Log y to the base x square plus log x to the base y square equal to 1. Y equal to x square minus 13 the value of x square plus y square is a very interesting question. See how we go about this. Log y to the base x square. This is sometimes tricky to do. All these questions what I prefer to do is call it as log y by log x square.
Boat to the base x. log x square to the base x is 2 log x to the base x which is 2 or this is nothing but 2 times log y to the base x which makes this 2 times log x to the base y not 2 1 by 2. Half log y to the base x this is 6 log x square to the base x is the denominator this is 2 the half log y to the basic. this is half log x to the base y. That makes it very simple.
This tells us that log y to the base x plus log x to the base y is equal to 2. Half of this plus half of this is equal to 1, or log x to the base y plus log y to the base x is 2. log x to the base y and log y to the base x are reciprocals of each other. We have k plus 1 by k is equal to 2. You know that anything of the form k and 1 by k, k plus 1 by k is greater than or equal to 2 or less than or equal to minus 2. It will be 2 only if k is 1. Only 1 plus 1 by 1 is 2. It is more than 1 or less than 1, positive and more than 1, positive and less than 1. Then it will be more than 2. 1 plus 1 by 1 is the only instance where k plus 1 by k can be 2. That means we know that k is 1. That means y is equal to x. Effectively, this whole thing is an elaborate way of saying x equal to y or we are solving for x equals x square minus 30 or x square minus x minus 30 is equal to 0. Notice, so This can be broken as 6 and 5. x minus 6 into x plus 5 is equal to 0. Minus 6 plus 5 is minus 1, minus 6 into 5 is minus 30. x is 6 or minus 5. x cannot be minus 5, not defined. x is 6, x is 6, that means y is 6. We have already established that x is equal to y. If looking at x equal to y equal to 6 or x square plus y square, 6 square plus 6 square, 36 plus 36, 72. When Geeta increases the speed from 12 kilometers per hour to 20 kilometers per hour speed from 12 kilometers per hour to 20 kilometers per hour she takes one hour lesser fewer or lesser than the usual time to cover the distance between her home and her office and the distance between her home and office very very straightforward question speed 12 to 20 The ratio 3 is to 5. Time taken.
Therefore, should be in the ratio 5 is to 3. This is 1 hour more. This is 1 hour more. The difference between the time taken is more. This is 5x.
These questions work really well if we think of it as a ratio. This is 5x. This is 3x. The difference is 2x.
This 2x is 1 hour or this 1.5 hours, this is 2.5 hours. Distance traveled is either 12 kilometers per hour into 2 and half hour or 20 kilometers per hour into 1 and half hour. 3 by 2 into 20 which is 30 or 5 by 2 into 12 which is also 30. Obviously, the distance has to be the same.
If distance is the same, speed into time is speed into time. That means speeds are in the ratio A is to B, times have to be in the ratio B is to A. Ratio of the speed is 3 is to 5 ratio of the time should be the other way around 5 is to 3 This is 5 is to 3 5 X and 3 X.
That means the difference is 2 X the difference is 1 R That is 1 R. This is 1.5 R. This is 2.5 R.
Solve it distance is 20 into 3 by 2 R 30 kilometers 20 kilometers per hour Into 3 by 2 R 30 kilometers super straightforward question should be looking at these killing them The number of triangles that can be formed by choosing points from seven points in a line and five points in another parallel line Seven points in this line one two three four five six seven one two three four five I have to form triangles. I cannot have three points from the same line forming a triangle. So, I should either have two points here and one here or two points here and one here.
So, I am looking at either 7c2 into 5c1 or 7c1 into 5c2. 7c2 is 7 into 6 by 2, which is 21. 5c1 is 5. 21 into 5. 7c1 is 7. 5c2 is 10. 5 into 4 by 2. 7 into 10. 7 into 10 is 70. 21 into 5 is 105. 105 plus 70 is 175. Easy as pie. Very straightforward question. Aruna purchased a certain number of apples for rupees 20 each and number of mangoes for rupees 125 each. Apples, mangoes.
price per unit rupees 20 rupees 25 she sells all the apples at 10% profit all the mangoes at a 20% loss overall she makes neither profit nor loss all the apples at a 10% profit 10% profit profit rupees 2 per unit mangoes at a 20% loss losses 20% of this minus 5 per unit She makes no profit no loss. That means this profit from all the apples offsets the loss from all the mangoes Everything she makes two rupees profit here. She makes five rupees loss Two into whatever this is will be equal to five into whatever the unit This is R the number of units in the ratio five is to two per unit number is two and minus five Whatever she makes here is offset here R She sells 5n number of apples and 2n number of mangoes.
