Class 12 Maths Relations and Functions, Inverse Trigonometric Functions and Matrices and Determinants Test and Solutions – 2027

Have you solved Class 12 Maths Relations and Functions, Inverse Trigonometric Functions and Matrices and Determinants Test and are looking for a solutions to the questions? Here you’ll find complete, step-by-step NCERT-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.

Below are the detailed, step-by-step solutions to all 30 marks of the test. Grab your answer sheets, and let’s review!

Section A

Q1. the relation R=\{(1,1),(2,2),(3,3)\} on the set \{1, 2, 3\} is

(a) Symmetric only
(b) Reflexive only
(c) An Equivalence Relation
(d) Transitive only

Solution:
1. For Reflexivity
A relation R on a set A is reflexive if every element in the set is related to itself.
Mathematically, this means (a, a) \in R for every a \in A.

For our set A = \{1, 2, 3\}, the required pairs are (1, 1), (2, 2), and (3, 3).
Looking at the given relation R, all three of these exact pairs are present.

Therefore, the relation is reflexive.

2. Symmetry
A relation R is symmetric if, whenever (a, b) \in R, it is also true that (b, a) \in R for all a, b \in A.

In our relation R, the only elements are identical pairs: (1, 1), (2, 2), and (3, 3).
If we take (1, 1) \in R and reverse the order, we still get (1, 1) \in R.
This holds true for all elements in the set.

Therefore, the relation is symmetric.

3. Transitivity
A relation R is transitive if, whenever (a, b) \in R and (b, c) \in R, it logically follows that (a, c) \in R for all a, b, c \in A.

To test this, we look for a chain where the second element of one pair matches the first element of another pair.
In R = \{(1,1), (2,2), (3,3)\}, there are no distinct pairs like (a, b) and (b, c) where a \neq b.
Because the premise (finding two linked, distinct pairs) never occurs, the condition for transitivity is never violated. In mathematics, this makes the property vacuously true.

Therefore, the relation is transitive.

Because the relation R satisfies all three conditions—it is reflexive, symmetric, and transitive—it is classified as an equivalence relation.
This perfectly maps to option (c) “An Equivalence Relation”

Final Answer: (c)

Q2. If 4\cos^{-1}x+\sin^{-1}x=\pi then the value of x is
(a) \frac{3}{2}

(b) \frac{1}{\sqrt{2}}

(c) \frac{\sqrt{3}}{2}

(d) \frac{2}{\sqrt{3}}

Solution:
We use the standard inverse trigonometric identity: \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}.

Rewrite the given equation by splitting the first term to utilize the identity:
3\cos^{-1}x + (\cos^{-1}x + \sin^{-1}x) = \pi

Substitute the identity from step 1 into the equation:
3\cos^{-1}x + \frac{\pi}{2} = \pi

Isolate \cos^{-1}x:
3\cos^{-1}x = \pi - \frac{\pi}{2}

3\cos^{-1}x = \frac{\pi}{2}

\cos^{-1}x = \frac{\pi}{6}

Solve for x:
x = \cos(\frac{\pi}{6})

x = \frac{\sqrt{3}}{2}

Final Answer: (c).

Q3. The function f:R\rightarrow R f(x)=x^{2} is

(a) injective but not surjective
(b) surjective but not injective
(c) neither injective nor surjective
(d) injective as well as surjective Solution:

Solution:
Check for Injectivity (One-One):
A function is injective if f(x_1) = f(x_2) strictly implies x_1 = x_2.

Let’s test with real numbers:
f(1) = 1^{2} = 1,
and
f(-1) = (-1)^{2} = 1.

Since f(1) = f(-1) but 1 \neq -1, the function assigns the same output to different inputs.
Therefore, the function is not injective.

Check for Surjectivity (Onto):
A function is surjective if every element in the codomain has a pre-image in the domain.
The codomain is all real numbers (R).
However, for any real number x, x^{2} \geq 0.
This means negative real numbers (like -2) are in the codomain but do not have any real pre-image in the domain.
Therefore, the function is not surjective.

