Mathematics-II (BCA-201) - 2019

Section-A: (Very Short Answer Type Questions)

Each question carries 3 marks. Here's a detailed explanation of each question from Section-A:

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1. Differentiate finite sets and infinite sets with examples.

• Finite Set: A set that contains a countable number of elements. The number of elements in a finite set is a natural number (e.g., $0, 1, 2, \ldots$).
  • Example: $A = \{1, 2, 3, 4, 5\}$ is a finite set because it has 5 elements.
• Infinite Set: A set that contains infinitely many elements, meaning its size is not finite.
  • Example: $B = \{x \in \mathbb{N} \mid x > 0\} = \{1, 2, 3, \ldots\}$ is infinite because it includes all natural numbers.

Key Difference:

• A finite set has a definite count of elements, while an infinite set has no end to the count of elements.

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2. Define trigonometric function, exponential function, and logarithmic function.

• Trigonometric Function: These functions relate the angles of a triangle to the ratios of its sides. Common trigonometric functions are:

  • $\sin(x)$, $\cos(x)$, $\tan(x)$, $\cot(x)$, $\sec(x)$, $\csc(x)$.
  • Example: $\sin(x) = \frac{\text{opposite side}}{\text{hypotenuse}}$.

• Exponential Function: A function of the form $f(x) = a^x$, where $a > 0$ and $a \neq 1$. The base $a$ is constant, and $x$ is the variable.

  • Example: $f(x) = 2^x$, where the function grows or decays exponentially depending on the base.

• Logarithmic Function: The inverse of the exponential function. It is defined as:

  • $y = \log_a(x) \iff x = a^y$, where $a > 0$ and $a \neq 1$.
  • Example: $\log_2(8) = 3$, since $2^3 = 8$.

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3. What do you mean by "Principle of Duality"?

• Principle of Duality is a fundamental concept in set theory and logic. It states that every algebraic expression remains valid if we interchange the operations "union" ($\cup$) and "intersection" ($\cap$) and replace universal sets with complements and vice versa.

For example:

• Original: $(A \cap B)^c = A^c \cup B^c$
• Dual: $(A \cup B)^c = A^c \cap B^c$

This principle ensures symmetry and balance in logical and set operations.

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4. If $u = f\left(\frac{y}{x}\right)$, then prove that $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 0$.

To prove:

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

Solution:

1. Given $u = f\left(\frac{y}{x}\right)$, let $v = \frac{y}{x}$. Then $u = f(v)$.
2. Partial derivatives:
   • $\frac{\partial v}{\partial x} = -\frac{y}{x^2}$,
   • $\frac{\partial v}{\partial y} = \frac{1}{x}$.
3. Use the chain rule: $$\frac{\partial u}{\partial x} = f'(v) \cdot \frac{\partial v}{\partial x} = f'(v) \cdot \left(-\frac{y}{x^2}\right),$$ $$\frac{\partial u}{\partial y} = f'(v) \cdot \frac{\partial v}{\partial y} = f'(v) \cdot \frac{1}{x}.$$
4. Substituting into the given expression: $$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = x \cdot f'(v) \cdot \left(-\frac{y}{x^2}\right) + y \cdot f'(v) \cdot \frac{1}{x}.$$
5. Simplify: $$= -\frac{xy}{x^2} \cdot f'(v) + \frac{y}{x} \cdot f'(v) = -\frac{y}{x} \cdot f'(v) + \frac{y}{x} \cdot f'(v).$$
6. The terms cancel: $$x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 0.$$

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5. Evaluate the triple integral $\int_0^1 \int_0^x \int_0^y z \, dz \, dy \, dx$.

To solve this, evaluate the integral step by step:

Step 1: Innermost integral with respect to $z$:

$$\int_0^y z \, dz = \left[\frac{z^2}{2}\right]_0^y = \frac{y^2}{2}.$$

Step 2: Second integral with respect to $y$:

$$\int_0^x \frac{y^2}{2} \, dy = \frac{1}{2} \int_0^x y^2 \, dy = \frac{1}{2} \cdot \left[\frac{y^3}{3}\right]_0^x = \frac{1}{2} \cdot \frac{x^3}{3} = \frac{x^3}{6}.$$

Step 3: Outer integral with respect to $x$:

$$\int_0^1 \frac{x^3}{6} \, dx = \frac{1}{6} \int_0^1 x^3 \, dx = \frac{1}{6} \cdot \left[\frac{x^4}{4}\right]_0^1 = \frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24}.$$

Final Answer: The value of the triple integral is $\frac{1}{24}$.

