# IGNOU MMT-004 Solved Assignment 2024 | M.Sc. MACS

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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## IGNOU MMT-004 Assignment Question Paper 2024

mmt-003-solved-assignment-2024-qp-2afdf4f2-c576-4553-8e87-8dfd8225b98e

# mmt-003-solved-assignment-2024-qp-2afdf4f2-c576-4553-8e87-8dfd8225b98e

MMT-003 Solved Assignment 2024 QP
1. a) Let $\mathrm{X}=\mathrm{C}\left[0,1\right]$$\mathrm{X}=\mathrm{C}\left[0,1\right]$X=C[0,1]\mathrm{X}=\mathrm{C}[0,1]$\mathrm{X}=\mathrm{C}\left[0,1\right]$. Define $d:X×X\to \mathbf{R}$$d:X×X\to \mathbf{R}$d:X xx X rarrRd: X \times X \rightarrow \mathbf{R}$d:X×X\to \mathbf{R}$ by $\mathrm{d}\left(\mathrm{f},\mathrm{g}\right)={\int }_{0}^{1}|\mathrm{f}\left(\mathrm{t}\right)-\mathrm{g}\left(\mathrm{t}\right)|\mathrm{d}\mathrm{t},\mathrm{f},\mathrm{g}\in \mathrm{X}$$\mathrm{d}\left(\mathrm{f},\mathrm{g}\right)={\int }_{0}^{1} |\mathrm{f}\left(\mathrm{t}\right)-\mathrm{g}\left(\mathrm{t}\right)|\mathrm{d}\mathrm{t},\mathrm{f},\mathrm{g}\in \mathrm{X}$d(f,g)=int_(0)^(1)|f(t)-g(t)|dt,f,ginX\mathrm{d}(\mathrm{f}, \mathrm{g})=\int_0^1|\mathrm{f}(\mathrm{t})-\mathrm{g}(\mathrm{t})| \mathrm{dt}, \mathrm{f}, \mathrm{g} \in \mathrm{X}$\mathrm{d}\left(\mathrm{f},\mathrm{g}\right)={\int }_{0}^{1}|\mathrm{f}\left(\mathrm{t}\right)-\mathrm{g}\left(\mathrm{t}\right)|\mathrm{d}\mathrm{t},\mathrm{f},\mathrm{g}\in \mathrm{X}$ where the integral is the Riemann integral. Show that $d$$d$dd$d$ is a metric on $X$$X$XX$X$. Find $d\left(f,g\right)$$d\left(f,g\right)$d(f,g)d(f, g)$d\left(f,g\right)$ where $f\left(x\right)=4x$$f\left(x\right)=4x$f(x)=4xf(x)=4 x$f\left(x\right)=4x$ and $g\left(x\right)={x}^{3},x\in \left[0,1\right]$$g\left(x\right)={x}^{3},x\in \left[0,1\right]$g(x)=x^(3),x in[0,1]g(x)=x^3, x \in[0,1]$g\left(x\right)={x}^{3},x\in \left[0,1\right]$.
b) Let ( $X,d\right)$$X,d\right)$X,d)X, d)$X,d\right)$ be a metric space and $a\in X$$a\in X$a in Xa \in X$a\in X$ be a fixed point of $X$$X$XX$X$. Show that the function ${f}_{a}:X\to \mathbf{R}$${f}_{a}:X\to \mathbf{R}$f_(a):X rarrRf_a: X \rightarrow \mathbf{R}${f}_{a}:X\to \mathbf{R}$ given by ${\mathrm{f}}_{\mathrm{a}}\left(\mathrm{x}\right)=\mathrm{d}\left(\mathrm{x},\mathrm{a}\right)$${\mathrm{f}}_{\mathrm{a}}\left(\mathrm{x}\right)=\mathrm{d}\left(\mathrm{x},\mathrm{a}\right)$f_(a)(x)=d(x,a)\mathrm{f}_{\mathrm{a}}(\mathrm{x})=\mathrm{d}(\mathrm{x}, \mathrm{a})${\mathrm{f}}_{\mathrm{a}}\left(\mathrm{x}\right)=\mathrm{d}\left(\mathrm{x},\mathrm{a}\right)$ is continuous. Is it uniformly continuous? Justify you answer.
