CSIR NET MATHEMATICS JUNE 2019 | Real Analysis | Complete Solutions of Section B

In this video, we are going to discuss the problem of real analysis from CSINET June 2019. Example number 21 from booklet C, Marks 3. Let f defined from real number to real number be a continuous and one-one function, which of the following is true? f is onto, f is either strictly decreasing or strictly increasing, third option there exists x belongs to R such that f of x is equal to 1 and f is unbounded. we will solve this problem using counter example so first example f of x is equal to e power x so f of x is equal to e power x is continuous as well as one one but it is not onto why because f of x is equal to zero is not possible for any real number f of x is equal to e power x and it is zero when x has value minus infinity and minus infinity is not in real number therefore f of x never be 0. Therefore we cannot get any inverse image of 0. Therefore f is not onto. Now the second example f of x is equal to 1 plus e power x. So f of x is continuous as well as 1 1. Now f of x never be 1 because e power x never be 0 which is discussed in the example number 1. So the similar way f of x never be 1. So the third option that exists x belongs to R such that f of x is equal to 1 for all continuous and 1 1 function on real number to real number which is not possible because f of x is equal to 1 plus e power x does not have value f of x is equal to 1 for real number. So third option is also wrong. Now the example number 3 f of x is equal to 1 upon 1 plus e power x. It is also continuous as well as 1 1 since 0 less than e power x less than infinity. If we add 1 then the inequality will be 1 less than 1 plus e power x less than infinity. Now if we take the reciprocal then 1 greater than 1 upon 1 plus e power x greater than 1 upon infinity which is 0. Now rewrite 0 less than 1 upon 1 plus e power x less than 1 which shows that f of x is bounded. So f is bounded when f is continuous and 1, 1. So, third option that f is unbounded if it is continuous and 1, 1 on real values which is wrong. So, there is only one correct option which is second one that f is either strictly decreasing or increasing. Example number 22 from booklet C marks 3. Let g n x is equal to n x divided by 1 plus n square x square. x belongs to closed open interval 0 to infinity. Which of the following is true for n approach to infinity? First option, gn converges to 0 pointwise but not uniform. Second option, gn converges to 0 uniform. Third option, gn converges to x for all x belongs to close interval 0 to infinity. And the fourth option, gnx converges to x divided by 1 plus x square for all x belongs to 0 to infinity. So, here we will recall two important concepts. convergence of sequence of function pointwise and uniform. So, first we recall the definition of pointwise convergence. Let D be a subset of R and let fn be a sequence of functions defined on D. We say that the convergence pointwise on D if limit n tends to infinity fnx exists for each point x in D. So, this limit limit n tends to infinity fn x exists then we can say that fn is a pointwise convergent and note that limit n tends to infinity fn x is equal to fx this is pointwise limit of sequence fn now uniform convergent limit n tends to infinity fn equal to f uniformly convergent means limit n tends to infinity supremum of modulus fn x minus fx where fx is a pointwise limit of fn equal to 0. To check the uniform convergent we will use MN test for uniform convergent. First step find limit n tends to infinity fnx is equal to fx that means point wise limit then the modulus fnx minus fx say hx calculate maximum of hx to find the maximum of hx just find extremum point by solving h dash x equal to 0. Suppose x0 is a point of extremum then calculate h double dash x at x0 if it is less than then hx has maximum value if limit n tends to infinity at the point of maximum this hx has maximum value 0 as n approaches to infinity then it is uniform convergent otherwise it is not uniformly convergent. So for our example g nx is equal to nx divided by 1 plus n square x square. So in the first step we are going to calculate the point wise limit limit n tends to infinity g n x is equal to limit n tends to infinity n x divided by 1 plus n square x square divide numerator and denominator by n square then it is limit n tends to infinity x divided by n 1 upon n square plus x square as n tends to infinity 1 upon n tends to 0 so numerator is 0 this is 0 plus x square so it is always 0 so g x which is a point wise limit has value 0 for all x and limit exists therefore it is a point wise convergent but the third option g n converges to x for all x belongs to closed open interval 0 to infinity it is wrong as well as fourth option is also wrong which says that g n x converges to x divided by 1 plus x square for all x because it will give the non-zero value third option and fourth option is cancelled now we are going to check the uniform convergent so for uniform convergent we are going to calculate the maximum of hx where hx is equal to gnx minus gx here gx is equal to 0 so its value is gnx and gnx is equal to nx