# IGNOU MMT-005 Solved Assignment 2024 | M.Sc. MACS

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

365.00

Access via our Android App Only

Details For MMT-005 Solved Assignment

## IGNOU MMT-05 Assignment Question Paper 2024

mmt-005-solved-assignment-2024-qp-b5864dd3-9e6a-403d-817d-3f61b693b950

# mmt-005-solved-assignment-2024-qp-b5864dd3-9e6a-403d-817d-3f61b693b950

1. Determine whether each of the following statement is true or false. Justify your answer with a short proof or a counter example.
i) If $z=a+ib$$z=a+ib$z=a+ibz=a+i b$z=a+ib$, where $a$$a$aa$a$ and $b$$b$bb$b$ are integers, then $|1+z+{z}^{2}+\cdots +{z}^{n}|\ge |z{|}^{n}$$\left|1+z+{z}^{2}+\cdots +{z}^{n}\right|\ge |z{|}^{n}$|1+z+z^(2)+cdots+z^(n)| >= |z|^(n)\left|1+z+z^2+\cdots+z^n\right| \geq|z|^n$|1+z+{z}^{2}+\cdots +{z}^{n}|\ge |z{|}^{n}$ if $a>0$$a>0$a > 0a>0$a>0$.
ii) If $f\left(z\right)$$f\left(z\right)$f(z)f(z)$f\left(z\right)$ and $\overline{f\left(z\right)}$$\overline{f\left(z\right)}$bar(f(z))\overline{f(z)}$\overline{f\left(z\right)}$ are analytic functions in $a$$a$aa$a$ domain, then $f$$f$ff$f$ is necessarily a constant.
iii) A real-valued function $u\left(x,y\right)$$u\left(x,y\right)$u(x,y)u(x, y)$u\left(x,y\right)$ is harmonic in $D$$D$DD$D$ iff $u\left(x,-y\right)$$u\left(x,-y\right)$u(x,-y)u(x,-y)$u\left(x,-y\right)$ is harmonic in $D$$D$DD$D$.
iv) $\underset{n\to \mathrm{\infty }}{lim}\left(n!{\right)}^{1/n}=\mathrm{\infty }$$\underset{n\to \mathrm{\infty }}{lim} \left(n!{\right)}^{1/n}=\mathrm{\infty }$lim_(n rarr oo)(n!)^(1//n)=oo\lim _{n \rightarrow \infty}(n !)^{1 / n}=\infty$\underset{n\to \mathrm{\infty }}{lim}\left(n!{\right)}^{1/n}=\mathrm{\infty }$.
v) The inequality $|{e}^{a}-{e}^{b}|\le |a-b|$$\left|{e}^{a}-{e}^{b}\right|\le |a-b|$|e^(a)-e^(b)| <= |a-b|\left|e^a-e^b\right| \leq|a-b|$|{e}^{a}-{e}^{b}|\le |a-b|$ holds for $a,b\in D=\left\{w:\mathrm{Re}w\le 0\right\}$$a,b\in D=\left\{w:\mathrm{Re}w\le 0\right\}$a,b in D={w:Re w <= 0}a, b \in D=\{w: \operatorname{Re} w \leq 0\}$a,b\in D=\left\{w:\mathrm{Re}w\le 0\right\}$.
vi) If $f\left(z\right)=\sum _{n=0}^{\mathrm{\infty }}{a}_{n}\left(z-a{\right)}^{n}$$f\left(z\right)=\sum _{n=0}^{\mathrm{\infty }} {a}_{n}\left(z-a{\right)}^{n}$f(z)=sum_(n=0)^(oo)a_(n)(z-a)^(n)f(z)=\sum_{n=0}^{\infty} a_n(z-a)^n$f\left(z\right)=\sum _{n=0}^{\mathrm{\infty }}{a}_{n}\left(z-a{\right)}^{n}$ has the property that $\sum _{n=0}^{\mathrm{\infty }}{f}^{\left(n\right)}\left(a\right)$$\sum _{n=0}^{\mathrm{\infty }} {f}^{\left(n\right)}\left(a\right)$sum_(n=0)^(oo)f^((n))(a)\sum_{n=0}^{\infty} f^{(n)}(a)$\sum _{n=0}^{\mathrm{\infty }}{f}^{\left(n\right)}\left(a\right)$ converges, then $f$$f$ff$f$ is necessarily an entire function.
