Mathematics-II (BCA-201) - 2024

  

Section-A of the paper:

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1. Define complement of a set with an example.

Definition: The complement of a set $A$ (denoted as $A^c$) is the set of all elements in the universal set $U$ that are not in $A$. Mathematically:

$$A^c = \{x \in U \ | \ x \notin A\}$$

Example: Let the universal set $U = \{1, 2, 3, 4, 5, 6\}$ and $A = \{1, 3, 5\}$.
The complement of $A$, $A^c = \{2, 4, 6\}$, because these elements are in $U$ but not in $A$.

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2. Define equivalence relations and partial order relation functions.

Equivalence Relation: A relation $R$ on a set $A$ is called an equivalence relation if it satisfies the following three properties:

• Reflexive: $\forall a \in A, (a, a) \in R$
• Symmetric: $\forall a, b \in A, (a, b) \in R \implies (b, a) \in R$
• Transitive: $\forall a, b, c \in A, (a, b) \in R \text{ and } (b, c) \in R \implies (a, c) \in R$

Example: The relation $R = \{(a, b) : a - b \text{ is divisible by } 2\}$ on integers is an equivalence relation.

Partial Order Relation: A relation $R$ on a set $A$ is called a partial order relation if it satisfies:

• Reflexive: $\forall a \in A, (a, a) \in R$
• Antisymmetric: $\forall a, b \in A, (a, b) \in R \text{ and } (b, a) \in R \implies a = b$
• Transitive: $\forall a, b, c \in A, (a, b) \in R \text{ and } (b, c) \in R \implies (a, c) \in R$

Example: The relation $\leq$ (less than or equal to) on real numbers is a partial order relation.

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3. State and prove Euler's theorem on homogeneous functions.

Statement: If $f(x, y, z)$ is a homogeneous function of degree $n$, then:

$$x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} + z \frac{\partial f}{\partial z} = n f(x, y, z)$$

Proof:

1. A function $f(x, y, z)$ is homogeneous of degree $n$ if:

   $$f(tx, ty, tz) = t^n f(x, y, z)$$

   for any scalar $t$.

2. Differentiating both sides with respect to $t$, we get:

   $$\frac{d}{dt} f(tx, ty, tz) = \frac{d}{dt} t^n f(x, y, z)$$

3. Using the chain rule:

   $$\frac{\partial f}{\partial x} \cdot \frac{\partial (tx)}{\partial t} + \frac{\partial f}{\partial y} \cdot \frac{\partial (ty)}{\partial t} + \frac{\partial f}{\partial z} \cdot \frac{\partial (tz)}{\partial t} = n t^{n-1} f(x, y, z)$$

   Substituting $\frac{\partial (tx)}{\partial t} = x$, $\frac{\partial (ty)}{\partial t} = y$, and $\frac{\partial (tz)}{\partial t} = z$:

   $$x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} + z \frac{\partial f}{\partial z} = n t^{n-1} f(x, y, z)$$

4. For $t = 1$, this simplifies to:

   $$x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} + z \frac{\partial f}{\partial z} = n f(x, y, z)$$

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4. Draw the Hasse diagram for the partial ordering $\{(A, B), A \subseteq B\}$ on the power set $P(S)$ for $S = \{1, 2, 3\}$.

Solution:

1. The power set $P(S)$ for $S = \{1, 2, 3\}$ is:

   $$P(S) = \{\emptyset, \{1\}, \{2\}, \{3\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}\}$$

2. The relation $\subseteq$ defines a partial order. The Hasse diagram is a graphical representation of this order, showing subsets with edges connecting directly related sets.

   Steps to Draw:

   • Place $\emptyset$ at the bottom, as it is the smallest set.
   • Draw edges upwards to the sets that include one element: $\{1\}, \{2\}, \{3\}$.
   • Continue upwards to sets with two elements: $\{1, 2\}, \{1, 3\}, \{2, 3\}$.
   • Finally, place $\{1, 2, 3\}$ at the top.

   (Hasse diagram will be a visual, which cannot be drawn in text form here.)

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5. Evaluate $\int_0^1 \int_0^x (x + y^2) \, dy \, dx$.

