Class 12 Maths Full Syllabus minus calculus Test 1 – 2026 and Solution

Have you solved Class 12 Maths Full Syllabus minus Calculus Test – 1  and are looking for a solutions to the questions? Here you’ll find complete, step-by-step BOARD Exam-style solutions for every question. Designed for both students and teachers, these answers make board exam revision easier with clear explanations, formula highlights, and downloadable PDF options.

Q1: If A = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, then A^{-1} is:

(A) \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}

(B) \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}

(C) \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

(D) \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

Solution:
If a matrix A is a diagonal matrix where A = \text{diag}(d_1, d_2, d_3), provided that no diagonal element is zero, the inverse A^{-1} is simply the diagonal matrix formed by the reciprocals of the diagonal elements.

Mathematically: A^{-1} = \text{diag} \left(\frac{1}{d_1}, \frac{1}{d_2}, \frac{1}{d_3} \right)
Step 1: Identify the Matrix Type
Given matrix A:
A = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, then A^{-1}

Since all non-diagonal elements are zero, A is a Diagonal Matrix.

Step 2: Identify the Diagonal Elements
The diagonal elements are:
d_1 = -1
d_2 = 1
d_3 = 1

Step 3: Calculate the Reciprocals
We take the reciprocal of each diagonal element to find the elements of A^{-1}:

1. \frac{1}{d_1} = \frac{1}{-1} = -1
2. \frac{1}{d_2} = \frac{1}{1} = 1
3. \frac{1}{d_3} = \frac{1}{1} = 1

Step 4: Construct the Inverse Matrix
Placing these values back into the diagonal matrix form:
A^{-1} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, then A^{-1}

note: In this specific case, notice that A^{-1} = A. This is a special type of matrix known as an Involutory Matrix (where A^2 = I).

Correct Option: (D)

Q2: If vector \vec{a}=3\hat{i}+2\hat{j}-\hat{k} and vector \vec{b}=\hat{i}-\hat{j}+\hat{k}, then which of the following is correct?

(A) \vec{a} || \vec{b}

(B) \vec{a} \perp \vec{b}

(C) |\vec{b}| > |\vec{a}|

(D) |\vec{a}| = |\vec{b}|

Solution:
To determine the relationship between two vectors, we first check their dot product (scalar product). If the dot product is zero, the vectors are perpendicular (orthogonal).

Step 1: Calculate the Dot Product
We calculate \vec{a} \cdot \vec{b} using components:
\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3

Substituting values:
\vec{a} \cdot \vec{b} = (3)(1) + (2)(-1) + (-1)(1)

Step 2: Simplify
\vec{a} \cdot \vec{b} = 3 - 2 -1
\vec{a} \cdot \vec{b} = 0

Conclusion: Since the dot product is zero, the vectors are perpendicular to each other.

Correct Option: (B)

Q3: If A is a square matrix of order 2 such that \text{det}(A)=4, then \text{det}(4 \text{adj} A) is equal to:

(A) 16
(B) 64
(C) 256
(D) 512

Solution:
We use two key properties of determinants for a square matrix of order n:

1. |kA| = k^n|A| (where k is a scalar)
2. |\text{adj} A| = |A|^{n-1}

Step 1: Apply Scalar Property
Here, the order n = 2.
We need to find |4 \text{adj} A|.
Using property 1:
|4 \text{adj} A| = 4^2 |\text{adj} A| = 16 |\text{adj} A|

Step 2: Apply Adjoint Property
Using property 2 with n=2:
|\text{adj} A| = |A|^{2-1} = |A|^1 = 4

Step 3: Final Calculation
Substitute the value back into the equation from Step 1:
|4 \text{adj} A| = 16 \times 4
|4 \text{adj} A| = 64

Correct Option: (B)

Q4: If E and F are two independent events such that P(E)=\frac{2}{3} and P(F)=\frac{3}{7}, then P(E/\overline{F}) is equal to:

(A) \frac{1}{6}

(B) \frac{1}{2}

(C) \frac{2}{3}

(D) \frac{7}{9}

Solution:
The question asks for the conditional probability of E given not F.

Concept: If E and F are independent events, then the occurrence of one does not affect the probability of the other.
Consequently, E and \overline{F} (not F) are also independent events.

Step 1: Apply Independence Property
Since E and \overline{F} are independent:
P(E/ \overline{F}) = P(E)

Step 2: Substitute Value
Given P(E) = \frac{2}{3}, therefore:

P(E/ \overline{F}) = \frac{2}{3}

Correct Option: (C)

Q5: Let A=\begin{bmatrix}1&-2&-1\\ 0&4&-1\\ -3&2&1\end{bmatrix}, B=\begin{bmatrix}-2\\ -5\\ -7\end{bmatrix}, C=\begin{bmatrix}9&8&7\end{bmatrix}, which of the following is defined?

(A) Only AB

(B) Only AC

(C) Only BA

(D) All AB, AC and BA

Solution:
Matrix multiplication XY is defined only if the number of columns in the first matrix X is equal to the number of rows in the second matrix Y.

Step 1: Analyze Dimensions
Matrix A is of order 3 \times 3.
Matrix B is of order 3 \times 1.
Matrix C is of order 1 \times 3.

Step 2: Check Compatibility

• Check AB: (3 \times 3) \times (3 \times 1). Cols of A (3) = Rows of B (3). Defined.
• Check AC: (3 \times 3) \times (1 \times 3). Cols of A (3) \neq Rows of C (1). Not Defined.
• Check BA: (3 \times 1) \times (3 \times 3). Cols of B (1) \neq Rows of A (3). Not Defined.

Conclusion: Only the product AB is possible.

Correct Option: (A)

Q6: If \vec{a}+\vec{b}+\vec{c}=\vec{0}, |\vec{a}|=\sqrt{37}, |\vec{b}|=3 and |\vec{c}|=4, then angle between \vec{b} and \vec{c} is:

(A) \frac{\pi}{6}

(B) \frac{\pi}{4}

(C) \frac{\pi}{3}

(D) \frac{\pi}{2}

Solution: To find the angle between \vec{b} and \vec{c}, we isolate these vectors on one side of the equation .

