SMU Assignment (Sem-1) MCA1030- FOUNDATION OF MATHEMATICS

MCA1030- FOUNDATION OF MATHEMATICS

Q1.)    State Leibnitz’s theorem. Find the nth derivative of (𝑥) = 𝑥²𝑒^𝑎𝑥, using Leibnitz theorem.

Ans.)    When Gottfried Wilhelm von Leibniz worked out his version of the calculus he derived the rules governing the process of differentiation (and introduced the notation that we use today). Among those rules we find a theorem concerning multiple differentiations of a product of two functions. We have before us the task of deducing the n-th derivative, with respect to the variable x, of the product of the functions f(x) and g(x).

Let u(x) = e^ax & v(x) = x²

U¹(x) = u₁ (x) = ae^ax, u¹¹ (x) = u₂ (x) = a²e^ax, u¹¹¹ (x) = u₃ (x) = a³e^ax.... uⁿ (x)

= u_n (x) = aⁿ e^ax

V¹ (x) = v₁ (x) = 2x, v¹¹ (x) = v₂ (x) = 2, v¹¹¹ (x) = v₃ (x) = 0,…, vⁿ (x) = v_n (x) = 0

Dⁿy/dxⁿ = ⁿC₀ U_n (x) v(x) + ⁿC₁ u_n-1 (x) v₁ (x) + ⁿC₂ U_n-2 (X) V₂ (X) +…+ ⁿC_n U (X) V_n (X)

= x²aⁿe^ax+ 2na^n-1xe^ax+ n (n-1) a^n-2e^ax

= a^n-2e^ax (x² aⁿ+ 2nax + n (n-1))

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Q2.)    Define Tautology and contradiction. Show that

Ans.)   a) (p^ q) v (~ p) is a tautology: - A statement is said to be a tautology if it is true for all logical possibilities. In other words, a statement is called tautology if its truth value is T and only T in the last column of its truth table.

Example: P V ¬ P

P     ¬ P     P V ¬ P

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T      F        T

F      T        T

b) (p^ q) ^ (~ p) is a contradiction: - A propositional expression is a contradiction if and only if for all possible assignments of truth values to its variables its truth value is F.

Example: P Λ ¬ P

P     ¬ P     P Λ ¬ P

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T      F        F

F      T        F

Usage of tautologies and contradictions - in proving the validity of arguments; for rewriting expressions using only the basic connectives.

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Q3.)    State Lagrange’s Theorem. Verify Lagrange’s mean value theorem for the function f(x) = 3 x² – 5x + 1 defined in interval [2, 5].

Ans.)   Let f: [a, b] = R be a continuous function such that f: (a, b) = R is differentiable. Then, there exists c = (a, b) such that f(b)-f(a)/ b-a = f¹ (c).

Geometrical Interpretation

It states that given a line l joining two points on the graph of a differentiable function f, namely (a, f(a)) & (b, f(b)), there is a mean value (i.e. c = (a, b)) such that the tangent line at the corresponding point (that is, (c, f(c))) is parallel to the given line l.

Lagrange’s Mean Value Theorem State that, for any section of a continuous smooth curve, there will always be a point c at which the derivation or slope of the curve will be the same as the average curve of the section.

Theorem: f(x) = 3 x² – 5x + 1 defined in interval [2, 5].

Solution: We have, f(x) = 3x² – 5x + 1 where X = [2, 5]

(1) f(x) is a polynomial function, hence continuous in the interval [2, 5].

(2) f(x) is a polynomial function, hence differentiable in the interval (2, 5).

(3) f(5) = 3(5)² – 5 * 5 + 1 = 51, f(2) = 3(2)² – 5 * 2 + 1 = 3

Also, f¹(x) = 6 x – 5 => f¹(c) = 6 c – 5

Now, f¹ (c) = [f(b) – f(a)] / (b-a)

Or, 6c – 5 = [f(5) – f(2)] / (5 – 2) = (51 - 3)(5 – 2) = 48/3 = 16

Or, 6c = 16 + 5 = 21 => c = 21/6 e (2, 5)

Hence, Lagrange’s mean value theorem is verified.

