, the zero is in the second position from the left.Therefore, it seems that regardless of the value of k, the zero is always in the second position from the left in the original number.So, the answer is that the zero is located in the second position from the left in the decimal notation of the number.

💡To find the best approximation of (log_4 20) using the given logarithmic values, we can use the change of base formula. First, recall the change of base formula:[log_b a = frac{log_c a}{log_c b}]Using base 10, we have:[log_4 20 = frac{log_{10} 20}{log_{10} 4}]Next, we need to express (log_{10} 20) and (log_{10} 4) in terms of the given logarithms. For (log_{10} 20):[log_{10} 20 = log_{10} (2 times 10) = log_{10} 2 + log_{10} 10 = 0.300 + 1 = 1.300]For (log_{10} 4):[log_{10} 4 = log_{10} (2^2) = 2 log_{10} 2 = 2 times 0.300 = 0.600]Now, substitute these values back into the change of base formula:[log_4 20 = frac{1.300}{0.600} = frac{13}{6}]Thus, the best approximation of (log_4 20) is (frac{13}{6}).

💡Okay, so I've got this problem here where Sheela deposits Rs. 3800 in her bank savings account, and this amount is 22% of her monthly income. The question is asking for her monthly income. Hmm, okay, let's break this down step by step.First, I need to understand what the problem is saying. Sheela is depositing a certain amount into her savings account, and that amount is a percentage of her monthly income. Specifically, it's 22%. So, if I think about percentages, 22% means 22 per 100, or 0.22 in decimal form.Now, the problem is giving me the amount that represents 22% of her income, which is Rs. 3800. So, I need to find out what the total income is, given that 22% of it is Rs. 3800.Let me think about how percentages work. If I have a percentage of a number, I can find the whole number by dividing the given amount by the percentage. In this case, I have 22% of her income equals Rs. 3800, so to find her total income, I should divide Rs. 3800 by 22%.But wait, percentages can be tricky sometimes. I need to make sure I'm using the right operation. Since 22% is a part of her total income, and I know the part, I need to find the whole. That means I should divide the part by the percentage to get the whole.So, mathematically, I can write this as:Total Income = (Amount Deposited) / (Percentage as a decimal)Plugging in the numbers:Total Income = 3800 / 0.22Now, let me do the division. 3800 divided by 0.22. Hmm, I think I can do this by moving the decimal places to make it easier. If I multiply both numerator and denominator by 100, it becomes 380000 divided by 22. That might be easier to handle.So, 380000 divided by 22. Let's see, 22 times 17000 is 374000, right? Because 22 times 1000 is 22000, so 22 times 17000 is 22000 times 17, which is 374000.Subtracting that from 380000, I get 6000 left. Now, 22 times 272 is 5984, which is close to 6000. So, adding that to 17000, I get 17272.7272...So, approximately, her total income is Rs. 17272.73.Wait a minute, let me double-check my calculations to make sure I didn't make any mistakes. If I take 22% of 17272.73, does it equal 3800?Calculating 22% of 17272.73:0.22 * 17272.73 = 3800Yes, that seems to check out. So, my calculation seems correct.But just to be thorough, let me try another method. Maybe using fractions instead of decimals to see if I get the same result.22% is the same as 22/100, so:Total Income = 3800 / (22/100) = 3800 * (100/22)Calculating that:3800 * 100 = 380000380000 / 22 = ?Again, we're back to the same division problem. So, I think my previous calculation is solid.Another way to think about it is to set up an equation where 22% of the total income equals 3800.Let me denote the total income as 'I'.So, 0.22 * I = 3800To solve for I, I divide both sides by 0.22:I = 3800 / 0.22Which is the same as before, leading to I = 17272.73Alright, so all these methods are converging to the same answer, which gives me more confidence that Rs. 17272.73 is indeed her monthly income.I think I've covered all the bases here. I've approached the problem using different methods—decimal division, fraction multiplication, and setting up an equation—and all of them lead me to the same result. I've also double-checked my calculations to ensure accuracy. So, I'm pretty confident that Sheela's monthly income is Rs. 17272.73.