Step 1: Rearrange Equation
\vec{b} + \vec{c} = -\vec{a}

Step 2: Square Both Sides
Squaring the magnitudes:
|\vec{b} + \vec{c}|^2 = |-\vec{a}|^2

Using the vector identity
|\vec{x}+\vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 + 2|\vec{x}||\vec{y}|\cos\theta

|\vec{x}|^2 + |\vec{y}|^2 + 2|\vec{x}||\vec{y}|\cos\theta = |-\vec{a}|^2

Step 3: Substitute Values
Given |\vec{a}|=\sqrt{37}, |\vec{b}|=3, |\vec{c}|=4:
3^2 + 4^2 + 2(3)(4)\cos\theta = (\sqrt{37})^2
9 + 16 + 24\cos\theta = 37
25 + 24 \cos\theta = 37

Step 4: Solve for \theta
24 \cos\theta = 37 - 25
24\cos\theta = 12
\cos\theta = \frac{1}{2}

Since \cos\frac{\pi}{3} = \frac{1}{2},
the angle \theta = \frac{\pi}{3}.

Correct Option: (C)

Question 7: If A=\begin{bmatrix}7&0&x \\ 0&7&0 \\ 0&0&y\end{bmatrix} is a scalar matrix, then y^{x} is equal to

(A) 0

(B) 1

(C) 7

(D) \pm 7

Solution:
We must apply the definition of a Scalar Matrix to find the values of x and y .

Step 1: Definition of Scalar Matrix
A diagonal matrix is a scalar matrix if all its diagonal elements are equal.
Therefore, a_{11} = a_{22} = a_{33} = 7.
This implies y = 7.

Step 2: Definition of Diagonal Elements
In a diagonal matrix (and thus a scalar matrix), all non-diagonal elements must be zero.
The element x is in position a_{13} (non-diagonal).
Therefore, x = 0.

Step 3: Calculate y^x Substitute y=7 and x=0:
y^x = 7^0 = 1

Correct Option: (B)

Q8: If A and B are invertible matrices, then which of the following is not correct?

(A) (A+B)^{-1}=B^{-1}+A^{-1}

(B) (AB)^{-1}=B^{-1}A^{-1}

(C) \text{adj}(A)=|A|A^{-1}

(D) |A|^{-1}=|A^{-1}|

Solution: We evaluate each property of invertible matrices .

• Option (B): This is the Reversal Law for inverses. It is Correct.
• Option (C): We know A(\text{adj} A) = |A|I. Multiplying by A^{-1}, we get \text{adj} A = |A|A^{-1}. This is Correct.
• Option (D): The determinant of an inverse is the reciprocal of the determinant. |A^{-1}| = \frac{1}{|A|} = |A|^{-1}. This is Correct.
• Option (A): Matrix addition does not distribute over inversion in this manner. Generally, (A+B)^{-1} \neq A^{-1} + B^{-1}. This is Incorrect.

Correct Option: (A)

Question 9: The graph of a trigonometric function is as shown. Which of the following will represent graph of its inverse?

(A) Graph passing through origin, defined on [-1, 1] with range [-\pi/2, \pi/2] (B) Graph passing through origin, different curvature (C) Graph defined on range [0, \pi] (D) Discontinuous graph

Solution: Step 1: Identify the Original Function The provided graph in the question shows a wave passing through the origin latex[/latex], reaching 1 at \pi/2 and -1 at -\pi/2. This is the graph of y = \sin x restricted to the interval [-\pi/2, \pi/2].

Step 2: Identify the Inverse Function The inverse of this function is y = \sin^{-1}x (arcsin). The properties of \sin^{-1}x are:

• Domain: [-1, 1] (x-axis)
• Range: [-\pi/2, \pi/2] (y-axis)
• Nature: It is an increasing function passing through the origin.
• Concavity: It is concave up (convex) for x > 0.

Step 3: Match with Options

• Option (A) shows a graph with domain [-1, 1] and range [-\pi/2, \pi/2], with the correct shape for arcsin.
• Option (C) shows a range of [0, \pi], which corresponds to \cos^{-1}x.

Correct Option: (A)

Q10: The projection vector of vector \vec{a} on vector \vec{b} is

(A) \left(\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|^{2}}\right)\vec{b}

(B) \frac{\vec{a}\cdot\vec{b}}{|\vec{a}|}

(C) \frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}

(D) \left(\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|^{2}}\right)\vec{b}

Solution:
The projection of a vector \vec{a} on \vec{b} is a vector that points in the direction of \vec{b} with a magnitude equal to the scalar projection.

Formula:
\text{Vector Projection} = (\text{Scalar Projection}) \times (\text{Unit Vector of } \vec{b})

= \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \right) \hat{b}

= \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \frac{\vec{b}}{|\vec{b}|} \right)

= \frac{(\vec{a} \cdot \vec{b})\vec{b}}{|\vec{b}|^2}

Correct Option: (A)

Question 11: If a line makes angles of \frac{3\pi}{4} and \frac{\pi}{3} with the positive directions of x, y and z-axis respectively, then \theta is

(A) \frac{-\pi}{3} \text{only}

(B) \frac{\pi}{3} \text{only}

(C) \frac{\pi}{6}

(D) \pm\frac{\pi}{3}

Solution:
We use the property of direction cosines: l^2 + m^2 + n^2 = 1,
where l = \cos \alpha, m = \cos \beta, n = \cos \gamma .

Step 1: Identify Known Angles
\alpha = \frac{3\pi}{4} \Rightarrow l = \cos\left(\frac{3\pi}{4} \right) = -\frac{1}{\sqrt{2}}

\beta = \frac{\pi}{3} \Rightarrow m = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}

\gamma = \theta

Step 2: Apply Formula
\left( -\frac{1}{\sqrt{2}} \right)^2 + \left( \frac{1}{2} \right)^2 + \cos^2\theta = 1

\frac{1}{2} + \frac{1}{4} + \cos^2\theta = 1

\frac{3}{4} + \cos^2\theta = 1

Step 3: Solve for \theta
\cos^2\theta = 1 - \frac{3}{4}

\cos^2\theta = \frac{1}{4}

\cos\theta = \pm \frac{1}{2}

The possible angles with the axis are \frac{\pi}{3} (for +1/2) and \frac{2\pi}{3} (for -1/2).