divided by 1 plus n square x square so differentiation of h h dash x n minus n cube x square divided by 1 plus n square x square whole square now for the point of extremum h dash will be 0. So evaluate with 0 then we have n minus n cube x square is equal to 0. So after this simplification we get the value x is equal to 1 by n. So x is equal to 1 by n is a point of extremum. Now calculate h double dash x by differentiation h dash x one more time and we have h double dash x is equal to minus 6 n cube x plus 2 n power 5 x cube divided by 1 plus n square x square power 3 now evaluate h double dash x at 1 by n then we have minus n square by 2 which is less than 0 therefore this point of extremum is a point of maximum so x equal to 1 by n is a point of maximum evaluate hx at point of maximum x is equal to 1 by n then limit n tends to infinity h of 1 by n substitute x is equal to 1 by n so n into 1 by n then numerator is 1 and the denominator of 1 plus n square into 1 upon n power 2 so n square n square cancel we have 1 so 1 upon 1 plus 1 which is 1 by 2 not equal to 0 therefore this limit has non-zero value therefore it is not uniformly convergent so it is pointwise convergent but not uniform which is option number one so option two is cancel and option one is correct answer example number 29 from booklet c mass 3 which of the following sets is uncountable first option x belongs to r such that log x is equal to p by q for some pq belongs to natural number second option x belongs to r such that cos x power n plus sin x power n is equal to 1 for some n belongs to n third option x belongs to r x is equal to log p by q for some pq belongs to N and the fourth option x belongs to R, x is equal to log p by q for some pq belongs to N. Now first we recall the concept of countable and uncountable set so a set is countable either if it is finite or infinite and we can find one-to-one correspondence between the elements of set and set of natural numbers so if we are able to find one-to-one correspondence between the set S and natural number then we can say that set is countable set. Two important result: set of rational number is countable and set of real number is uncountable that we are going to use option 1 log x is equal to p by q where pq belongs to N some natural numbers now this is a graph of log x, log x function increasing function and the given set contain only positive rational number so we can easily find the one-to-one correspondence between the rational number and the given set the rational number is countable therefore the given set is countable so first option is wrong now option number two cos x power n plus sin x power n equal to one if we take n equal to two then we have trigonometric result cos x square plus sin x square is equal to one and it is true for all real numbers and set of real number is uncountable therefore the given set is uncountable and hence it is a correct answer now we are going to check the option number three which is x is equal to log p by q where p, q are the natural numbers so you can rewrite this set e power x is equal to p by q and graph of exponential function is increasing so we can easily find there is a one-to-one correspondence between the set and the set of rational numbers and set of rational number is countable therefore the given set is countable so option number three is wrong now option number four where cos x is equal to p by q now cos x and sine x are the periodic function so within the period they are one-to-one correspondence between the rational numbers and hence we can extend it to the real line So, we have 1 to 1 correspondence between the set and the set of rational number and the rational number is countable. So, the given set is countable and the fourth option is also wrong. So, there is only one correct answer second. Example number 30 from booklet C marks 3. Consider a sequence a n, a n equal to minus 1 power n 1 by 2 minus 1 by n. Let b n equal to sigma k equal to 1 to n ak for all n belongs to n then which of the following is true first option limit n tends to infinity bn equal to 0 second option limit supremum n tends to infinity bn greater than 1 by 2 third option lim inf n tends to infinity bn less than minus 1 by 2 and the fourth option 0 less than equal to lim inf bn less than equal to lim sup bn less than equal to 1 by 2 so once First we recall important result of sequence, if a sequence of real numbers converge to real number X then any subsequence of Xn also converge to X. In other words, if a sequence Xn of real numbers has two subsequence that converge to different limits then it diverges. That means sequence Xn is diverges if we are able to find two subsequence which converge to different limits. So given sequence An is equal to minus 1 power n 1 by 2 minus 1 by n. So first term of the sequence is n equal to 1 so minus 1 power 1 1 by 2 minus 1 by 1. So we have minus 1 by 2 plus 1. Now a2, so for n equal to 2, minus 1 power 2, 1 by 2, minus 1 by 2. Now n equal to 3, so minus 1 by 3, 1 by 2, minus 1 by 3. So we have minus 1 by 2, plus 1 by 3. a4, 1 by 2, minus 1 by 4. a5, minus 1 by 2, plus 1 by 5. a6, 1 by 2, minus 1 by 6 