CSIR NET MATHEMATICS JUNE 2019 | UNIT-I | Real Analysis & Linear Algebra | 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 will 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 T. so this limit limit n tends to infinity fn x exists then we can say that fn is a pointwise convergent 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 n tends to infinity is 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 M test for uniform convergent. First step find limit n tends to infinity fn x is equal to fx that means pointwise limit then the modulus fn x 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 x 0 is a point of extremum then calculate h double dash x at x 0 if it is less than then h x has maximum value if limit n tends to infinity at the point of maximum this h x has maximum value 0 as n approach to infinity then it is uniform convergent otherwise it is not uniformly convergent so for our example g n x is equal to n x 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 gx which is a pointwise limit has value 0 for all x and limit is exist therefore it is a pointwise convergent but the third option gn converts 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 gnx converts 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 x will be 0. So, evaluate with 0. Then, we have n minus n cube x square is equal to 0. So, after the 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 23 booklet c marks three consider the vector space pn of real polynomial in x of degree less than or equal to n define transformation t from p2 to p3 by T of fx is equal to integration 0 to x ft dt plus f dash x. Then the matrix representation of T with respect to the basis 1x x square and 1x x square x cube and we have given four matrices associated the given linear transformation from P2 to P3. Here 1, x, x square is a basis for P2 which is the domain space and 1, x, x square, x cube is a basis for P3 which is the codomain space. So it is a standard problem and which is continuously asked in the net exam. So we should know how to find the matrix associated the linear transformation. So we begin with the transformation T of f(x) is equal to integration 0 to x f t dt plus f dash x. We first find the images of basis vector of domain space with respect to T so basis vector 1, f of x is equal to 1 and f dash x is 0 so T of 1 is equal to integration 0 to t, f(t) is equal to 1 and f dash x is equal to 0 now after the integration we have T evaluated from 0 to x then its value is x now it can be rewritten in the form of linear combination of basis vector of codomain space basis so codomain space basis is 1, x, x square, x cube and x can be written in the form of 0 plus 1 into x plus 0 into x square plus 0 into x cube and so the coordinates of T1 with respect to the basis of codomain space is 0 1 0 and now similarly basis vector 2 which is f(x) is equal to x as derivative is 1 therefore its image integration 0 to x f(t) is equal to t and f x is equal to 1 now after the integration is t square by 2 evaluate between 0 to x then we have x square plus x square by 2 plus 1 it can be rewrite as a linear combination of basis vector of codomain basis which is 1 plus 0 into x plus 1 by 2 into x square plus 0 into x cube and hence the coordinate of tx with respect to the basis of co-domain spaces 1 0 1 by 2 and 0 now similarly for the basis vector 3 f of x is equal to x square its derivative is 2x then we have t of x square is equal to integration 0 to x t square dt plus f dash x is equal to 2x after the integration we have t cube by 3 evaluate between the limit 0 to x plus 2x then we have x cube by 3 plus 2x now we write this image as a linear combination of basis vector of codomain space which is 0 plus 2 into x plus 0 into x square plus 1 by 3 into x cube and hence its coordinates with respect to the basis of co-domain space is 0 2 0 1 by 3. Now we construct matrix associated linear transformation which is nothing but the coordinates of images with respect to the co-domain space are the columns therefore t1 tx and tx square which is 0 1 0 1 0 1 by 2 0 and 0 2 0 1 by 3 and which is a option number 2 and its correct answer example number 24 booklet c marks 3 let pax denote the characteristic polynomial of matrix a then for which of the matrices pax minus P A inverse X is constant. In the option we have given 2 cross 2 matrices. So first we begin with the generalized matrix A equal to A B C D and its characteristic polynomial P A X is equal to determinant X I minus A. So determinant X minus A minus B minus C X minus D. Then X square minus A plus D X plus A D minus B C. You can rewrite in the form of X square minus trace of a where a plus d is trace of a which is sum of diagonals and ad minus bc represent the determinant of a. So this is generalized form of characteristic polynomial of 2 cross 2 matrices. Now a inverse is equal to 1 upon determinant a d minus b minus ca which is a shortcut for finding 2 cross 2 matrix is inverse. PA inverse X is equal to determinant XI minus A inverse. Then we have X square minus trace of A inverse into X plus determinant A inverse. In the example we want the difference of PAX minus PA inverse X. So its difference is X square cancel with the X square then we have trace of A inverse minus trace of A into X plus determinant A minus determinant A inverse but we want constant so this difference should be constant it is constant when the coefficient of x should be 0 so trace of a inverse minus trace of a equal to 0 in other words trace of a inverse is equal to trace of a trace