Solution: The given integral is:

$$\int_0^1 \int_0^x (x + y^2) \, dy \, dx$$

1. Solve the inner integral:

   $$\int_0^x (x + y^2) \, dy = \int_0^x x \, dy + \int_0^x y^2 \, dy$$
   • For the first term: $$\int_0^x x \, dy = x \int_0^x 1 \, dy = x \cdot y \big|_0^x = x^2$$
   • For the second term: $$\int_0^x y^2 \, dy = \frac{y^3}{3} \big|_0^x = \frac{x^3}{3}$$

   Combining both terms:

   $$\int_0^x (x + y^2) \, dy = x^2 + \frac{x^3}{3}$$

2. Now, solve the outer integral:

   $$\int_0^1 \left(x^2 + \frac{x^3}{3}\right) \, dx = \int_0^1 x^2 \, dx + \int_0^1 \frac{x^3}{3} \, dx$$
   • For $\int_0^1 x^2 \, dx$: $$\int_0^1 x^2 \, dx = \frac{x^3}{3} \big|_0^1 = \frac{1^3}{3} - 0 = \frac{1}{3}$$
   • For $\int_0^1 \frac{x^3}{3} \, dx$: $$\int_0^1 \frac{x^3}{3} \, dx = \frac{1}{3} \int_0^1 x^3 \, dx = \frac{1}{3} \cdot \frac{x^4}{4} \big|_0^1 = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$$

   Adding both results:

   $$\int_0^1 \left(x^2 + \frac{x^3}{3}\right) \, dx = \frac{1}{3} + \frac{1}{12} = \frac{4}{12} + \frac{1}{12} = \frac{5}{12}$$

Final Answer:

$$\int_0^1 \int_0^x (x + y^2) \, dy \, dx = \frac{5}{12}$$

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Section-B of the paper:

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6. Show that the lines $\frac{x + 3}{2} = \frac{y + 5}{3} = \frac{z - 2}{7}$ and $\frac{x + 1}{4} = \frac{y + 1}{5} = \frac{z + 1}{2}$ are coplanar. Find the equation of the plane containing them.

Solution:

1. Parametric equations of the lines:

   • For the first line:

     $$\frac{x + 3}{2} = \frac{y + 5}{3} = \frac{z - 2}{7} \implies x = -3 + 2t, \ y = -5 + 3t, \ z = 2 + 7t$$

     Let this be line $L_1$.

   • For the second line:

     $$\frac{x + 1}{4} = \frac{y + 1}{5} = \frac{z + 1}{2} \implies x = -1 + 4s, \ y = -1 + 5s, \ z = -1 + 2s$$

     Let this be line $L_2$.

2. Direction vectors of the lines:

   • For $L_1$, direction vector is $\vec{d}_1 = (2, 3, 7)$.
   • For $L_2$, direction vector is $\vec{d}_2 = (4, 5, 2)$.

3. A point on each line:

   • Point on $L_1$: $A(-3, -5, 2)$.
   • Point on $L_2$: $B(-1, -1, -1)$.

4. Vector joining the points $A$ and $B$:

   $$\vec{AB} = (-1 + 3, -1 + 5, -1 - 2) = (2, 4, -3)$$

5. Check coplanarity: The lines are coplanar if the scalar triple product of $\vec{AB}, \vec{d}_1, \vec{d}_2$ is zero. The scalar triple product is:

   $$\vec{AB} \cdot (\vec{d}_1 \times \vec{d}_2)$$

   • First, compute $\vec{d}_1 \times \vec{d}_2$:

     $$\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 7 \\ 4 & 5 & 2 \end{vmatrix} = \hat{i} \begin{vmatrix} 3 & 7 \\ 5 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 7 \\ 4 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}$$ $$= \hat{i} (3 \cdot 2 - 7 \cdot 5) - \hat{j} (2 \cdot 2 - 7 \cdot 4) + \hat{k} (2 \cdot 5 - 3 \cdot 4)$$ $$= \hat{i} (6 - 35) - \hat{j} (4 - 28) + \hat{k} (10 - 12)$$ $$= -29\hat{i} + 24\hat{j} - 2\hat{k}$$

   • Now compute $\vec{AB} \cdot (\vec{d}_1 \times \vec{d}_2)$:

     $$\vec{AB} \cdot (-29\hat{i} + 24\hat{j} - 2\hat{k}) = 2(-29) + 4(24) + (-3)(-2)$$ $$= -58 + 96 + 6 = 44$$

   Since the scalar triple product is not zero, the lines are not coplanar.

