💡To determine the total number of cats at CleverCat Academy, we'll use the principle of inclusion-exclusion for three sets. Let's define:- ( J ) as the set of cats that can jump.- ( S ) as the set of cats that can spin.- ( F ) as the set of cats that can fetch.Given data:- ( |J| = 60 )- ( |S| = 35 )- ( |F| = 40 )- ( |J cap S| = 25 )- ( |S cap F| = 20 )- ( |J cap F| = 22 )- ( |J cap S cap F| = 12 )- Cats that can perform none of the tricks: 10First, calculate the number of cats that can perform exactly two tricks:- Only Jump and Spin: ( 25 - 12 = 13 )- Only Spin and Fetch: ( 20 - 12 = 8 )- Only Jump and Fetch: ( 22 - 12 = 10 )Next, calculate the number of cats that can perform only one trick:- Only Jump: ( 60 - (13 + 10 + 12) = 25 )- Only Spin: ( 35 - (13 + 8 + 12) = 2 )- Only Fetch: ( 40 - (10 + 8 + 12) = 10 )Finally, sum all these numbers and add the cats that can perform none of the tricks:[ 25 + 13 + 10 + 2 + 10 + 12 + 12 + 10 = 92 ]Therefore, the total number of cats in the academy is 92.

💡First, I need to determine the ratio of the profits shared by x and y. Given that the total profit is Rs. 1600 and x's share is Rs. 400, y's share would be Rs. 1200.Next, I recognize that the profit is distributed in proportion to their investments. Therefore, the ratio of x's profit to y's profit is the same as the ratio of their investments.I can set up the proportion as follows:400 / X = 1200 / 15000To find the value of X, I'll cross-multiply:400 * 15000 = 1200 * XThis simplifies to:6000000 = 1200XFinally, I'll solve for X by dividing both sides by 1200:X = 6000000 / 1200X = 5000Therefore, x invested Rs. 5000 in the business.

💡To determine the range of values for ( a ) such that the function ( f(x) = x^2 + 2ax + 1 ) does not have any good points, we start by setting ( f(x) = x ). This gives us the equation:[x = x^2 + 2ax + 1]Rearranging the equation, we get:[x^2 + (2a - 1)x + 1 = 0]For the function to have no good points, this quadratic equation must have no real roots. This occurs when the discriminant ( Delta ) is less than zero. The discriminant of a quadratic equation ( ax^2 + bx + c = 0 ) is given by:[Delta = b^2 - 4ac]Applying this to our equation, we have:[Delta = (2a - 1)^2 - 4 times 1 times 1 = (2a - 1)^2 - 4]Setting the discriminant to be less than zero:[(2a - 1)^2 - 4 < 0]Solving this inequality:[(2a - 1)^2 < 4]Taking square roots on both sides:[|2a - 1| < 2]This leads to:[-2 < 2a - 1 < 2]Adding 1 to all parts of the inequality:[-1 < 2a < 3]Dividing by 2:[-frac{1}{2} < a < frac{3}{2}]Thus, the range of values for ( a ) is ( left(-frac{1}{2}, frac{3}{2}right) ).

💡First, I need to find out how much of the ribbon was used to create the pieces. The original length of the ribbon is 51 meters, and after cutting, 36 meters remain. So, the length of the ribbon that was cut into pieces is:51 meters - 36 meters = 15 metersNext, I should convert this length from meters to centimeters because the pieces are measured in centimeters. Since 1 meter equals 100 centimeters:15 meters * 100 centimeters/meter = 1500 centimetersNow, I can determine how many 15-centimeter pieces can be cut from 1500 centimeters of ribbon by dividing the total length of the cut ribbon by the length of each piece:1500 centimeters / 15 centimeters/piece = 100 piecesTherefore, the number of pieces cut from the ribbon is 100.

