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Csir net june 2018: Real analysis Question (1)

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General Mathematics Question : Given $(x_n)$ is convergent sequence in $\mathbb{R}$ and $(y_n)$ is bounded sequence in $\mathbb{R}$ then we can conclude that, (a) $(x_n+y_n)$ is convergent. (b) $(x_n+y_n)$ is bounded. (c) $(x_n+y_n)$ has no convergent subsequence. (d)$(x_n+y_n)$ has no bounded subsequence. Solution : we discuss options one by one (a) take $x_n=1$ for all $n\in\mathbb{N}$ and $y_n=(-1)^n$ for all $n\in\mathbb{N}$ then, sequence $(x_n)$ is convergent and $(y_n)$ is bounded. But, $(x_n+y_n)=(0,2,0,2,0,2,....)$ which is not convergent. Hence (a) is false. (b) given that sequence $(x_n)$ is convergent and hence it is bounded. Also given that, $(y_n)$ is bounded. So that sum of two bounded sequence $(x_n+y_n)$ is also bounded. Hence (b) is true. (c), (d): By option (b) as sequence $(x_n+y_n)$ is bounded sequence and hence by "Bolzano Weierstrass theorem for sequences", it has convergent (hence bounded) subsequence.  So that, (c),(d) are fa...
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IIT-JAM 2018 Abstract Algebra Question (3)

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General Mathematics Question Solution : we know that,  Number of homomorphisms from $\mathbb{Z}_n$ to $\mathbb{Z}_m$ is $gcd(m,n)$ Now, as $221=13\times 17$. Hence, $gcd(51,221)≠1$, $gcd(91,221)≠1$, $gcd(119,221)≠1$, But $gcd(21,221)=1$ Hence number of homomorphisms from $\mathbb{Z}_{21}$ to $\mathbb{Z}_{221}$ is $1$ that is, there is only one homomorphism from $\mathbb{Z}_{21}$ to $\mathbb{Z}_{221}$. So that this must be trivial homomorphism( zero map).  Hence if $f:\mathbb{Z}_{21}\rightarrow\mathbb{Z}_{221}$ is homomorphism then it is trivial homomorphism ( zero map) i.e. $f(a)=0$ for all $a\in\mathbb{Z}_{21}$. Hence option (A) is correct.  (Note other options are not correct because in those case there exists non-trivial homomorphisms, because gcd of numbers $51,91,119$ with $221$ is not $1$.) Hence answer is "option (A)" only. Book recommend : see below, Must buy it. Nice offer must vist! 
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Sequence Question Real analysis IIT JAM 2018

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General Mathematics Question :  Solution : Note that,   An element $m$ is maximum of set $S$ iff $s≤m$ for all $s\in S$ and $m\in S$. Further, if a set $S$ has maximum element then it is supremum of $S$. Similarly, an element $l$ is minimum or least element of set $S$ if $l≤s$ all $s\in S$ and $l\in S$. Further, if a set $S$ has minimum or least element then it is infimum of $S$. Hence, $sup\{a_n: n\in\mathbb{N}\}$ $=sup\{a_1,a_2,a_3,...\}$ $=sup\{3,\frac{5}{4},\frac{5}{3},\frac{17}{16},...\}$ $=3$ (since maximum  of set $\{a_1,a_2,...\}$ is 3.   ) Similarly,  $inf\{a_n: n\in\mathbb{N}\}$ $=inf\{a_1,a_2,a_3,...\}$ $=inf\{3,\frac{5}{4},\frac{5}{3},\frac{17}{16},...\}$ $=1$ (since, $\{a_1,a_2,...\}$ has minimum or least element 1)  Hence option (A) is true, (C) is false.  Now to discuss options (B) and (C): we note that,  $\text{lim inf} (a_n)$ is infimum of set of all subsequential limits of sequence $(a_n)$ and similarly, $\te...
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