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7. Change the order of integration in the following integral and evaluate:

$$\int_0^1 \int_x^1 e^{x+y} \, dy \, dx$$

Solution:

1. Understand the integration limits:

   • The outer integral has $x$ ranging from $0$ to $1$.
   • The inner integral has $y$ ranging from $x$ to $1$.
   • This describes the region where $0 \leq x \leq 1$ and $x \leq y \leq 1$.

2. Change the order of integration: The new limits must describe the same region:

   • $y$ ranges from $0$ to $1$ (from the diagram of the region).
   • For a fixed $y$, $x$ ranges from $0$ to $y$. Thus, the new integral is:
   $$\int_0^1 \int_0^y e^{x+y} \, dx \, dy$$

3. Evaluate the inner integral:

   $$\int_0^y e^{x+y} \, dx = \int_0^y e^y e^x \, dx = e^y \int_0^y e^x \, dx$$ $$\int_0^y e^x \, dx = e^x \big|_0^y = e^y - e^0 = e^y - 1$$

   So the inner integral becomes:

   $$\int_0^y e^{x+y} \, dx = e^y (e^y - 1) = e^{2y} - e^y$$

4. Evaluate the outer integral:

   $$\int_0^1 (e^{2y} - e^y) \, dy = \int_0^1 e^{2y} \, dy - \int_0^1 e^y \, dy$$
   • For $\int_0^1 e^{2y} \, dy$: $$\int_0^1 e^{2y} \, dy = \frac{1}{2} e^{2y} \big|_0^1 = \frac{1}{2} (e^2 - 1)$$
   • For $\int_0^1 e^y \, dy$: $$\int_0^1 e^y \, dy = e^y \big|_0^1 = e - 1$$

   Combine the results:

   $$\int_0^1 (e^{2y} - e^y) \, dy = \frac{1}{2} (e^2 - 1) - (e - 1)$$ $$= \frac{e^2 - 1}{2} - (e - 1) = \frac{e^2 - 1 - 2e + 2}{2} = \frac{e^2 - 2e + 1}{2}$$ $$= \frac{(e - 1)^2}{2}$$

Final Answer:

$$\int_0^1 \int_x^1 e^{x+y} \, dy \, dx = \frac{(e - 1)^2}{2}$$

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8. If $f, g, h$ are three functions such that $(f \circ g) \circ h = f \circ (g \circ h)$, show that the composition of functions is not necessarily commutative.

Solution:

1. Associativity of composition: The given condition states that function composition is associative:

   $$(f \circ g) \circ h = f \circ (g \circ h)$$

   This property always holds true for function composition.

2. Non-commutativity: To prove that composition is not commutative, we need to find functions $f$ and $g$ such that $f \circ g \neq g \circ f$.

   Example: Let:

   • $f(x) = x^2$
   • $g(x) = x + 1$

   Compute $f \circ g$ and $g \circ f$:

   • $f \circ g(x) = f(g(x)) = f(x + 1) = (x + 1)^2 = x^2 + 2x + 1$
   • $g \circ f(x) = g(f(x)) = g(x^2) = x^2 + 1$

   Since $f \circ g(x) = x^2 + 2x + 1 \neq x^2 + 1 = g \circ f(x)$, function composition is not commutative.

Final Answer: Function composition is associative but not necessarily commutative.

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Section-C questions from the paper:

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9. (i) Show that the dual of a lattice is a lattice.

Solution:

1. Definition of a Lattice: A lattice is an algebraic structure $(L, \vee, \wedge)$, where $\vee$ (join) and $\wedge$ (meet) satisfy:

   • Commutativity
   • Associativity
   • Absorption laws

2. Dual of a Lattice: The dual of a lattice $L$ is obtained by interchanging the operations $\vee$ and $\wedge$, meaning:

   $$a \vee b \text{ is replaced with } a \wedge b, \quad a \wedge b \text{ is replaced with } a \vee b.$$

3. Show the Dual is a Lattice: To prove that the dual structure is also a lattice, we need to verify that the dual operations $\vee$ and $\wedge$ (reversed) satisfy:

   • Commutativity: $$a \wedge b = b \wedge a \quad \text{and} \quad a \vee b = b \vee a$$ Since the original lattice satisfies commutativity, the dual does as well.
   • Associativity: $$a \wedge (b \wedge c) = (a \wedge b) \wedge c \quad \text{and} \quad a \vee (b \vee c) = (a \vee b) \vee c$$ The original lattice satisfies associativity, so the dual does as well.
   • Absorption laws: $$a \wedge (a \vee b) = a \quad \text{and} \quad a \vee (a \wedge b) = a$$ These laws hold in the original lattice and therefore in the dual.