💡Okay, let's tackle this problem step by step. I have the function ( f(x) = sin(2omega x + frac{pi}{4}) ) where ( omega > 0 ). The problem states that the smallest positive period ( T ) of this function satisfies ( frac{pi}{3} < T < frac{pi}{2} ). Additionally, the graph of ( y = f(x) ) is symmetric about the point ( (frac{3pi}{4}, 0) ). I need to determine which of the statements A, B, C, or D is true.First, I remember that the period ( T ) of a sine function ( sin(kx + phi) ) is given by ( T = frac{2pi}{|k|} ). In this case, the function is ( sin(2omega x + frac{pi}{4}) ), so the coefficient ( k ) is ( 2omega ). Therefore, the period ( T ) is ( frac{2pi}{2omega} = frac{pi}{omega} ).Given that ( frac{pi}{3} < T < frac{pi}{2} ), I can substitute ( T ) with ( frac{pi}{omega} ) to get:[frac{pi}{3} < frac{pi}{omega} < frac{pi}{2}]To solve for ( omega ), I can invert the inequalities, remembering to reverse the inequality signs when taking reciprocals:[3 > omega > 2]So, ( omega ) is between 2 and 3.Next, the graph is symmetric about the point ( (frac{3pi}{4}, 0) ). For a function to be symmetric about a point ( (a, b) ), it must satisfy the condition:[f(a + h) - b = -(f(a - h) - b)]In this case, ( a = frac{3pi}{4} ) and ( b = 0 ), so the condition simplifies to:[fleft(frac{3pi}{4} + hright) = -fleft(frac{3pi}{4} - hright)]Let me plug ( x = frac{3pi}{4} + h ) into the function:[fleft(frac{3pi}{4} + hright) = sinleft(2omegaleft(frac{3pi}{4} + hright) + frac{pi}{4}right)]Simplify the argument of the sine function:[2omega cdot frac{3pi}{4} + 2omega h + frac{pi}{4} = frac{3piomega}{2} + 2omega h + frac{pi}{4}]Similarly, for ( x = frac{3pi}{4} - h ):[fleft(frac{3pi}{4} - hright) = sinleft(2omegaleft(frac{3pi}{4} - hright) + frac{pi}{4}right)]Simplify:[frac{3piomega}{2} - 2omega h + frac{pi}{4}]The condition for symmetry is:[sinleft(frac{3piomega}{2} + 2omega h + frac{pi}{4}right) = -sinleft(frac{3piomega}{2} - 2omega h + frac{pi}{4}right)]Using the identity ( sin(A) = -sin(-A) ), the right side becomes:[-sinleft(frac{3piomega}{2} - 2omega h + frac{pi}{4}right) = sinleft(-frac{3piomega}{2} + 2omega h - frac{pi}{4}right)]For the equality to hold for all ( h ), the arguments inside the sine functions must differ by an integer multiple of ( 2pi ). Therefore:[frac{3piomega}{2} + 2omega h + frac{pi}{4} = -frac{3piomega}{2} + 2omega h - frac{pi}{4} + 2pi k]Where ( k ) is an integer. Let me simplify this equation:[frac{3piomega}{2} + frac{pi}{4} = -frac{3piomega}{2} - frac{pi}{4} + 2pi k]Bring like terms to one side:[frac{3piomega}{2} + frac{pi}{4} + frac{3piomega}{2} + frac{pi}{4} = 2pi k]Combine terms:[3piomega + frac{pi}{2} = 2pi k]Divide both sides by ( pi ):[3omega + frac{1}{2} = 2k]Solve for ( omega ):[3omega = 2k - frac{1}{2} implies omega = frac{2k - frac{1}{2}}{3} = frac{4k - 1}{6}]We know from earlier that ( 2 < omega < 3 ). Let's find the integer ( k ) such that ( omega ) falls within this range.Let me compute ( omega ) for different integer values of ( k ):- For ( k = 2 ): [ omega = frac{4(2) - 1}{6} = frac{8 - 1}{6} = frac{7}{6} approx 1.1667 ] This is less than 2, so not in our desired range.- For ( k = 3 ): [ omega = frac{4(3) - 1}{6} = frac{12 - 1}{6} = frac{11}{6} approx 1.8333 ] Still less than 2.- For ( k = 4 ): [ omega = frac{4(4) - 1}{6} = frac{16 - 1}{6} = frac{15}{6} = 2.5 ] This is within the range ( 2 < omega < 3 ).- For ( k = 5 ): [ omega = frac{4(5) - 1}{6} = frac{20 - 1}{6} = frac{19}{6} approx 3.1667 ] This is greater than 3, so outside our desired range.Therefore, the only valid value is ( k = 4 ), giving ( omega = 2.5 ). So, the function is ( f(x) = sin(5x + frac{pi}{4}) ).Now, let's analyze each option:**Option A: ( f(frac{pi}{2}) = 1 )**Compute ( f(frac{pi}{2}) ):[fleft(frac{pi}{2}right) = sinleft(5 cdot frac{pi}{2} + frac{pi}{4}right) = sinleft(frac{5pi}{2} + frac{pi}{4}right) = sinleft(frac{11pi}{4}right)]Simplify ( frac{11pi}{4} ):[frac{11pi}{4} = 2pi + frac{3pi}{4}]So,[sinleft(frac{11pi}{4}right) = sinleft(frac{3pi}{4}right) = frac{sqrt{2}}{2}]Therefore, ( f(frac{pi}{2}) = frac{sqrt{2}}{2} ), not 1. So, option A is false.