Direction angles with positive axes are typically given in [0, \pi]. Negative angles are not standard.
Between the options provided, only \frac{\pi}{3} is valid and listed correctly.

Correct Option: (B)

Question 12: Which of the following can be both a symmetric and skew-symmetric matrix?

(A) Unit Matrix

(B) Diagonal Matrix

(C) Null Matrix

(D) Row Matrix

Solution:
We check the conditions for a matrix A to be both symmetric and skew-symmetric .
Condition 1: Symmetric \Rightarrow A^T = A
Condition 2: Skew-symmetric \Rightarrow A^T = -A

Step 1: Equate Conditions
From the above, A = -A. A + A = O (Zero Matrix)
2A = 0
2A = 0
A = O

Thus, the matrix must be a Null (Zero) Matrix.

Correct Option: (C)

Section B

Question 13: Evaluate \tan^{-1}[2 \sin(2 \cos^{-1}\frac{\sqrt{3}}{2})]

Solution: We solve this expression from the innermost bracket outward.
Step 1: Evaluate the innermost inverse cosine
We know that \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}.

Therefore, the principal value is:
\cos^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{6}

Step 2: Substitute and solve the sine argument
Substitute this back into the expression:
2 \cos^{-1}(\frac{\sqrt{3}}{2}) = 2\times \frac{\pi}{6}

2 \cos^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{3}

Now the expression becomes:
\tan^{-1}[2 \sin(\frac{\pi}{3})]

Step 3: Evaluate the sine function
We know that \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}.

Substitute this into the expression:
\tan^{-1}[2 \times \frac{\sqrt{3}}{2}]

= \tan^{-1}[\sqrt{3}]

Step 4: Final Evaluation
We know that \tan(\frac{\pi}{3}) = \sqrt{3}.

Therefore:
\tan^{-1}[\sqrt{3}]= \frac{\pi}{3}

Final Answer: \boxed{ \tan^{-1}[2 \sin(2 \cos^{-1}\frac{\sqrt{3}}{2})]= \frac{\pi}{3} }

Question 14: The diagonals of a parallelogram are given by \vec{a}=2\hat{i}-\hat{j}+\hat{k} and \vec{b}=\hat{i}+3\hat{j}-\hat{k}. Find the area of the parallelogram.

Solution:
When the diagonals \vec{d_1} and \vec{d_2} of a parallelogram are given, the area is calculated using the cross product.

Formula:
\text{Area} = \frac{1}{2} |\vec{d_1} \times \vec{d_2}|

Here,
\vec{d_1} = \vec{a} = 2\hat{i}-\hat{j}+\hat{k} and \vec{d_2} = \vec{b} = \hat{i}+3\hat{j}-\hat{k}.

Step 1: Calculate the Cross Product (\vec{a} \times \vec{b})
\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 3 & -1 \end{vmatrix}

Expanding along the first row:
\vec{a} \times \vec{b} = \hat{i}[(-1)(-1) - (1)(3)] - \hat{j}[(2)(-1) - (1)(1)] + \hat{k}[(2)(3) - (-1)(1)]

\vec{a} \times \vec{b} = \hat{i}[1 - 3] - \hat{j}[-2 - 1] + \hat{k}[6 + 1]

\vec{a} \times \vec{b} = -2\hat{i} + 3\hat{j} + 7\hat{k}

Step 2: Calculate the Magnitude (|\vec{a} \times \vec{b}|)

|\vec{a} \times \vec{b}| = \sqrt{(-2)^2 + (3)^2 + (7)^2}
= \sqrt{4 + 9 + 49}
= \sqrt{62}

Step 3: Calculate the Area \text{Area} = \frac{1}{2} \sqrt{62} sq. units.

Final Answer: \boxed{Area = \frac{\sqrt{62}}{2}} sq. units.

Q15 (a): Two friends while flying kites from different locations, find the strings of their kites crossing each other. The strings can be represented by vectors \vec{a}=3\hat{i}+\hat{j}+2\hat{k} and \vec{b}=2\hat{i}-2\hat{j}+4\hat{k}. Determine the angle formed between the kite strings.

Solution:
To find the angle \theta between two vectors \vec{a} and \vec{b}, we use the dot product formula:
\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}

Step 1: Calculate the Dot Product (\vec{a} \cdot \vec{b})
\vec{a} \cdot \vec{b} = a_1 a_2 + b_1 b_2 + c_1 c_2
\vec{a} \cdot \vec{b} = (3)(2) + (1)(-2) + (2)(4)
\vec{a} \cdot \vec{b} = 6 - 2 + 8
\vec{a} \cdot \vec{b}= 12

Step 2: Calculate the Magnitudes (|\vec{a}| and |\vec{b}|)
Magnitude of \vec{a}:
|\vec{a}| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{9 +1 + 4} = \sqrt{14}

Magnitude of \vec{b}:
|\vec{b}| = \sqrt{2^2 + (-2)^2 + 4^2} = \sqrt{4 +4 + 16} = \sqrt{24} = 2\sqrt{6}

Step 3: Substitute Values into Formula
\cos \theta = \frac{12}{\sqrt{14} \cdot 2\sqrt{6}}

\cos \theta = \frac{6}{\sqrt{84}}

\cos \theta = \frac{3}{\sqrt{21}}

Rationalizing the denominator:
\cos \theta = \frac{\sqrt{21}}{7}

Final Answer: \boxed{ \theta = \cos^{-1}\left(\frac{\sqrt{21}}{7}\right) }

Q15(b): If \vec{a} and \vec{b} are two non-collinear vectors, then find x, such that \vec{\alpha}=(x-2)\vec{a}+\vec{b} and \vec{\beta}=(3+2x)\vec{a}-2\vec{b} are collinear.

Solution:
Two vectors \vec{\alpha} and \vec{\beta} are collinear if one is a scalar multiple of the other, or if their corresponding components are proportional.

Given:
\vec{\alpha}=(x-2)\vec{a}+1\vec{b}

\vec{\beta} = (3+2x)\vec{a} - 2\vec{b}

Step 1: Set Condition for Collinearity
Since \vec{a} and \vec{b} are non-collinear vectors,
for \vec{\alpha} and \vec{\beta} to be collinear, the ratio of the coefficients of \vec{a} must equal the ratio of the coefficients of \vec{b}.

