 So, good morning to everyone. Today, we will take exercise seven of our sequence and city chapter. So, in this exercise, we are going to see the applications of arithmetic mean geometric mean inequalities. And we will find the value of a given function like we will find the maximum or minimum value of a given function with the help of this inequalities. Right. So, let's begin with question number one, question number one of the exercise number seven. So, question is saying, question is saying the minimum value of four raised to power X plus four to the power two minus X, and X belongs to real number, right. So, we have to find the minimum value of this given function. So, I'm taking it as four raised to power X plus four raised to power two minus X, we have to find the minimum value of this, right. So, let's consider two number like this four raised to power X and for this we can write it as four is squared upon four couple X, right. So, let's consider these two numbers. And now we know that the value of a to the power X for X belonging to for X belonging to any real number this function always gives a positive value right. So, this is our graph of a raise to power X, right. This is our Y coordinate this is our X coordinate so this a raise to power X always it's a positive value for all the real values of X. So, these two numbers are positive. So, four raised to power X and the four square upon four raised to power X is always positive. So, since these two numbers are positive, we can apply the arithmetic mean and geometric mean inequalities which says that arithmetic mean is always greater than or equal to geometric mean. So, let's apply this inequality on these two numbers. So, it will be four raised to power X plus four raised to power two minus X. This is nothing but four raised to power two minus X this divided by two. This must be greater than or equal to four X into four raised to power two minus X whole raised to power one by two. So, this will become four raised to power X plus four raised to power two minus X greater than or equal to two into this will become four raised to power two four raised to power two and whole raised to power one by two that will be four only. So, this thing the minimum so the minimum value of this four raised to power X plus four raised to power two minus X minimum we can say the minimum value of this should be eight because this thing will be always greater than or equal to eight. And we are asked to find the minimum value so the minimum value of four for this will be eight. Hope this is clear to all. So, option D is correct. Now let's move to the next question, question number two. Now it is saying if theta lies between zero to pi, then the minimum value of sine cube theta plus Cossack cube theta plus two is we have to find the minimum value like of sine cube theta plus Cossack cube theta plus two when theta lies between zero to pi. So, theta lies between zero to pi. Okay. And we have to find the minimum value of sine cube theta plus Cossack cube theta plus two, we have to find the minimum value of this. So, we can write this as sine cube theta plus one upon sine cube theta plus two. Right. This is a function in theta. So we have to find the minimum value of this function. Now, within the interval of zero to pi, we know that sine cube theta means sine theta is positive between zero to pi between zero to pi sine theta is always positive. Right. So, this sine cube theta will also be positive and this one by sine cube theta will also be positive. Okay. So, let me tell you one thing, the value of this sine cube theta plus one by sine cube theta, the value of this thing, this will be always greater than or equal to two. Okay, now how it comes actually, if you consider, let me show you the explanation of this, why this value will be always greater than or equal to two. So, let me take two numbers, x and one by x, okay, where x is positive. So, the value of this x plus one by x means arithmetic mean is always greater than or equal to geometric mean, applying this inequality on these two numbers, x and one by x, we can see x plus one by x by two will be always greater than or equal to x into one by x raise to power half. So, from here, we can see that the x plus one by x will always be greater than or equal to two since this x and this x will be cancelled out. So, this, if x is a positive number, so the sum of x plus one by x will always be greater than or equal to two. So, applying same thing here, we can see that the sine cube theta, let's take it as x and this one by x and since x is greater than zero, like x is always positive here, sine theta is always positive between zero to pi, so sine cube theta will always be positive. So, we can say that this will be greater than or equal to two. So, what will be the minimum value of this thing? So, the minimum value sine cube theta plus cosine cube theta plus two, the minimum value of this will be, we will take minimum value of these two things that will be equal to two, two plus two, that will be four. The minimum value of the given function will be