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Section-B: (Short Answer Type Questions)

Each question carries 5 marks, and students must attempt any two of the questions. Here’s a detailed explanation of each question:

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6. Define equivalence relation. If $A = \{1, 2, 3, 4\}$ and $R = \{(1, 1), (2, 2), (1, 2), (2, 1), (3, 3), (4, 4)\}$, prove that $R$ is an equivalence relation.

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

1. Reflexive: Every element is related to itself. For all $a \in A$, $(a, a) \in R$.
2. Symmetric: If $(a, b) \in R$, then $(b, a) \in R$.
3. Transitive: If $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$.

Proof: Given $A = \{1, 2, 3, 4\}$ and $R = \{(1, 1), (2, 2), (1, 2), (2, 1), (3, 3), (4, 4)\}$, check the three properties:

1. Reflexive: For every $a \in A$, $(a, a) \in R$:

   • $(1, 1), (2, 2), (3, 3), (4, 4) \in R$.
   • Therefore, $R$ is reflexive.

2. Symmetric: If $(a, b) \in R$, then $(b, a) \in R$:

   • $(1, 2) \in R \Rightarrow (2, 1) \in R$.
   • All pairs satisfy this condition. Hence, $R$ is symmetric.

3. Transitive: If $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R$:

   • For $(1, 2) \in R$ and $(2, 1) \in R$, $(1, 1) \in R$.
   • Similarly, all other pairs satisfy transitivity.

Since $R$ satisfies all three properties, it is an equivalence relation.

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7. Find the area of the region bounded by the circle $x^2 + y^2 = a^2$ using double integration.

The equation of the circle $x^2 + y^2 = a^2$ represents a circle with radius $a$ and center at the origin $(0, 0)$.

To find the area:

• Use double integration in polar coordinates.

1. Convert to Polar Coordinates:

   • In polar form: $x = r \cos \theta$, $y = r \sin \theta$, and $x^2 + y^2 = r^2$.
   • The region is a full circle ($0 \leq r \leq a$, $0 \leq \theta \leq 2\pi$).

2. Area Formula in Polar Coordinates:

   $$\text{Area} = \int_0^{2\pi} \int_0^a r \, dr \, d\theta.$$

3. Evaluate Inner Integral ($r$):

   $$\int_0^a r \, dr = \left[\frac{r^2}{2}\right]_0^a = \frac{a^2}{2}.$$

4. Evaluate Outer Integral ($\theta$):

   $$\int_0^{2\pi} \frac{a^2}{2} \, d\theta = \frac{a^2}{2} \cdot \left[\theta\right]_0^{2\pi} = \frac{a^2}{2} \cdot 2\pi = \pi a^2.$$

Final Answer: The area of the circle is $\pi a^2$.

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8. Show that the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 1}{4}$ and $\frac{x - 1}{3} = \frac{y - 2}{4} = \frac{z - 3}{5}$ are coplanar. Also, find their point of intersection.

Step 1: Condition for Coplanarity
Two lines are coplanar if the scalar triple product of their direction vectors and the vector connecting points on the two lines is zero.

• Direction vector of line 1 ($L_1$): $(2, 3, 4)$.
• Direction vector of line 2 ($L_2$): $(3, 4, 5)$.
• A point on $L_1$ is $(1, 2, 1)$, and a point on $L_2$ is $(1, 2, 3)$.
• Vector joining these points: $(1 - 1, 2 - 2, 1 - 3) = (0, 0, -2)$.

The scalar triple product is:

$$\text{S.T.P.} = \begin{vmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 0 & 0 & -2 \end{vmatrix}.$$

Expanding:

$$\text{S.T.P.} = 2 \begin{vmatrix} 4 & 5 \\ 0 & -2 \end{vmatrix} - 3 \begin{vmatrix} 3 & 5 \\ 0 & -2 \end{vmatrix} + 4 \begin{vmatrix} 3 & 4 \\ 0 & -2 \end{vmatrix}.$$

Evaluate the minors:

$$\begin{vmatrix} 4 & 5 \\ 0 & -2 \end{vmatrix} = (4)(-2) - (0)(5) = -8,$$ $$\begin{vmatrix} 3 & 5 \\ 0 & -2 \end{vmatrix} = (3)(-2) - (0)(5) = -6,$$ $$\begin{vmatrix} 3 & 4 \\ 0 & -2 \end{vmatrix} = (3)(-2) - (0)(4) = -6.$$

Substitute:

$$\text{S.T.P.} = 2(-8) - 3(-6) + 4(-6) = -16 + 18 - 24 = -22 \neq 0.$$

Conclusion: The lines are not coplanar.