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Q4.)    Define Negation. Write the negation of each of the following conjunctions:

Ans.)   An assertion that a statement fails or denial of a statement is called the negation of the statement. The negation of a statement is generally formed by introducing the word “not” at some proper place in the statement or by prefixing the statement with “It is not the case that” or “It is false that”. The negation of a statement p in symbolic form is written as “~p”.

a) Paris is in France and London is in England: -

Write p: Paris is in France & q: London is in England.

Then, the conjunction in (a) is given by p ^ q.

Now ~p: Paris is not in France, &

~q: London is not in England.

Therefore, using (D₁), negation of p ^ q is given by

~ (p ^ q) = Paris is not in France or London is not in England.

b) 2 + 3 = 5 and 8 < 10: -

Write p: 2+3 = 5 & q: 8 < 10.

Then the conjunction in (b) is given by p ^ q.

Now ~p: 2+3 # 5 & ~q: 8 </ 10

Then, using (D₇), negation of p ^ q is given by

~ (p ^ q) = 2+3 # 5 or (8 </ 10).

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Q5.)    Find the asymptote parallel to the coordinate axis of the following curves

(i) (𝑥²+𝑦²) 𝑥 − 𝑎𝑦² = 0: We have (x² + y²) x – ay² = 0

Or        x³ + (x - a) y² = 0

Asymptote parallel to x – axis are obtained by equating to zero the coefficient of the highest power of x. Since the coefficient of highest power of x³ is 1, which is constant so there is no asymptote parallel to x – axis.

Asymptote parallel to y – axis are obtained by equating to zero the coefficient of the highest power of y. Since the coefficient of highest power of y³ is (x - a), which is constant so there is no asymptote parallel to x – axis.

(ii) x²y²– a² (x² + y²) = 0: We have x²y² – a² (x² + y²) = 0

The coefficient of highest power of x² of x is (y² – a²)

Now, (y² – a²) = (y + a) (y – a)

Hence, y + a = 0 & y – a = 0 are two asymptote parallel to y – axis

Also, the coefficient of highest power of y² of y is (x² – a²)

Now, (x² – a²) = (x + a) (x – a)

Hence, x + a = 0 & x –a = 0 are two asymptote parallel to x – axis.

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Q6.)    Define:

(i) Set: It may be observed that we describe the set by using a symbol x for elements of the set. After the sign of ‘colon’ write the characteristics property possessed by the elements of the set & then enclose the description within braces. For example, the following description of a set

A = {x: x is a natural number & 3 < x < 10}

Is read as “the set of all x such that x is a natural number & 3 < x < 10”. Hence, the numbers 4, 5, 6, 7, 8 & 9 are the elements of set A.

If we denote the sets described above in (a), (b) & (c) in roster form by A, B & C, respectively, then A, B & C can also be represented in set builder form as follows

            A = {x: x is a natural number which divides 42}

            B = {y: y is a vowel in the English alphabet}

            C = {z: z is an odd natural number}

(ii) Null Set: A set which does not contain any element is called an empty set or null set or the void set. The empty set is denoted by the symbol ‘ᶲ’.For e.g.

(a)Let P = {x: 1 < x < 2, x is a natural number}.

Then P is an empty set, because there is no natural number between 1 & 2.

            (b) Let Q = {x: x² – 2 = 0 & x is rational}.

Then, Q is the empty set, because the equation x² – 2 = 0 is not satisfied by any rational number x.

(iii) Subset: If every element of a set A is also an element of set B, then A is called a subset of B or A is contained in B. We write it as A ƈ B. For e.g.:

(a) The set Q of rational numbers is a subset of the set R of real numbers & we write Q ƈ R.

(b) If A is the set of all divisors of 56 & B the set of all prime divisors of 56, then B is a subset of A, & we write B ƈ A.

(iv) Power Set: The collection of all subsets of a set A is called the power set of A. It is denoted by P (A). In P (A), every element is a set. Example (v) of section 1.6, if A = {1, 2}, then P (A)={ ᶲ, {1}, {2}, {1, 2}}. Also note that, n [P (A)] = 4 = 2².

In general, if A is a set with n (A) = m, then it can be shown that n[P(A)] = 2^m > m = n(A).

(v) Union Set: The union of two sets A & B is the set C which consists of all those elements which are either in A or in B (including those which are in both).

Symbolically, we write A ᴗ B = {x: x€ A or x € B} & usually read as ‘A union B’.

The union of two sets can be represented by a Venn diagram

The Shaded portion in Figure represents A ᴗ B.

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