Profits will be 5n into 2, 10n. Losses will be 2n into minus 5, minus 10n. Goes off, comes to 0. Now what does she do? So she sells 5n and 2n. Now instead if she sells all the apples at a 20% loss, new scenario, loss per apple is minus 4. Now, Mang goes at a 10% profit, 25 into 10, 10% of this is 2.5.
Overall, she makes a loss of 150 rupees. So, minus 4 into 5n plus 2.5 into 2n equals rupees 150, but a loss. I am putting profit as positive, loss as negative.
Minus 4, atrocious. Minus 4 into 5n. I am sorry about this.
Minus 20n plus 5n which is minus 15n is minus 150 or n is 10 or she buys 50 apples and 20 mangoes. Nice. What is the question we have then? The number of apples purchased by Aruna is 50. 50 apples and 20 mangoes.
The key part here is to look at the first statement and then quote. The losses, the number of apples in the ratio, apples to mangoes in the ratio 5 is to 2. Remember, we do that way through. 5n and 2n, then plug it into this, find n. The sum of coefficients of all the terms in the expansion of 5x minus 9 whole power 4. Very nicely expanded. We figure out what needs to happen.
Then we say, hey, 5x minus 9 into 5x minus 9 into 5x minus 9 into 5x minus 9. Find it, multiply. Remember, every term. Barring the constant term will have an x unit Every term is going to have a coefficient and it's going to be an x power 4 term an x cubed term x square term x term unit term Let's say this expansion a x power 4 plus b x cube plus c x square plus d x plus e That's what it becomes.
I don't care what a b c d are this is how it will become What do I need to find I need to find a plus b plus c plus d plus e That is what I want to find sum of all coefficient in this expansion. Now if I call this as say polynomial of x. This is a polynomial in x and I want to find a plus b plus c plus d plus c. What do I do? I say nice polynomial in x.
Let me find p of 1. That will be nothing but a plus b plus c plus d. Or I can expand this and find all the coefficients. Or I can just substitute x equal to 1. Whatever that value is, that is my answer. If I put x equal to 1, 5 into 1 minus 9, 5 minus 9 minus 4 whole power 4, which is 4 power 4, 16 square. 256. Sum of all coefficients is probably the easiest question in this whole lot in binomial theorem.
Planck in x equal to 1, which is expanded expression, find the numerical value here too. A new sequence is obtained from the sequence of positive integers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 by deleting all the perfect squares. In the 2022nd term of this new sequence, 2022nd is awfully large number.
We need to find that number. Let's simplify this. 1, 2, 3, 4, 5. Let's say I consider it till 100. New sequence would be 2, 3, 5, 6, 7 till 99. If I go till number 100, I won't even have 100. 100 won't be there in my sequence. Therefore, I'm going to include 101. If we go and take all numbers till 101, till then there are 10 squares sitting there. How many numbers get used?
100 minus 10, 90 numbers get used. 101 is the 91st number. 101 is the 91st number.
Square plus 1. I can simplify like that. I'm going to look at this 2022nd term. I need to think about 2022nd term here. This is going from here is difficult. I'm going to think about a square plus one there.
Close to 2022. 2025 is a square. 45 square is 2025. So what I'm going to do? Nice. I'm going to think about 1, 2, 3 up till 2025 and then 2026 number.
Of which 1, 4, 9 all of these will disappear. In my new sequence, we will have 2, 3, 5, 2024, 2026. 2026, up to this number, I will have 2025 minus 45 terms. Why minus 45? 45 square is 2025. So, if I subtract this, it is 1980 terms. So, 2026 is the one and… 1 9 8 2th term is 2027, 1 9 8 3, 1983rd term is 2028, 84th term is 2029 and so on.