vii) If a power series $\sum _{n=0}^{\mathrm{\infty }}{a}_{n}{z}^{n}$$\sum _{n=0}^{\mathrm{\infty }} {a}_{n}{z}^{n}$sum_(n=0)^(oo)a_(n)z^(n)\sum_{n=0}^{\infty} a_n z^n$\sum _{n=0}^{\mathrm{\infty }}{a}_{n}{z}^{n}$ converges for $|z|<1$$|z|<1$|z| < 1|z|<1$|z|<1$ and if ${b}_{n}\in \mathbb{C}$${b}_{n}\in \mathbb{C}$b_(n)inCb_n \in \mathbb{C}${b}_{n}\in \mathbb{C}$ is such that $|{b}_{n}|<{n}^{2}|{a}_{n}|$$\left|{b}_{n}\right|<{n}^{2}\left|{a}_{n}\right|$|b_(n)| < n^(2)|a_(n)|\left|b_n\right|<n^2\left|a_n\right|$|{b}_{n}|<{n}^{2}|{a}_{n}|$ for all $n\ge 0$$n\ge 0$n >= 0n \geq 0$n\ge 0$, then $\sum _{n=0}^{\mathrm{\infty }}{b}_{n}{z}^{n}$$\sum _{n=0}^{\mathrm{\infty }} {b}_{n}{z}^{n}$sum_(n=0)^(oo)b_(n)z^(n)\sum_{n=0}^{\infty} b_n z^n$\sum _{n=0}^{\mathrm{\infty }}{b}_{n}{z}^{n}$ converges for $|z|<1$$|z|<1$|z| < 1|z|<1$|z|<1$.
viii) If $f$$f$ff$f$ is entire and $f\left(z\right)=f\left(-z\right)$$f\left(z\right)=f\left(-z\right)$f(z)=f(-z)f(z)=f(-z)$f\left(z\right)=f\left(-z\right)$ for all $z$$z$zz$z$, then there exists an entire function $g$$g$gg$g$ such that $f\left(z\right)=g\left({z}^{2}\right)$$f\left(z\right)=g\left({z}^{2}\right)$f(z)=g(z^(2))f(z)=g\left(z^2\right)$f\left(z\right)=g\left({z}^{2}\right)$ for all $z\in \mathbb{C}$$z\in \mathbb{C}$z inCz \in \mathbb{C}$z\in \mathbb{C}$.
ix) A mobius transformation which maps the upper half plane $\left\{z:\mathrm{Im}z>0\right\}$$\left\{z:\mathrm{Im}z>0\right\}${z:Im z > 0}\{z: \operatorname{Im} z>0\}$\left\{z:\mathrm{Im}z>0\right\}$ onto itself and fixing $0,\mathrm{\infty }$$0,\mathrm{\infty }$0,oo0, \infty$0,\mathrm{\infty }$ and no other points, must be of the form $Tz=\alpha z$$Tz=\alpha z$Tz=alpha zT z=\alpha z$Tz=\alpha z$ for some $\alpha >0$$\alpha >0$alpha > 0\alpha>0$\alpha >0$ and $\alpha \ne 1$$\alpha \ne 1$alpha!=1\alpha \neq 1$\alpha \ne 1$.
x) If $f$$f$ff$f$ is entire and $\mathrm{Re}f\left(z\right)$$\mathrm{Re}f\left(z\right)$Re f(z)\operatorname{Re} f(z)$\mathrm{Re}f\left(z\right)$ is bounded as $|z|\to \mathrm{\infty }$$|z|\to \mathrm{\infty }$|z|rarr oo|z| \rightarrow \infty$|z|\to \mathrm{\infty }$, then $f$$f$ff$f$ is constant.