💡Alright, so I have this function here: ( f(x) = sqrt{2} sinleft(2x + frac{pi}{4}right) ). I need to tackle two parts. First, find the smallest positive period and the monotonic intervals of the function. Second, when ( x ) is in the interval (left[-frac{pi}{4}, frac{pi}{4}right]), I have to find the maximum and minimum values of ( f(x) ) along with the corresponding ( x ) values.Starting with part (1). I remember that for sine functions of the form ( sin(Bx + C) ), the period is ( frac{2pi}{|B|} ). In this case, ( B = 2 ), so the period should be ( frac{2pi}{2} = pi ). That seems straightforward. So the smallest positive period is ( pi ).Now, for the monotonic intervals. Monotonicity refers to where the function is increasing or decreasing. Since it's a sine function, it should have intervals where it increases and decreases periodically. To find these intervals, I might need to look at the derivative of ( f(x) ).Let me compute the derivative. The derivative of ( sin(u) ) is ( cos(u) cdot u' ). So, ( f'(x) = sqrt{2} cosleft(2x + frac{pi}{4}right) cdot 2 ). Simplifying, that's ( 2sqrt{2} cosleft(2x + frac{pi}{4}right) ).To find where the function is increasing or decreasing, I need to determine where the derivative is positive or negative. The function is increasing when ( f'(x) > 0 ) and decreasing when ( f'(x) < 0 ).So, ( f'(x) > 0 ) when ( cosleft(2x + frac{pi}{4}right) > 0 ), and ( f'(x) < 0 ) when ( cosleft(2x + frac{pi}{4}right) < 0 ).Let me solve ( cosleft(2x + frac{pi}{4}right) > 0 ). The cosine function is positive in the intervals ( (-frac{pi}{2} + 2kpi, frac{pi}{2} + 2kpi) ) for any integer ( k ).So, ( 2x + frac{pi}{4} ) must lie within ( (-frac{pi}{2} + 2kpi, frac{pi}{2} + 2kpi) ).Let me solve for ( x ):( -frac{pi}{2} + 2kpi < 2x + frac{pi}{4} < frac{pi}{2} + 2kpi )Subtract ( frac{pi}{4} ) from all parts:( -frac{3pi}{4} + 2kpi < 2x < frac{pi}{4} + 2kpi )Divide all parts by 2:( -frac{3pi}{8} + kpi < x < frac{pi}{8} + kpi )So, the function is increasing on intervals ( left(-frac{3pi}{8} + kpi, frac{pi}{8} + kpiright) ) for any integer ( k ).Similarly, when ( cosleft(2x + frac{pi}{4}right) < 0 ), the function is decreasing. The cosine function is negative in ( (frac{pi}{2} + 2kpi, frac{3pi}{2} + 2kpi) ).So, ( 2x + frac{pi}{4} ) must lie within ( (frac{pi}{2} + 2kpi, frac{3pi}{2} + 2kpi) ).Solving for ( x ):( frac{pi}{2} + 2kpi < 2x + frac{pi}{4} < frac{3pi}{2} + 2kpi )Subtract ( frac{pi}{4} ):( frac{pi}{4} + 2kpi < 2x < frac{5pi}{4} + 2kpi )Divide by 2:( frac{pi}{8} + kpi < x < frac{5pi}{8} + kpi )So, the function is decreasing on intervals ( left(frac{pi}{8} + kpi, frac{5pi}{8} + kpiright) ) for any integer ( k ).Okay, so that's part (1) done. The period is ( pi ), and the function increases on ( left(-frac{3pi}{8} + kpi, frac{pi}{8} + kpiright) ) and decreases on ( left(frac{pi}{8} + kpi, frac{5pi}{8} + kpiright) ).Moving on to part (2). I need to find the maximum and minimum values of ( f(x) ) when ( x ) is in ( left[-frac{pi}{4}, frac{pi}{4}right] ).First, let me understand the interval. ( x ) ranges from ( -frac{pi}{4} ) to ( frac{pi}{4} ). Let me see what the argument of the sine function does in this interval.The argument is ( 2x + frac{pi}{4} ). So when ( x = -frac{pi}{4} ), the argument is ( 2(-frac{pi}{4}) + frac{pi}{4} = -frac{pi}{2} + frac{pi}{4} = -frac{pi}{4} ).When ( x = frac{pi}{4} ), the argument is ( 2(frac{pi}{4}) + frac{pi}{4} = frac{pi}{2} + frac{pi}{4} = frac{3pi}{4} ).So, the argument ( 2x + frac{pi}{4} ) ranges from ( -frac{pi}{4} ) to ( frac{3pi}{4} ).Now, let's think about the sine function over this interval. The sine function reaches its maximum at ( frac{pi}{2} ) and its minimum at ( -frac{pi}{2} ). But in our interval ( -frac{pi}{4} ) to ( frac{3pi}{4} ), the sine function increases from ( -frac{pi}{4} ) to ( frac{pi}{2} ), and then decreases from ( frac{pi}{2} ) to ( frac{3pi}{4} ).So, the maximum value of ( sin(theta) ) in this interval occurs at ( theta = frac{pi}{2} ), which is 1. The minimum value occurs at the left end, ( theta = -frac{pi}{4} ), which is ( sin(-frac{pi}{4}) = -frac{sqrt{2}}{2} ).But wait, hold on. The function is ( sqrt{2} sin(theta) ), so the maximum value would be ( sqrt{2} times 1 = sqrt{2} ), and the minimum would be ( sqrt{2} times (-frac{sqrt{2}}{2}) = -1 ).So, the maximum value of ( f(x) ) is ( sqrt{2} ), and the minimum is ( -1 ).Now, I need