2. a) Let $A$$A$AA$A$ and $B$$B$BB$B$ be any two subsets of a metric space (X, d), then show that
i) int $\mathrm{A}=\cup \left\{\mathrm{E}$$\mathrm{A}=\cup \left\{\mathrm{E}$A=uu{E\mathrm{A}=\cup\{\mathrm{E}$\mathrm{A}=\cup \left\{\mathrm{E}$ : is open and $\mathrm{E}\subseteq \mathrm{A}\right\}$$\mathrm{E}\subseteq \mathrm{A}\right\}$EsubeA}\mathrm{E} \subseteq \mathrm{A}\}$\mathrm{E}\subseteq \mathrm{A}\right\}$
ii) $\mathrm{int}\left(\mathrm{A}\cap \mathrm{B}\right)=\mathrm{int}\mathrm{A}\cap \mathrm{int}\mathrm{B}$$\mathrm{int}\left(\mathrm{A}\cap \mathrm{B}\right)=\mathrm{int}\mathrm{A}\cap \mathrm{int}\mathrm{B}$int(AnnB)=int Ann int B\operatorname{int}(\mathrm{A} \cap \mathrm{B})=\operatorname{int} \mathrm{A} \cap \operatorname{int} \mathrm{B}$\mathrm{int}\left(\mathrm{A}\cap \mathrm{B}\right)=\mathrm{int}\mathrm{A}\cap \mathrm{int}\mathrm{B}$
iii) $\phantom{\rule{1em}{0ex}}\mathrm{int}\left(A\cup B\right)\supseteq$$\phantom{\rule{1em}{0ex}}\mathrm{int}\left(A\cup B\right)\supseteq$quad int(A uu B)supe\quad \operatorname{int}(A \cup B) \supseteq$\phantom{\rule{1em}{0ex}}\mathrm{int}\left(A\cup B\right)\supseteq$ int $A\cap$$A\cap$A nnA \cap$A\cap$ int $B$$B$BB$B$
iv) $\overline{\mathrm{A}\cap \mathrm{B}}\subseteq \overline{\mathrm{A}}\cap \overline{\mathrm{B}}$$\overline{\mathrm{A}\cap \mathrm{B}}\subseteq \overline{\mathrm{A}}\cap \overline{\mathrm{B}}$bar(AnnB)sube bar(A)nn bar(B)\overline{\mathrm{A} \cap \mathrm{B}} \subseteq \overline{\mathrm{A}} \cap \overline{\mathrm{B}}$\overline{\mathrm{A}\cap \mathrm{B}}\subseteq \overline{\mathrm{A}}\cap \overline{\mathrm{B}}$.
b) Find the interior, boundary and closure of the following sets $\mathbf{A}$$\mathbf{A}$A\mathbf{A}$\mathbf{A}$ in $\mathbf{R}$$\mathbf{R}$R\mathbf{R}$\mathbf{R}$ with the usual metric and discrete metric.
i) $A=\mathbf{Q}$$A=\mathbf{Q}$A=QA=\mathbf{Q}$A=\mathbf{Q}$, the set of rationals in $\mathbf{R}$$\mathbf{R}$R\mathbf{R}$\mathbf{R}$
ii) $\mathrm{A}=\right]1,2\right]\cup \right]2,4\left[$$\mathrm{A}=\right]1,2\right]\cup \right]2,4\left[$A=]1,2]uu]2,4[\mathrm{A}=] 1,2] \cup] 2,4[$\mathrm{A}=\right]1,2\right]\cup \right]2,4\left[$
3. a) Let $\left(X,{d}_{1}\right)$$\left(X,{d}_{1}\right)$(X,d_(1))\left(X, d_1\right)$\left(X,{d}_{1}\right)$ and $\left(Y,{d}_{2}\right)$$\left(Y,{d}_{2}\right)$(Y,d_(2))\left(Y, d_2\right)$\left(Y,{d}_{2}\right)$ be metric spaces. Show that $f:X\to Y$$f:X\to Y$f:X rarr Yf: X \rightarrow Y$f:X\to Y$ is continuous if and only if $f\left(\overline{A}\right)\subseteq \overline{f\left(A\right)}$$f\left(\overline{A}\right)\subseteq \overline{f\left(A\right)}$f( bar(A))sube bar(f(A))f(\bar{A}) \subseteq \overline{f(A)}$f\left(\overline{A}\right)\subseteq \overline{f\left(A\right)}$ where $A$$A$AA$A$ is any subset of $X$$X$XX$X$
b) Let $\left({\mathrm{X}}_{1},{\text{}\mathrm{d}}_{1}\right)$$\left({\mathrm{X}}_{1},{\text{}\mathrm{d}}_{1}\right)$(X_(1),d_(1))\left(\mathrm{X}_1, \mathrm{~d}_1\right)$\left({\mathrm{X}}_{1},{\text{}\mathrm{d}}_{1}\right)$ and $\left({\mathrm{X}}_{2},{\text{}\mathrm{d}}_{2}\right)$$\left({\mathrm{X}}_{2},{\text{}\mathrm{d}}_{2}\right)$(X_(2),d_(2))\left(\mathrm{X}_2, \mathrm{~d}_2\right)$\left({\mathrm{X}}_{2},{\text{}\mathrm{d}}_{2}\right)$ be two discrete metric spaces. Verify that the product metric on ${\mathrm{X}}_{1}×{\mathrm{X}}_{2}$${\mathrm{X}}_{1}×{\mathrm{X}}_{2}$X_(1)xxX_(2)\mathrm{X}_1 \times \mathrm{X}_2${\mathrm{X}}_{1}×{\mathrm{X}}_{2}$ is discrete.
c) Show that an infinite discrete metric space $\mathrm{X}$$\mathrm{X}$X\mathrm{X}$\mathrm{X}$ is bounded but not totally bounded.