2. a) If $f=u+iv$$f=u+iv$f=u+ivf=u+i v$f=u+iv$ is entire such that ${u}_{x}+{v}_{y}=0$${u}_{x}+{v}_{y}=0$u_(x)+v_(y)=0u_x+v_y=0${u}_{x}+{v}_{y}=0$ in $\mathbb{C}$$\mathbb{C}$C\mathbb{C}$\mathbb{C}$ then show that $f$$f$ff$f$ has the form $f\left(z\right)=az+b$$f\left(z\right)=az+b$f(z)=az+bf(z)=a z+b$f\left(z\right)=az+b$ where $a,\mathbf{b}$$a,\mathbf{b}$a,ba, \mathbf{b}$a,\mathbf{b}$ are constants with $\mathrm{Re}a=0$$\mathrm{Re}a=0$Re a=0\operatorname{Re} a=0$\mathrm{Re}a=0$.
b) Consider $f\left(z\right)={z}^{2}-z$$f\left(z\right)={z}^{2}-z$f(z)=z^(2)-zf(z)=z^2-z$f\left(z\right)={z}^{2}-z$ and the closed circular region $R=\left\{z:|z|\le 1\right\}$$R=\left\{z:|z|\le 1\right\}$R={z:|z| <= 1}R=\{z:|z| \leq 1\}$R=\left\{z:|z|\le 1\right\}$. Find points in $R$$R$RR$R$ where $|f\left(z\right)|$$|f\left(z\right)|$|f(z)||f(z)|$|f\left(z\right)|$ has its maximum and minimum values.
c) Find the points where the function $f\left(z\right)=\frac{\mathrm{log}\left(z+4\right)}{{z}^{2}+i}$$f\left(z\right)=\frac{\mathrm{log}\left(z+4\right)}{{z}^{2}+i}$f(z)=(log(z+4))/(z^(2)+i)f(z)=\frac{\log (z+4)}{z^2+i}$f\left(z\right)=\frac{\mathrm{log}\left(z+4\right)}{{z}^{2}+i}$ is not analytic.
3. a) Evaluate the following integrals:
i) $I={\int }_{0}^{2\pi }f\left({e}^{i\theta }\right){\mathrm{cos}}^{2}\left(\theta /2\right)d\theta$$I={\int }_{0}^{2\pi } f\left({e}^{i\theta }\right){\mathrm{cos}}^{2}\left(\theta /2\right)d\theta$I=int_(0)^(2pi)f(e^(i theta))cos^(2)(theta//2)d thetaI=\int_0^{2 \pi} f\left(e^{i \theta}\right) \cos ^2(\theta / 2) d \theta$I={\int }_{0}^{2\pi }f\left({e}^{i\theta }\right){\mathrm{cos}}^{2}\left(\theta /2\right)d\theta$.
ii) $\phantom{\rule{1em}{0ex}}I={\int }_{0}^{2\pi }f\left({e}^{i\theta }\right){\mathrm{sin}}^{2}\theta /2d\theta$$\phantom{\rule{1em}{0ex}}I={\int }_{0}^{2\pi } f\left({e}^{i\theta }\right){\mathrm{sin}}^{2}\theta /2d\theta$quad I=int_(0)^(2pi)f(e^(i theta))sin^(2)theta//2d theta\quad I=\int_0^{2 \pi} f\left(e^{i \theta}\right) \sin ^2 \theta / 2 d \theta$\phantom{\rule{1em}{0ex}}I={\int }_{0}^{2\pi }f\left({e}^{i\theta }\right){\mathrm{sin}}^{2}\theta /2d\theta$.
b) Find the image of the circle $|z|=r\left(r\ne 1\right)$$|z|=r\left(r\ne 1\right)$|z|=r(r!=1)|z|=r(r \neq 1)$|z|=r\left(r\ne 1\right)$ under the mapping $w=f\left(z\right)=\frac{z-i}{z+i}$$w=f\left(z\right)=\frac{z-i}{z+i}$w=f(z)=(z-i)/(z+i)w=f(z)=\frac{z-i}{z+i}$w=f\left(z\right)=\frac{z-i}{z+i}$. What happens when $r=1$$r=1$r=1r=1$r=1$ ?