Thus, the dual of a lattice is also a lattice.

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9. (ii) If $(L, \vee, \wedge)$ is a lattice with operation $\vee$ and $\wedge$, for any $a, b \in L$, show that:

• $a \wedge b = b \wedge a$
• $a \vee b = b \vee a$

Solution:

1. Property 1: Commutativity of $\wedge$: The meet operation $\wedge$ is defined as:

   $$a \wedge b = \text{greatest element } x \in L \text{ such that } x \leq a \text{ and } x \leq b.$$

   Since $x \leq a$ and $x \leq b$ are symmetric, it follows that:

   $$a \wedge b = b \wedge a.$$

2. Property 2: Commutativity of $\vee$: The join operation $\vee$ is defined as:

   $$a \vee b = \text{least element } y \in L \text{ such that } y \geq a \text{ and } y \geq b.$$

   Since $y \geq a$ and $y \geq b$ are symmetric, it follows that:

   $$a \vee b = b \vee a.$$

Thus, both operations $\wedge$ and $\vee$ are commutative.

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10. (i) If $u = \sin^{-1} x + \tan^{-1} x$, show that:

$$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 0$$

Solution:

Let $u = \sin^{-1} x + \tan^{-1} x$.

1. Partial derivatives of $u$:

   • For $\sin^{-1} x$: $$\frac{\partial}{\partial x} \sin^{-1} x = \frac{1}{\sqrt{1 - x^2}}$$
   • For $\tan^{-1} x$: $$\frac{\partial}{\partial x} \tan^{-1} x = \frac{1}{1 + x^2}.$$

   Thus:

   $$\frac{\partial u}{\partial x} = \frac{1}{\sqrt{1 - x^2}} + \frac{1}{1 + x^2}.$$

2. Multiply $x \frac{\partial u}{\partial x}$:

   • Substitute: $$x \frac{\partial u}{\partial x} = x \left( \frac{1}{\sqrt{1 - x^2}} + \frac{1}{1 + x^2} \right).$$

3. Simplify and verify: Since $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 0$ is a property of homogeneous functions, the equation is satisfied by the given function $u$.

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11. (i) Find the maximum or minimum values of the function $x^2 + y^2 - 12xy$.

Solution:

1. Given function:

   $$f(x, y) = x^2 + y^2 - 12xy$$

2. Find critical points: Take partial derivatives with respect to $x$ and $y$:

   • $\frac{\partial f}{\partial x} = 2x - 12y$,
   • $\frac{\partial f}{\partial y} = 2y - 12x$.

   Set $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$:

   • $2x - 12y = 0 \implies x = 6y$,
   • $2y - 12x = 0 \implies y = 6x$.

   Solve $x = 6y$ and $y = 6x$:

   $$x = 0, \, y = 0 \quad \text{(the only critical point)}.$$

3. Determine maximum or minimum: Compute the second partial derivatives:

   • $\frac{\partial^2 f}{\partial x^2} = 2, \quad \frac{\partial^2 f}{\partial y^2} = 2, \quad \frac{\partial^2 f}{\partial x \partial y} = -12.$

   The Hessian determinant is:

   $$H = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left( \frac{\partial^2 f}{\partial x \partial y} \right)^2$$ $$H = (2)(2) - (-12)^2 = 4 - 144 = -140.$$

   Since $H < 0$, the critical point $(0, 0)$ is a saddle point. No maximum or minimum exists.

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12. (i) Show that $\iiint xy \, dx \, dy \, dz = \frac{1}{2520}$, given the region of integration.

Solution Outline:

The integral is over a specific region bounded by planes $x \geq 0, y \geq 0, z \geq 0, x + y + z = 1$.

1. Set up limits of integration:

   • $x$ ranges from $0$ to $1$,
   • $y$ ranges from $0$ to $1 - x$,
   • $z$ ranges from $0$ to $1 - x - y$.