\frac{\text{Coeff of } \vec{a} \text{ in } \vec{\alpha}}{\text{Coeff of } \vec{a} \text{ in } \vec{\beta}} = \frac{\text{Coeff of } \vec{b} \text{ in } \vec{\alpha}}{\text{Coeff of } \vec{b} \text{ in } \vec{\beta}}

Step 2: Substitute Coefficients
\frac{x-2}{3+2x} = \frac{1}{-2}

Step 3: Solve for x:
-2(x-2) = 1(3+2x)
-2x + 4 = 3 + 2x

Rearrange terms:
4 - 3 = 2x + 2x
1 = 4x
x = \frac{1}{4}

Final Answer: \boxed{x = \frac{1}{4}}

Q16: Solve the following linear programming problem Graphically:
Maximise Z = x + 2y

Subject to constraints:
x - y \ge 0
x + 2y \le 6
x \ge 0, y \ge 0

Solution:
Step 1: Convert Inequalities to Equations
To determine the feasible region, we first write the associated equations for the boundary lines:
Line 1: x - y = 0 \Rightarrow y = x
This line passes through the origin (0,0) and (2,2).

Constraint Check (x - y \ge 0): Let’s test a point on the x-axis, e.g., (1, 0). 1 - 0 \ge 0 is True. Therefore, the region lies below the line y = x.

Line 2: x + 2y = 6
If x = 0, then 2y = 6 \Rightarrow y = 3. Point: (0, 3).
If y = 0, then x = 6. Point: (6, 0).

Constraint Check (x + 2y \le 6): Test origin (0, 0). 0 + 0 \le 6 is True.
Therefore, the region lies towards the origin.

Non-negativity: x \ge 0, y \ge 0 implies the region is in the First Quadrant.

Step 2: Determine the Intersection Point
We need to find the intersection of the lines y = x and x + 2y = 6.

Substitute x = y into the second equation:
y + 2y = 6
3y = 6 \Rightarrow y = 2
Since x = y, then x = 2.
The intersection point is A(2, 2).

Step 3: Identify the Feasible Region Corner Points
Based on the graph intersections and constraints:
The region is bounded by the x-axis, the line y=x, and the line x+2y=6.
The corner points are:

1. O (0, 0) (Origin)
2. B (6, 0) (x-intercept of Line 2)
3. A (2, 2) (Intersection point)

(Note: The point (0,3) is excluded because it violates the condition x \ge y)

Step 4: Evaluate Z at Corner Points
We substitute the coordinates of the corner points into the objective function Z = x + 2y:

Final Answer: The maximum value of Z is 6.
Interestingly, this maximum value occurs at two corner points, A(2, 2) and B(6, 0).
This implies that Z is maximum at every point on the line segment joining A and B.

Q17: Find shortest distance between the lines given by \vec{r}=(1-\lambda)\hat{i}+(\lambda-2)\hat{j}+(3-2\lambda)\hat{k} and \vec{r}=(\mu+1)\hat{i}+(2\mu-1)\hat{j}-(2\mu+1)\hat{k} are skew lines.

Solution:
Given Lines:
\vec{r} = (1-\lambda)\hat{i} + (\lambda-2)\hat{j} + (3-2\lambda)\hat{k}

Rearranging to vector form \vec{a_1} + \lambda\vec{b_1}: \vec{r} = (\hat{i} - 2\hat{j} + 3\hat{k}) + \lambda(-\hat{i} + \hat{j} - 2\hat{k})

Here, \vec{a_1} = \hat{i} - 2\hat{j} + 3\hat{k} and

\vec{b_1} = -\hat{i} + \hat{j} - 2\hat{k}

\vec{r} = (\mu+1)\hat{i} + (2\mu-1)\hat{j} - (2\mu+1)\hat{k}

Rearranging to vector form \vec{a_2} + \mu\vec{b_2}: \vec{r} = (\hat{i} - \hat{j} - \hat{k}) + \mu(\hat{i} + 2\hat{j} - 2\hat{k})

Here, \vec{a_2} = \hat{i} - \hat{j} - \hat{k} and

\vec{b_2} = \hat{i} + 2\hat{j} - 2\hat{k}

Step 1: Find \vec{a_2} - \vec{a_1}
\vec{a_2} - \vec{a_1} = (\hat{i} - \hat{j} - \hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k})

\vec{a_2} - \vec{a_1} = (1-1)\hat{i} + (-1+2)\hat{j} + (-1-3)\hat{k}

\vec{a_2} - \vec{a_1} = 0\hat{i} + 1\hat{j} - 4\hat{k}

Step 2: Find Cross Product \vec{b_1} \times \vec{b_2}
\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix}

\vec{b_1} \times \vec{b_2} = \hat{i}(-2 - (-4)) - \hat{j}(2 - (-2)) + \hat{k}(-2 - 1)

\vec{b_1} \times \vec{b_2} = 2\hat{i} - 4\hat{j} - 3\hat{k}

Step 3: Magnitude of Cross Product |\vec{b_1} \times \vec{b_2}|
|\vec{b_1} \times \vec{b_2}| = \sqrt{2^2 + (-4)^2 + (-3)^2}

|\vec{b_1} \times \vec{b_2}| = \sqrt{4 + 16 + 9} = \sqrt{29}

Step 4: Calculate Shortest Distance (SD)
SD = \left| \frac{(\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1})}{ |\vec{b_1} \times \vec{b_2}|} \right|

SD = \left| \frac{(2\hat{i} - 4\hat{j} - 3\hat{k}) \cdot (0\hat{i} + \hat{j} - 4\hat{k})}{\sqrt{29}} \right| = \left| \frac{0 - 4 + 12}{\sqrt{29}} \right| = \frac{8}{\sqrt{29}} units.

Answer: \boxed{ \text{The shortest distance is } \frac{8}{\sqrt{29}} \text{units}}.

Q18 (a): The probability distribution for the number of students being absent in a class on a Saturday is as follows:

(i) Calculate p.
(ii) Calculate the mean of the number of absent students on Saturday.