four. So, hence, option C is correct for this question. So, let's take the third question. It is saying if A, B, C and D are four real numbers of the same sign, then the value of A by B plus B by C plus C by D plus D by A lies in the interval. Lies in the interval, we have to say in which interval the value of this should lie. Okay, so let me consider four numbers A by B, B by C, C by D and D by A. Now, it is saying that the A, B, C, D are four real numbers of the same sign. Let's take like A, B, C, D are positive numbers. So, anyhow, these four numbers, what I have written here will be positive and let's take like A, B, C, D are negative. Then also this ratio, this A by B, B by C, C by D and D by A will be always positive. Okay, so we can say these four numbers are greater than zero. So, here we can apply the concept of arithmetic mean greater than or equal to geometric mean, right? So, applying this inequality on the above four numbers, we will get A by B plus B by C plus C by D plus D by A upon four. This will be greater than or equal to A by B into B by C into C by D into B by A raise to the power one by four. So, here you can see it all gets cancelled, B, B, C, C, D, D. So, we are left with A by B plus B by C plus C by D plus D by A. That should be greater than or equal to four, one raise to power one by four. This will be nothing but one only. So, this will be greater than or equal to four, right? So, we have to, what was asked in the question, the value of this lies in the interval. So, the value of this thing will be lie in the interval four to infinity. So, option D is correct. The value of this will lie between four to infinity. It will be greater than or equal to four. So, option D is correct. And now let's take the next question, question number four. It is saying x lies between zero to pi by two. Then the minimum value of two times sin x plus cos x plus cos x two x whole q is. So, we have to find the minimum value of this thing. And range of x is given, x lies between zero to pi by two. Okay. And we have to find the minimum value of two times sin x plus cos x plus cos x two x whole raise to power three. We have to find the minimum value of this function. So, within an interval of zero to pi by two, sin x will be always positive. So, sin x is positive between zero to pi by two. Within zero to pi by two, cos x will also be positive. Okay. And this cos x two x, cos x two x, like when x lies between zero to pi by two. So, two x will lie between zero to pi. So, within zero to pi this cos x function is also positive. So, we have considered these three functions, these three functions sin x cos x and cos x two x and they both, sorry, all the three are positive. Right. So, here also we can apply the concept of arithmetic mean greater than or equal to geometric mean. What we will get, we will get sin x plus cos x plus cos x two x upon three, this should be greater than or equal to sin x into cos x into cos x two x. I am writing cos x two x as one upon sin two x. Right. This whole raise to power one by three. So, it will be sin x plus cos x plus cos x two x. This should be greater than or equal to three into this sin x into cos x and I am writing this sin two x as two sin x cos x. Whole raise to power one by three. So, this sin x, this sin x will be cancelled cos x cos x will be cancelled. So, it will finally become this thing. Let us write it as a. Okay. So, this a is greater than or equal to three into one by what is this one by two raise to power one by three. Or we can do one thing like we could have taken cube on both the sides earlier itself. Okay, not an issue. Let me consider. Let me take cube now. So, taking cube taking cube on both sides, it will become greater than or equal to 27 into one by two. Okay. So, our question was two times a cube. Right. Two times in our function it is given that two times of this thing sin x plus cos x plus cos x cos x plus cos x cos x to x cube. So, multiplying by two, we will get two into a cube, should be greater than or equal to 27. So, our function the the minimum value of our required function will be 27. So, we can write 2 into is sin x plus cos x plus cos x 2 x, cos x 2 x whole raise to power 3 should be greater than or equal to 27. So, minimum value will be 27. So, option A is correct. Now, let us take question number 5. In this question, it is given A plus B plus C is equal to 3, where A is greater than 0, B is greater than 0 and C is greater than 0. Then the greatest value of A square into B cube into C square is, we have to find the greatest value of A square B cube C square and the given information is A plus B plus C is equal to 3. So, this is given in the question A plus B plus C is equal to 3 and all the 3 numbers this A, B and C are positive, okay. And we have to find the greatest value of A square B cube C square, we have to find the maximum value of this. So, let me do one thing. Let me consider number A by 2, A by 2, then B by 3, B by 3, B by 3 and C by 2 and C by 2. Now, look, this A, B, C are positive. So, anyhow, these value A by 2, B by 3 and C by 2 will be