of a inverse is the sum of the diagonals of a inverse so d plus a divided by determinant of a and trace of a is a plus d since in our example all the matrices have non-zero trace therefore we cancel d plus a with the d plus a and determinant a is equal to 1 that means the difference is constant when determinant a is 1 so we will check this condition for the given options so first option 3 3 2 4 is determinant is 12 minus 6 which is 6 and which is not 1 so first option is wrong for the second option 4 3 2 3 is determinant value is 6 so which is also wrong third option 3 2 4 3 is determinant is 9 minus 8 which is 1 so it is the correct option we will check fourth option also so a is equal to 8 minus 9 which is minus 1 so which is not also 1 and the correct answer is third option example number 25 from booklet c marks 3 which of the following matrices is not diagonalizable so again four option we will first recall some important concept and then important result important concepts algebraic multiplicity the algebraic multiplicity of an eigenvalue lambda is the number of occurrence of lambda as a root of characteristic polynomial and the geometric multiplicity the geometric multiplicity of an eigenvalue lambda is the number of independent eigenvector associated with the lambda. Now some important result an eigenvalue of upper or lower triangular matrices are diagonal values. The second important result if algebraic and geometric multiplicity of each eigenvalue are equal then it is diagonalizable matrix. Third result geometric multiplicity is always less than equal to algebraic multiplicity if algebraic multiplicity is one then geometric multiplicity is also one if a matrix has distinct eigenvalue then it is diagonalizable matrix so we will use this result to check the diagonalization of four matrices the first option a is equal to 2 0 1 2 3 0 0 0 2 it is upper triangular matrix therefore this eigenvalue are diagonal values therefore the eigenvalues of upper triangular matrix a is 2, 2, 3 now algebraic multiplicity of eigenvalue lambda equal to 2 is 2 because it repeated two times and algebraic multiplicity of 3 is 1 algebraic multiplicity of 3 is 1 so its geometric multiplicity is 1 so we don't have problem with the algebraic multiplicity 1 now we are going to calculate the eigenvector associated with the so eigenvector is the solution of a minus lambda into x is equal to 0 so lambda equal to 2 so a minus 2i into x is equal to 0 subtract 2 from the diagonal entries of matrix a then we have 2 minus 2 0 3 minus 2 1 and 2 minus 2 0 so if we find the equation associated this system then we have two equal two solutions first z is equal to zero second y equal to zero we don't get any value of x so x belongs to r so solution x is equal to x 0 0 x into 1 0 0 therefore the eigenvector associated with the lambda equal to 2 eigenvalue is vector v equal to 1 0 0 so only one eigenvector associated with the lambda equal to 2 therefore algebraic multiplicity is 2 but the geometric multiplicity is 1 and therefore they are not equal therefore a is not diagonalizable matrix so the first answer is correct now we are going to check the other answers a equal to 1 1 1 1 so first find is eigenvalues determinant A minus lambda equal to 0 characteristic equation its root will give the eigenvalue of A then after the calculation lambda minus 2 into lambda equal to 0 therefore we have two eigenvalues lambda equal to 0 and lambda equal to 2 distinct eigenvalues of A therefore A is diagonalizable matrix so second option is wrong option number 3 A equal to 2 1 0 0 3 0 0 0 3 it is also upper triangular matrix so diagonals value will give the eigenvalue therefore A has 2 3 3 the algebraic multiplicity of 2 is 1 therefore its geometric multiplicity is also one so we are not going to calculate eigenvector associated with the two algebraic multiplicity of three is two because it repeated two times so we are going to calculate eigenvector associated with the three eigenvectors are the solution of A minus lambda into x is equal to zero lambda equal to three now subtract three from the diagonal values so we have two minus three is minus one one zero three minus three is zero three minus three is 0 now corresponding equation minus x plus y equal to 0 so y equal to x z belongs to R so x is equal to x y z here y equal to x so you can write down x so x x z we can separate two vectors one is multiple of x second is multiple of z so 1 1 0 is a multiple of x and 0 0 1 is a multiple of z therefore corresponding eigenvectors v1 equal to 1 1 0 and v2 equal to 1 0 0 so two eigenvectors associated with the lambda equal to 3 so the algebraic multiplicity of lambda equal to 3 is 2 multiplicity of lambda equal to 3 is also 2 so algebraic and geometric multiplicity are equal therefore it is diagonalizable matrix so third option is also wrong and the fourth option we are going to calculate the eigenvalues determinant A minus lambda I equal to 0 and after the calculation you have two distinct eigenvalues for a lambda equal to 2 and lambda equal to 3 therefore a is diagonalizable matrix so fourth option is also wrong there is only first correct answer example number 26 from booklet c marks 3 what is the rank of following matrix so we have done 5 cross 5 matrix and the options are first option 2 second option 3 third option 4 fourth option 5 so important result the rank of matrix a is the number of non-zero rows in row echelon form of a so first we convert matrix a into its row echelon form so we apply row operation so first row operation