4. a) If $p\left(z\right)={a}_{0}+{a}_{1}z+\cdots +{a}_{n-1}{z}^{n-1}+{z}^{n}\left(n\ge 1\right)$$p\left(z\right)={a}_{0}+{a}_{1}z+\cdots +{a}_{n-1}{z}^{n-1}+{z}^{n}\left(n\ge 1\right)$p(z)=a_(0)+a_(1)z+cdots+a_(n-1)z^(n-1)+z^(n)(n >= 1)p(z)=a_0+a_1 z+\cdots+a_{n-1} z^{n-1}+z^n(n \geq 1)$p\left(z\right)={a}_{0}+{a}_{1}z+\cdots +{a}_{n-1}{z}^{n-1}+{z}^{n}\left(n\ge 1\right)$, then show that there exists a real $R>0$$R>0$R > 0R>0$R>0$ such that ${2}^{-1}|z{|}^{n}\le |p\left(z\right)|\le 2|z{|}^{n}$${2}^{-1}|z{|}^{n}\le |p\left(z\right)|\le 2|z{|}^{n}$2^(-1)|z|^(n) <= |p(z)| <= 2|z|^(n)2^{-1}|z|^n \leq|p(z)| \leq 2|z|^n${2}^{-1}|z{|}^{n}\le |p\left(z\right)|\le 2|z{|}^{n}$ for $|z|\ge R$$|z|\ge R$|z| >= R|z| \geq R$|z|\ge R$.
b) Find all solutions to the equation $\mathrm{sin}z=5$$\mathrm{sin}z=5$sin z=5\sin z=5$\mathrm{sin}z=5$.
5. a) Find the constant $c$$c$cc$c$ such that $f\left(z\right)=\frac{1}{{z}^{n}+{z}^{n-1}+\cdots +{z}^{2}+{z}^{-n}}+\frac{c}{z-1}$$f\left(z\right)=\frac{1}{{z}^{n}+{z}^{n-1}+\cdots +{z}^{2}+{z}^{-n}}+\frac{c}{z-1}$f(z)=(1)/(z^(n)+z^(n-1)+cdots+z^(2)+z^(-n))+(c)/(z-1)f(z)=\frac{1}{z^n+z^{n-1}+\cdots+z^2+z^{-n}}+\frac{c}{z-1}$f\left(z\right)=\frac{1}{{z}^{n}+{z}^{n-1}+\cdots +{z}^{2}+{z}^{-n}}+\frac{c}{z-1}$ can be extended to be analytic at $z=1$$z=1$z=1z=1$z=1$, when $n\in \mathbb{N}$$n\in \mathbb{N}$n inNn \in \mathbb{N}$n\in \mathbb{N}$ is fixed.
b) Find all the singularities of the function $f\left(z\right)=\mathrm{exp}\left(\frac{z}{\mathrm{sin}z}\right)$$f\left(z\right)=\mathrm{exp}\left(\frac{z}{\mathrm{sin}z}\right)$f(z)=exp((z)/(sin z))f(z)=\exp \left(\frac{z}{\sin z}\right)$f\left(z\right)=\mathrm{exp}\left(\frac{z}{\mathrm{sin}z}\right)$.
c) Evaluate ${\text{∮}\phantom{\rule{thinmathspace}{0ex}}}_{C}\frac{dz}{{z}^{2}+1}$${\text{∮}\phantom{\rule{thinmathspace}{0ex}}}_{C}\frac{dz}{{z}^{2}+1}$oint_(C)(dz)/(z^(2)+1)\oint_C \frac{d z}{z^2+1}${\text{∮}\phantom{\rule{thinmathspace}{0ex}}}_{C}\frac{dz}{{z}^{2}+1}$ where $c$$c$cc$c$ is the circle $|z|=4$$|z|=4$|z|=4|z|=4$|z|=4$.
6. a) Find the maximum modulus of $f\left(z\right)=2z+5i$$f\left(z\right)=2z+5i$f(z)=2z+5if(z)=2 z+5 i$f\left(z\right)=2z+5i$ on the closed circular region defined by $|z|\le 2$$|z|\le 2$|z| <= 2|z| \leq 2$|z|\le 2$.
b) Evaluate ${\int }_{C}\frac{{z}^{3}+3}{z\left(z-i{\right)}^{2}}dz$${\int }_{C} \frac{{z}^{3}+3}{z\left(z-i{\right)}^{2}}dz$int _(C)(z^(3)+3)/(z(z-i)^(2))dz\int_C \frac{z^3+3}{z(z-i)^2} d z${\int }_{C}\frac{{z}^{3}+3}{z\left(z-i{\right)}^{2}}dz$, where $c$$c$cc$c$ is the eight like figure shown in Fig. 1.