From here till 2022, I won't get the next square. 46 square is far away. After 45 square, I'm not finding anything else. So, 1981, the 198 first term is 2026. So, to this, I am adding 45. 2022 plus 45 is my answer, 2067. 2067 will be my 2022nd term. I know that squares get knocked off.
Finding the nth term is difficult. Finding the rank of n square plus 1, that is easy. n square plus 1, 45 square plus 1. So, 45 square plus 1, I am looking at 45 square plus 1. I know before this I have kicked out 45 numbers. So, 45 square plus 1 will be the 45 square plus 1 minus 45, that term. This number, before that 45 terms would have got kicked out.
that is 1981. I find the closest square to this 2022 such that this number is below 2022 with a square number. So I don't, after this I don't skip any other number till I go to the 2022 term. 46 square is very large. 45 square plus 1 will be here.
46 square plus 1 will be another 90 or 91, so far away, not before 2022. So I'm good. 2022 term will be 2067. The third, fourteenth and sixty-ninth terms of an arithmetic progression form three distinct and consecutive terms of a geometric progression. The next term of the geometric progression is the nth term of the arithmetic progression. Then n equals t3, t14, t69. Geometric progression.
You know I have to figure out the next term of this geometric progression and then bring it back to the arithmetic progression method. I know this is a plus 2d, a plus 13d, a plus 68d. I am not going to do that. I am going to define this as b. Geometric progressions get defined very well with the middle term.
From here 11 terms before, this is b minus 11d. From here 14 to 69, 69 minus 14 is 55, this would be b plus 55d. Why am I defining like this? Because for my geometric progression, I like it if I am anchored around the middle term.
a, b, c, I like it as b by r, b, b plus r, easier. I can say b square is equal to ac and then simplify and then find d. So my life is easier if I can anchor around a middle term.
So I say hey nice interesting so I am going to anchor around this term as a middle term. From here I have to add common difference 55 times to go here. 14th term 69th term.
Backward I have to subtract common difference 11 times to get here. So b square equals b minus 11d into b plus 55d. something to go with, something that I can simplify it and get some value of d.
Let us see if we can get that b minus 11 d into b plus 55 d equal to b square. So, b square minus 11 bd plus 55 bd minus 11 into 55 d square equals b square. 44BD equals 11 into 55D square.
We are told this is distinct and consecutive terms. That means the common difference is not 0. I can cancel off a D or I get that 4B equals 55D. I am cancelling 1D, 1D remains. 4B equals 55D nice or B 4B equals 55D, D is 4 by 55 times B nice wonderful.
What is the next term of the geometric progression nice. Next term of the geometric progression this this this and the term after that It is B plus 55 D whole square by B looking for B plus 55 B whole square by B fourth term of the geometric progression We need to find this and then bring it back to this form b plus 55 d the whole square d is 4 by 55 b which is nice nice now 4 by 55 into 55 makes it 4 b this is 5 b the whole square by b is the fourth term of the g b 25 b square by b which is 25 b is the fourth term of the b square this is nth term of the AP. AP was anchored around B.
B plus some number of times D equals 25B or some number of times D is 24B. D is 4 by 55B. This into 4 by 55B is 24B. This is 6. B gets cancelled this number is 55 into 6 which is 330. So from B we add common difference 330 times to get the next term.
It's brilliant. We need to remember that for us B is not the first term, it's the 14th term. From here we add 330 times common difference to get the next term 330 times. So what term are we looking for from here 330 times the third term to the right or this term?
The nth term is 14 plus 3.30 which is 3.44. Super tricky, very heavy duty algebraic, something where we need to anchor around one term, use the idea that if we have a, b, c in g, p, then my sequence is b square equal to a, c, solving for b square equal to a, c. Other idea that we use here, we have three terms in a, b, g, p, a, b, c, the fourth term is d.
Then we know BD equal to C square, D equals C square by B. We use that to derive that term. Then come back to the AP and then find how many times common difference we need to add.
344 is the answer. Properly algebraic heavy duty solving involved. The number is minus 16, 2 power x plus 3 minus 2 power 2x minus 1. minus 16 and 2 power 2x minus 1 plus 16 are in AP, are in arithmetic progression, what is x equal to?