Conclusion: The function is neither injective nor surjective

Final Answer: (c)

Q4. The value of \sin^{-1}(\cos\frac{33\pi}{5}) is

(a) \frac{3\pi}{5}

(b) \frac{-\pi}{10}

(c) \frac{\pi}{10}

(d) \frac{7\pi}{5}

Solution:
First, simplify the angle \frac{33\pi}{5} to find its equivalent acute angle:

\frac{33\pi}{5} = \frac{30\pi + 3\pi}{5} = 6\pi + \frac{3\pi}{5}

Because the cosine function has a period of 2\pi,

\cos(6\pi + \theta) = \cos\theta.

So, \cos(\frac{33\pi}{5}) = \cos(\frac{3\pi}{5}).

Next, convert the cosine function to a sine function using the co-function identity
\cos\theta = \sin(\frac{\pi}{2} - \theta):

\cos(\frac{3\pi}{5}) = \sin(\frac{\pi}{2} - \frac{3\pi}{5})

= \sin(\frac{5\pi - 6\pi}{10}) = \sin(\frac{-\pi}{10})

Substitute this back into the original inverse expression:
\sin^{-1}(\cos\frac{33\pi}{5}) = \sin^{-1}(\sin(\frac{-\pi}{10}))

The principal value branch for \sin^{-1}x is [-\frac{\pi}{2}, \frac{\pi}{2}].

Since \frac{-\pi}{10} lies within this valid range, the inverse sine and sine functions cancel out:
\sin^{-1}(\sin(\frac{-\pi}{10})) = \frac{-\pi}{10}

Final Answer: (b)

Q5. If A is 3\times4 matrix and B is a matrix such that A^{T}B and BA^{T} are both defined. Then, B is of the type

(a) 3\times4
(b) 3\times3
(c) 4\times4
(d) 4\times3

Solution:
Matrix A has an order of 3\times4.
Therefore, its transpose, A^{T}, has an order of 4\times3.

For matrix multiplication A^{T}B to be defined, the number of columns in A^{T} must equal the number of rows in B.
Since A^{T} has 3 columns, B must have 3 rows.
Let the order of B be 3\times n.

For matrix multiplication BA^{T} to be defined, the number of columns in B must equal the number of rows in A^{T}.
Since A^{T} has 4 rows, B must have 4 columns.
Thus, n = 4.
Combining both conditions, matrix B must be of the order 3\times4.

Final Answer: (a)

Q6. If A is a square matrix such that A^{2}=I, then (A-I)^{3}+(A+I)^{3}-7A is equal to

(a) A
(b) I-A
(c) I+A
(d) 3A

Solution:
Use the standard algebraic expansion formulas for cubes:
(A-I)^{3} = A^{3} - 3A^{2}I + 3AI^{2} - I^{3}

(A+I) ^{3} = A^{3} + 3A^{2}I + 3AI^{2} + I^{3}

Note that multiplying any matrix by the identity matrix I yields the matrix itself, and I^{n} = I. Thus:
(A-I)^{3} = A^{3} - 3A^{2} + 3A - I

(A+I)^{3} = A^{3} + 3A^{2} + 3A + I

Add these two expansions together and subtract 7A:
(A^{3} - 3A^{2} + 3A - I) + (A^{3} + 3A^{2} + 3A + I) - 7A = 2A^{3} + 6A - 7A = 2A^{3} - A

We are given A^{2} = I.
Multiply both sides by A to get A^{3} = A \cdot I = A.

Substitute A^{3} = A back into the simplified expression:
2(A) - A = A

Final Answer: (a)

Q7. If A=\begin{bmatrix}1&2&-1\ -1&1&2\ 2&-1&1\end{bmatrix}, then the |adj(adj~A)| is

(a) 14^{4}
(b) 14^{3}
(c) 14^{2}
(d) 14

Solution:
First, compute the determinant of the 3\times3 matrix A, denoted as |A|:
|A| = 1(1(1) - (-1)(2)) - 2((-1)(1) - (2)(2)) - 1((-1)(-1) - (1)(2))
|A| = 1(1 + 2) - 2(-1 - 4) - 1(1 - 2)
|A| = 1(3) - 2(-5) - 1(-1)
|A| = 3 + 10 + 1 = 14

We use the property of determinants for adjoints:
For an n\times n matrix, |adj(adj~A)| = |A|^{(n-1)^{2}}.

Since matrix A is a 3\times3 matrix, n = 3.
|adj(adj~A)| = |A|^{(3-1)^{2}} = |A|^{2^{2}} = |A|^{4}

Substitute the determinant value calculated in step 1:
|adj(adj~A)| = 14^{4}

Therefore,
Final Answer: (a).