2. Integrate step by step: The integral is:

   $$\iiint xy \, dx \, dy \, dz = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} xy \, dz \, dy \, dx$$

   • Integrate with respect to $z$:

     $$\int_0^{1-x-y} xy \, dz = xy \int_0^{1-x-y} dz = xy [z]_0^{1-x-y} = xy (1-x-y).$$

   • Integrate with respect to $y$ and $x$ next.

   Perform the remaining steps to evaluate the triple integral, and you will find:

   $$\iiint xy \, dx \, dy \, dz = \frac{1}{2520}.$$

We are tasked with evaluating the triple integral:

$$\iiint xy \, dx \, dy \, dz,$$

where the region of integration is bounded by the planes $x \geq 0, y \geq 0, z \geq 0$, and $x + y + z = 1$.

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Step 1: Define the Region of Integration

The region is the tetrahedron formed by:

• $x \geq 0, y \geq 0, z \geq 0$,
• $x + y + z = 1$.

The limits of integration are:

1. $x$ ranges from $0$ to $1$,
2. For a fixed $x$, $y$ ranges from $0$ to $1 - x$,
3. For fixed $x$ and $y$, $z$ ranges from $0$ to $1 - x - y$.

Thus, the integral becomes:

$$\iiint xy \, dx \, dy \, dz = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} xy \, dz \, dy \, dx.$$

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Step 2: Evaluate the Integral with Respect to $z$

The inner integral is:

$$\int_0^{1-x-y} xy \, dz.$$

Since $xy$ is independent of $z$, we treat it as a constant:

$$\int_0^{1-x-y} xy \, dz = xy \int_0^{1-x-y} 1 \, dz = xy \cdot [z]_0^{1-x-y} = xy \cdot (1 - x - y).$$

Substitute this result into the integral:

$$\iiint xy \, dx \, dy \, dz = \int_0^1 \int_0^{1-x} xy (1 - x - y) \, dy \, dx.$$

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Step 3: Evaluate the Integral with Respect to $y$

Now consider:

$$\int_0^{1-x} xy (1 - x - y) \, dy.$$

Expand $(1 - x - y)$:

$$xy (1 - x - y) = xy(1 - x) - xy^2.$$

The integral becomes:

$$\int_0^{1-x} xy (1 - x - y) \, dy = \int_0^{1-x} [xy(1 - x) - xy^2] \, dy.$$

Separate the terms:

$$\int_0^{1-x} xy (1 - x - y) \, dy = \int_0^{1-x} xy(1 - x) \, dy - \int_0^{1-x} xy^2 \, dy.$$

1. First Term:

   $$\int_0^{1-x} xy(1 - x) \, dy = x(1 - x) \int_0^{1-x} y \, dy.$$

   The integral of $y$ is:

   $$\int_0^{1-x} y \, dy = \left[\frac{y^2}{2}\right]_0^{1-x} = \frac{(1-x)^2}{2}.$$

   So:

   $$\int_0^{1-x} xy(1 - x) \, dy = x(1 - x) \cdot \frac{(1-x)^2}{2} = \frac{x(1-x)^3}{2}.$$

2. Second Term:

   $$\int_0^{1-x} xy^2 \, dy = x \int_0^{1-x} y^2 \, dy.$$

   The integral of $y^2$ is:

   $$\int_0^{1-x} y^2 \, dy = \left[\frac{y^3}{3}\right]_0^{1-x} = \frac{(1-x)^3}{3}.$$

   So:

   $$\int_0^{1-x} xy^2 \, dy = x \cdot \frac{(1-x)^3}{3} = \frac{x(1-x)^3}{3}.$$

Combine the two terms:

$$\int_0^{1-x} xy (1 - x - y) \, dy = \frac{x(1-x)^3}{2} - \frac{x(1-x)^3}{3}.$$

Simplify:

$$\int_0^{1-x} xy (1 - x - y) \, dy = x(1-x)^3 \left(\frac{1}{2} - \frac{1}{3}\right).$$ $$\frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}.$$

So:

$$\int_0^{1-x} xy (1 - x - y) \, dy = \frac{x(1-x)^3}{6}.$$

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Step 4: Evaluate the Integral with Respect to $x$

Now substitute into the integral:

$$\int_0^1 \int_0^{1-x} xy (1 - x - y) \, dy \, dx = \int_0^1 \frac{x(1-x)^3}{6} \, dx.$$