Solution:
(i) Calculate p:
The sum of all probabilities in a probability distribution must equal 1.
\Sigma P(X) = 1

Substitute the given values:
p + 2p + 3p + p = 1
7p = 1
p = \frac{1}{7}

(ii) Calculate the Mean (\mu):
The mean (expected value) is calculated using the formula:
\text{Mean} = \Sigma [X \cdot P(X)]

Construct the table of products:

• For X=0: 0 \times p = 0
• For X=2: 2 \times 2p = 4p
• For X=4: 4 \times 3p = 12p
• For X=5: 5 \times p = 5p

Sum the products:
\text{Mean} = 0 + 4p + 12p + 5p
\text{Mean}= 21p

Substitute the value of p = \frac{1}{7}:
\text{Mean} = 21 \times \frac{1}{7}
\text{Mean} = 3

Final Answer: (i) \boxed{ p = \frac{1}{7}} (ii) \boxed{ Mean = 3}

Q18 (b): For the vacancy advertised in the newspaper, 3000 candidates submitted their applications. From the data it was revealed that two third of the total applicants were females and other were males. The performance of the applicants indicates that the probability of a male getting a distinction in written test is 0.4 and that a female getting a distinction is 0.35. Find the probability that the candidate chosen at random will have a distinction in the written test.

Solution:
Step 1: Define Events

• Let E_1 be the event that the candidate is Male.
• Let E_2 be the event that the candidate is Female.
• Let A be the event that the candidate gets a Distinction.

Step 2: Assign Probabilities
Given:

• P(E_2) = \frac{2}{3} (Females)
• P(E_1) = 1 - \frac{2}{3} = \frac{1}{3} (Males)
• P(A|E_1) = 0.4 (Prob. of distinction given Male)
• P(A|E_2) = 0.35 (Prob. of distinction given Female)

Step 3: Apply Law of Total Probability
P(A) = P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)

Substitute the values:
P(A) = \left( \frac{1}{3} \cdot 0.4 \right) + \left( \frac{2}{3} \cdot 0.35 \right)

P(A) = \frac{2}{15} + \frac{14}{60}

P(A) = \frac{22}{60}

P(A) = \frac{11}{30}

Final Answer: The probability is \boxed{ \frac{11}{30}}.

Q19: Let R be a relation defined over N, where N is set of natural numbers, defined as “mRn if and only if m is a multiple of n, m, n \in N.” Find whether R is reflexive, symmetric and transitive or not.

Solution:
We check each property for the relation R = {(m, n) : m = kn, k \in N}.

1. Reflexive: A relation is reflexive if mRm for all m \in N.

• Check: Is m a multiple of m?
• Yes, because m = 1 \times m.
• Conclusion: R is Reflexive.

2. Symmetric: A relation is symmetric if mRn \Rightarrow nRm.

• Let m = 4 and n = 2.
• 4R2 is true because 4 is a multiple of 2 (4 = 2 \times 2).
• Check 2R4: Is 2 a multiple of 4? No, 2 is a factor of 4, not a multiple.
• Conclusion: R is Not Symmetric.

3. Transitive: A relation is transitive if mRn and nRp \Rightarrow mRp.

• Given mRn \Rightarrow m = k_1 n (where k_1 \in N).
• Given nRp \Rightarrow n = k_2 p (where k_2 \in N).
• Substitute n into the first equation: m = k_1 (k_2 p)
• m = (k_1 k_2) p
• Since k_1 and k_2 are natural numbers, their product K = k_1 k_2 is also a natural number.
• Thus, m = Kp, which implies m is a multiple of p.
• Conclusion: R is Transitive.

Final Answer: The relation R is Reflexive and Transitive but not Symmetric.

Q20(a): Find the image A' of the point A(1,6,3) in the line \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}. Also, find the equation of the line joining A and A'.

Solution:
Step 1: Coordinates of General Point on the Line
Let the given line be L. \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} = \lambda (say)

Any general point M on the line can be written as:
M(\lambda, 2\lambda+1, 3\lambda+2)

Step 2: Find the Foot of the Perpendicular (M)
Let M be the foot of the perpendicular drawn from point A(1,6,3) to the line.
The direction ratios of the line AM are:
x_2 - x_1 = \lambda - 1
y_2 - y_1 = (2\lambda + 1) - 6 = 2\lambda - 5
z_2 - z_1 = (3\lambda + 2) - 3 = 3\lambda - 1

Direction ratios of AM: (\lambda-1, 2\lambda-5, 3\lambda-1)

Since AM is perpendicular to the given line (whose direction ratios are 1, 2, 3), their dot product is zero:
a_1a_2 + b_1b_2 + c_1c_2 = 0
1(\lambda - 1) + 2(2\lambda - 5) + 3(3\lambda - 1) = 0
\lambda - 1 + 4\lambda - 10 + 9\lambda - 3 = 0
14\lambda - 14 = 0
\lambda = 1

Substitute \lambda = 1 into the coordinates of M:
M(1, 2(1)+1, 3(1)+2) = M(1, 3, 5)
So, the foot of the perpendicular is (1, 3, 5).

Step 3: Find the Image A’
Let the image of A be A'(x', y', z').
Since M is the midpoint of AA':
\frac{1 + x'}{2} = 1 \Rightarrow 1 + x' = 2 \Rightarrow x' = 1
\frac{6 + y'}{2} = 3 \Rightarrow 6 + y' = 6 \Rightarrow y' = 0
\frac{3 + z'}{2} = 5 \Rightarrow 3 + z' = 10 \Rightarrow z' = 7

Image A' is (1, 0, 7).

Step 4: Equation of Line Joining A and A’
The line passes through A(1, 6, 3) and A'(1, 0, 7).
Direction ratios: 1-1, 0-6, 7-3 \Rightarrow 0, -6, 4
Simplifying direction ratios (dividing by -2): 0, 3, -2

Equation of the line:
\frac{x-1}{0} = \frac{y-6}{3} = \frac{z-3}{-2}

Final Answer: \boxed{ Image: A'(1, 0, 7)}

\boxed{ \text{Equation of AA': } \frac{x-1}{0} = \frac{y-6}{3} = \frac{z-3}{-2} }

Q20(b): Find a point P on the line \frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9} such that its distance from point Q(2,4,-1) is 7 units. Also, find the equation of the line joining P and Q.