c) Find the radius of convergence of the following series.
i) $\phantom{\rule{1em}{0ex}}\sum _{k=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{k+1}}{k!}\left(z-1-i{\right)}^{k}$$\phantom{\rule{1em}{0ex}}\sum _{k=1}^{\mathrm{\infty }} \frac{\left(-1{\right)}^{k+1}}{k!}\left(z-1-i{\right)}^{k}$quadsum_(k=1)^(oo)((-1)^(k+1))/(k!)(z-1-i)^(k)\quad \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k !}(z-1-i)^k$\phantom{\rule{1em}{0ex}}\sum _{k=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{k+1}}{k!}\left(z-1-i{\right)}^{k}$
ii) $\phantom{\rule{1em}{0ex}}\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{6k+1}{2k+5}\right)}^{k}\left(z-2i{\right)}^{k}$$\phantom{\rule{1em}{0ex}}\sum _{k=1}^{\mathrm{\infty }} {\left(\frac{6k+1}{2k+5}\right)}^{k}\left(z-2i{\right)}^{k}$quadsum_(k=1)^(oo)((6k+1)/(2k+5))^(k)(z-2i)^(k)\quad \sum_{k=1}^{\infty}\left(\frac{6 k+1}{2 k+5}\right)^k(z-2 i)^k$\phantom{\rule{1em}{0ex}}\sum _{k=1}^{\mathrm{\infty }}{\left(\frac{6k+1}{2k+5}\right)}^{k}\left(z-2i{\right)}^{k}$
7. a) Expand $f\left(z\right)=\frac{1}{\left(z-1{\right)}^{2}\left(z-3\right)}$$f\left(z\right)=\frac{1}{\left(z-1{\right)}^{2}\left(z-3\right)}$f(z)=(1)/((z-1)^(2)(z-3))f(z)=\frac{1}{(z-1)^2(z-3)}$f\left(z\right)=\frac{1}{\left(z-1{\right)}^{2}\left(z-3\right)}$ in a Laurent series valid for
i) $0<|z-1|<2$$0<|z-1|<2$0 < |z-1| < 20<|z-1|<2$0<|z-1|<2$ and
ii) $\phantom{\rule{1em}{0ex}}0<|z-3|<2$$\phantom{\rule{1em}{0ex}}0<|z-3|<2$quad0 < |z-3| < 2\quad 0<|z-3|<2$\phantom{\rule{1em}{0ex}}0<|z-3|<2$.
b) Find the zeros and singularities of the function $f\left(z\right)=\frac{z}{4{\mathrm{cos}}^{2}z-1}$$f\left(z\right)=\frac{z}{4{\mathrm{cos}}^{2}z-1}$f(z)=(z)/(4cos^(2)z-1)f(z)=\frac{z}{4 \cos ^2 z-1}$f\left(z\right)=\frac{z}{4{\mathrm{cos}}^{2}z-1}$ in $|z|\le 1$$|z|\le 1$|z| <= 1|z| \leq 1$|z|\le 1$. Also find the residue at the poles.
c) Prove that the linear fractional transformation $\varphi \left(z\right)=\frac{2z-1}{2-z}$$\varphi \left(z\right)=\frac{2z-1}{2-z}$phi(z)=(2z-1)/(2-z)\phi(z)=\frac{2 z-1}{2-z}$\varphi \left(z\right)=\frac{2z-1}{2-z}$ maps the circle $c:|z|=1$$c:|z|=1$c:|z|=1c:|z|=1$c:|z|=1$ into itself. Also prove that $f\left(z\right)$$f\left(z\right)$f(z)f(z)$f\left(z\right)$ is conformal in $\overline{D}=\left\{z:|z|\le 1\right\}$$\overline{D}=\left\{z:|z|\le 1\right\}$bar(D)={z:|z| <= 1}\bar{D}=\{z:|z| \leq 1\}$\overline{D}=\left\{z:|z|\le 1\right\}$.
8. a) Find the image of the semi-infinite strip $x>0,0$x>0,0x > 0,0 < y < 1x>0,0<y<1$x>0,0 when $w=i/z$$w=i/z$w=i//zw=i / z$w=i/z$. Sketch the strip and its image.
b) Show that there is only one linear fractional transformation that maps three given distinct points ${z}_{1},{z}_{2}$${z}_{1},{z}_{2}$z_(1),z_(2)z_1, z_2${z}_{1},{z}_{2}$ and ${z}_{3}$${z}_{3}$z_(3)z_3${z}_{3}$ in the extended $z$$z$zz$z$ plane onto three specified distinct points ${w}_{1},{w}_{2}$${w}_{1},{w}_{2}$w_(1),w_(2)w_1, w_2${w}_{1},{w}_{2}$ and ${w}_{3}$${w}_{3}$w_(3)w_3${w}_{3}$ in the extended $w$$w$ww$w$ plane.
9. Evaluate the following integrals
a) ${\int }_{0}^{\mathrm{\infty }}\frac{{x}^{2}+2}{\left({x}^{2}+1\right)\left({x}^{2}+4\right)}dx$${\int }_{0}^{\mathrm{\infty }} \frac{{x}^{2}+2}{\left({x}^{2}+1\right)\left({x}^{2}+4\right)}dx$int_(0)^(oo)(x^(2)+2)/((x^(2)+1)(x^(2)+4))dx\int_0^{\infty} \frac{x^2+2}{\left(x^2+1\right)\left(x^2+4\right)} d x${\int }_{0}^{\mathrm{\infty }}\frac{{x}^{2}+2}{\left({x}^{2}+1\right)\left({x}^{2}+4\right)}dx$.
b) ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{2}2x}{1+{x}^{2}}dx$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }} \frac{{\mathrm{sin}}^{2}2x}{1+{x}^{2}}dx$int_(-oo)^(oo)(sin^(2)2x)/(1+x^(2))dx\int_{-\infty}^{\infty} \frac{\sin ^2 2 x}{1+x^2} d x${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{\mathrm{sin}}^{2}2x}{1+{x}^{2}}dx$.
$$a^2=b^2+c^2-2bc\:Cos\left(A\right)$$