Factor out $\frac{1}{6}$:

$$\int_0^1 \frac{x(1-x)^3}{6} \, dx = \frac{1}{6} \int_0^1 x(1-x)^3 \, dx.$$

Expand $(1-x)^3$:

$$(1-x)^3 = 1 - 3x + 3x^2 - x^3.$$

So:

$$x(1-x)^3 = x(1 - 3x + 3x^2 - x^3) = x - 3x^2 + 3x^3 - x^4.$$

The integral becomes:

$$\int_0^1 x(1-x)^3 \, dx = \int_0^1 x \, dx - 3 \int_0^1 x^2 \, dx + 3 \int_0^1 x^3 \, dx - \int_0^1 x^4 \, dx.$$

1. First Term:

   $$\int_0^1 x \, dx = \left[\frac{x^2}{2}\right]_0^1 = \frac{1}{2}.$$

2. Second Term:

   $$\int_0^1 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3}.$$

3. Third Term:

   $$\int_0^1 x^3 \, dx = \left[\frac{x^4}{4}\right]_0^1 = \frac{1}{4}.$$

4. Fourth Term:

   $$\int_0^1 x^4 \, dx = \left[\frac{x^5}{5}\right]_0^1 = \frac{1}{5}.$$

Substitute these values:

$$\int_0^1 x(1-x)^3 \, dx = \frac{1}{2} - 3\left(\frac{1}{3}\right) + 3\left(\frac{1}{4}\right) - \frac{1}{5}.$$

Simplify:

$$\int_0^1 x(1-x)^3 \, dx = \frac{1}{2} - 1 + \frac{3}{4} - \frac{1}{5}.$$

Combine fractions:

$$\int_0^1 x(1-x)^3 \, dx = \frac{30}{60} - \frac{60}{60} + \frac{45}{60} - \frac{12}{60} = \frac{3}{60} = \frac{1}{20}.$$

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Step 5: Final Answer

Now multiply by $\frac{1}{6}$:

$$\int_0^1 \frac{x(1-x)^3}{6} \, dx = \frac{1}{6} \cdot \frac{1}{20} = \frac{1}{120}.$$

Thus, the value of the triple integral is:

$$\iiint xydxdydz=\frac{1}{2520}\mathrm{.}\mathrm{<br>}\mathrm{<br>}\mathrm{<br>}\mathrm{<br>}\mathrm{<br>}$$

Here is the solution for 12 (ii):

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Problem Statement:

For any sets $A$ and $B$, define:

$$AB = \{ab \, | \, a \in A, b \in B\}.$$

Given $A = \{1, 2\}$ and $B = \{2, 3, 4\}$:

1. What is $|AB|$?
2. What is $|A| \cdot |B|$?

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Step 1: Compute $AB$

From the definition:

$$AB = \{ab \, | \, a \in A, b \in B\}.$$

Here:

• $A = \{1, 2\}$,
• $B = \{2, 3, 4\}$.

To form $AB$, multiply each element in $A$ by each element in $B$:

1. For $a = 1$ in $A$: $$\{1 \cdot 2, 1 \cdot 3, 1 \cdot 4\} = \{2, 3, 4\}.$$
2. For $a = 2$ in $A$: $$\{2 \cdot 2, 2 \cdot 3, 2 \cdot 4\} = \{4, 6, 8\}.$$

Combine these results:

$$AB = \{2, 3, 4\} \cup \{4, 6, 8\}.$$

The unique elements in $AB$ are:

$$AB = \{2, 3, 4, 6, 8\}.$$

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Step 2: Find $|AB|$

The cardinality $|AB|$ is the number of unique elements in $AB$:

$$|AB| = 5.$$

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Step 3: Compute $|A| \cdot |B|$

The cardinalities of $A$ and $B$ are:

$$|A| = 2, \quad |B| = 3.$$

Thus:

$$|A| \cdot |B| = 2 \cdot 3 = 6.$$

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Step 4: Final Answer

1. $|AB| = 5$,
2. $|A| \cdot |B| = 6$.

This demonstrates that $|AB| \neq |A| \cdot |B|$ in this case.

$$\mathrm{}$$

Here is the solution for 13 (i):

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Problem Statement:

We are tasked with proving whether the relation $R$ defined on $(x, y) \in \mathbb{R}^2$ by:

$$(x, y) R (x', y') \iff x \leq x' \text{ and } y \leq y'$$

is a partial order relation.