positive, okay. Now, the reason behind taking, breaking this A, B, C in these value is like, I will be applying the inequality AM is greater than or equal to GM on these, how much number, on this 7 numbers. So, while applying this, I will be having, while writing GM, we used to take the product of all these numbers. So, it will give me A square from here, B cube from these three numbers and C square from these two numbers, right. So, that's why I have dropped these, so I break these numbers in this fashion, right. Hope this is clear to everyone. So, now I am applying this AM is greater than or equal to GM on these 7 numbers. So, it will be basically, we will add all these 7 numbers. So, after adding, this will become A plus B plus C only and the value of this A plus B plus C is known to us, right, as in the value of A plus B plus C is given in the question. So, this will be A plus B plus C divided upon 7, this should be greater than or equal to geometric mean of all these numbers. So, what will be the geometric mean of all these numbers? It will be A square B cube C square divided upon 2 raised to power 4 into 3 raised to power 3, right, and whole raised to power 1 by 7. So, hope it is now clear, like why I have broken these numbers in this fashion. So, here we are getting A plus B plus C whose value is known to us and in GM, we are getting A square B cube and C square whose maximum value we have to calculate. So, now the putting the value of A plus B plus C here, we will get. 3 upon 7, let's raise it to power 7. So, 3 upon 7, whole to the power 7, it will be equal to A square B cube C square upon 2 raised to power 4 into 3 raised to power 3. So, it will become 3 raised to power 7 into 2 raised to power 4 into 3 raised to power 3 divided upon 7 raised to power 7. This value will always be greater than or equal to A square B cube C square, right. So, we can say the maximum value of this thing, the maximum value of this, the maximum value of A square B cube C square maximum will be equal to this thing. So, what is the 3 raised to power 7 plus 3, 3 raised to power 10 into 2 raised to power 4 whole divided upon 7 raised to power 7. So, this will be our answer. 3 raised to power 10 into 2 raised to power 4, 7 raised to power 7. This is option number B. So, hence, this will be our correct answer. Now, let's move to question number 6. It is saying x plus y plus z is equal to A and the minimum value of A by x plus A by y plus A by z is given as 81 to the power alpha. Then the value of alpha or lambda, lambda, 81 raised to power lambda, then the value of lambda is, we have to find the value of lambda. And given information is, x plus y plus z is equal to A, right. And minimum value of A upon x plus A upon y plus A upon z, right. So, it is saying the minimum value of this is 81 to the power lambda. So, let me take this, let me put the value of A here. So, we can write it as A by x plus A by y plus A upon z should be always greater than or equal to 81 raised to power lambda, right. Because the minimum value is 81. So, it will, 81 raised to power lambda, so either it will be more than 81 to the power lambda or it will be equal to 81 raised to power lambda. Now, what I will do, I will put the value of A here. So, putting the value of A here, it will be x, x plus y plus z upon x plus x plus y plus z upon y plus x plus y plus z upon z, this should be greater than or equal to 81 to the power lambda. Now, let me split this. We can write it as 1 plus y upon x plus z upon x plus x upon y plus 1 plus z upon y plus x upon z plus y upon z plus 1, this should be greater than or equal to 81 raised to power lambda. Now, if I consider this y by x as A, right, let me consider this y by x as A, then what will be our x by y? x by y here. If we consider it as A, it is nothing but it is reciprocal of this. So, it will become A by, sorry, 1 by A and we know the value of A plus 1 by A, this value, this A plus 1 by A, this should be always greater than or equal to 2 when A is positive number. Okay, so similarly, this z by x, let's consider it as B. So, this x by z will be also reciprocal of this z by x, so these two numbers considering z by x and x by z, this should also, the summation of these two numbers will also be greater than or equal to 2. So, we can write it as, we can further write it as this 1 plus 1 plus 1, this 3, considering these two numbers, we can write it as 2 plus 2 plus this z by y and y upon z. The minimum value we can write, the minimum value of this will be 2. So, this will be equal to, now we are taking the equality sign here, right. So, this I am equating to 81 to the power lambda, okay. Now, I am equating because we have taken this 2 here, no. So, this will be the minimum value will be 2. So, this 3 plus 2 plus 2 plus 2, this 9 is equal to 81 raise to the power lambda. So, what does it mean? 9 is equal to 9 raise to the power 2 lambda. So, 2 lambda will be equal to 1, from here we get lambda is equal to 1 by 2, right. We are talking for the minimum value, right. The minimum value of this function is 81 raise to the power lambda. So, hence