4. a) Find the first derivative ${\mathrm{f}}^{\mathrm{\prime }}\left(\mathbf{a}\right)$${\mathrm{f}}^{\mathrm{\prime }}\left(\mathbf{a}\right)$f^(‘)(a)\mathrm{f}^{\prime}(\mathbf{a})${\mathrm{f}}^{\mathrm{\prime }}\left(\mathbf{a}\right)$ of the function $\mathrm{f}$$\mathrm{f}$f\mathrm{f}$\mathrm{f}$ defined by $f:{\mathbf{R}}^{3}\to {\mathbf{R}}^{2}$$f:{\mathbf{R}}^{3}\to {\mathbf{R}}^{2}$f:R^(3)rarrR^(2)f: \mathbf{R}^3 \rightarrow \mathbf{R}^2$f:{\mathbf{R}}^{3}\to {\mathbf{R}}^{2}$ given by $f\left(x,y,z\right)=\left(xyz,x+y+{z}^{2}\right)$$f\left(x,y,z\right)=\left(xyz,x+y+{z}^{2}\right)$f(x,y,z)=(xyz,x+y+z^(2))f(x, y, z)=\left(x y z, x+y+z^2\right)$f\left(x,y,z\right)=\left(xyz,x+y+{z}^{2}\right)$ where $\mathbf{a}=\left(1.-1,2\right)$$\mathbf{a}=\left(1.-1,2\right)$a=(1.-1,2)\mathbf{a}=(1 .-1,2)$\mathbf{a}=\left(1.-1,2\right)$.
b) Let $\mathrm{E}$$\mathrm{E}$E\mathrm{E}$\mathrm{E}$ be an open subset of ${\mathbf{R}}^{n}$${\mathbf{R}}^{n}$R^(n)\mathbf{R}^n${\mathbf{R}}^{n}$ and $f:E\to {\mathbf{R}}^{m}$$f:E\to {\mathbf{R}}^{m}$f:E rarrR^(m)f: E \rightarrow \mathbf{R}^m$f:E\to {\mathbf{R}}^{m}$ be a function such that each of its components function ${f}_{i}$${f}_{i}$f_(i)f_i${f}_{i}$ are differentiable, then show that $f$$f$ff$f$ is differentiable. Is the converse of this result true? Justify your answer.
c) Near what points may the surface ${z}^{2}+xz+y=0$${z}^{2}+xz+y=0$z^(2)+xz+y=0z^2+x z+y=0${z}^{2}+xz+y=0$ be represented uniquely as a graph of a differentiable function $\mathrm{z}=\mathrm{k}\left(\mathrm{x},\mathrm{y}\right)$$\mathrm{z}=\mathrm{k}\left(\mathrm{x},\mathrm{y}\right)$z=k(x,y)\mathrm{z}=\mathrm{k}(\mathrm{x}, \mathrm{y})$\mathrm{z}=\mathrm{k}\left(\mathrm{x},\mathrm{y}\right)$ ? Locate such a point.
5. a) Use the method of Lagrange’s multiplier method to find the shortest possible distance from the ellipse ${x}^{2}+2{y}^{2}=2$${x}^{2}+2{y}^{2}=2$x^(2)+2y^(2)=2x^2+2 y^2=2${x}^{2}+2{y}^{2}=2$ to the line $x+y=2$$x+y=2$x+y=2x+y=2$x+y=2$.
b) Find the directional derivative of the function $f:{\mathbf{R}}^{4}\to {\mathbf{R}}^{3}$$f:{\mathbf{R}}^{4}\to {\mathbf{R}}^{3}$f:R^(4)rarrR^(3)f: \mathbf{R}^4 \rightarrow \mathbf{R}^3$f:{\mathbf{R}}^{4}\to {\mathbf{R}}^{3}$ defined by
$f\left(x,y,z,w\right)=\left({x}^{2}y,xyz,{x}^{2}+{y}^{2}+{z}^{2}\right)$$f\left(x,y,z,w\right)=\left({x}^{2}y,xyz,{x}^{2}+{y}^{2}+{z}^{2}\right)$f(x,y,z,w)=(x^(2)y,xyz,x^(2)+y^(2)+z^(2))f(x, y, z, w)=\left(x^2 y, x y z, x^2+y^2+z^2\right)$f\left(x,y,z,w\right)=\left({x}^{2}y,xyz,{x}^{2}+{y}^{2}+{z}^{2}\right)$
at $\mathrm{a}=\left(1,2,-1,-2\right)$$\mathrm{a}=\left(1,2,-1,-2\right)$a=(1,2,-1,-2)\mathrm{a}=(1,2,-1,-2)$\mathrm{a}=\left(1,2,-1,-2\right)$ in the direction $\mathrm{v}=\left(0,1,2,-2\right)$$\mathrm{v}=\left(0,1,2,-2\right)$v=(0,1,2,-2)\mathrm{v}=(0,1,2,-2)$\mathrm{v}=\left(0,1,2,-2\right)$.
6. a) Let A be a compact non-empty subset of a metric space (X, d) and let F be a closed subset of $X$$X$XX$X$ such that $A\cap F=\varphi$$A\cap F=\varphi$A nn F=phiA \cap F=\phi$A\cap F=\varphi$, then show that $d\left(A,F\right)>0$$d\left(A,F\right)>0$d(A,F) > 0d(A, F)>0$d\left(A,F\right)>0$ where $d\left(A,F\right)=inf\left\{d\left(a,b\right):a\in A,b\in F\right\}$$d\left(A,F\right)=inf\left\{d\left(a,b\right):a\in A,b\in F\right\}$d(A,F)=i n f{d(a,b):a in A,b in F}d(A, F)=\inf \{d(a, b): a \in A, b \in F\}$d\left(A,F\right)=inf\left\{d\left(a,b\right):a\in A,b\in F\right\}$.
b) Give an example of the following with justification
i) A vector-valued function $f:{\mathbf{R}}^{3}\to {\mathbf{R}}^{3}$$f:{\mathbf{R}}^{3}\to {\mathbf{R}}^{3}$f:R^(3)rarrR^(3)f: \mathbf{R}^3 \rightarrow \mathbf{R}^3$f:{\mathbf{R}}^{3}\to {\mathbf{R}}^{3}$ which is not differentiable at $\left(0,0,0\right)$$\left(0,0,0\right)$(0,0,0)(0,0,0)$\left(0,0,0\right)$.
ii) A function which is Legesgue measurable on $\mathbf{R}$$\mathbf{R}$R\mathbf{R}$\mathbf{R}$.
c) Show that the components of a metric space is either identical or pairwise disjoint.