and the remaining terms. Now bn is equal to sigma k equal to 1 to n ak. therefore b1 is equal to a1 and a1 has value 1 minus 1 by 2 b2 is sum of first two sequence of an a1 plus a2 so if we take the summation of a1 plus a2 so minus 1 by 2 minus plus 1 by 2 cancel then we have remaining part 1 minus 1 by 2 b3 sum of first three terms of sequence an a1 plus a2 plus a3 so minus 1 by 2 cancel with the plus 1 by 2 and the remaining part 1 minus 1 by 2 plus 1 by 3 and minus 1 by 2 similarly b4 so in b4 minus 1 by 2 cancel with the plus 1 by 2 1 minus 1 by 2 cancel with the plus 1 by 2 then we have 1 minus 1 by 2 plus 1 by 3 minus 1 by 4 similarly b5 so we have minus 1 by 2 plus 1 by 2 cancel 1 more minus 1 by 2 plus 1 by 2 cancel and the remaining part 1 minus 1 by 2 plus 1 by 3 minus 1 by 4 plus 1 by 5 and minus 1 by 2 similarly b6 1 minus 1 by 2 plus 1 by 3 minus 1 by 4 plus 1 by 5 minus 1 by 6 so if you observe the terms then for even terms just like b2 we have 1 minus 1 by 2 for b4 1 minus 1 by 2 plus 1 by 3 minus 1 by 4 for b6 1 minus 1 by 2 plus 1 by 3 minus 1 by 4 plus 1 by 5 minus 1 by 6 and for odd terms we have 1 minus 1 by 2 for b3 1 minus 1 by 2 plus 1 by 3 and one more term which is minus 1 by 2 similarly b5 we have one more term minus 1 by 2 so if we generalize then bn is equal to 1 minus 1 by 2 up to 1 by n minus 1 by 2 if n is odd and if n is even then we have 1 minus 1 by 2 up to 1 upon n. Now if we recall this series of log 1 plus x equal to x minus x square by 2 plus x cube by 3 minus x power 4 by 4. If we take x is equal to 1 then log 1 plus 1 is equal to log 2 and 1 minus 1 by 2 plus 1 by 3 minus 1 by 4 up to infinity terms if n is odd then we have odd term sequence b1 b3 b5 b7 and if limit n tends to infinity b2n plus 1 approach to this bracket approach to log 2 and minus 1 by 2 is constant so it has limit log 2 minus 1 by 2 if n is even then the even term 2, 4, 8 converges to log 2. So, we are able to find two subsequences. One is odd term sequence, second one is even term sequence have two different limits. Therefore, by the theorem bn has two different sequence which converges to different limits and therefore it is diverges. So, the first option limit n tends to infinity bn equal to 0 which is wrong. Now, we recall b1 equal to 1 minus 1 by 2 so which has positive value b2 equal to a1 plus a2 which is also 1 minus 1 by 2 positive b3 1 minus 1 by 2 plus 1 by 3 minus 1 by 2 so you can see 1 minus 1 by 2 is plus 1 by 2 cancel the minus 1 by 2 so it has value 1 by 3 so it is also positive b4 1 minus 1 by 2 positive and 1 by 3 minus 1 by 4 is positive so sum of positive number is positive similarly b5 positive b6 positive so for all n bn is positive so lim inf bn which says is less than minus 1 by 2 so third option is wrong now as we know bn tends to infinity limit n tends to infinity b to n plus 1 has value log 2 minus 1 by 2. So the log 2 has value 0.6931 minus 0.5 which has value 0.1931 and log 2 has value 0.6931. So lim sup represent the maximum value or largest value of the sequence and in the fourth option we have given that lim sup b n less than equal to 1 by 2 but we are able to find the value which is bigger than 1 by 2 which is 0.6931. So the fourth option says that it is less than 1 by 2 which is wrong and the second option which says that lim sup is greater than 1 by 2 it is correct. Example number 31 from booklet C maths 3 which of the following is true. First option sigma n equal to 1 to infinity minus 1 power n divided by n does not converge. Second option sigma n equal to 1 to infinity 1 upon n converges. Third option sigma m equal to 1 to infinity sigma n equal to 1 to infinity 1 upon m plus n power 2 converges. And the fourth option same series diverges. So it's a problem of convergence and divergence of series. So first we recall important test which we are going to use to check the convergence of each series first is comparison test let sigma an and sigma bn are positive term series if limit n tends to infinity an upon bn is finite and non-zero then both the series converge and diverge simultaneously second test p test series for the positive term series series sigma n equal to 1 to infinity 1 upon n power p is convergent series if p greater than 1 otherwise that means p less than equal to 1 it is divergent series third test alternating series states or Leibniz test if the given series is alternating series that means sigma n equal to 1 to infinity minus 1 power n a n is convergent if modulus a n is monotonically decreasing that means modulus a n plus 1 less than equal to modulus a n and limit n tends to infinity a n equal to 0 if these two conditions satisfy by alternating series then series is convergent series option 1 which is minus 1 power n divided by n which is alternating series it's compared with the standard form then a n is equal to 1 by n therefore a n plus 1 is equal to 1 upon n plus 1 since n less than n plus 1 1 upon a n plus 1 and modulus a n plus 1 less than a n will satisfy the first condition monotonically decreasing limit n tends to infinity a n equal to limit n tends to infinity 1 upon n as n tends to infinity 1 upon n tends to 0 