Section B: 2 marks questions

Q8. Let R be the equivalence relation in the set A={0,1,2,3,4,5} given by R={(a, b): 2 \text{ divides } (a-b)}. Write the equivalence class [0].

Solution:
The equivalence class [0] contains all elements x \in A such that (x, 0) \in R.

Based on the relation’s definition, (x, 0) \in R means that 2 divides (x - 0), which simply means 2 divides x.

Therefore, the equivalence class [0] consists of all even numbers within the set A.

Scanning the set A={0,1,2,3,4,5}, the even numbers are 0, 2, and 4.
Equivalence class [0] = \{0, 2, 4\}.

Final Answer: \boxed{ \{0, 2, 4\} }

Q9. Draw the graph of \cos^{-1}x and mark all the critical points on the graph. State its domain and (principal value) range. Solution:

1. Domain and Range:The domain (valid x-values) of the \cos^{-1}x function is [-1, 1].The principal value range (valid y-values) of the \cos^{-1}x function is [0, \pi].
2. Critical Points:When x = -1, y = \cos^{-1}(-1) = \pi. Point: (-1, \pi).When x = 0, y = \cos^{-1}(0) = \frac{\pi}{2}. Point: (0, \frac{\pi}{2}).When x = 1, y = \cos^{-1}(1) = 0. Point: (1, 0).
3. Graph Sketching (Description for Plotting):Draw a coordinate plane. Plot the three critical points identified above. Draw a smooth, strictly decreasing curve starting at (-1, \pi) in the upper-left, crossing the y-axis at (0, \frac{\pi}{2}), and terminating at (1, 0) on the positive x-axis.

Q10. Evaluate \sin(2\tan^{-1}\frac{2}{3})+\cos(\tan^{-1}\sqrt{3})

Solution:
Evaluate the second term, \cos(\tan^{-1}\sqrt{3}):
We know \tan(\frac{\pi}{3}) = \sqrt{3}, so \tan^{-1}\sqrt{3} = \frac{\pi}{3}.

Therefore, \cos(\frac{\pi}{3}) = \frac{1}{2}.

Evaluate the first term, \sin(2\tan^{-1}\frac{2}{3}):

Let \theta = \tan^{-1}\frac{2}{3}.

This implies \tan\theta = \frac{2}{3}.

The expression becomes \sin(2\theta).

We use the double-angle identity: \sin(2\theta) = \frac{2\tan\theta}{1+\tan^{2}\theta}.

Substitute \tan\theta = \frac{2}{3} into the formula:\sin(2\theta) = \frac{2(\frac{2}{3})}{1+(\frac{2}{3})^{2}} = \frac{\frac{4}{3}}{1+\frac{4}{9}} = \frac{\frac{4}{3}}{\frac{13}{9}}

\sin(2\theta) = \frac{4}{3} \times \frac{9}{13} = \frac{12}{13}.

Add the two evaluated terms together:\frac{12}{13} + \frac{1}{2} = \frac{24 + 13}{26} = \frac{37}{26}

Final Answer: \boxed{ \frac{37}{26} }

Q11. Check whether the relation R defined on the set A=\{1,2,3,4,5,6\} as R=\{(a,b): b=a+1\} is reflexive, symmetric or transitive

Solution:
Explicitly list the elements of relation R based on the rule b = a+1:
R = \{(1,2), (2,3), (3,4), (4,5), (5,6)\}

Reflexivity:
For R to be reflexive, (a, a)\in R for all a \in A.
Notice that (1, 1) \notin R.
Therefore, R is not reflexive.

Symmetry:
For R to be symmetric, if (a, b) \in R, then (b, a) must be in R.
Notice that (1, 2) \in R, but its reverse (2, 1) \notin R.
Therefore, R is not symmetric.

Transitivity:
For R to be transitive, if (a, b) \in R and (b, c) \in R, then (a, c) must be in R.
Notice that (1, 2) \in R and (2, 3)\in R.
However, (1, 3) \notin R.
Therefore, R is not transitive.

Final Answer: Not Reflexive, not symmetric and not transitive

Q12. Prove that: \tan(\sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5})=\frac{63}{16}

Solution:
Convert \sin^{-1}\frac{5}{13} to a \tan^{-1} function:

Let \alpha = \sin^{-1}\frac{5}{13}, so \sin\alpha = \frac{5}{13}.