**Option B: The graph of ( f(x) ) is symmetric about the line ( x = frac{pi}{8} )**For a function to be symmetric about the line ( x = a ), it must satisfy:[f(a + h) = f(a - h)]Let me check if ( fleft(frac{pi}{8} + hright) = fleft(frac{pi}{8} - hright) ).Compute ( fleft(frac{pi}{8} + hright) ):[sinleft(5left(frac{pi}{8} + hright) + frac{pi}{4}right) = sinleft(frac{5pi}{8} + 5h + frac{pi}{4}right) = sinleft(frac{7pi}{8} + 5hright)]Compute ( fleft(frac{pi}{8} - hright) ):[sinleft(5left(frac{pi}{8} - hright) + frac{pi}{4}right) = sinleft(frac{5pi}{8} - 5h + frac{pi}{4}right) = sinleft(frac{7pi}{8} - 5hright)]For symmetry, we need:[sinleft(frac{7pi}{8} + 5hright) = sinleft(frac{7pi}{8} - 5hright)]Using the identity ( sin(A) = sin(pi - A) ), we can see that:[sinleft(frac{7pi}{8} + 5hright) = sinleft(pi - left(frac{7pi}{8} + 5hright)right) = sinleft(frac{pi}{8} - 5hright)]But this is not equal to ( sinleft(frac{7pi}{8} - 5hright) ) unless ( h = 0 ). Therefore, the function is not symmetric about ( x = frac{pi}{8} ). So, option B is false.**Option C: ( f(x) ) is decreasing on the interval ( (frac{pi}{6}, frac{pi}{4}) )**To determine if ( f(x) ) is decreasing on this interval, we can look at its derivative:[f'(x) = 5cosleft(5x + frac{pi}{4}right)]A function is decreasing where its derivative is negative. So, we need to check the sign of ( f'(x) ) on ( (frac{pi}{6}, frac{pi}{4}) ).Compute the argument ( 5x + frac{pi}{4} ) for ( x ) in ( (frac{pi}{6}, frac{pi}{4}) ):- When ( x = frac{pi}{6} ): [ 5 cdot frac{pi}{6} + frac{pi}{4} = frac{5pi}{6} + frac{pi}{4} = frac{10pi}{12} + frac{3pi}{12} = frac{13pi}{12} ]- When ( x = frac{pi}{4} ): [ 5 cdot frac{pi}{4} + frac{pi}{4} = frac{5pi}{4} + frac{pi}{4} = frac{6pi}{4} = frac{3pi}{2} ]So, the argument ( 5x + frac{pi}{4} ) ranges from ( frac{13pi}{12} ) to ( frac{3pi}{2} ).Let's analyze ( cos(theta) ) in this interval:- ( frac{13pi}{12} ) is in the third quadrant where cosine is negative.- ( frac{3pi}{2} ) is where cosine is zero.- Between ( frac{13pi}{12} ) and ( frac{3pi}{2} ), cosine remains negative.Therefore, ( f'(x) = 5cos(theta) ) is negative throughout the interval ( (frac{pi}{6}, frac{pi}{4}) ), meaning ( f(x) ) is decreasing on this interval. So, option C is true.**Option D: ( f(x) ) has exactly two extreme points on the interval ( (0, frac{pi}{4}) )**Extreme points occur where the derivative is zero, i.e., where ( cos(5x + frac{pi}{4}) = 0 ). Let's solve for ( x ):[5x + frac{pi}{4} = frac{pi}{2} + kpi quad text{for integer } k]Solving for ( x ):[x = frac{frac{pi}{2} + kpi - frac{pi}{4}}{5} = frac{frac{pi}{4} + kpi}{5} = frac{pi}{20} + frac{kpi}{5}]Now, let's find all ( x ) in ( (0, frac{pi}{4}) ):- For ( k = 0 ): [ x = frac{pi}{20} approx 0.157pi ] Which is within ( (0, frac{pi}{4}) ) since ( frac{pi}{4} approx 0.25pi ).- For ( k = 1 ): [ x = frac{pi}{20} + frac{pi}{5} = frac{pi}{20} + frac{4pi}{20} = frac{5pi}{20} = frac{pi}{4} ] But ( x = frac{pi}{4} ) is the endpoint, not included in the open interval ( (0, frac{pi}{4}) ).- For ( k = -1 ): [ x = frac{pi}{20} - frac{pi}{5} = frac{pi}{20} - frac{4pi}{20} = -frac{3pi}{20} ] Negative, so not in our interval.Thus, the only critical point in ( (0, frac{pi}{4}) ) is ( x = frac{pi}{20} ). Therefore, ( f(x) ) has exactly one extreme point in this interval, not two. So, option D is false.**Conclusion**After evaluating all options, only option C is true.**Final Answer**boxed{C}