so a n approach to 0 as n tends to So, it satisfies the second condition so by the limitless test this series is convergent and the first option says that does not converge so first option is wrong. Now for option 2 we will use the p-series test here a n equal to 1 by n so it compared with the p-test is n power p so p is equal to 1 if p is equal to 1 then by the p-test it is divergent series but in the option it is given convergent series so it is wrong. Option 3 and 4 have the same series sigma m equal to 1 to infinity sigma n equal to 1 to infinity 1 upon m plus n power 2. So first we open the first summation by putting the value of n. So if we put n equal to 1, first term 1 upon m plus 1 power 2, second term 1 upon m plus 2 power 2 third term 1 upon m plus 3 power 2 now apply summation on each term now after open the each summation if you put m equal to 1 then we have first term 1 upon 1 plus 1 power 2 so 2 square second term m equal to 2 so 1 upon 3 square third term m equal to 3 so 1 upon 4 square and continue if you open the second summation then m equal to 1 so first term is 1 upon 1 plus 2 which is 3 so 3 square then second term if you put 2 then 1 upon 4 square and continue if you open the third summation then it begin with m equal to 1 so it is it begin with the 4 1 upon 4 square plus 1 upon 5 square plus 1 upon 6 square continue now 1 upon 2 square comes only single time in all the terms so first term is 1 upon 2 square now 1 upon 3 square comes only two times one upon four square comes only three times one upon five square comes only four times and continue. If we rewrite then we have one upon two square plus two upon three square plus three upon four square plus four upon five square and in another way in the series from sigma n equal to one to infinity it is n upon n plus one power two. Now this series convert into the single summation n equal to 1 to infinity n divided by n plus 1 power 2 so here an equal to n upon n plus 1 power 2 now we consider bn is equal to 1 upon n now by taking the ratio limit n tends to infinity an upon bn then we have n square divided by n plus 1 whole square divide numerator and denominator by n square and apply as n tends to infinity we have 1 upon 1 plus 0 because as n tends to infinity 1 upon n tends to 0 then its value is 1 which is finite and non-zero by the comparison test an and bn converges and diverges simultaneously since bn is divergent series by the p test we already discussed in the second option to bn is divergent so an must be divergent series the first option is wrong and the fourth option is correct example number 32 from booklet c maths 3 for n belongs to capital n which of the following is true square root n plus 1 minus square root n greater than 1 upon square root n for all except the possible finitely many n second option square root n plus 1 minus square root n less than 1 upon square root n third option square root n plus 1 minus square root n greater than 1 and the fourth option square root n plus 1 minus square root n greater than 2 now we begin with the first and second option there is a difference of greater than and less than sign so we have two terms left hand side term and right hand side term square root n plus 1 minus square root n and 1 upon square root n both the terms are positive we are not beginning with the sign and multiply square root n on the left hand side then square root n into n plus 1 minus n 1 by taking minus n on the other side we have square out n into n plus 1 1 plus n squaring both side then opening the brackets we have n square n and n square plus 2n plus 1 n square cancel with n square and 2n minus n is equal to n then 0 and n plus 1 now n is a natural number so n plus 1 is always greater than 0 so that is a less than sign 0 less than n plus 1 and all the terms are positive terms so it preserve these signs so you apply the sign on the above terms then finally we have square root n plus 1 minus square root n less than 1 upon square root n and the first option has greater than 1 upon square root n the first option is wrong but the second option is correct now we are going to check the remaining option option number three in option number four so we begin with left and right side term square root n plus one minus square root n and one now taking square root n on the other side we have one plus square root n by squaring both side now squaring this term one plus square root n n plus one n plus two root n plus one n cancel with n one cancel with one then we have 0 root n as n is natural number then the last term is 0 less than 2 root n and all the terms are positive it preserves the sign so apply this sign on all the above terms then finally we have square root n plus 1 minus square root n less than 1 and the third option we have greater than one so third option is wrong if it is less than one so it must less than two so fourth option is greater than two which is also wrong hence there is only one option is correct second option square root n plus one minus square root n less than one upon square root n thanks for watching for more videos like share subscribe press the bell icon of next gen maths