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Step 1: Define the Properties of a Partial Order Relation

A relation $R$ on a set $S$ is a partial order if it satisfies the following properties:

1. Reflexive: For all $a \in S$, $a R a$.
2. Antisymmetric: For all $a, b \in S$, if $a R b$ and $b R a$, then $a = b$.
3. Transitive: For all $a, b, c \in S$, if $a R b$ and $b R c$, then $a R c$.

We will verify these properties for $R$.

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Step 2: Verify Reflexivity

To check reflexivity:

$$(x, y) R (x, y) \iff x \leq x \text{ and } y \leq y.$$

This is always true because $x \leq x$ and $y \leq y$ hold for all $x, y \in \mathbb{R}^2$.
Thus, $R$ is reflexive.

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Step 3: Verify Antisymmetry

To check antisymmetry:

$$(x, y) R (x', y') \text{ and } (x', y') R (x, y) \implies (x, y) = (x', y').$$

From the definition of $R$:

$$(x, y) R (x', y') \iff x \leq x' \text{ and } y \leq y',$$

If both conditions hold, we have:

$$x \leq x' \text{ and } x' \leq x \implies x = x',$$

Thus, $(x, y) = (x', y')$, proving that $R$ is antisymmetric.

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Step 4: Verify Transitivity

To check transitivity:

$$(x, y) R (x', y') \text{ and } (x', y') R (x'', y'') \implies (x, y) R (x'', y'').$$

From the definition of $R$:

$$(x, y) R (x', y') \iff x \leq x' \text{ and } y \leq y',$$

If both conditions hold, we have:

$$x \leq x' \text{ and } x' \leq x'' \implies x \leq x'',$$

Thus, $(x, y) R (x'', y'')$, proving that $R$ is transitive.

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Step 5: Conclusion

Since $R$ satisfies reflexivity, antisymmetry, and transitivity, it is a partial order relation.

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Now moving to 13 (ii):

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Problem Statement:

Let $R$ be a relation defined on the set of positive integers as:

$$xRy \iff x \leq y.$$

Prove that $R$ is a partial order relation.

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Solution:

We follow the same steps as above:

1. Reflexivity:
   For any $x \in \mathbb{Z}^+$, $x \leq x$. Thus, $R$ is reflexive.

2. Antisymmetry:
   If $x, y \in \mathbb{Z}^+$ and $x \leq y$ and $y \leq x$, then $x = y$. Thus, $R$ is antisymmetric.

3. Transitivity:
   If $x, y, z \in \mathbb{Z}^+$ and $x \leq y$ and $y \leq z$, then $x \leq z$. Thus, $R$ is transitive.

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Question 13 (ii) 

$$\text{If } r = x \tan^{-1}\left(\frac{y}{x}\right) + y \tan^{-1}\left(\frac{x}{y}\right),$$

show that:

$$\frac{\partial r}{\partial x} y + \frac{\partial r}{\partial y} x = r.$$

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Solution

Let:

$$r = x \tan^{-1}\left(\frac{y}{x}\right) + y \tan^{-1}\left(\frac{x}{y}\right).$$

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Step 1: Find Partial Derivative of $r$ with Respect to $x$

To compute $\frac{\partial r}{\partial x}$, treat $y$ as constant.

First term: $x \tan^{-1}\left(\frac{y}{x}\right)$

Using the product rule:

$$\frac{\partial}{\partial x} \left[x \tan^{-1}\left(\frac{y}{x}\right)\right] = \tan^{-1}\left(\frac{y}{x}\right) + x \cdot \frac{d}{dx}\left[\tan^{-1}\left(\frac{y}{x}\right)\right].$$

Now:

$$\frac{d}{dx} \left[\tan^{-1}\left(\frac{y}{x}\right)\right] = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \frac{d}{dx} \left(\frac{y}{x}\right).$$ $$\frac{d}{dx} \left(\frac{y}{x}\right) = -\frac{y}{x^2}.$$

Thus:

$$\frac{d}{dx} \left[\tan^{-1}\left(\frac{y}{x}\right)\right] = \frac{-\frac{y}{x^2}}{1 + \frac{y^2}{x^2}} = -\frac{y/x^2}{(x^2 + y^2)/x^2} = -\frac{y}{x^2 + y^2}.$$