Any term in arithmetic progression, this plus this is equal to 2 times this. So, 2 times 2 power x plus 3 minus 2 power 2x minus 1 minus 16. equals minus 16 plus 2 power 2x minus 1 plus 16. Lovely. This worked rather well for us. So, this is 2 power x plus 4 minus 2 power 2x minus 32 equals 2 power 2x minus 1, which is good.
What we are going to do, this is 2 power 2x, this is 2 power 2x minus 1. I am going to say rather nicely 2 power x equal to y, we should get a quadratic. This is 2 power x into 2 power 4, 2 power x is y, 2 power 4 is 16, 16y minus y square minus 32. This number is 2 power 2x by 2 is equal to y square by 2. bring the minus y square this side 16 y minus 32 3 by 2 y square 32 y minus 64 3 y square 3 y square minus 32 y plus 64 is equal to 0. we have to break this 32 into two parts to get the product 3 into 64 So, 3 into 64 is 6 into 32, this is 12 into 16, 24 into 8, 48 into 4, 96 into 2, we are not doing there. Some two of these numbers need to add to 32, 24 into 8 works, 3y square.
minus 24y minus 8y plus 64 equal to 0. 3y into y minus 8 minus 8 into y minus 8 is 0. 3y minus 8 into y minus 8 is 0. y is 8 or 8 by 3. y is 2 power x. So, we know that 2 power x is equal to 8 or 8 by 3. There are two possible values, both should work. We have asked this for and they are saying x equal to what?
I can obviously not say x is such that 2 power x is 8 by 3. x is such that 2 power x is 8 gives us x equal to 3. 3 works, we are through. 2 cube is 8, x is 3. In all of these questions, frame the arithmetic progression property first. Then, if you put 2 power x equal to y or k, then 2 power 2x becomes y square or k square.
we generate a quadratic in y solve that and then come back to x the key thing is to go from here to here by making this substitution and we are through. Mrs. Sharma and Mr. Sharma, Mrs. and Mr. Ahuja along with four other persons are to be seated at the round table for dinner nice If Mrs and Mr Sharma are to be seated next to each other and Mrs and Mr Ahuja are not to be seated next to each other then the total number of seating arrangements are lovely lovely for eight people around a circle eight people around a circle is seven factorial that we know okay but this is not that we are told that there are two couples one couple have to be seated next to each other Another couple, Mrs. and Mr. Sharma are to be seated next to each other. And Mrs. and Mr. Zakuja are not to be seated next to each other.
Our problem for circular arrangement, we say n people are to be arranged around a circle. Only relative positions are there. So that's why the answer is not n factorial, but n minus 1 factorial.
Because a circle does not have a starting step and an end step. The whole idea of circular arrangement kicks only relatively. So we place one person sit there. Around that person or next to that person next to next to that person this side and that side we seat everybody I want to follow the same approach Mr. Sun, Mrs. And Mr. Sharma are to be seated next to each other.
So I say Mrs. And Mr. Sharma two seats taken by them. It's the Sharma family Then we have one two, three, four, five six seats left Around these two, these six can be placed. The second condition were not there. Mr. and Mr. Ahuja are not to be seated next to each other.
Assume that condition is not there. Then we can say, six people, six seats. Now the circular arrangement idea is not there because the Sharmas are together. They anchor my positioning. I'll have one, two, three, four, five, six.
One position to the left, two positions to the left, three to the left, four to the left, five to the left, six to the left. Out of this block, I am through 6 factorial. To keep one thing in mind, Mrs. and Mr. Sharma are together.
They could be Mrs. and Mr., Mr. and Mrs., both ways. So, my answer would be 2 into 6 factorial. I could have Mrs. and wife and husband, husband and wife, both will work. But this is not enough. This is not my answer.
It is not 2 into 6 factorial because the 2 into 6 factorial does not account for the factor. Mrs. and Mr. Ahuja are not to be seated next to each other. How do we handle that not to be seated next to each other? You say, okay, let me count all possibilities where they are seated next to each other. I'll subtract that.
My answer should have been 2 into 6 factorial. From this, I'm going to subtract all possibilities where they are not seated next to each other. So, all possibilities where they are indeed seated next to each other. Sorry.