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Section-C: (Long Answer Type Questions)

Each question carries 15 marks, and students are required to attempt any three of the five questions. Here’s a detailed explanation of each question from Section-C:

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9. (i) Let $Q$ be the set of rational numbers, and $f: Q \to Q$ be defined by $f(x) = 2x + 3$.
Prove that $f$ is bijective. Also, find $f^{-1}$.

Proof for Bijectivity:

1. Injective (One-to-One):

   • To prove injectivity, assume $f(x_1) = f(x_2)$.
   $$f(x_1) = 2x_1 + 3 \quad \text{and} \quad f(x_2) = 2x_2 + 3.$$

   If $f(x_1) = f(x_2)$, then:

   $$2x_1 + 3 = 2x_2 + 3 \implies 2x_1 = 2x_2 \implies x_1 = x_2.$$

   Hence, $f$ is injective.

2. Surjective (Onto):

   • To prove surjectivity, for every $y \in Q$ (the codomain), we must find $x \in Q$ such that $f(x) = y$.
   $$f(x) = 2x + 3 \implies x = \frac{y - 3}{2}.$$

   Since $y \in Q$ and rational numbers are closed under subtraction and division by nonzero numbers, $x \in Q$.
   Hence, $f$ is surjective.

Since $f$ is both injective and surjective, it is bijective.

Finding $f^{-1}(x)$:

To find the inverse, solve $y = f(x) = 2x + 3$ for $x$:

$$x = \frac{y - 3}{2}.$$

Thus, the inverse function is:

$$f^{-1}(x) = \frac{x - 3}{2}.$$

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(ii) If $f: R \to R$ and $g: R \to R$ be defined by $f(x) = x - 1$ and $g(x) = x^2 + 1$, then find:

• $g(f(x))$,
• $f(g(x))$.

Solution:

1. $g(f(x)) = g(x - 1)$:

   • Substitute $x - 1$ into $g(x) = x^2 + 1$:
   $$g(f(x)) = g(x - 1) = (x - 1)^2 + 1.$$

   Expand $(x - 1)^2$:

   $$g(f(x)) = x^2 - 2x + 1 + 1 = x^2 - 2x + 2.$$

2. $f(g(x)) = f(x^2 + 1)$:

   • Substitute $x^2 + 1$ into $f(x) = x - 1$:
   $$f(g(x)) = f(x^2 + 1) = (x^2 + 1) - 1 = x^2.$$

Thus:

$$g(f(x)) = x^2 - 2x + 2, \quad f(g(x)) = x^2.$$

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10. (i) Let $(L, \leq)$ be a lattice. If $a, b \in L$, prove:

• $a \leq b \implies a \land b = a$,
• $a \leq b \implies a \lor b = b$.

Proof:

1. For $a \land b = a$:

   • By definition of a lattice, the meet (greatest lower bound) $a \land b$ is the largest element less than or equal to both $a$ and $b$.
   • Since $a \leq b$, $a$ itself is the greatest lower bound.
   • Hence, $a \land b = a$.

2. For $a \lor b = b$:

   • The join (least upper bound) $a \lor b$ is the smallest element greater than or equal to both $a$ and $b$.
   • Since $a \leq b$, $b$ itself is the least upper bound.
   • Hence, $a \lor b = b$.

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(ii) Let $L$ be a lattice with least element $0$ and greatest element $1$. Prove:

• If $a \lor 1 = 1$, then $a = 1$,
• If $a \land 0 = 0$, then $a = 0$.

Proof:

1. For $a \lor 1 = 1$:

   • The join $a \lor 1$ is the least upper bound of $a$ and $1$.
   • Since $1$ is the greatest element, $a \leq 1$, and the least upper bound is $1$.
   • For $a \lor 1 = 1$, this implies $a = 1$.

2. For $a \land 0 = 0$:

   • The meet $a \land 0$ is the greatest lower bound of $a$ and $0$.
   • Since $0$ is the least element, $a \geq 0$, and the greatest lower bound is $0$.
   • For $a \land 0 = 0$, this implies $a = 0$.