7. a) Let $\mathbf{Q}$$\mathbf{Q}$Q\mathbf{Q}$\mathbf{Q}$ be the set of rationals with the metric defined on $\mathbf{Q}$$\mathbf{Q}$Q\mathbf{Q}$\mathbf{Q}$ by $d:\mathbf{Q}×\mathbf{Q}\to \mathbf{R}$$d:\mathbf{Q}×\mathbf{Q}\to \mathbf{R}$d:QxxQrarrRd: \mathbf{Q} \times \mathbf{Q} \rightarrow \mathbf{R}$d:\mathbf{Q}×\mathbf{Q}\to \mathbf{R}$, defined by $d\left(x,y\right)=|x-y|,\mathrm{\forall }x,y\in \mathbf{R}$$d\left(x,y\right)=|x-y|,\mathrm{\forall }x,y\in \mathbf{R}$d(x,y)=|x-y|,AA x,y inRd(x, y)=|x-y|, \forall x, y \in \mathbf{R}$d\left(x,y\right)=|x-y|,\mathrm{\forall }x,y\in \mathbf{R}$.
Show that $\left\{{\left(1+\frac{1}{\mathrm{n}}\right)}^{\mathrm{n}}\right\}$$\left\{{\left(1+\frac{1}{\mathrm{n}}\right)}^{\mathrm{n}}\right\}${(1+(1)/(n))^(n)}\left\{\left(1+\frac{1}{\mathrm{n}}\right)^{\mathrm{n}}\right\}$\left\{{\left(1+\frac{1}{\mathrm{n}}\right)}^{\mathrm{n}}\right\}$ is Cauchy sequence in $\mathbf{Q}$$\mathbf{Q}$Q\mathbf{Q}$\mathbf{Q}$, but does not converge in $\mathbf{Q}$$\mathbf{Q}$Q\mathbf{Q}$\mathbf{Q}$ and $\left\{\frac{1}{{3}^{n}}\right\}$$\left\{\frac{1}{{3}^{n}}\right\}${(1)/(3^(n))}\left\{\frac{1}{3^n}\right\}$\left\{\frac{1}{{3}^{n}}\right\}$ is a Cauchy sequence $\mathbf{Q}$$\mathbf{Q}$Q\mathbf{Q}$\mathbf{Q}$ which converges in $\mathbf{Q}$$\mathbf{Q}$Q\mathbf{Q}$\mathbf{Q}$ to the limit 0 .
b) Which of the following sets are totally bounded? Give reasons for your answer. Are they compact?
i) $\phantom{\rule{1em}{0ex}}2\mathbf{N}$$\phantom{\rule{1em}{0ex}}2\mathbf{N}$quad2N\quad 2 \mathbf{N}$\phantom{\rule{1em}{0ex}}2\mathbf{N}$ in $\left(\mathbf{N},d\right)$$\left(\mathbf{N},d\right)$(N,d)(\mathbf{N}, d)$\left(\mathbf{N},d\right)$ where $d$$d$dd$d$ is the discrete metric.
ii) $\phantom{\rule{1em}{0ex}}\left[0,2\right]\cup \left[5,10\right]$$\phantom{\rule{1em}{0ex}}\left[0,2\right]\cup \left[5,10\right]$quad[0,2]uu[5,10]\quad[0,2] \cup[5,10]$\phantom{\rule{1em}{0ex}}\left[0,2\right]\cup \left[5,10\right]$ in $\left(\mathbf{R},d\right)$$\left(\mathbf{R},d\right)$(R,d)(\mathbf{R}, d)$\left(\mathbf{R},d\right)$ where $d$$d$dd$d$ is the Euclidean metric.
c) Which of the following sets are connected sets in ${\mathbf{R}}^{2}$${\mathbf{R}}^{2}$R^(2)\mathbf{R}^2${\mathbf{R}}^{2}$ with the metric given against it? Justify your answer.
i) $\phantom{\rule{1em}{0ex}}\mathrm{A}=\left\{\left(\mathrm{x},\mathrm{y}\right):0\le \mathrm{x}\le 1,0\le \mathrm{y}\le 2\right\}$$\phantom{\rule{1em}{0ex}}\mathrm{A}=\left\{\left(\mathrm{x},\mathrm{y}\right):0\le \mathrm{x}\le 1,0\le \mathrm{y}\le 2\right\}$quadA={(x,y):0 <= x <= 1,0 <= y <= 2}\quad \mathrm{A}=\{(\mathrm{x}, \mathrm{y}): 0 \leq \mathrm{x} \leq 1,0 \leq \mathrm{y} \leq 2\}$\phantom{\rule{1em}{0ex}}\mathrm{A}=\left\{\left(\mathrm{x},\mathrm{y}\right):0\le \mathrm{x}\le 1,0\le \mathrm{y}\le 2\right\}$ under the standard metric.
ii) $\mathrm{A}=\left\{\left(\mathrm{x},\mathrm{y}\right):{\mathrm{x}}^{2}+{\mathrm{y}}^{2}=1\right\}$$\mathrm{A}=\left\{\left(\mathrm{x},\mathrm{y}\right):{\mathrm{x}}^{2}+{\mathrm{y}}^{2}=1\right\}$A={(x,y):x^(2)+y^(2)=1}\mathrm{A}=\left\{(\mathrm{x}, \mathrm{y}): \mathrm{x}^2+\mathrm{y}^2=1\right\}$\mathrm{A}=\left\{\left(\mathrm{x},\mathrm{y}\right):{\mathrm{x}}^{2}+{\mathrm{y}}^{2}=1\right\}$ under the discrete metric.