That means... Mrs. and Mr. Ahuja could have been in these two seats or these two or these two or these two or these two Combination 1, combination 2, combination 3, combination 4, combination 5 Assume they are in these two, then there are four seats remaining, four people, four factorial ways They were in these two 4 seats remaining, 4 factorial ways are the number of ways that I should not count is 4 factorial into 5. Why 4 factorial? After I have put Mrs. and Mr. Ahuja, 4 seats are remaining, 4 people, 4 factorial ways.
Why 5? Mrs. and Mr. Ahuja could be seated next to each other in 5 different combinations. 1 and 2, 2 and 3, 3 and 4, 4 and 5, 5 and 6. One other thing that I need to remember here which I... kind of overlooked when I did this question originally is the fact that Mrs and Mr Ahuja in each of these combination could be husband and wife or husband and wife, husband and wife, husband and wife, husband and wife, husband and wife. So it is not just 4 factorial into 5, it is 4 factorial into 5 into 2. So for the husband and wife being, Mrs and Mrs Sharma being there in one position, They could be husband and wife here in taking these two slots and then I'll have to subtract this much.
So if I had them here, I have six factorial combinations. From that, I'll have to subtract this. And then I could have them the other way around.
From that, I'll have to subtract this. So husband and wife, husband and wife could happen both ways. It could be H and W, W and H for Sharma's.
H and W, W and H for Ahuja's. This is for the Ahuja part, this is for the Sharma part. Nice. Now, how do I combine this? I have to subtract one from the other.
I have to keep one thing very careful in mind. Given one sequence of Sharma's, that is 6 factorial, I will have to subtract 4 factorial into 5 into 2. The whole thing into 2, for all of these arrangements, I will have to add one by one. 6 factorial minus 4 factorial into 5 into 2 whole thing into 2 husband and wife, wife and husband each of those combinations 6 factorial is 720 5 to 2 is 10, 4 factorial is 24, this is 240. 2 times 720 minus 240 is 2 times 480, which is 960. Excellent, excellent, excellent question.
I'm going to go over the thought process one more time. I'm going to first think in terms of, hey, what do we do if we have the first condition alone in mind, Mrs. and Mr. Sharma together, both of them together, and 6 slots, 6 people, 6 factorial. Both of them together could be husband and wife, wife and husband, 2 into 6 factorial, total possibilities Now for each of those possibilities, one sequence of husband and wife here, I could have I am subtracting the scenarios where the Ahujas are also sitting together So I am sitting the Ahujas together, here, here, here, here or here, 5 ways Each of those 5 ways, there are 4 factorial ways for the remaining 4 people to sit And each of those 5 ways, I could have HNW W and H, H and W, W and H, each of them into 2. 4 into 5, 4 factorial into 5 into 2, 240 ways in which we can seat the Ahuja such that they are seated together for each of those 6 factorial ways. For the husband and wife like this, then husband and wife like this.
So, my final answer is 2 times 6 factorial minus this number. Assuming that husband and wife are, wife and husband are here, I will have to subtract this. Assuming wife and husband are like this, I have to subtract this. So, twice of 6 factorial minus this, 720 minus 240, 480, whole thing into 2, 960. Beautiful question, super challenging question.
Let 50 distinct positive integers be sorted such that the highest among them is 100. The average of the 25, largest 25 integers among them exceeds the average of the remaining integers by 50. the maximum possible value for the sum of all the 50 integers so nice 100 integers sorry 50 integers distinct all of them top 25 are here bottom 25 are here this will be some average this will have some average this difference is 50 largest among these is 100 The average of the largest 25 integers among them exceeds the average of the remaining by 50. We want to find the maximum possible value of sum of all. We want these to be as high as possible. They are distinct.
Forget about this 25. If these 25 are high, if the average is 50, this will also be as high as possible. What is the scenario for these 25 being high? I have numbers that are close to each other starting from 100. 99, 98, 97, 96, 95, 94, 93, 92, 91. and so on.
That means I take 25 integers starting from 100 or 76 to 100. Those are my 25 largest integers. What is the average among these? Will be the middle term 100 plus 76 by 2, 176 by 2. This average will be 88. Beyond 88, 89 to 100 are 12 numbers.
