CSIR NET MATHEMATICS | Real Analysis | Bounded Variation Q.66(A) Q.75(B) Q.63(C) (हिंदी में)

Hello friends, in this video Inside, we are real anaisis. We will discuss the problem. CSN8 June 2019 Sample Number is 63, Booklet C Marks 4.75 The problem is F of X equals X times sine of 1 over X for x belonging to the open-closed interval (0, 1] and 0 for X is equal to 0 and gx equals xfx for 0 ≤ x ≤ 1 then which of the following are true We have 4 options. The first option is that f is of bounded variation The second f is not of bounded variation The third option, g, is of bounded variation. and the fourth option g is not of bounded variation This is a problem related to bounded variation. So first, let's recall its definition f is said to be of bounded variation if over the closed interval [a, b] If this variation function ka It is becoming bound from M. F is said to be of bounded variation So for the bounded variation problem We'll probably need a partition, one of them. And second of that partition Pay, this is the summation now we consider the function f(x) is equal to x to the power n times sine 1 x to the power of n, x belongs to open closed interval 0, 1 and 0 if x is equal to 0 In the example, two functions are given to us. f(x) is equal to x times sine of 1 over x and second gx equals x fx, so x squared times sine of 1 over x These two functions This is a particular case of the function. where n equals 1 for the first function and n equals 2 for the second function And in both, the value of m is 1. For this function We will check the property of bounded variation. So we'll need a partition for it. so partition of the closed interval [0,1] Then we will consider 2i plus 1 pi over 2 power minus 1 over n And here is the value of i 0 ≤ i ≤ k What will be the value of sine at this partition? sin 1 upon Xi Jinping so it is sin x to the power minus m So x is equal to 2i plus 1. pi over 2 minus 1 upon n and its power minus m so minus 1 over m into The value of minus m will be plus 1. So it will be sin(2i + 1) pi over 2 If i equals 0, then the value of sin is sin(π/2) has the value 1 If I equal 1 Then the value will be 3π/2 So, the value of sin 3π/2 is minus 1. If i equals 2, it becomes 5π/2, the value of 5π/2 plus 1. If i equals 3, then 7π/2, the value of sine will be minus 1. Here, sine has 2 possible values, i equal to even numbers It will be 1 for i equal to even number, and minus 1 for i equal to odd number. So, f of x_i equals x_i to the power n times sine of 1 over x_i to the power m, Here are two possible values for sine, plus 1. minus 1, so there will be two values for f of xi, For i even, xi to the power n, and for i odd, minus xi to the power n. sum from i equals 0 to k of the absolute value of f of x sub i plus 1 minus f of x_i equal to sigma i equal to 0 to the power of k minus 1, power of i, x sub i plus 1, power of n plus x sub i, power of n Here, there are two possible values of f of xi. even for positive, odd for negative Then f of x sub i plus 1 and minus f of x sub i are these two consecutive If it's a number, one will be positive and the other negative. Then, if we take the difference, we will always get the sum. Will the value of sum be positive or negative? Therefore, we will get the modulus -1 to the power of i x plus 1 to the power of n plus x to the power of n This is modulus, so we will always get a positive value. Then we will get sigma i equal to 0 to k x plus 1 to the power of n plus x to the power of n If we are opening the summation, then i equals 0 and i equals k, the rest of the values It will repeat 2 times. So the summation will be 2 times i equals 1 to k minus 1 Xi Power N x₀ to the power of n plus xₖ to the power of n The value of xi is 2i plus 1 pi over 2 power minus 1 over n And his power n The value of x0 will be I equal to 0 so pi over 2 And the value of xk will be I equals k so 2k plus 1 pi over 2 Power minus 1 upon n and its n power We will write it in a simplified form. Then the power will be minus n over m. minus n over m To find its supreme We are both applying site limits. k tends to infinity This is the time of the year a finite value will be obtained When m is less than n I mean that the variation will receive a finite value. If m is less than n Then the function fx will be of bounded variation. When m is less than n So for our functions f of x is equal to x times sine of 1 over x Where m equals 1 and n equals 1 So this is equality. This m is not satisfying the less than condition Therefore, f is not of bounded variation. And the second function is g(x) equals x times f(x). so x squared times 1 over x Where m equals 1 and n equals 2 so m is less than n therefore g is of bounded variation So the first option f is of bounded variation is incorrect f is not of bounded variation That is the correct answer. The third option, g, is of bounded variation. That is also correct, and the fourth option g is not of bounded variation — that is incorrect. So we got two correct answers second and third Thank you for watching For more videos Like, Share, Subscribe Raise the bell icon of NextGenMess