IGNOU BMTC 133 Solved Assignment 2023-24- IGNOU BMTC 133 stands for Indira Gandhi National Open University’s Bachelor of Science (B.Sc.) Discipline Specific Core Course (DSC) in Real Analysis. It’s a foundational course that covers essential concepts of mathematical analysis, including limits, continuity, differentiation, and integration.
- Course Code: BMTC-133
- Course Name: Real Analysis
- Programme: Bachelor of Science (B.Sc.)
- School: School of Sciences (SOS)
1. State whether the following statements are true or false. Give a short proof or a counterexample in support of your answer:
a) A real-valued function of three variables which is continuous everywhere is
differentiable.
Ans. The five statements in the image are all true. Here are the reasons why:
1. Every infinite set is an open set. This is not true. An infinite set can be either open or
closed. For example, the set of all real numbers is infinite and closed, while the set of
all positive integers is infinite and open.
2. The negation of (p\wedge\sim q \rightarrow q) is true. This is true using De Morgan’s
Laws. The negation of the statement is (\(\lnot p\) \vee q) \wedge (\(\lnot \sim q\) \vee
\(\lnot p\)), which simplifies to (q \vee \lnot p) \wedge (q \vee p). This statement is
always true regardless of the truth values of p and q, because it is the disjunction of
two tautologies (statements that are always true).
3. -1 is a limit point of the interval ]-2,1[. This is true. A limit point of an interval is a
point that is arbitrarily close to infinitely many points in the interval. In this case, for
any positive epsilon value, we can find a number within the interval ]-2,1[ that is less
than epsilon away from -1. For example, if epsilon is 0.1, then the number -0.9 is
within the interval and less than 0.1 away from -1.
4. The necessary condition for a function f to be integrable is that it is continuous. This
is not true. A function can be integrable even if it is not continuous at some points.
For example, the function f(x) = |x| is not continuous at x = 0, but it is integrable on
the interval [-1, 1]. However, if a function is not integrable, then it must be
discontinuous at some points.
5. The function (f:\mathbb{R}\rightarrow\mathbb{R}) defined by (f(x)=|x-2|+|3-x|;)
differentiable at (x=5) . This is true. A function is differentiable at a point if the limit
of its difference quotient exists at that point. In this case, the difference quotient for
f(x) is: \lim_{h \to 0} \frac{f(x + h) – f(x)}{h} = \lim_{h \to 0} \frac{|x+h-2|+|3-x-h| –
(|x-2|+|3-x|)}{h} IGNOU BMTC 133 Solved Assignment 2023-24
This can be further simplified and evaluated at x = 5 to show that the limit exists and is equal
to 2, which means that f(x) is differentiable at x = 5.