we are able to take this value as this y by x and plus x by y as 2. So, from here we are getting the value of lambda as 1 by 2. So, option A is correct. Option A is correct. Now, let's take the next question, question number 7. It is saying ABC are 3 positive numbers and ABC square has the greatest value 1 by 64. So, it's given that ABC are positive numbers, okay. And the greatest value like this, ABC square, the greatest value of this is 1 by 64. Then we have to find the value of A, B and C, okay. So, since these 3 numbers ABC are positive, we can apply arithmetic mean is greater than or equal to geometric mean. So, we can write it as A plus B plus C by 3, this should be greater than equal to ABC raise to power 1 by 3. But here this given is ABC square, not ABC, right. So, we have to split this ABC accordingly, what we have done earlier. So, let me split, let me consider, let me consider numbers as A, B and I will split this C as C by 2 and C by 2. Why am I splitting? Because while multiplying it will become AB into C square. So, now I will apply arithmetic mean greater than geometric mean on these 4 numbers. So, it will become A plus B plus C by 4 greater than or equal to ABC square raise to power 1 by 4. Now, checking the power 4 on both sides, it will be A plus B plus C by 4 raise to power 4. This should be greater than or equal to ABC square, okay. Now, the maximum value of ABC square is 1 by 64. So, this A plus B plus C upon 4 whole raise to power 4, right, this will be equal to ABC square means this equality will hold only when these 4 numbers what we have considered here. What we have considered here, these 4 numbers must be equal if we are, if this equality holds true, right, if this equality holds true, then these the considered 4 numbers must be equal to must all the 4 numbers must be equal. So, for these 2 hold, we can say A is equal to B is equal to C by 2, right, anyhow this C by 2 and C by 2 will be equal to one another. So, A must be equal to B and that must be equal to C by 2. So, let us take it as X, right. So, our A will become 2X, B will become, sorry, A will become X, B will become, B will also be equal to X and C will become 2X, okay. So, we can write it as this, no, this X plus ABC, we have considered it, we have equated all these 4 terms. So, we are getting A equal to B equal to C by 2 and that is equal to X, okay. Now, I will be putting this value, maximum value of AB square here. So, we will get A plus B plus C by 4 whole raise to power 4 is equal to 1 upon 64, right, 1 upon 64, the maximum value of AB square is 1 upon 64. So, from here, we will get A plus B plus C whole raise to power 4 is equal to 4 raise to power 4 upon 64. So, it will become 4, right. A plus B plus C whole raise to power 4 will become equal to 4. So, now put the value of ABC here. So, what was the condition? For that condition to be true, like that equality to be true, we can say this A plus B plus C X plus X plus 2X that will be 4X raise to power 4 must be equal to 4, okay. So, this will be 4 raise to power 4, whether I am doing something wrong or what, let me check once. I feel I have done some mistakes, I feel I have done some mistakes, I think. So, these four numbers we have considered, now we have applied arithmetic mean greater than geometric mean here and the value of ABC square is okay, okay, okay. I have done mistake here actually. This ABC square upon 4 will come no, these four numbers will get multiplied. So, this will be ABC square upon 4, right. So, here one more 4 will come. So, it will become like 64 into 4 here, right. So, it will basically become 1 here. It will be equal to 1 here, right. So, what we can say, we can say this 4X raise to power 4, this is A plus B plus C is 4X, 4X raise to power 4 that must be equal to 1, this must be equal to 1, right. So, from here we can say X raise to power 4 will be equal to 1 by 4X raise to power 4 will be equal to 1 by 4 raise to power 4, 4 raise to power 4, okay. So, from here we can say X must be equal to 1 by 4, okay. X must be equal to 1 by 4. So, this A what was X? This was equal to A. A was nothing but X, we have considered A as X. So, the value of A will be 1 by 4 and A and B both are equal. So, A is equal to B is equal to 1 by 4 and the value of C will be 2 times of X that is 2 times of 1 by 4 that will be nothing but 1 by 2, right. So, we got the answer as A and B, the value of A and B will be 1 by 4 and the value of C will be 1 by 2. So, let me check the option 1 by 2, 1 by 4, 1 by 4 and 1 by 2. Yes, it's here in the option number C. So, this is the correct answer. So, this exercise is done. There were basically seven questions only in this exercise, but we have only applied the concept of arithmetic mean greater than or equal to geometric mean inequalities here. We have not applied any concept of calculers in calculating the maximum and minimum value of a given function. So, some good questions based on the inequalities was there in the exercise which we have covered already. So, this is all for today. We will meet you once again shortly. Till then, Tata, goodbye.