Before 88, 87 to 76 are 12 numbers. 88 is the middle term. That's where the average will properly offset.
This side slightly lesser, slightly lower. Greater lower, greater lower. Perfectly offsetting each other, average will be 88. Average of the 25 largest integers is 88. Average of the 25 smallest integers is 50 less than that.
38 what is the overall average or what is the sum of all this sum of all this is 38 into 25 sum of all this is 88 and 25 or the total is 38 into 25 plus 88 and 25 or 25 times 88 plus 38 16 1 9 plus 3 12. 25 x 126 or 50 x 63 or 100 x 31.5, 3150. 25 x 126 is 3150, that's maximum. Maximize the largest 25 integers. that will automatically maximize that average automatically maximize that set that is maximized the second one is automatically maximized because the average the differences between averages is 50 the 25 largest integers have an average of 88 smallest integers have the average of 38 it's the best case scenario overall total is 25 times 38 plus 88 88 plus 38 find that number the set of real values for x which satisfy this inequality log 8 base 27 is not equal to log x to the base 3. We will think of it as log 8 by log 27 less than or equal to log x to the base 3. We will choose both as base 3. log 8 to the base 3, this is 3, is less than or equal to 3 times log x to the base 3. 8 is 2 cube, so it is log 2 to the base 3, 3 times over. Log 2 to the base 3 is less than or equal to log x to the base 3, or x greater than or equal to.
x can be 2. These two will be equal. Anything more than 2, this will hold good. x is greater than or equal to 2. I am saying this is out, this is out. Z is only lesser than.
I think this should be the answer. We will try. This time, we are solving for log x to the base 3 is less than, say, 9 to the power 1 by log 3 to the base 2. Simplifying this, 9 to the power log 2 to the base 3. Think about it, I want it to be on the 9 as the base.
I can write this as 3 square and simplify this into this 3 square whole power log 2 to the base 3 or 3 power log 4 to the base 3. It becomes 2 log 2 to the base 3 which is log of 2 square to the base 3 log 4 to the base 3. 3 par log 4 to the base 3. This is a par log x to the base a. Log x to the a par log x to the base a is x. The very definition of logarithms.
Take log on both sides to the base a. This will become clear. a power log x to the base a is x.
How do I prove that? Take log to the base x, log to the base a. Log of this is log of this.
Boot to the base a, this will become equal to this. Or this side, 3 power log 4 to the base 3 is 4. This side, log x to the base 3 less than 4. 3 power 4 is 81, x is less than 81. We already got x is greater than or equal to 2. x can go from 2 to 81. 2 included, 81 not included. Solved methodically, we should get this. This funda, a power log x to the base a, y is equal to x. Take log on both sides, this is log of a to the power log x to the base a, this is log x.
take both sides log to the base a. So, base a to the base a. First, apply the exponent rule log of x to the base a times log a equals log x to the base a.
Log a to the base a is 1. Log x to the base a equal to log x to the base a, which means this whole thing. x is equal to log a power log x to the base a, very definition of logarithm. We use that here.
The value of k for which the following lines are concurrent, three lines go through the same point. What is the value of k? There is no k here, no k, there is only k here.
Solve these two. What do I do? I know x minus y is 1. You know 2x plus 3y.
equals 12, multiply this by 2, 2x minus 2y equals 2, subtract one from the other, 5y equals 10, mean y is 2, this is 2, this is 3, x is 3, y is 2. Substitute that here, it should be satisfied, x is 3, y is 2, what we get, 2 into 3 minus 3 into 2. plus k is equal to 0. 6 minus 6 plus k equal to 0. k should be 0. If three lines are concurrent, they should have a common point. All three paths to the same point. Find the point these two paths to substitute that here should work. Find k. k is 0. 3B.
The lengths of the sides of a triangle are x, 21 and 40 where x is the shortest side. A possible value of it. Sum of any two sides of a triangle are greater than the third. x is the shortest side.
x plus 21 greater than 40 is our equation. Or x greater than 40 minus 21, 19. 19 plus 21 won't form a triangle. These two won't.
3B. We're basically saying take this mark. I want everybody to get this right.