8. a) Consider $\mathbf{Z}$$\mathbf{Z}$Z\mathbf{Z}$\mathbf{Z}$ and let ${\mathcal{F}}_{1}$${\mathcal{F}}_{1}$F_(1)\mathcal{F}_1${\mathcal{F}}_{1}$ denote the class of subsets of $\mathbf{Z}$$\mathbf{Z}$Z\mathbf{Z}$\mathbf{Z}$, given by ${\mathcal{F}}_{1}=\left\{\mathrm{A}\subset \mathbf{Z}$${\mathcal{F}}_{1}=\left\{\mathrm{A}\subset \mathbf{Z}$F_(1)={AsubZ\mathcal{F}_1=\{\mathrm{A} \subset \mathbf{Z}${\mathcal{F}}_{1}=\left\{\mathrm{A}\subset \mathbf{Z}$ : either $\mathrm{A}$$\mathrm{A}$A\mathrm{A}$\mathrm{A}$ is finite or ${\mathrm{A}}^{\mathrm{c}}$${\mathrm{A}}^{\mathrm{c}}$A^(c)\mathrm{A}^{\mathrm{c}}${\mathrm{A}}^{\mathrm{c}}$ is finite }. Check whether ${\mathcal{F}}_{1}$${\mathcal{F}}_{1}$F_(1)\mathcal{F}_1${\mathcal{F}}_{1}$ is a $\sigma$$\sigma$sigma\sigma$\sigma$ algebra or not.
b) Let A be any set in $\mathbf{R}$$\mathbf{R}$R\mathbf{R}$\mathbf{R}$, show that ${m}^{\ast }\left(A\right)={m}^{\ast }\left(A+x\right)$${m}^{\ast }\left(A\right)={m}^{\ast }\left(A+x\right)$m^(**)(A)=m^(**)(A+x)m^*(A)=m^*(A+x)${m}^{\ast }\left(A\right)={m}^{\ast }\left(A+x\right)$ where ${m}^{\ast }$${m}^{\ast }$m^(**)m^*${m}^{\ast }$ denotes the outer measure.
c) Find the measure of the following sets.
i) $\phantom{\rule{1em}{0ex}}E=\bigcap _{n=1}^{\mathrm{\infty }}\left(a-\frac{1}{n},b\right)$$\phantom{\rule{1em}{0ex}}E=\bigcap _{n=1}^{\mathrm{\infty }} \left(a-\frac{1}{n},b\right)$quad E=nnn_(n=1)^(oo)(a-(1)/(n),b)\quad E=\bigcap_{n=1}^{\infty}\left(a-\frac{1}{n}, b\right)$\phantom{\rule{1em}{0ex}}E=\bigcap _{n=1}^{\mathrm{\infty }}\left(a-\frac{1}{n},b\right)$
ii) $\mathrm{E}=\mathbf{Q}\cup \left\{1,2,3,4\right\}$$\mathrm{E}=\mathbf{Q}\cup \left\{1,2,3,4\right\}$E=Quu{1,2,3,4}\mathrm{E}=\mathbf{Q} \cup\{1,2,3,4\}$\mathrm{E}=\mathbf{Q}\cup \left\{1,2,3,4\right\}$
iii) $\mathrm{E}=\right]5,7\left[\cup \left[7,7.5\right]$$\mathrm{E}=\right]5,7\left[\cup \left[7,7.5\right]$E=]5,7[uu[7,7.5]\mathrm{E}=] 5,7[\cup[7,7.5]$\mathrm{E}=\right]5,7\left[\cup \left[7,7.5\right]$.
9. a) Show that if $f$$f$ff$f$ is measurable, then the function ${f}^{a}\left(x\right)$${f}^{a}\left(x\right)$f^(a)(x)f^a(x)${f}^{a}\left(x\right)$ given by
${f}^{a}\left(x\right)=\left\{\begin{array}{cc}a& \text{if}f\left(x\right)>a\\ f\left(x\right)& \text{if}f\left(x\right)\le a\end{array}$${f}^{a}\left(x\right)=\left\{\begin{array}{cc}a& \text{if}f\left(x\right)>a\\ f\left(x\right)& \text{if}f\left(x\right)\le a\end{array}\right\$f^(a)(x)={[a,” if “f(x) > a],[f(x),” if “f(x) <= a]:}f^a(x)=\left\{\begin{array}{cc} a & \text { if } f(x)>a \\ f(x) & \text { if } f(x) \leq a \end{array}\right.${f}^{a}\left(x\right)=\left\{\begin{array}{cc}a& \text{if}f\left(x\right)>a\\ f\left(x\right)& \text{if}f\left(x\right)\le a\end{array}$
is also measurable.