multiplied fourth row by minus 1 added to the fifth row then we have 0 0 0 0 1 now multiply third row by minus 1 added to the corresponding entries of fourth row then we have 0 0 0 1 1 then next operation multiply second row by minus 1 and add it to the third row then we have 0 0 1 1 1 in the third row multiply first row by minus 1 add it to the second row then we have 0 1 1 1 1 1 in the second row this is a row echelon form of matrix A and it has 5 non-zero row therefore rank of matrix A is 5 and which is the option number 4. Example number 27 from booklet C marks 3. Let V denote the vector space of real valued closed interval 0, 1. Let W be the subspace span by the set sin x cos x tan x then the dimension of w over r is first option 1 second option 2 third option 3 fourth option infinite so first the dimension of w is the number of independent vectors in the set sine x cos x and tan x now important result the set f1 x f2 x f3 x these are the functions of function is independent set of function if the following wronskian determinant is non-zero for at least 1x so first we try to find the wronskian of sin x cos x and tan x if these three are independent then the rank of matrix is 3 if it is not independent then we find the number of independent vectors out of these three vectors so we will have possible answer 1 2 or 3 not infinite the fourth option is already cancel now wronskian first row is function second row is first derivative third row is a second derivative here the functions are f1 equal to sin x f2 equal to cos x f3 equal to tan x sin x is derivative cos x and derivative cos x is minus sin x in the second column cos x is derivative minus sin x that is minus sin x is minus cos x and the third column tan x is derivative sec square x and derivative of sec square is 2 sec x tan x Now evaluate the determinant sin x multiply the minus sin x to sec square x tan x minus minus plus sec square x multiply is the cos x so we have sec x. Now similarly for second value minus cos x multiplied cos x multiplied to sec square x tan x minus minus plus sin x into sec square x and the last term tan x into minus cos square x minus tan square x now after the simplification we have value of Wronskian is minus tan x to tan square x plus 3 now if we take x is equal to pi by 4 then value of Wronskian is minus tan pi by 4 into 2 tan square pi by 4 value of tan pi by 4 is 1 therefore we have minus 5 which is non-zero. So at x is equal to pi by 4 once again is non-zero. Therefore the given set of function sin x cos x and tan x are independent and hence dimension of w is the number of vectors in the set which is 3. So dimension of w is 3. Example number 28 booklet c marks 3. Let v be the vector space of polynomials in the variable t of degree at most two over R and inner product on V is defined by inner product f with g is equal to integration 0 to 1 f(t) g(t) dt for f, g belongs to V. Let W span by the vectors 1 minus t square, 1 plus t square and W perp be the orthogonal complement of W in V which of the following conditions satisfy for all h belongs to W perp first option h is an even function that means h(t) is equal to h of minus t second option h is an odd function that means h(t) is equal to minus times h of minus t third option h(t) is equal to 0 has real solution and the fourth option h of 0 is equal to 0 now first we recall two concept orthogonal vectors and orthogonal complement so orthogonal vectors two vectors u and v are orthogonal if their inner product is zero orthogonal complement orthogonal complement of a subspace W of a vector space V is the set W perp of all vectors in V that are orthogonal to every vector in W. That means orthogonal complement W perp is a set of all those vectors which are perpendicular to every vector in W. That means if W belongs to W perp then W inner product with U is 0 for every U in W. Now suppose h(t) belongs to W perp so H3 belongs to V and V contains the polynomials in variable t of degree at most 2. So, h(t) is in the form of a plus bt plus ct square. Now, it belongs to W perp. So, it is orthogonal to every vector of W. So, it is orthogonal to 1 minus t square, 1 plus t square. And hence, we have inner product h with 1 minus t square is 0 and inner product h with 1 plus t square is 0. Now we have given the definition of inner product f with g is equal to integration 0 to 1 ft gt dt. So inner product 1 minus t square is equal to 0. So by the definition it is integration 0 to 1 ht 1 minus t square dt where ht is equal to a plus bt plus ct square into 1 minus t square dt equal to 0. Now just we open the brackets and integrate. after the integration we evaluate between the limit 0 to 1 if we put 0 then lower limit is 0 and if we put 1 then we have a minus a by 3 plus b by 2 minus b by 4 plus c by 3 minus c by 5 equal to 0 now after the simplification we have equation 40 a plus 15 b plus 8 c is equal to 0 now we have another condition h in a product 1 plus t square is equal to 0 and we apply the definition then we have the second equation after the integration 8a plus 45 b plus 32 c equal to 0 now we have two equation and three unknowns by simple elimination we are able to find the value of a and b a is equal to 1 by 5 c and b is equal to minus 16 by 15 c since h t is equal to a plus b t plus c t square then substitute the value of a and b then we have h t is equal to c by 15 3 minus 16 t plus 15 t square now we are going to find the h of minus t just replace t by minus t in the definition of h t then we have c by 15, 3 minus 16t plus 15t square, so we have 3 plus 