IGNOU MMT-004 Solved Assignment 2024 | M.Sc. MACS
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IGNOU MMT-004 Assignment Question Paper 2024
mmt-003-solved-assignment-2024-qp-2afdf4f2-c576-4553-8e87-8dfd8225b98e
- a) Let
X=C[0,1] \mathrm{X}=\mathrm{C}[0,1] . Defined:X xx X rarrR d: X \times X \rightarrow \mathbf{R} byd(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} where the integral is the Riemann integral. Show thatd d is a metric onX X . Findd(f,g) d(f, g) wheref(x)=4x f(x)=4 x andg(x)=x^(3),x in[0,1] g(x)=x^3, x \in[0,1] .
2. a) Let
i) int
ii)
iii)
iv)
b) Find the interior, boundary and closure of the following sets
i)
ii)
3. a) Let
b) Let
c) Show that an infinite discrete metric space
4. a) Find the first derivative
b) Let
c) Near what points may the surface
5. a) Use the method of Lagrange’s multiplier method to find the shortest possible distance from the ellipse
b) Find the directional derivative of the function
6. a) Let A be a compact non-empty subset of a metric space (X, d) and let F be a closed subset of
b) Give an example of the following with justification
i) A vector-valued function
ii) A function which is Legesgue measurable on
c) Show that the components of a metric space is either identical or pairwise disjoint.
7. a) Let
Show that
i)
ii)
c) Which of the following sets are connected sets in
i)
ii)
8. a) Consider
b) Let A be any set in
c) Find the measure of the following sets.
ii)
iii)
9. a) Show that if
b) Verify Bounded Convergence Theorem for the sequence of functions
- State whether the following statements are True or False. Justify your answers.
a) The sequence{((1)/(n),(1)/(n)):ninN} \left\{\left(\frac{1}{\mathrm{n}}, \frac{1}{\mathrm{n}}\right): \mathrm{n} \in \mathbf{N}\right\} is convergent inR^(2) \mathbf{R}^2 under the discrete metric onR^(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 fromR^(n) \mathbf{R}^n toR^(m) \mathbf{R}^m never vanishes.
e) Ifint_(A)fdm=int_(A)gdm \int_{\mathrm{A}} \mathrm{fdm}=\int_{\mathrm{A}} \mathrm{gdm} for allAin M \mathrm{A} \in \boldsymbol{M} , thenf=g \mathrm{f}=\mathrm{g} .
MMT-004 Sample Solution 2024
mmt-004-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee
- a) Let
X=C[0,1] \mathrm{X}=\mathrm{C}[0,1] . Defined:X xx X rarrR d: X \times X \rightarrow \mathbf{R} byd(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} where the integral is the Riemann integral. Show thatd d is a metric onX X . Findd(f,g) d(f, g) wheref(x)=4x f(x)=4 x andg(x)=x^(3),x in[0,1] g(x)=x^3, x \in[0,1] .
- Non-negativity:
d(f,g) >= 0 d(f, g) \geq 0 - Identity of indiscernibles:
d(f,g)=0 d(f, g) = 0 if and only iff=g f = g - Symmetry:
d(f,g)=d(g,f) d(f, g) = d(g, f) - Triangle inequality:
d(f,h) <= d(f,g)+d(g,h) d(f, h) \leq d(f, g) + d(g, h)
-
Non-negativity:
For anyf,g in X f, g \in X , the absolute value function|*| |\cdot| ensures that|f(t)-g(t)| >= 0 |f(t) – g(t)| \geq 0 for allt in[0,1] t \in [0, 1] . Therefore, the integralint_(0)^(1)|f(t)-g(t)|dt \int_0^1 |f(t) – g(t)| \, dt is also non-negative. Hence,d(f,g) >= 0 d(f, g) \geq 0 . -
Identity of indiscernibles:
Iff=g f = g , thenf(t)-g(t)=0 f(t) – g(t) = 0 for allt in[0,1] t \in [0, 1] , sod(f,g)=int_(0)^(1)|f(t)-g(t)|dt=int_(0)^(1)0dt=0 d(f, g) = \int_0^1 |f(t) – g(t)| \, dt = \int_0^1 0 \, dt = 0 .
Conversely, ifd(f,g)=0 d(f, g) = 0 , thenint_(0)^(1)|f(t)-g(t)|dt=0 \int_0^1 |f(t) – g(t)| \, dt = 0 . Since the integrand is non-negative, it must be zero almost everywhere, implying thatf(t)=g(t) f(t) = g(t) for almost allt in[0,1] t \in [0, 1] . Sincef f andg g are continuous, they must be equal everywhere on[0,1] [0, 1] , sof=g f = g . -
Symmetry:
By the properties of the absolute value function,|f(t)-g(t)|=|g(t)-f(t)| |f(t) – g(t)| = |g(t) – f(t)| for allt in[0,1] t \in [0, 1] . Therefore,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) . -
Triangle inequality:
For anyf,g,h in X f, g, h \in X and for allt in[0,1] t \in [0, 1] , we have|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)| by the triangle inequality for real numbers. Integrating both sides over[0,1] [0, 1] gives:int_(0)^(1)|f(t)-h(t)|dt <= int_(0)^(1)(|f(t)-g(t)|+|g(t)-h(t)|)dt=int_(0)^(1)|f(t)-g(t)|dt+int_(0)^(1)|g(t)-h(t)|dt=d(f,g)+d(g,h) \int_0^1 |f(t) – h(t)| \, dt \leq \int_0^1 (|f(t) – g(t)| + |g(t) – h(t)|) \, dt = \int_0^1 |f(t) – g(t)| \, dt + \int_0^1 |g(t) – h(t)| \, dt = d(f, g) + d(g, h) Thus,d(f,h) <= d(f,g)+d(g,h) d(f, h) \leq d(f, g) + d(g, h) .
- For
0 <= x <= 2 0 \leq x \leq 2 ,4x >= x^(3) 4x \geq x^3 , so|4x-x^(3)|=4x-x^(3) |4x – x^3| = 4x – x^3 . - For
2 < x <= 1 2 < x \leq 1 ,4x < x^(3) 4x < x^3 , so|4x-x^(3)|=x^(3)-4x |4x – x^3| = x^3 – 4x .
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