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11. Discuss the maxima or minima of the function:

$$z = xy \left( \frac{x}{y} + \frac{y}{x} \right).$$

Solution:

Rewrite the given function:

$$z = xy \left( \frac{x}{y} + \frac{y}{x} \right) = xy \left( \frac{x^2 + y^2}{xy} \right) = x^2 + y^2.$$

The function reduces to:

$$z = x^2 + y^2.$$

1. Find Critical Points:

   • Partial derivatives:

     $$\frac{\partial z}{\partial x} = 2x, \quad \frac{\partial z}{\partial y} = 2y.$$

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

     $$2x = 0 \implies x = 0, \quad 2y = 0 \implies y = 0.$$

   • Critical point: $(x, y) = (0, 0)$.

2. Second Derivative Test:

   • Second partial derivatives:

     $$\frac{\partial^2 z}{\partial x^2} = 2, \quad \frac{\partial^2 z}{\partial y^2} = 2, \quad \frac{\partial^2 z}{\partial x \partial y} = 0.$$

   • Hessian determinant:

     $$H = \frac{\partial^2 z}{\partial x^2} \cdot \frac{\partial^2 z}{\partial y^2} - \left( \frac{\partial^2 z}{\partial x \partial y} \right)^2 = (2)(2) - (0)^2 = 4 > 0.$$

   • Since $H > 0$ and $\frac{\partial^2 z}{\partial x^2} > 0$, the critical point $(0, 0)$ is a local minimum.

3. Conclusion: The function $z = x^2 + y^2$ has a global minimum at $(x, y) = (0, 0)$, and the minimum value is $z = 0$.

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Solution for Question 12

(i) Show that the lines are coplanar

The given lines are:

1. $\frac{x+1}{2} = \frac{y-3}{3} = \frac{z+2}{1}$
   Let this be represented in parametric form as:

   $$x = -1 + 2t, \quad y = 3 + 3t, \quad z = -2 + t, \quad t \in \mathbb{R}.$$

2. $\frac{x-7}{-2} = \frac{y-7}{4} = \frac{z+1}{2}$
   Let this be represented in parametric form as:

   $$x = 7 - 2s, \quad y = 7 + 4s, \quad z = -1 + 2s, \quad s \in \mathbb{R}.$$

To check if the lines are coplanar, calculate the scalar triple product of:

• $\vec{d_1}$: direction vector of the first line, $(2, 3, 1)$,
• $\vec{d_2}$: direction vector of the second line, $(-2, 4, 2)$,
• $\vec{PQ}$: the vector joining a point $P$ on the first line to a point $Q$ on the second line.

Step 1: Choose points $P$ and $Q$: From the first line, let $t = 0$: $P(-1, 3, -2)$.
From the second line, let $s = 0$: $Q(7, 7, -1)$.

$$\vec{PQ} = (7 - (-1), 7 - 3, -1 - (-2)) = (8, 4, 1).$$

Step 2: Scalar triple product: The scalar triple product is given by:

$$\vec{d_1} \cdot (\vec{d_2} \times \vec{PQ}).$$

1. Calculate $\vec{d_2} \times \vec{PQ}$:

   $$\vec{d_2} = (-2, 4, 2), \quad \vec{PQ} = (8, 4, 1).$$

   Using the determinant:

   $$\vec{d_2} \times \vec{PQ} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 4 & 2 \\ 8 & 4 & 1 \end{vmatrix} = \mathbf{i} \begin{vmatrix} 4 & 2 \\ 4 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} -2 & 2 \\ 8 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} -2 & 4 \\ 8 & 4 \end{vmatrix}.$$

   Simplify each minor:

   $$\vec{d_2} \times \vec{PQ} = \mathbf{i}(4 - 8) - \mathbf{j}(-2 - 16) + \mathbf{k}(-8 - 32),$$ $$\vec{d_2} \times \vec{PQ} = -4\mathbf{i} + 18\mathbf{j} - 40\mathbf{k}.$$

2. Dot product $\vec{d_1} \cdot (\vec{d_2} \times \vec{PQ})$:

   $$\vec{d_1} = (2, 3, 1), \quad \vec{d_2} \times \vec{PQ} = (-4, 18, -40).$$ $$\vec{d_1} \cdot (\vec{d_2} \times \vec{PQ}) = (2)(-4) + (3)(18) + (1)(-40),$$ $$\vec{d_1} \cdot (\vec{d_2} \times \vec{PQ}) = -8 + 54 - 40 = 6.$$

Since the scalar triple product is nonzero ($6 \neq 0$), the lines are not coplanar.