Solution:
Step 1: Coordinates of General Point P
Let the given line be: \frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9} = \lambda

Any point P on this line can be written in terms of \lambda:
x = \lambda - 5
y = 4\lambda - 3
z = -9\lambda + 6

So, P(\lambda - 5, 4\lambda - 3, -9\lambda + 6).

Step 2: Apply the Distance Formula
We are given that the distance PQ = 7 units.
Using the distance formula PQ^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2
PQ^2 = (\lambda - 5 - 2)^2 + (4\lambda - 3 - 4)^2 + (-9\lambda + 6 + 1)^2
PQ^2 = (\lambda - 7)^2 + (4\lambda - 7)^2 + (-9\lambda + 7)^2 = 49

Notice that (-9\lambda + 7)^2 is the same as (9\lambda - 7)^2.
Expanding the terms:
(\lambda^2 - 14\lambda + 49) + (16\lambda^2 - 56\lambda + 49 ) + (81\lambda^2 -126\lambda + 49) = 49
(1+ 16 + 81)\lambda^2 + (-14 - 56 -126)\lambda + (49 + 49 + 49) = 49
98\lambda^2 - 196\lambda + 147 = 49
98\lambda^2 - 196\lambda + 98 = 0

Divide the entire equation by 98:
\lambda^2 - 2\lambda + 1 = 0
(\lambda - 1)^2 = 0 \Rightarrow \lambda = 1

Step 3: Find Coordinates of P
Substitute \lambda = 1 into the coordinates of P:
P(1 - 5, 4(1) - 3 , -9(1) + 6)
P(-4, 1, -3)

Step 4: Equation of Line PQ
Now we find the equation of the line passing through P(-4, 1, -3) and Q(2, 4, -1).
Direction ratios (DRs) of PQ:
2 - (-4) = 6
4 - 1 = 3
-1 - (-3) = 2

DRs are 6, 3, 2.

The equation of the line is:
\frac{x-2}{6} = \frac{y-4}{3} = \frac{z+1}{2}

Final Answer: \boxed{\text{Point } P(-4, 1, -3)}

\boxed{\text{Equation of the line joining PQ: } \frac{x - 2}{6} = \frac{y - 4}{3} = \frac{z + 1}{2}}.

Q21(A): A school wants to allocate students into three clubs Sports, Music and Drama. The number of students in Sports club should be equal to the sum of the number of students in Music and Drama club. The number of students in Music club should be 20 more than half the number of students in Sports club. The total number of students to be allocated in all three clubs are 180. Find the number of students allocated to different clubs, using matrix method.

Solution:
Step 1: Formulation of Linear Equations
Let the number of students in the Sports, Music, and Drama clubs be denoted by x, y, and z respectively.

According to the first condition: x = y + z \Rightarrow x - y - z = 0 …(i)
According to the second condition: y = \frac{x}{2} + 20 \Rightarrow 2y = x + 40 \Rightarrow -x + 2y + 0z = 40 …(ii)
According to the third condition, : x + y + z = 180 …(iii)

Step 2: Matrix Form
The system of linear equations can be written in the form AX = B, where:
A = \begin{bmatrix} 1 & -1 & -1 \\ -1 & 2 & 0 \\ 1 & 1 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 0 \\ 40 \\ 180 \end{bmatrix}

Step 3: Determinant of Matrix A
First, we must check if the matrix is non-singular (i.e., |A| \neq 0) to ensure a unique solution exists.
|A| = 1(2 - 0) - (-1)(-1 - 0) + (-1)(-1 - 2)
|A| = 1(2) + (1)(-1) -1(-3)
|A| = 2 - 1 + 3 = 4

Since |A| \neq 0, the system has a unique solution given by X = A^{-1}B.

Step 4: Finding the Adjoint of A
We calculate the cofactors for each element A_{ij}:

• A_{11} = + (2 - 0) = 2
• A_{12} = - (-1 - 0) = 1
• A_{13} = + (-1 - 2) = -3
• A_{21} = - (-1 + 1) = 0
• A_{22} = + (1 + 1) = 2
• A_{23} = - (1 + 1) = -2
• A_{31} = + (0 + 2) = 2
• A_{32} = - (0 - 1) = 1
• A_{33} = + (2 - 1) = 1

The adjoint matrix is the transpose of the cofactor matrix:
adj A = \begin{bmatrix} 2 & 1 & -3 \\ 0 & 2 & -2 \\ 2 & 1 & 1 \end{bmatrix}^T = \begin{bmatrix} 2 & 0 & 2 \\ 1 & 2 & 1 \\ -3 & -2 & 1 \end{bmatrix}

Step 5: Finding the Inverse Matrix
The inverse is given by A^{-1} = \frac{1}{|A|} (adj A):
A^{-1} = \frac{1}{4} \begin{bmatrix} 2 & 0 & 2 \\ 1 & 2 & 1 \\ -3 & -2 & 1 \end{bmatrix}

Step 6: Calculating the Solution
X = A^{-1}B = \frac{1}{4} \begin{bmatrix} 2 & 0 & 2 \\ 1 & 2 & 1 \\ -3 & -2 & 1 \end{bmatrix} \begin{bmatrix} 0 \\ 40 \\ 180 \end{bmatrix}

Multiplying the matrices:
X = \frac{1}{4} \begin{bmatrix} (2)(0) + (0)(40) + (2)(180) \\ (1)(0) + (2)(40) + (1)(180) \\ (-3)(0) + (-2)(40) + (1)(180) \end{bmatrix}

X = \frac{1}{4} \begin{bmatrix} 0 + 0 + 360 \\ 0 + 80 + 180 \\ 0 - 80 + 180 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 360 \\ 260 \\ 100 \end{bmatrix}

X = \begin{bmatrix} 90 \\ 65 \\ 25 \end{bmatrix} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}

Final Answer:

• Number of students in Sports club (x): 90
• Number of students in Music club (y): 65
• Number of students in Drama club (z): 25

Q21(B): Given A=\begin{bmatrix}-4&4&4\\ -7&1&3\\ 5&-3&-1\end{bmatrix} and B=\begin{bmatrix}1&-1&1\\ 1&-2&-2\\ 2&1&3\end{bmatrix}, find AB. Hence solve the system of linear equations:
x-y+z=4
x-2y-2z=9
2x+y+3z=1 +1