b) Verify Bounded Convergence Theorem for the sequence of functions $\left\{{f}_{n}\right\}$$\left\{{f}_{n}\right\}${f_(n)}\left\{f_n\right\}$\left\{{f}_{n}\right\}$ where
${\mathrm{f}}_{\mathrm{n}}\left(\mathrm{x}\right)=\frac{1}{\left(1+\mathrm{x}/\mathrm{n}{\right)}^{\mathrm{n}}},0\le \mathrm{x}\le 1,\mathrm{n}\in \mathbf{N}$${\mathrm{f}}_{\mathrm{n}}\left(\mathrm{x}\right)=\frac{1}{\left(1+\mathrm{x}/\mathrm{n}{\right)}^{\mathrm{n}}},0\le \mathrm{x}\le 1,\mathrm{n}\in \mathbf{N}$f_(n)(x)=(1)/((1+x//n)^(n)),0 <= x <= 1,ninN\mathrm{f}_{\mathrm{n}}(\mathrm{x})=\frac{1}{(1+\mathrm{x} / \mathrm{n})^{\mathrm{n}}}, 0 \leq \mathrm{x} \leq 1, \mathrm{n} \in \mathbf{N}${\mathrm{f}}_{\mathrm{n}}\left(\mathrm{x}\right)=\frac{1}{\left(1+\mathrm{x}/\mathrm{n}{\right)}^{\mathrm{n}}},0\le \mathrm{x}\le 1,\mathrm{n}\in \mathbf{N}$
c) Find the fourier series of the function $f$$f$ff$f$ defined by
$f\left(x\right)=\left\{\begin{array}{c}-{x}^{2},-\pi $f\left(x\right)=\left\{\begin{array}{c}-{x}^{2},-\pi f(x)={[-x^(2)”,”-pi < x <= 0],[x^(2)”,”0 < x < pi]:}f(x)=\left\{\begin{array}{c} -x^2,-\pi<x \leq 0 \\ x^2, 0<x<\pi \end{array}\right.$f\left(x\right)=\left\{\begin{array}{c}-{x}^{2},-\pi
1. State whether the following statements are True or False. Justify your answers.
a) The sequence $\left\{\left(\frac{1}{\mathrm{n}},\frac{1}{\mathrm{n}}\right):\mathrm{n}\in \mathbf{N}\right\}$$\left\{\left(\frac{1}{\mathrm{n}},\frac{1}{\mathrm{n}}\right):\mathrm{n}\in \mathbf{N}\right\}${((1)/(n),(1)/(n)):ninN}\left\{\left(\frac{1}{\mathrm{n}}, \frac{1}{\mathrm{n}}\right): \mathrm{n} \in \mathbf{N}\right\}$\left\{\left(\frac{1}{\mathrm{n}},\frac{1}{\mathrm{n}}\right):\mathrm{n}\in \mathbf{N}\right\}$ is convergent in ${\mathbf{R}}^{2}$${\mathbf{R}}^{2}$R^(2)\mathbf{R}^2${\mathbf{R}}^{2}$ under the discrete metric on ${\mathbf{R}}^{2}$${\mathbf{R}}^{2}$R^(2)\mathbf{R}^2${\mathbf{R}}^{2}$.
b) A subset in a metric space is compact if it is closed.
c) Continuous image of a path connected space is path connected.
d) The second derivative of a linear map from ${\mathbf{R}}^{n}$${\mathbf{R}}^{n}$R^(n)\mathbf{R}^n${\mathbf{R}}^{n}$ to ${\mathbf{R}}^{m}$${\mathbf{R}}^{m}$R^(m)\mathbf{R}^m${\mathbf{R}}^{m}$ never vanishes.
e) If ${\int }_{\mathrm{A}}\mathrm{f}\mathrm{d}\mathrm{m}={\int }_{\mathrm{A}}\mathrm{g}\mathrm{d}\mathrm{m}$${\int }_{\mathrm{A}} \mathrm{f}\mathrm{d}\mathrm{m}={\int }_{\mathrm{A}} \mathrm{g}\mathrm{d}\mathrm{m}$int_(A)fdm=int_(A)gdm\int_{\mathrm{A}} \mathrm{fdm}=\int_{\mathrm{A}} \mathrm{gdm}${\int }_{\mathrm{A}}\mathrm{f}\mathrm{d}\mathrm{m}={\int }_{\mathrm{A}}\mathrm{g}\mathrm{d}\mathrm{m}$ for all $\mathrm{A}\in \mathbit{M}$$\mathrm{A}\in \mathbit{M}$Ain M\mathrm{A} \in \boldsymbol{M}$\mathrm{A}\in \mathbit{M}$, then $\mathrm{f}=\mathrm{g}$$\mathrm{f}=\mathrm{g}$f=g\mathrm{f}=\mathrm{g}$\mathrm{f}=\mathrm{g}$.
