Mathematics:Teaching, Learning and Exploring.
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Find Notes,Questions and Answers for Different Topics on the following link.
http://mathskthm.me.pn/SETNET.html
(Part B Question: NET June-2015) Let \(V \) be the space of twice differentiable functions on \(\mathbb{R} \) satisfying \( f''-2f'+f =0\). Define \( T: V \to \mathbb{R}^2 \) by \(T(f)=(f'(0),f(0)) \). Then T is
A. one-one and onto
B. one-one but not onto
C. onto but not one-one
D. neither one--one nor onto
\(\textbf{Option C}\)
The general solution of differential equation is \(f(x) = c_1e^x+c_2xe^x \).
So the function \(T \) is
\(T(f) = (f'(0),f(0)) = (c_1+c_2,c_1) \)
This function is clearly onto but not one-one.
(Part B Question: NET June-2015) Let \(f: X \to X \) be such that \(f(x) = x \) for all \(x \in X \). Then
A. \(f \) is one-one and onto
B. \(f \) is one-one but not onto
C. \(f \) is onto but not one-one
D. \(f \) need not be either one-one or onto
\(\textbf{Option A}\)
\(f \) is onto as for \(x \in X \) there is \(y = f(x) \in X \) such that \(f(y) = x \). \(f \) is one -one as
\(f(x) = f(y) \implies f((f(x))=f(f(y)) \implies x=y \)
(Part C Question: NET June-2015) Which of the following sets of functions are uncountable?
A. \(\{f: f:\mathbb{N} \to \{1,2\}\}\)
B. \(\{f: f:\{1,2\} \to \mathbb{N}\}\)
C. \(\{f: f:\{1,2\} \to \mathbb{N}, f(1) \leq f(2)\}\)
D. \(\{f: f:\mathbb{N} \to \{1,2\}, f(1) \leq f(2)\}\)
\(\textbf{Options A and D}\)
Options A and D: The cardinality of \(\{f: f:\mathbb{N} \to \{1,2\}\}\) and \(\{f: f:\mathbb{N} \to \{1,2\}, f(1) \leq f(2)\}\)
is \( 2^\mathbb{N}= c\) and so are uncountable.
Options B and C: The cardinality of \(\{f: f:\{1,2\} \to \mathbb{N}, f(1) \leq f(2)\}\) and \(\{f: f:\{1,2\} \to \mathbb{N}, f(1) \leq f(2)\}\) is
\(\mathbb{N} \times \mathbb{N} \) and so are countable.
(Part C Question: NET June-2015) Let \(p_n(x) = x^n \) for \(x \in \mathbb{R} \) and let \(\mathcal{P} = \) span \(\{p_0,p_1,\ldots\} \). Then
A. \(\mathcal{P} \) is the vector space of all real valued continuous functions on \(\mathbb{R} \).
B. \(\mathcal{P} \) is a subspace of all real valued continuous functions on \(\mathbb{R} \).
C. \(\{p_0,p_1, \ldots \} \) is a linearly independent set in the vector space of all continuous functions on \(\mathbb{R} \)
D. Trigonometric functions belongs to \(\mathcal{P} \).
\(\textbf{Options B and C}\)
Option A: Not every continuous real valued function on \(\mathbb{R} \) is a polynomial.
Option B: All polynomials in \(\mathbb{R}[x] \) are continuous and they are closed under addition and scalar multiplication.
Option C: \(a_0+a_1x+\cdots+a_nx^n = 0 \iff a_0=a_1=\ldots=a_n=0 \)
Option D: Trigonometric functions are not polynomials!
(Part C Question: NET June-2015)Consider the set \(\mathbb{Z} \) of integers, with the topology \(\tau \) in which a subset is closed if and only if it is empty, or \(\mathbb{Z} \) , or finite. Which of the following statements are true?
A. \(\tau \) is the subspace topology induced from the usual topology on \(\mathbb{R} \).
B. \(\mathbb{Z} \) is compact in the topology \(\tau \)
C. \(\mathbb{Z} \) is Hausdorff in the topology \(\tau \).
D. Every infinite subset of \(\mathbb{Z}\) is dense in the topology \(\tau \).
\(\textbf{Options B and D}\)
\(U \) is open in \(\tau \iff |\mathbb{Z}-U| < \infty \)
Option A: The set \((-1,1) \cap \mathbb{Z} =\{0\} \) is open in induced topology from the usual topology on \(\mathbb{R} \) but not in the topology \(\tau \).
Option B: Let \(\mathcal C\) be an open cover of \(\mathbb{Z}\).Let \(U \in \mathcal C\) be any set in \(C\). It covers all but a finite number of points of \(\mathbb{Z} \).So for each of those points we pick an element of \(\mathcal{C}\) which covers each of those points.
Hence we have a finite subcover of \(\mathbb{Z} \)
Option C: \(\mathbb{Z} \) is not Hausdorff in \(\tau \) as no two open sets in \(\tau \) are disjoint.
Option D: Since complement of any open set in \(\tau \) is finite, every infinite set in \(\mathbb{Z} \) intersects any open set in \(\tau \).
\(\textbf{Remark:} \) These results can be generalised to co-finite topology on any infinite set.
(Part B Question: NET June-2015)Which of the following subsets of \(\mathbb{R}^n \) is compact (w.r.t. usual topology of \(\mathbb{R}^n)\) ?
A. \(\{(x_1,x_2,\ldots,x_n):|x_i|<1,1\leq i\leq n\} \)
B. \(\{(x_1,x_2,\ldots,x_n):x_1+x_2+\cdots+x_n = 0\} \)
C. \(\{(x_1,x_2,\ldots,x_n):x_i\geq 0,1\leq i\leq n\} \)
D. \(\{(x_1,x_2,\ldots,x_n):1\leq i\leq 2^i, 1 \leq i \leq n\} \)
\(\textbf{Option D}\)
The set \(\{(x_1,x_2,\ldots,x_n):|x_i|<1,1\leq i\leq n\} \) is not closed.
The set \(\{(x_1,x_2,\ldots,x_n):x_1+x_2+\cdots+x_n = 0\} \) is not bounded.
The set \(\{(x_1,x_2,\ldots,x_n):x_i\geq 0,1\leq i\leq n\} \) is not bounded.
The set \(\{(x_1,x_2,\ldots,x_n):1\leq i\leq 2^i, 1 \leq i \leq n\} \) is closed and bounded.
(\(\textbf{Hint:} \) Draw sets in \(\mathbb{R}^2) \)
(Part B Question: NET June-2015) A polynomial of odd degree with real coefficients must have
A. at least one real root
B. no real root
C. only real roots
D. at least one root which is not real
\(\textbf{Option A}\)
Complex roots of polynomials with real coefficients comes in conjugate pairs (\(a+ ib \) and \(a-ib \)). So polynomials of odd degrees with real coefficients must have at least one real root.
(Part B Question: NET June-2015) The number of subfields of a field \(F \) of cardinality \(2^{100} \) is
A. 2
B. 4
C. 9
D. 100
\(\textbf{Option C}\)
No. of subfields of \(F \) = no. of divisors of \(100 (= 2^2 \times 5^2) = (2 +1) \times (2+1) = 9\)
(SET Feb-2005) How many solutions are there in \(\mathbb{R} \) for the equation \(\sin(x) = x+1 \)?
A. Finitely many solutions
B. Countably infinite solutions
C. Uncountably many solution
D. No solution
\( \textbf{Option A}\)
The graph of \(\sin(x) \) and \(x+1 \) intersects at a point. So there is only one solution.
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