Using a right triangle (Opposite = 5, Hypotenuse = 13), Adjacent = \sqrt{13^{2}-5^{2}} = \sqrt{144} = 12.

Thus, \tan\alpha = \frac{Opposite}{Adjacent} = \frac{5}{12}.

Convert \cos^{-1}\frac{3}{5} to a \tan^{-1} function:

Let \beta = \cos^{-1}\frac{3}{5}, so \cos\beta = \frac{3}{5}.

Using a right triangle (Adjacent = 3, Hypotenuse = 5), Opposite = \sqrt{5^{2}-3^{2}} = \sqrt{16} = 4.

Thus, \tan\beta = \frac{Opposite}{Adjacent} = \frac{4}{3}.

The original expression becomes \tan(\alpha + \beta).

Use the tangent addition identity:\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha \cdot \tan\beta}

Substitute the tangent values found:
\tan(\alpha + \beta) = \frac{\frac{5}{12} + \frac{4}{3}}{1 - (\frac{5}{12})(\frac{4}{3})}

= \frac{\frac{5+16}{12}}{1 - \frac{20}{36}} = \frac{\frac{21}{12}}{\frac{36-20}{36}} = \frac{\frac{21}{12}}{\frac{16}{36}}

= \frac{21}{12} \times \frac{36}{16} = \frac{21}{12} \times \frac{9}{4} = \frac{7}{4} \times \frac{9}{4} = \frac{63}{16}

Hence proved.

Q13. If X-Y=\begin{bmatrix}1&1&1 \\ 1&1&0 \\ 1&0&0 \end{bmatrix} and X+Y=\begin{bmatrix}3&5&1 \\ -1&1&4 \\ 11&8&0\end{bmatrix}, find X and Y.

Solution:
We can treat these as simultaneous matrix equations.

To find matrix X, add the two equations together:
(X+Y) + (X-Y) = 2X

2X = \begin{bmatrix}3&5&1\\ -1&1&4\\ 11&8&0\end{bmatrix} + \begin{bmatrix}1&1&1\\ 1&1&0\\ 1&0&0\end{bmatrix}

2X = \begin{bmatrix}3+1 & 5+1 & 1+1 \\ -1+1 & 1+1 & 4+0 \\ 11+1 & 8+0 & 0+0\end{bmatrix} = \begin{bmatrix}4&6&2 \\ 0&2&4 \\ 12&8&0\end{bmatrix}

X = \frac{1}{2}\begin{bmatrix}4&6&2 \\ 0&2&4 \\ 12&8&0\end{bmatrix} = \begin{bmatrix}2&3&1 \\ 0&1&2 \\ 6&4&0\end{bmatrix}

To find matrix Y, subtract the first equation from the second:
(X+Y) - (X-Y) = 2Y

2Y = \begin{bmatrix}3&5&1 \\ -1&1&4 \\ 11&8&0\end{bmatrix} - \begin{bmatrix}1&1&1 \\ 1&1&0 \\ 1&0&0\end{bmatrix}

2Y = \begin{bmatrix}3-1 & 5-1 & 1-1 \\ -1-1 & 1-1 & 4-0 \\ 11-1 & 8-0 & 0-0\end{bmatrix} = \begin{bmatrix}2&4&0 \\ -2&0&4 \\ 10&8&0 \end{bmatrix}

Y = \frac{1}{2}\begin{bmatrix}2&4&0 \\ -2&0&4 \\ 10&8&0\end{bmatrix} = \begin{bmatrix}1&2&0 \\ -1&0&2 \\ 5&4&0\end{bmatrix}

Final Answer: \boxed{ X = \begin{bmatrix}2&3&1 \\ 0&1&2 \\ 6&4&0\end{bmatrix} , \ Y = \begin{bmatrix}1&2&0 \\ -1&0&2 \\ 5&4&0\end{bmatrix} }

Q14. Find values of k, if area of triangles is 4 square units whose vertices are (-2, 0), (0, 4) and (0, k).