16 t plus 15 t square but which is not equal to h of t therefore ht is not an even function and which is the first option so first option is wrong which is also not equal to minus times ht therefore it is not an odd function so second option is also wrong now ht is equal to 0 but c is not 0 for all the polynomials so this bracket is 0 then we have 3 minus 16t plus 15t square equal to 0 Now it is a quadratic equation so we have delta b square minus 4ac b is equal to minus 16 a is equal to 15 and c is equal to 3 We have value of delta which is 76 greater than 0 and hence this quadratic equation has real solutions and therefore the third option that h of t equal to 0 has real solution is correct and the last option h of 0 so if we put t equal to 0 then 16 t term will be 0 15 t square term will be 0 then we have 3 c by 15 and c is not 0 for all the polynomials therefore h of 0 equal to 0 is not always and hence the fourth option is also wrong example number 29 from booklet C mastery 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 cos x is equal to 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 logarithm 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 1 if we take n equal to 2 then we have trigonometric result cos x square plus sin x square is equal to 1 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 pq are the natural numbers so you can rewrite this set e power x is equal to pq 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 sin 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 one to one correspondence between the set and the set of rational number and the rational number is countable so the given set is countable as 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, a k for all n belongs to n. Then which of the following is true? First option, limit n tends to infinity b n 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 sub bn less than equal to 1 by 2 so once we recall important result of sequence if a sequence of real numbers converges to real number x then any subsequence of xn also converges to x. In other words if a sequence xn of real numbers has two subsequences that converge to different limits then it diverges. That means sequence xn is diverges if we are able to find two subsequences which converge to different limits. So given sequence a n 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 a 2 for n equal to 2 minus 1 power to 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 a n 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 a n a1 plus a2 plus a3 so minus 1 by 2 cancel 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 4 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 means 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 b2n 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 bn 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 March 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 test comparison test. Let sigma a n and sigma b n 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 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 test 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 satisfied by alternating series then series is convergent series. Solution 1 which is minus 1 power n divided by n which is alternating series it's compared with the standard form then an is equal to 1 by n therefore an plus 1 is equal to 1 upon n plus 1 since n less than n plus 1 1 upon an plus 1 and modulus an plus 1 less than an will satisfy the first condition monotonically decreasing limit n tends to infinity an equal to limit n tends to infinity 1 upon n as n tends to infinity 1 upon n tends to 0 so an approach to 0 as n tends to infinity so it satisfy the second condition so by the Leibniz 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 is given convergent series so it is wrong option 3 & 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 2 times 1 upon 4 square come only 3 times 1 upon 5 square comes only four times and continue If we rewrite then we have 1 upon 2 square plus 2 upon 3 square plus 3 upon 4 square plus 4 upon 5 square and in another way in the series from sigma n equal to 1 to infinity it is n upon n plus 1 power 2 Now this series convert into the single summation n equal to 1 to infinity n divided by n plus 1 power 2. So here a n 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 so bn is divergent so an must be divergent series option is wrong and the fourth option is correct example number 32 from booklet C March 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 side term square root n plus 1 minus square root n and 1 upon square root n both the terms are positive cannot begin 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 root 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 then 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 3 and option number 4. So we begin with left hand right hand side term square root n plus 1 minus square root n and 1. Now taking square root n on the other side we have 1 plus square root n by squaring both sides. Now squaring this term 1 plus square root n n plus 1 and plus 2 root n plus 1 n cancel with n 1 cancel with 1 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 1 so third option is wrong if it is less than 1 so it must less than 2 so fourth option is greater than 2 which is also wrong and there is only one option is correct second option square root n plus 1 minus square root n less than 1 upon square root n thanks for watching for more videos like share subscribe press the bell icon of next gen maths