Solution:
Step 1: Calculate the Product AB
We multiply matrix A by matrix B:
AB=\begin{bmatrix}-4&4&4\\ -7&1&3\\ 5&-3&-1\end{bmatrix}\begin{bmatrix}1&-1&1\\ 1&-2&-2\\ 2&1&3\end{bmatrix}

Multiplying row by column:
Row 1:
c_{11} = (-4)(1) + (4)(1) + (4)(2) = -4 + 4 + 8 = 8
c_{12} = (-4)(-1) + (4)(-2) + (4)(1) = 4 - 8 + 4 = 0
c_{12} = (-4)(1) + (4)(-2) + (4)(3) = -4 - 8 + 12 = 0

Row 2:
c_{21} = (-7)(1) + (1)(1) + (3)(2) = -7 + 1 + 6 = 0
c_{22} = (-7)(-1) + (1)(-2) + (3)(1) = 7 - 2 + 3 = 8
c_{22} = (-7)(1) + (1)(-2) + (3)(3) = -7 - 2 + 9 = 0

Row 3:
c_{31} = (5)(1) + (-3)(1) + (-1)(2) = 5 - 3 - 2 = 0
c_{32} = (5)(-1) + (-3)(-2) + (-1)(1) = -5 + 6 - 1 = 0
c_{32} = (5)(1) + (-3)(-2) + (-1)(3) = 5 + 6 - 3 = 8

Thus, AB = \begin{bmatrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{bmatrix} = 8I.

Step 2: Relate Matrix B to the System of Equations
The given system of linear equations is:
x - y +z = 4
x - 2y - 2z = 9
2x + y + 3z = 1

This can be written in matrix form CX = D,
where: C = \begin{bmatrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, and D = \begin{bmatrix} 4 \\ 9 \\ 1 \end{bmatrix}.

Notice that the coefficient matrix C is exactly the same as matrix B given in the problem.
So, the system is BX = D.
To find X, we need B^{-1}:
X = B^{-1}D

Step 3: Use the Result from Step 1 to Find Inverse
From Step 1, we know that AB = 8I.
Multiplying both sides by B^{-1}, we get:
A(BB^{-1}) = 8IB^{-1}
AI = 8B^{-1}
A = 8B^{-1} \implies B^{-1} = \frac{1}{8}A

Step 4: Solve for X
Substitute B^{-1} = \frac{1}{8}A into X = B^{-1}D:
X = \frac{1}{8} \begin{bmatrix}-4&4&4\\ -7&1&3\\ 5&-3&-1\end{bmatrix} \begin{bmatrix} 4 \\ 9 \\ 1 \end{bmatrix}

Perform matrix multiplication:
X = \frac{1}{8} \begin{bmatrix} (-4)(4) + (4)(9) + (4)(1) \\ (-7)(4) + (1)(9) + (3)(1) \\ (5)(4) + (-3)(9) + (-1)(1) \end{bmatrix}

X = \frac{1}{8} \begin{bmatrix} -16 + 36 + 4 \\ -28 + 9 + 3 \\ 20 - 27 - 1 \end{bmatrix}

\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{8} \begin{bmatrix} 24 \\ -16 \\ -8 \end{bmatrix} = \begin{bmatrix} 3 \\ -2 \\ -1 \end{bmatrix}

Final Answer: The solution to the system of equations is: \boxed{ x = 3, y = -2, z = -1}

Q22: If A=R-{3} and B=R-{1}, consider the function f:A\rightarrow B defined by f(x)=\frac{x-2}{x-3} for all x\in A. Then show that f is bijective and find f^{-1}(x).

Solution:

Step 1: Check for One-One (Injectivity)
Let x_1, x_2 \in A such that f(x_1) = f(x_2).
\frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3}
\frac{x_1-2}{x_2-3}={x_2-2}{x_1-3}
x_1x_2 - 3x_1 - 2x_2 + 6 = x_1x_2 - 3x_2 - 2x_1 + 6

Subtracting x_1x_2 and 6 from both sides:
- 3x_1 - 2x_2 = - 3x_2 - 2x_1
-3x_1 + 2x_1 = -3x_2 + 2x_2
-x_1 = -x_2 \implies x_1 = x_2

Since f(x_1) = f(x_2) \implies x_1 = x_2, the function f is one-one.

Step 2: Check for Onto (Surjectivity)
Let y \in B be an arbitrary element.
We need to find x \in A such that f(x) = y.
y=\frac{x-2}{x-3}
y(x-3)=x-2
xy - 3y = x - 2
xy - x = 3y - 2
x(y - 1) = 3y - 2
x = \frac{3y - 2}{y - 1}

Since y \in B = R - {1}, y \neq 1, so the denominator is non-zero.
Also, we must verify if x \neq 3 (since domain A = R - {3}).
If x = 3, then 3 = \frac{3y - 2}{y - 1} \implies 3y - 3 = 3y - 2 \implies -3 = -2, which is impossible.

Thus, for every y \in B, there exists a valid pre-image x \in A.
Therefore, f is onto.

Since f is both one-one and onto, it is bijective.

Step 3: Find f^{-1}(x)
From Step 2, we found x in terms of y:
x = \frac{3y - 2}{y - 1}

By definition of inverse function, f^{-1}(y) = x.
Therefore, f^{-1}(y) = \frac{3y - 2}{y - 1}.

Replacing y with x, we get:
Therefore, f^{-1}(x) = \frac{3x - 2}{x - 1}.

Final Answer: The function is bijective. The inverse function is: \boxed{f^{-1}(x) = \frac{3x - 2}{x - 1}}

Q23: Show that the points with position vectors 2\hat{i}, -\hat{i}-4\hat{j} and -\hat{i}+4\hat{j} form an isosceles triangle. Also find the area of the triangle.