$$cos\:3\theta =4\:cos^3\:\theta -3\:cos\:\theta$$

## MMT-004 Sample Solution 2024

mmt-004-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee

# mmt-004-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee

MMT-004 Solved Assignment 2024 SS
1. a) Let $\mathrm{X}=\mathrm{C}\left[0,1\right]$$\mathrm{X}=\mathrm{C}\left[0,1\right]$X=C[0,1]\mathrm{X}=\mathrm{C}[0,1]$\mathrm{X}=\mathrm{C}\left[0,1\right]$. Define $d:X×X\to \mathbf{R}$$d:X×X\to \mathbf{R}$d:X xx X rarrRd: X \times X \rightarrow \mathbf{R}$d:X×X\to \mathbf{R}$ by $\mathrm{d}\left(\mathrm{f},\mathrm{g}\right)={\int }_{0}^{1}|\mathrm{f}\left(\mathrm{t}\right)-\mathrm{g}\left(\mathrm{t}\right)|\mathrm{d}\mathrm{t},\mathrm{f},\mathrm{g}\in \mathrm{X}$$\mathrm{d}\left(\mathrm{f},\mathrm{g}\right)={\int }_{0}^{1} |\mathrm{f}\left(\mathrm{t}\right)-\mathrm{g}\left(\mathrm{t}\right)|\mathrm{d}\mathrm{t},\mathrm{f},\mathrm{g}\in \mathrm{X}$d(f,g)=int_(0)^(1)|f(t)-g(t)|dt,f,ginX\mathrm{d}(\mathrm{f}, \mathrm{g})=\int_0^1|\mathrm{f}(\mathrm{t})-\mathrm{g}(\mathrm{t})| \mathrm{dt}, \mathrm{f}, \mathrm{g} \in \mathrm{X}$\mathrm{d}\left(\mathrm{f},\mathrm{g}\right)={\int }_{0}^{1}|\mathrm{f}\left(\mathrm{t}\right)-\mathrm{g}\left(\mathrm{t}\right)|\mathrm{d}\mathrm{t},\mathrm{f},\mathrm{g}\in \mathrm{X}$ where the integral is the Riemann integral. Show that $d$$d$dd$d$ is a metric on $X$$X$XX$X$. Find $d\left(f,g\right)$$d\left(f,g\right)$d(f,g)d(f, g)$d\left(f,g\right)$ where $f\left(x\right)=4x$$f\left(x\right)=4x$f(x)=4xf(x)=4 x$f\left(x\right)=4x$ and $g\left(x\right)={x}^{3},x\in \left[0,1\right]$$g\left(x\right)={x}^{3},x\in \left[0,1\right]$g(x)=x^(3),x in[0,1]g(x)=x^3, x \in[0,1]$g\left(x\right)={x}^{3},x\in \left[0,1\right]$.
To show that $d$$d$dd$d$ is a metric on $X$$X$XX$X$, we need to verify the following properties for all $f,g,h\in X$$f,g,h\in X$f,g,h in Xf, g, h \in X$f,g,h\in X$:
1. Non-negativity: $d\left(f,g\right)\ge 0$$d\left(f,g\right)\ge 0$d(f,g) >= 0d(f, g) \geq 0$d\left(f,g\right)\ge 0$
2. Identity of indiscernibles: $d\left(f,g\right)=0$$d\left(f,g\right)=0$d(f,g)=0d(f, g) = 0$d\left(f,g\right)=0$ if and only if $f=g$$f=g$f=gf = g$f=g$
3. Symmetry: $d\left(f,g\right)=d\left(g,f\right)$$d\left(f,g\right)=d\left(g,f\right)$d(f,g)=d(g,f)d(f, g) = d(g, f)$d\left(f,g\right)=d\left(g,f\right)$
4. Triangle inequality: $d\left(f,h\right)\le d\left(f,g\right)+d\left(g,h\right)$$d\left(f,h\right)\le d\left(f,g\right)+d\left(g,h\right)$d(f,h) <= d(f,g)+d(g,h)d(f, h) \leq d(f, g) + d(g, h)$d\left(f,h\right)\le d\left(f,g\right)+d\left(g,h\right)$
Let’s verify each property:
1. Non-negativity:
For any $f,g\in X$$f,g\in X$f,g in Xf, g \in X$f,g\in X$, the absolute value function $|\cdot |$$|\cdot |$|*||\cdot|$|\cdot |$ ensures that $|f\left(t\right)-g\left(t\right)|\ge 0$$|f\left(t\right)-g\left(t\right)|\ge 0$|f(t)-g(t)| >= 0|f(t) – g(t)| \geq 0$|f\left(t\right)-g\left(t\right)|\ge 0$ for all $t\in \left[0,1\right]$$t\in \left[0,1\right]$t in[0,1]t \in [0, 1]$t\in \left[0,1\right]$. Therefore, the integral ${\int }_{0}^{1}|f\left(t\right)-g\left(t\right)|\phantom{\rule{thinmathspace}{0ex}}dt$${\int }_{0}^{1} |f\left(t\right)-g\left(t\right)|\phantom{\rule{thinmathspace}{0ex}}dt$int_(0)^(1)|f(t)-g(t)|dt\int_0^1 |f(t) – g(t)| \, dt${\int }_{0}^{1}|f\left(t\right)-g\left(t\right)|\phantom{\rule{thinmathspace}{0ex}}dt$ is also non-negative. Hence, $d\left(f,g\right)\ge 0$$d\left(f,g\right)\ge 0$d(f,g) >= 0d(f, g) \geq 0$d\left(f,g\right)\ge 0$.