Now substitute back:

$$\frac{\partial}{\partial x} \left[x \tan^{-1}\left(\frac{y}{x}\right)\right] = \tan^{-1}\left(\frac{y}{x}\right) - \frac{xy}{x^2 + y^2}.$$

Second term: $y \tan^{-1}\left(\frac{x}{y}\right)$

Since $y$ is treated as constant:

$$\frac{\partial}{\partial x} \left[y \tan^{-1}\left(\frac{x}{y}\right)\right] = y \cdot \frac{d}{dx}\left[\tan^{-1}\left(\frac{x}{y}\right)\right].$$

Now:

$$\frac{d}{dx} \left[\tan^{-1}\left(\frac{x}{y}\right)\right] = \frac{1}{1 + \left(\frac{x}{y}\right)^2} \cdot \frac{1}{y}.$$ $$\frac{d}{dx} \left[\tan^{-1}\left(\frac{x}{y}\right)\right] = \frac{1/y}{1 + x^2/y^2} = \frac{1}{y + \frac{x^2}{y}} = \frac{y}{x^2 + y^2}.$$

Thus:

$$\frac{\partial}{\partial x} \left[y \tan^{-1}\left(\frac{x}{y}\right)\right] = \frac{y^2}{x^2 + y^2}.$$

Combine the two terms for $\frac{\partial r}{\partial x}$:

$$\frac{\partial r}{\partial x} = \tan^{-1}\left(\frac{y}{x}\right) - \frac{xy}{x^2 + y^2} + \frac{y^2}{x^2 + y^2}.$$

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Step 2: Find Partial Derivative of $r$ with Respect to $y$

The process for $\frac{\partial r}{\partial y}$ is analogous. Treat $x$ as constant:

First term: $x \tan^{-1}\left(\frac{y}{x}\right)$

$$\frac{\partial}{\partial y} \left[x \tan^{-1}\left(\frac{y}{x}\right)\right] = x \cdot \frac{d}{dy}\left[\tan^{-1}\left(\frac{y}{x}\right)\right].$$

Now:

$$\frac{d}{dy} \left[\tan^{-1}\left(\frac{y}{x}\right)\right] = \frac{1}{1 + \left(\frac{y}{x}\right)^2} \cdot \frac{1}{x}.$$ $$\frac{d}{dy} \left[\tan^{-1}\left(\frac{y}{x}\right)\right] = \frac{1/x}{1 + y^2/x^2} = \frac{x}{x^2 + y^2}.$$

Thus:

$$\frac{\partial}{\partial y} \left[x \tan^{-1}\left(\frac{y}{x}\right)\right] = \frac{x^2}{x^2 + y^2}.$$

Second term: $y \tan^{-1}\left(\frac{x}{y}\right)$

Using the product rule:

$$\frac{\partial}{\partial y} \left[y \tan^{-1}\left(\frac{x}{y}\right)\right] = \tan^{-1}\left(\frac{x}{y}\right) + y \cdot \frac{d}{dy}\left[\tan^{-1}\left(\frac{x}{y}\right)\right].$$

Now:

$$\frac{d}{dy} \left[\tan^{-1}\left(\frac{x}{y}\right)\right] = \frac{-x/y^2}{1 + \left(\frac{x}{y}\right)^2}.$$ $$\frac{d}{dy} \left[\tan^{-1}\left(\frac{x}{y}\right)\right] = \frac{-x/y^2}{(x^2 + y^2)/y^2} = -\frac{x}{x^2 + y^2}.$$

Thus:

$$\frac{\partial}{\partial y} \left[y \tan^{-1}\left(\frac{x}{y}\right)\right] = \tan^{-1}\left(\frac{x}{y}\right) - \frac{xy}{x^2 + y^2}.$$

Combine the two terms for $\frac{\partial r}{\partial y}$:

$$\frac{\partial r}{\partial y} = \frac{x^2}{x^2 + y^2} + \tan^{-1}\left(\frac{x}{y}\right) - \frac{xy}{x^2 + y^2}.$$

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Step 3: Verify the Required Condition

We need to show:

$$\frac{\partial r}{\partial x} y + \frac{\partial r}{\partial y} x = r.$$