Think about this. It is a beautiful funda sitting here. I draw a number line and I have points p and q here. Distance between these two is q minus p.
More generally, the sum two points in the number line, distance between them is modulus of m minus n. Why am I saying modulus of m minus n? I do not even need to know whether it is m here and n here. or n here and m here, doesn't matter.
Two points in the number line, what is the distance between them? Modulus of p minus q, modulus of q minus p, modulus of m minus n. That idea is super powerful. That idea is going to crack open this question brilliantly.
Whenever you have modulus of x minus 1 plus modulus of x minus 4 plus modulus of x plus 3, all of that, then say, okay, the bunch of things involved here, I'm going to draw a nice number line. I have a modulus of x, so I'm going to put 0. I have models of x minus 1, I am going to put 1, x minus 2, 2, x minus 4, 4, x minus 6, 6, x minus 10 My diagram is looking ugly, I am going to erase everything, I thought there was some x plus as well I am going to put 0 here, 1, 2, 4, 6 and 10, nice f of x is modulus of x plus 2 times modulus of x minus 1 is 2 plus modulus of x minus 2 plus modulus of x minus 4 plus modulus of x minus 6 is 2 times modulus of x minus 10. Nice. And so, forget the 2 times part. So, suppose x were here, x minus 0, x minus 1, x minus 2, x minus 4, x minus 6, x minus 10. I am calculating all these distances. Suppose x were here.
x minus 10, x minus 6, x minus 4, x minus 2, x minus 1, x minus 0. I am adding all these distances. These are two types. I am adding all of them. It is better to add from here. I do not have to think about x here.
If I put x at 0, that will be better than putting x here. All these distances will not get added. First step.
Likewise, beyond 10, I don't need to worry. I'll worry about 10. But if I worry about 10, 10 to 6, 10 to 4, 10 to 2, 10 to 1, 10 to 0, instead of that, I had x in the middle. To 10, to 0, to 6, to 1, to 4, to 2. I can count like that. I can already intuitively set that it will be better if I put x in the middle, not on one side. That's a beautiful idea.
Digging deeper, if I put x here, this is modulus of x minus 2, this is modulus of x minus 4. Add these two distances, that's 2. Wherever x is in between, it doesn't matter. Likewise, x minus 0 is this distance. x minus 1, I'm putting x here, x minus 1 is this distance.
x minus 6 is this distance, x minus 10 is this distance. If I put x here, I'll have x minus 4, x minus 6, x minus 10, x minus 2. Just depending on where x moves, I'll have the numbers changing a little bit, right? I can sense that if x is more to the center, Then I can split these terms.
If I put x's here, then each of these keeps adding up. If I put x inside, then x to 1 and x to 0 gets shortened. If I put x of 0, then each of these distances gets added.
If I put x here, all of these lengths become shortened. x to 0 also will be, will not be non-zero, but it's going to get offset. I can sense that x is somewhere in between, if not in between 2 and 4. It is in between 1 and 10. Where in between? What I am saying is in between 10, this modulus of x minus 2 and modulus of x minus 4. These two add up to 2. Likewise, modulus of x minus 1 and x minus 10. x minus 1 and x minus 10. These two.
adapter 9. So let's pair them up and we will use our fact that we have color coding available and pairing these two. Then what do we pair? We pair up modulus of x and modulus of x minus 6. Modulus of x should go all the way to here. and then we pair up modulus of x minus 1 and x minus 10. If you pair them up like this, modulus of x minus 2 is modulus of x minus 4 is 2. This plus this is 6. This plus this is 9. While the distance between 1 to 10, if x is in between anywhere here, is adding up to 9. This distance of that distance adds up to 9. This distance plus this distance adds up to 2. This distance plus this distance adds up to 6. Or effectively, we have 2, we have 6, we have 9 into 2. 9 into 2 is 18 plus 6 is 24 plus 2 is 26. That is the minimum.