2. Identity of indiscernibles:
If $f=g$$f=g$f=gf = g$f=g$, then $f\left(t\right)-g\left(t\right)=0$$f\left(t\right)-g\left(t\right)=0$f(t)-g(t)=0f(t) – g(t) = 0$f\left(t\right)-g\left(t\right)=0$ for all $t\in \left[0,1\right]$$t\in \left[0,1\right]$t in[0,1]t \in [0, 1]$t\in \left[0,1\right]$, so $d\left(f,g\right)={\int }_{0}^{1}|f\left(t\right)-g\left(t\right)|\phantom{\rule{thinmathspace}{0ex}}dt={\int }_{0}^{1}0\phantom{\rule{thinmathspace}{0ex}}dt=0$$d\left(f,g\right)={\int }_{0}^{1} |f\left(t\right)-g\left(t\right)|\phantom{\rule{thinmathspace}{0ex}}dt={\int }_{0}^{1} 0\phantom{\rule{thinmathspace}{0ex}}dt=0$d(f,g)=int_(0)^(1)|f(t)-g(t)|dt=int_(0)^(1)0dt=0d(f, g) = \int_0^1 |f(t) – g(t)| \, dt = \int_0^1 0 \, dt = 0$d\left(f,g\right)={\int }_{0}^{1}|f\left(t\right)-g\left(t\right)|\phantom{\rule{thinmathspace}{0ex}}dt={\int }_{0}^{1}0\phantom{\rule{thinmathspace}{0ex}}dt=0$.
Conversely, if $d\left(f,g\right)=0$$d\left(f,g\right)=0$d(f,g)=0d(f, g) = 0$d\left(f,g\right)=0$, then ${\int }_{0}^{1}|f\left(t\right)-g\left(t\right)|\phantom{\rule{thinmathspace}{0ex}}dt=0$${\int }_{0}^{1} |f\left(t\right)-g\left(t\right)|\phantom{\rule{thinmathspace}{0ex}}dt=0$int_(0)^(1)|f(t)-g(t)|dt=0\int_0^1 |f(t) – g(t)| \, dt = 0${\int }_{0}^{1}|f\left(t\right)-g\left(t\right)|\phantom{\rule{thinmathspace}{0ex}}dt=0$. Since the integrand is non-negative, it must be zero almost everywhere, implying that $f\left(t\right)=g\left(t\right)$$f\left(t\right)=g\left(t\right)$f(t)=g(t)f(t) = g(t)$f\left(t\right)=g\left(t\right)$ for almost all $t\in \left[0,1\right]$$t\in \left[0,1\right]$t in[0,1]t \in [0, 1]$t\in \left[0,1\right]$. Since $f$$f$ff$f$ and $g$$g$gg$g$ are continuous, they must be equal everywhere on $\left[0,1\right]$$\left[0,1\right]$[0,1][0, 1]$\left[0,1\right]$, so $f=g$$f=g$f=gf = g$f=g$.
3. Symmetry:
By the properties of the absolute value function, $|f\left(t\right)-g\left(t\right)|=|g\left(t\right)-f\left(t\right)|$$|f\left(t\right)-g\left(t\right)|=|g\left(t\right)-f\left(t\right)|$|f(t)-g(t)|=|g(t)-f(t)||f(t) – g(t)| = |g(t) – f(t)|$|f\left(t\right)-g\left(t\right)|=|g\left(t\right)-f\left(t\right)|$ for all $t\in \left[0,1\right]$$t\in \left[0,1\right]$t in[0,1]t \in [0, 1]$t\in \left[0,1\right]$. Therefore, $d\left(f,g\right)={\int }_{0}^{1}|f\left(t\right)-g\left(t\right)|\phantom{\rule{thinmathspace}{0ex}}dt={\int }_{0}^{1}|g\left(t\right)-f\left(t\right)|\phantom{\rule{thinmathspace}{0ex}}dt=d\left(g,f\right)$$d\left(f,g\right)={\int }_{0}^{1} |f\left(t\right)-g\left(t\right)|\phantom{\rule{thinmathspace}{0ex}}dt={\int }_{0}^{1} |g\left(t\right)-f\left(t\right)|\phantom{\rule{thinmathspace}{0ex}}dt=d\left(g,f\right)$d(f,g)=int_(0)^(1)|f(t)-g(t)|dt=int_(0)^(1)|g(t)-f(t)|dt=d(g,f)d(f, g) = \int_0^1 |f(t) – g(t)| \, dt = \int_0^1 |g(t) – f(t)| \, dt = d(g, f)$d\left(f,g\right)={\int }_{0}^{1}|f\left(t\right)-g\left(t\right)|\phantom{\rule{thinmathspace}{0ex}}dt={\int }_{0}^{1}|g\left(t\right)-f\left(t\right)|\phantom{\rule{thinmathspace}{0ex}}dt=d\left(g,f\right)$.
4. Triangle inequality:
For any $f,g,h\in X$$f,g,h\in X$f,g,h in Xf, g, h \in X$f,g,h\in X$ and for all $t\in \left[0,1\right]$$t\in \left[0,1\right]$t in[0,1]t \in [0, 1]$t\in \left[0,1\right]$, we have $|f\left(t\right)-h\left(t\right)|=|f\left(t\right)-g\left(t\right)+g\left(t\right)-h\left(t\right)|\le |f\left(t\right)-g\left(t\right)|+|g\left(t\right)-h\left(t\right)|$$|f\left(t\right)-h\left(t\right)|=|f\left(t\right)-g\left(t\right)+g\left(t\right)-h\left(t\right)|\le |f\left(t\right)-g\left(t\right)|+|g\left(t\right)-h\left(t\right)|$|f(t)-h(t)|=|f(t)-g(t)+g(t)-h(t)| <= |f(t)-g(t)|+|g(t)-h(t)||f(t) – h(t)| = |f(t) – g(t) + g(t) – h(t)| \leq |f(t) – g(t)| + |g(t) – h(t)|