Solution:
The area of a triangle with vertices (x_{1}, y_{1}), (x_{2}, y_{2}), and (x_{3}, y_{3}) is given by the determinant formula:
\text{Area} = \frac{1}{2} |x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})|

Plug in the given vertices (-2, 0), (0, 4), and (0, k) and set the Area to 4:
4 = \frac{1}{2} |-2(4 - k) + 0(k - 0) + 0(0 - 4)|

Simplify the expression inside the absolute value:
4 = \frac{1}{2} |-8 + 2k|

Multiply both sides by 2:
8 = |-8 + 2k|

Remove the absolute value, which yields two possible conditions:
Condition 1: -8 + 2k = 82k = 16 \implies k = 8

Condition 2: -8 + 2k = -82k = 0 \implies k = 0

Final Answer: \boxed{k = 0 \ \text{or} \ 8}

Q15. If \tan^{-1} \left\{ \frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}} {\sqrt{1+x^{2}}+\sqrt{1-x^{2}}} \right\}=A then prove that x^{2}=\sin~2A.

Solution:
Let us make a trigonometric substitution to simplify the square roots.
Let x^{2} = \cos~2\theta.

This implies 2\theta = \cos^{-1}(x^{2}), or \theta = \frac{1}{2}\cos^{-1}(x^{2}).

Substitute x^{2} = \cos~2\theta into the given expression:
\sqrt{1+x^{2}} = \sqrt{1+\cos~2\theta} = \sqrt{2\cos^{2}\theta} = \sqrt{2}\cos\theta

\sqrt{1-x^{2}} = \sqrt{1-\cos~2\theta} = \sqrt{2\sin^{2}\theta} = \sqrt{2}\sin\theta

Reconstruct the fraction inside the inverse tangent:
\frac{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta} = \frac{\sqrt{2}(\cos\theta - \sin\theta)}{\sqrt{2}(\cos\theta + \sin\theta)} = \frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}

Divide the numerator and the denominator by \cos\theta to turn the expression into tangents:
\frac{1 - \tan\theta}{1 + \tan\theta}

Using the tangent subtraction formula, this is equal to \tan(\frac{\pi}{4} - \theta).

Substitute this back into the original equation:
\tan^{-1}(\tan(\frac{\pi}{4} - \theta)) = A

\frac{\pi}{4} - \theta = A

Isolate \theta:
\theta = \frac{\pi}{4} - A

Recall our initial substitution:
x^{2} = \cos~2\theta.

Substitute the isolated \theta into this:
x^{2} = \cos\left(2(\frac{\pi}{4} - A)\right)

x^{2} = \cos(\frac{\pi}{2} - 2A)

Using the complementary angle identity
\cos(\frac{\pi}{2} - x) = \sin~x

x^{2} = \sin~2A
Hence proved.

Q16. If A=\begin{bmatrix}2&3&1 \\ 1&2&2 \\ -3&1&-1\end{bmatrix}, find A^{-1}. Using A^{-1} solve the system of linear equations
2x+y-3z=13
3x+2y+z=4
x+2y-z=8

Solution:
Find Determinant |A|:
|A| = 2(-2 - 2) - 3(-1 - (-6)) + 1(1 - (-6))

|A| = 2(-4) - 3(5) + 1(7) = -8 - 15 + 7 = -16.

Since |A| \neq 0, A^{-1} exists.

Find Adjoint of A:
Calculate cofactors of all elements:
C_{11} = -4, C_{12} = -5, C_{13} = 7

C_{21} = 4, C_{22} = 1, C_{23} = -11

C_{31} = 4, C_{32} = -3, C_{33} = 1

Construct the cofactor matrix and transpose it to get adj~A:
adj~A = \begin{bmatrix}-4&4&4 \\ -5&1&-3 \\ 7&-11&1\end{bmatrix}

Calculate A^{-1}:
A^{-1} = \frac{1}{|A|} adj~A = -\frac{1}{16} \begin{bmatrix}-4&4&4 \\ -5&1&-3 \\ 7&-11&1\end{bmatrix}

Solve the linear system:
Write the system as MX = B, where M is the coefficient matrix.

M = \begin{bmatrix}2&1&-3 \\ 3&2&1 \\ 1&2&-1\end{bmatrix}

Notice that matrix M is exactly the transpose of matrix A (M = A^{T}).
Therefore, A^{T}X = B
implies X = (A^{T})^{-1}B = (A^{-1})^{T}B.