So, this whole question becomes far better if we understand the distance between x and y on the number line is modulus of x minus y. So we are doing modulus of x minus 2 plus modulus of x minus 4 distance to 2, distance to 4. Best or minimum that the number is between 2 and 4. Look at the number between 2 and 4. I can pair 2 and 4. I can pair 0 and 6. I can pair 1 and 10. Why am I pairing 1 and 10? Because this is getting doubled. x minus 1 and x minus 10 coming twice over. 9 into 2, 18 plus 6 plus 2. 18 plus 6, 24 plus 2, 26. Just to drive home that funda, that modulus of p minus q is the distance between p and q.
That idea is super powerful. So grab onto that idea, it will be very simple. It's an excellent question.
We're dealing with f of x here. Not really a functions question. It's a beautiful question based on one simple funda. Suppose I say 2 power y.
It is dealing with a 2 power y. y can take any value. 2 power y cannot be negative. Super important idea, very powerful idea.
2 power y cannot be negative. It's a very simple, very powerful idea. 2 power y cannot be negative.
It cannot be negative. 2 power 3 is positive, 2 power 1 by 3 is positive, 2 power minus 3 is positive, 2 power minus 1 by 3 is positive. 2 power y cannot be negative. Super important idea.
So, this question, Therefore, it's a wonderful question which is 81 power x plus 81 power f of x equal to 3. We want to find out all possible values of f of x. First of all, if this went to 0, it could go to 0. Assuming it can go to 0, x can be 81 power x equal to 3. 3 power 4 is 81. x can be 0.25. x cannot be more than 0.25, tallying point.
We are not worried about x, we are talking about f of x. Now, what we can notice here is 81 per x is less than or equal to 3. 81 per f of x is less than or equal to 3. This cannot be negative, that cannot be more than 3. This cannot be negative, this cannot be more than 3. 81 per 0.25 is 3. So x has to be less than 0.25, f of x has to be less than or equal to 0.25. If f of x is equal to 0.25, then this part has to go to 0. It should be minus infinity.
81 power minus infinity can be taken to be 0. So f of x cannot be 0.25, but can be anything up to 0.25. It should be 0.25 or lesser. It cannot be more than 0.25.
This is out. This is out. This is out.
Key point here is this one. 2 power y cannot be negative. 10 power y cannot be negative.
11 power y cannot be negative. 14 power y cannot be negative. Positive number power something cannot be negative. Very simple panda.
Very powerful panda. This cannot be negative. This cannot be negative. I mean this is less than or equal to 3, this is less than or equal to 3. What people less than or equal to 3? Less than 3. So that one cannot be 0, cannot go to 0. Only minus 3, it will go to 0. So if 81 power f of x is less than 3, f of x is less than 0.2.
So it cannot go above 0.2. Beautiful question. Particularly a very interesting question to deal with mod x and mod y.
So first of all, whenever mod is involved, We know if x equal to 2 satisfies it, x equal to minus 2 will satisfy it. In some form, whatever satisfying means, if x equal to 2 works, x equal to minus 2 will work. There is a mod sitting here as well. If y equal to 1.8 works, y equal to minus 1.8 will also work. So we say, I am going to forget about negatives.
I deal only with positives. I draw x axis, y axis. Then I think about. 2x plus 3y equal to 6. I understand this.
I get this. I want to think of it as 2x plus 3y equal to 6. When x is positive, y is positive. When x is positive, x is equal to mod x. When y is positive, y equal to mod x.
So I am going to forget these three quadrants. I am going to deal only with this quadrant. Then think about 2x plus 3y equal to 6. Then say, hey, nice.
Then x equal to 3, 3 comma 0. y is 0. That will work. 0 comma 2 will be, if you join this, this is the equation, 2x plus 3y equal to 6. Nice. And I say, hey nice, but this is 2x plus 3y equal to 6. I am dealing with modulus. Plus and minus both will work.
So if I put x is minus 3 also it should work. If you put 0 comma minus 2 also it should work. If a point here works, a point here will work.
Point here works, a point here will work. Instead of x put minus x, that will work. Instead of y put minus y, that will work.
R, I'm dealing with a shape. A nice little rhombus. This being 6, this being 4. Area is half into product of diagonals. Beautiful question to drive home the idea of thinking about modulus The mod x and mod y is involved draw in the first quadrant nice and simple take that shape Tend to it will be linear image here.
It will be symmetric all over the board linear through Same funda applies for x square y square also Best wishes Shh