## MMT-005 Sample Solution 2024

Simply click “Install” to download and install the app, and then follow the instructions to purchase the required assignment solution. Currently, the app is only available for Android devices. We are working on making the app available for iOS in the future, but it is not currently available for iOS devices.

Yes, It is Complete Solution, a comprehensive solution to the assignments for IGNOU. Valid from January 1, 2023 to December 31, 2023.

Yes, the Complete Solution is aligned with the IGNOU requirements and has been solved accordingly.

Yes, the Complete Solution is guaranteed to be error-free.The solutions are thoroughly researched and verified by subject matter experts to ensure their accuracy.

As of now, you have access to the Complete Solution for a period of 6 months after the date of purchase, which is sufficient to complete the assignment. However, we can extend the access period upon request. You can access the solution anytime through our app.

The app provides complete solutions for all assignment questions. If you still need help, you can contact the support team for assistance at Whatsapp +91-9958288900

No, access to the educational materials is limited to one device only, where you have first logged in. Logging in on multiple devices is not allowed and may result in the revocation of access to the educational materials.

Payments can be made through various secure online payment methods available in the app.Your payment information is protected with industry-standard security measures to ensure its confidentiality and safety. You will receive a receipt for your payment through email or within the app, depending on your preference.

The instructions for formatting your assignments are detailed in the Assignment Booklet, which includes details on paper size, margins, precision, and submission requirements. It is important to strictly follow these instructions to facilitate evaluation and avoid delays.

$$tan\:\theta =\frac{sin\:\theta }{cos\:\theta }$$

## Terms and Conditions

• The educational materials provided in the app are the sole property of the app owner and are protected by copyright laws.
• Reproduction, distribution, or sale of the educational materials without prior written consent from the app owner is strictly prohibited and may result in legal consequences.
• Any attempt to modify, alter, or use the educational materials for commercial purposes is strictly prohibited.
• The app owner reserves the right to revoke access to the educational materials at any time without notice for any violation of these terms and conditions.
• The app owner is not responsible for any damages or losses resulting from the use of the educational materials.
• The app owner reserves the right to modify these terms and conditions at any time without notice.
• By accessing and using the app, you agree to abide by these terms and conditions.
• Access to the educational materials is limited to one device only. Logging in to the app on multiple devices is not allowed and may result in the revocation of access to the educational materials.

Our educational materials are solely available on our website and application only. Users and students can report the dealing or selling of the copied version of our educational materials by any third party at our email address (abstract4math@gmail.com) or mobile no. (+91-9958288900).

In return, such users/students can expect free our educational materials/assignments and other benefits as a bonafide gesture which will be completely dependent upon our discretion.

Scroll to Top
Scroll to Top