to find the corresponding ( x ) values where these extrema occur.For the maximum value, ( sin(theta) = 1 ) when ( theta = frac{pi}{2} ). So,( 2x + frac{pi}{4} = frac{pi}{2} )Solving for ( x ):( 2x = frac{pi}{2} - frac{pi}{4} = frac{pi}{4} )( x = frac{pi}{8} )Now, checking if ( frac{pi}{8} ) is within our interval ( left[-frac{pi}{4}, frac{pi}{4}right] ). Since ( frac{pi}{8} approx 0.3927 ) radians and ( frac{pi}{4} approx 0.7854 ) radians, yes, ( frac{pi}{8} ) is within the interval.For the minimum value, ( sin(theta) = -frac{sqrt{2}}{2} ) occurs at ( theta = -frac{pi}{4} ). So,( 2x + frac{pi}{4} = -frac{pi}{4} )Solving for ( x ):( 2x = -frac{pi}{4} - frac{pi}{4} = -frac{pi}{2} )( x = -frac{pi}{4} )Which is exactly the left endpoint of our interval. So, the minimum occurs at ( x = -frac{pi}{4} ).Wait, but I should also check if there's a possibility of another minimum within the interval. Since the sine function is decreasing from ( frac{pi}{2} ) to ( frac{3pi}{4} ), but in our interval, the argument only goes up to ( frac{3pi}{4} ), which is less than ( pi ). So, the minimum should indeed be at the left endpoint.Just to be thorough, let me compute ( f(x) ) at the endpoints and at critical points within the interval.Critical points occur where the derivative is zero. From part (1), the derivative is ( 2sqrt{2} cosleft(2x + frac{pi}{4}right) ). Setting this equal to zero:( cosleft(2x + frac{pi}{4}right) = 0 )Which occurs when ( 2x + frac{pi}{4} = frac{pi}{2} + kpi ), for integer ( k ).Solving for ( x ):( 2x = frac{pi}{2} - frac{pi}{4} + kpi = frac{pi}{4} + kpi )( x = frac{pi}{8} + frac{kpi}{2} )Within our interval ( left[-frac{pi}{4}, frac{pi}{4}right] ), let's see which ( k ) values give ( x ) in this range.For ( k = 0 ): ( x = frac{pi}{8} ) which is within the interval.For ( k = -1 ): ( x = frac{pi}{8} - frac{pi}{2} = -frac{3pi}{8} ), which is less than ( -frac{pi}{4} ) (since ( -frac{3pi}{8} approx -1.178 ) and ( -frac{pi}{4} approx -0.785 )), so it's outside.For ( k = 1 ): ( x = frac{pi}{8} + frac{pi}{2} = frac{5pi}{8} ), which is greater than ( frac{pi}{4} ), so also outside.Therefore, the only critical point within the interval is ( x = frac{pi}{8} ), which we already considered for the maximum.So, evaluating ( f(x) ) at the endpoints and the critical point:At ( x = -frac{pi}{4} ):( f(-frac{pi}{4}) = sqrt{2} sinleft(2(-frac{pi}{4}) + frac{pi}{4}right) = sqrt{2} sinleft(-frac{pi}{2} + frac{pi}{4}right) = sqrt{2} sinleft(-frac{pi}{4}right) = sqrt{2} times (-frac{sqrt{2}}{2}) = -1 )At ( x = frac{pi}{8} ):( f(frac{pi}{8}) = sqrt{2} sinleft(2(frac{pi}{8}) + frac{pi}{4}right) = sqrt{2} sinleft(frac{pi}{4} + frac{pi}{4}right) = sqrt{2} sinleft(frac{pi}{2}right) = sqrt{2} times 1 = sqrt{2} )At ( x = frac{pi}{4} ):( f(frac{pi}{4}) = sqrt{2} sinleft(2(frac{pi}{4}) + frac{pi}{4}right) = sqrt{2} sinleft(frac{pi}{2} + frac{pi}{4}right) = sqrt{2} sinleft(frac{3pi}{4}right) = sqrt{2} times frac{sqrt{2}}{2} = 1 )So, comparing these values: ( f(-frac{pi}{4}) = -1 ), ( f(frac{pi}{8}) = sqrt{2} ), and ( f(frac{pi}{4}) = 1 ).Therefore, the maximum value is ( sqrt{2} ) at ( x = frac{pi}{8} ), and the minimum value is ( -1 ) at ( x = -frac{pi}{4} ).I think that covers everything. Let me just recap:1. The period is ( pi ).2. The function increases on intervals ( left(-frac{3pi}{8} + kpi, frac{pi}{8} + kpiright) ) and decreases on ( left(frac{pi}{8} + kpi, frac{5pi}{8} + kpiright) ).3. On the interval ( left[-frac{pi}{4}, frac{pi}{4}right] ), the maximum is ( sqrt{2} ) at ( x = frac{pi}{8} ), and the minimum is ( -1 ) at ( x = -frac{pi}{4} ).I don't see any mistakes in my reasoning, so I think this is correct.**Final Answer**(1) The smallest positive period is (boxed{pi}), and the function is increasing on intervals (left[-frac{3pi}{8} + kpi, frac{pi}{8} + kpiright]) and decreasing on intervals (left[frac{pi}{8} + kpi, frac{5pi}{8} + kpiright]) for any integer (k).(2) The maximum value of (f(x)) is (boxed{sqrt{2}}) at (x = boxed{frac{pi}{8}}), and the minimum value is (boxed{-1}) at (x = boxed{-frac{pi}{4}}).

💡To determine the greatest integer less than ( frac{11}{3} ), I start by converting the improper fraction into a mixed number. Dividing 11 by 3 gives ( 3 frac{2}{3} ). This value is slightly more than 3. Since the question asks for the greatest integer that is strictly less than ( frac{11}{3} ), the appropriate integer is 3.