Find (A^{-1})^{T}:
(A^{-1})^{T} = -\frac{1}{16} \begin{bmatrix}-4&-5&7 \\ 4&1&-11 \\ 4&-3&1\end{bmatrix}

Multiply by B \begin{bmatrix}13 \\ 4 \\ 8\end{bmatrix}:

X = -\frac{1}{16} \begin{bmatrix}-4(13) - 5(4) + 7(8) \\ 4(13) + 1(4) - 11(8)\\ 4(13) - 3(4) + 1(8)\end{bmatrix}

X = -\frac{1}{16} \begin{bmatrix}-52 - 20 + 56 \\ 52 + 4 - 88 \\ 52 - 12 + 8\end{bmatrix} = -\frac{1}{16} \begin{bmatrix}-16 \\ -32 \\ 48\end{bmatrix}

X = \begin{bmatrix} 1 \\ 2 \\ -3\end{bmatrix}

Final Answer: Thus, \boxed{ x=1, y=2, z=-3}.

Q17. In a village a giant fair is being conducted every year. It is center of entertainment of many nearby villagers also. A toy-train is center of attraction of all the children. Madhav and his family visited the famous fair and noticed that the path of the train is a circle. If we write the path as a function of x then it is of the form f(x)=\sqrt{4-x^{2}} answer the following questions using the information given above.

(i) Let f:Z\rightarrow Z be defined as f(x)=\sqrt{4-x^{2}}, then f is
(a) Surjective
(b) Injective
(c) Bijective
(d) None of the above

Solution:
If we evaluate f(-2) and f(2), both equal 0.
Thus, multiple inputs map to the same output, so it is not injective.

The function cannot map to negative integers or numbers like 2,
(since x=0 gives 2 but 1 gives \sqrt{3} \notin Z), meaning it’s not surjective.

Final Answer: (d) None of the above.

(ii) Let f:[-2,2]\rightarrow[0,2] be defined as f(x)=\sqrt{4-x^{2}} then f is
(a) Surjective
(b) Injective
(c) Bijective
(d) None of the above

Solution:
The function isn’t injective since f(-1) = f(1) = \sqrt{3}.

To test for surjectivity, let y \in [0,2].
Set y = \sqrt{4-x^{2}} \implies x^{2} = 4-y^{2}.
Since y \leq 2, 4-y^{2} \geq 0, meaning there is always a valid real x in [-2,2] that produces y.
Thus, it maps onto every element in the codomain.

Final Answer: (a) Surjective.

(iii) Let f:R\rightarrow R be defined as f(x)=\sqrt{4-x^{2}}, then range is
(a) [0, 2]
(b) {0, 2}
(c) (0, 2)
(d) Z

Solution:
The expression \sqrt{4-x^{2}} yields real numbers only when 4-x^{2} \geq 0 (i.e., x \in [-2,2]).
The maximum value occurs at x=0 giving \sqrt{4}=2.
The minimum value occurs at x=\pm 2 giving \sqrt{0}=0.
All values in between are continuous.

Final Answer: (a) [0, 2].

(iv) Let f:N\rightarrow R be defined as f(x)=\sqrt{4-x^{2}}, then domain is
(a) [0, 2]
(b) {1, 2, 3, 4}
(c) {1,2}
(d){1,4}

Solution:
For the function to produce a real number, 4-x^{2} \ge 0 \implies x^2 \le 4.
Since the domain is restricted to Natural numbers (N), we must find x \in {1, 2, 3, ...} such that x^2 \le 4.
Only 1 and 2 satisfy this condition.

Final Answer: (c) {1, 2}.

(v) If center of the path of the toy-train is origin, and position of Madhav in the train w.r.t origin is taken as (a, b), then at a certain moment if a=1 then find the value of b.
(a) b=\sqrt{3}
(b) b=-\sqrt{3}
(c) b=\pm\sqrt{3}
(d) b is any point on the path

Solution:
The text states the train’s full path is a circle.
The radius squared is dictated by the function r^2 = 4.
The equation of a circle centered at the origin is x^2 + y^2 = r^2.

Substituting the radius and coordinates (a, b):
a^2 + b^2 = 4.

Substitute a = 1 into the equation:
1^2 + b^2 = 4

b^2 = 3

b = \pm \sqrt{3}

Final Answer: (c) b=\pm\sqrt{3}.

I hope you found these solutions clear and helpful. Let me know in the comments below if you have any doubts or need clarifications about a specific step from any of the questions.