Solution:
Step 1: Define Position Vectors
Let the vertices of the triangle be A, B, and C with position vectors:
\vec{OA} = \vec{a} = 2\hat{i}

\vec{OB} = \vec{b} = -\hat{i} - 4\hat{j}

\vec{OC} = \vec{c} = -\hat{i} + 4\hat{j}

Step 2: Calculate Side Lengths
We find the vectors representing the sides:
\vec{AB} = \vec{b} - \vec{a} = (-\hat{i} - 4\hat{j}) - (2\hat{i}) = -3\hat{i} +-4 \hat{j}

\vec{BC} = \vec{c} - \vec{b} = (-\hat{i} + 4\hat{j}) - (-\hat{i} - 4\hat{j}) = 0\hat{i} + 8\hat{j}

\vec{CA} = \vec{a} - \vec{c} = (2\hat{i}) - (-\hat{i} - 4\hat{j}) = 3\hat{i} - 4\hat{j}

Now, calculate magnitudes:
|\vec{AB}| = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5

|\vec{BC}| = \sqrt{0^2 + 8^2} = \sqrt{64} = 8

|\vec{CA}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5

Since |\vec{AB}| = |\vec{CA}| = 5, the triangle has two equal sides. Therefore, the triangle is isosceles.

Step 3: Calculate Area of Triangle
The area of triangle ABC is given by \frac{1}{2} |\vec{AB} \times \vec{AC}|.
First, find \vec{AC} = -\vec{CA} = -3\hat{i} + 4\hat{j}.

\vec{AB} = -3\hat{i} - 4\hat{j} + 0\hat{k}

\vec{AC} = -3\hat{i} + 4\hat{j} + 0\hat{k}

Cross product \vec{AB} \times \vec{AC}:
\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & -4 & 0 \\ -3 & 4 & 0 \end{vmatrix}

\vec{AB} \times \vec{AC} = \hat{i}(0 - 0) - \hat{j}(0 - 0) + \hat{k}((-3)(4) - (-4)(-3))

\vec{AB} \times \vec{AC} = \hat{k}(-12 -12) = -24\hat{k}

Magnitude of cross product:
|\vec{AB} \times \vec{AC}| = |-24\hat{k}| = 24

Area = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} \times 24 = 12 sq units.

Final Answer: The triangle is isosceles and its area is: \boxed{12 \text{ sq units}}

Question 24 (Relations & Functions Case Study)

Context: A teacher writes five relations on the set A = {1, 2, 3}:

• R_1 = {(2,3), (3,2)}
• R_2 = {(1,2), (1,3), (3,2)}
• R_3 = {(1,2), (2,1), (1,1)}
• R_4 = {(1,1), (1,2), (3,3), (2,2)}
• R_5 = {(1,1), (1,2), (3,3), (2,2), (2,1), (2,3), (3,2)}

Solutions:

(i) Identify the relation which is reflexive, transitive but not symmetric.

• Reflexive: Must contain (1,1), (2,2), (3,3). Candidates are R_4 and R_5.
• Transitive Check:
  • R_4: Contains (1,2) and (2, 2) and (1,2) . The pairs (1,2), (2,2) and (1,2) satisfies the condition. It is transitive.
  • R_5: Contains (1,2) and (2,3), but does not contain (1,3). Thus, R_5 is not transitive.
• Symmetric Check:
  • R_4: Contains (1,2) but not (2,1). Thus, it is not symmetric.
• Answer: R_4

(ii) Identify the relation which is reflexive and symmetric but not transitive.

• Reflexive: Candidates are R_4 and R_5.
• Symmetric Check:
  • R_4: Not symmetric (missing (2,1)).
  • R_5: Contains (1,2)\leftrightarrow (2,1) and (2,3) \leftrightarrow (3,2). It is symmetric.
• Transitive Check:
  • R_5: We have (1,2) and (2,3) , but the pair (1,3) is missing. Therefore, it is not transitive.
• Answer: R_5

(iii) (a) Identify the relations which are symmetric but neither reflexive nor transitive.

• Symmetric Candidates: R_1 (has (2,3) and (3,2)),
  R_3 (has (1,2) and (2,1)), and
  R_5 (has (1,2) and (2,1) and has (2,3) and (3,2))
• Reflexive Check: R_1 and R_3 are not reflexive (missing diagonal pairs).
• Transitive Check:
  • R_1: Has (2,3) and (3,2), but missing (2,2). Not transitive.
  • R_3: Has (2,1) and (1,2), but missing (2,2). Not transitive.
• Answer: R_1 and R_3

(OR) (iii) What pairs should be added to the relation R_2 to make it an equivalence relation?

• Current R_2: {(1,2), (1,3), (3,2)}.
• Reflexivity: We must add (1,1), (2,2), (3,3).
• Symmetry:
  • For (1,2) , add (2,1).
  • For (1,3) , add (3,1) .
  • For (3,2) , add (2,3).
• Transitivity:
  With the pairs added above, 1 connects to 2 and 3, and 2 connects to 3. All elements communicate with each other.
  The smallest equivalence relation containing these pairs on the set {1,2,3} is the Universal Relation.
• Answer: We must add the pairs: \boxed{{(1,1), (2,2), (3,3), (2,1), (3,1), (2,3)}}.

Solution: Question 25 (Probability Case Study)

Context: A bank offers three types of loans with the following probabilities of customers choosing them and subsequent default rates:

• Fixed Rate (F): P(F) = 10% = 0.10. Default probability P(D|F) = 5% = 0.05.
• Floating Rate (L): P(L) = 20% = 0.20. Default probability P(D|L) = 3% = 0.03.
• Variable Rate (V): P(V) = 70% = 0.70. Default probability P(D|V) = 1% = 0.01.

(i) What is the probability that a customer after availing the loan will default on the loan repayment?

Solution:
We use the Law of Total Probability:
P(D) = P(F) \cdot P(D|F) + P(L) \cdot P(D|L) + P(V) \cdot P(D|V)

Substitute the values:
P(D) = (0.1 \times 0.05) + (0.2 \times 0.03) + (0.7 \times 0.01)

P(D) = 0.005 + 0.006 + 0.007

P(D) = 0.018

Answer: The probability of default is 0.018 (or 1.8%).

(ii) A customer after availing the loan, defaults on loan repayment. What is the probability that he availed the loan at a variable rate of interest?

Solution:
We use Bayes’ Theorem to find P(V|D):
P(V|D) = \frac{P(V) \cdot P(D|V)}{P(D)}

Using the result from part (i) for P(D):
P(V|D) = \frac{0.70 \times 0.01}{0.018}

P(V|D) = \frac{0.007}{0.018}

P(V|D) = \frac{7}{18}

Answer: \boxed{ \text{The probability is } \frac{7}{18}}.