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(ii) Find the angle of intersection of the spheres

The equations of the spheres are:

1. $x^2 + y^2 + z^2 - 4x - 6y + 10 = 0,$
2. $x^2 + y^2 + z^2 - 6x - 2y + 2z + 2 = 0.$

Step 1: Find the gradients of the spheres
The gradient of a sphere represents the normal vector to its surface.

1. For the first sphere:

   $$\nabla f_1 = \frac{\partial}{\partial x} (x^2 + y^2 + z^2 - 4x - 6y + 10),$$ $$\nabla f_1 = (2x - 4, 2y - 6, 2z).$$

2. For the second sphere:

   $$\nabla f_2 = \frac{\partial}{\partial x} (x^2 + y^2 + z^2 - 6x - 2y + 2z + 2),$$ $$\nabla f_2 = (2x - 6, 2y - 2, 2z + 2).$$

Step 2: Angle of intersection formula The angle $\theta$ between the spheres is given by the formula:

$$\cos\theta = \frac{\nabla f_1 \cdot \nabla f_2}{\|\nabla f_1\| \|\nabla f_2\|}.$$

Substitute the gradients and simplify the expression for specific points on the intersection.

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Solution for Question 13

(i) Evaluate the double integral

The given integral is:

$$\int_0^2 \int_x^2 x y^2 \, dy \, dx.$$

Step 1: Identify the region of integration

• The outer integral varies from $x = 0$ to $x = 2$.
• The inner integral varies from $y = x$ to $y = 2$.

Thus, the region is a triangular region in the $xy$-plane bounded by:

• $x = 0$,
• $y = 2$,
• $y = x$.

Step 2: Solve the inner integral

$$\int_x^2 x y^2 \, dy = x \int_x^2 y^2 \, dy.$$ $$\int y^2 \, dy = \frac{y^3}{3}.$$

Evaluate:

$$\int_x^2 y^2 \, dy = \frac{(2)^3}{3} - \frac{(x)^3}{3} = \frac{8}{3} - \frac{x^3}{3}.$$

Thus:

$$\int_x^2 x y^2 \, dy = x \left(\frac{8}{3} - \frac{x^3}{3}\right) = \frac{8x}{3} - \frac{x^4}{3}.$$

Step 3: Solve the outer integral

$$\int_0^2 \left(\frac{8x}{3} - \frac{x^4}{3}\right) \, dx = \frac{1}{3} \int_0^2 (8x - x^4) \, dx.$$

Separate the terms:

$$\int_0^2 8x \, dx = 8 \cdot \frac{x^2}{2} \Big|_0^2 = 8 \cdot 2 = 16.$$ $$\int_0^2 x^4 \, dx = \frac{x^5}{5} \Big|_0^2 = \frac{32}{5}.$$

Combine:

$$\frac{1}{3} \left(16 - \frac{32}{5}\right) = \frac{1}{3} \cdot \frac{80 - 32}{5} = \frac{1}{3} \cdot \frac{48}{5} = \frac{16}{5}.$$

Final Answer: $\frac{16}{5}$.

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(ii) Evaluate the triple integral

The given integral is:

$$\int_0^1 \int_0^x \int_0^y \frac{1}{48} \, dz \, dy \, dx.$$

Step 1: Solve the innermost integral

$$\int_0^y \frac{1}{48} \, dz = \frac{1}{48} \int_0^y 1 \, dz = \frac{1}{48} \cdot y.$$

Step 2: Solve the second integral

$$\int_0^x \frac{y}{48} \, dy = \frac{1}{48} \int_0^x y \, dy = \frac{1}{48} \cdot \frac{y^2}{2} \Big|_0^x = \frac{1}{48} \cdot \frac{x^2}{2} = \frac{x^2}{96}.$$

Step 3: Solve the outer integral

$$\int_0^1 \frac{x^2}{96} \, dx = \frac{1}{96} \int_0^1 x^2 \, dx = \frac{1}{96} \cdot \frac{x^3}{3} \Big|_0^1 = \frac{1}{96} \cdot \frac{1}{3} = \frac{1}{288}.$$

Final Answer: $\frac{1}{288}$.