NCERT Solutions for Class 12 Maths Chapter 1

A = {1, 2, 3 … 13, 14}
R = {(x, y): 3x − y = 0}
∴ R = {(1, 3), (2, 6), (3, 9), (4, 12)}
R is not reflexive since (1, 1), (2, 2) … (14, 14) ∉ R.
Also, R is not symmetric as (1, 3) ∈ R, but (3, 1) ∉ R. [3(3) − 1 ≠ 0]
Also, R is not transitive as (1, 3), (3, 9) ∈ R, but (1, 9) ∉ R. [3(1) − 9 ≠ 0] Hence, R is neither reflexive, nor symmetric, nor transitive.

R = {(a, b): a ≤ b²}
It can be observed that (1/2,1/2)∉R, since, 1/2>(1/2)²
∴ R is not reflexive.
Now, (1, 4) ∈ R as 1 < 42 But, 4 is not less than 12. ∴ (4, 1) ∉ R ∴ R is not symmetric. Now, (3, 2), (2, 1.5) ∈ R [as 3 < 2² = 4 and 2 < (1.5)² = 2.25] But, 3 > (1.5)² = 2.25
∴ (3, 1.5) ∉ R
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.

Let A = {1, 2, 3, 4, 5, 6}.
A relation R is defined on set A as: R = {(a, b): b = a + 1}
∴ R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
We can find (a, a) ∉ R, where a ∈ A.
For instance, (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) ∉ R
∴ R is not reflexive.
It can be observed that (1, 2) ∈ R, but (2, 1) ∉ R.
∴ R is not symmetric.
Now, (1, 2), (2, 3) ∈ R but, (1, 3) ∉ R
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.

Let A = {1, 2, 3}. A relation R on A is defined as R = {(1, 2), (2, 1)}.
It is clear that (1, 1), (2, 2), (3, 3) ∉ R,
∴ R is not reflexive.
Now, as (1, 2) ∈ R and (2, 1) ∈ R, then R is symmetric.
Now, (1, 2) and (2, 1) ∈ R, however, (1, 1) ∉ R,
∴ R is not transitive.
Hence, R is symmetric but neither reflexive nor transitive.

Set A is the set of all books in the library of a college.
R = {x, y): x and y have the same number of pages}
Now, R is reflexive since (x, x) ∈ R as x and x has the same number of pages.
Let (x, y) ∈ R
⇒ x and y have the same number of pages.
⇒ y and x have the same number of pages.
⇒ (y, x) ∈ R
∴ R is symmetric.
Now, let (x, y) ∈R and (y, z) ∈ R.
⇒ x and y and have the same number of pages and y and z have the same number of pages.
⇒ x and z have the same number of pages.
⇒ (x, z) ∈ R ∴ R is transitive.
Hence, R is an equivalence relation.

R = {(P, Q): Distance of point P from the origin is the same as the distance of point Q from the origin}
Clearly, (P, P) ∈ R since the distance of point P from the origin is always the same as the distance of the same point P from the origin.
∴ R is reflexive.
Now, Let (P, Q) ∈ R.
⇒ The distance of point P from the origin is the same as the distance of point Q from the origin.
⇒ The distance of point Q from the origin is the same as the distance of point P from the origin.
⇒ (Q, P) ∈ R, ∴ R is symmetric.
Now, Let (P, Q), (Q, S) ∈ R.
⇒ The distance of points P and Q from the origin is the same and also, the distance of points Q and S from the origin is the same.
⇒ The distance of points P and S from the origin is the same.
⇒ (P, S) ∈ R ∴ R is transitive.
Therefore, R is an equivalence relation.
The set of all points related to P ≠ (0, 0) will be those points whose distance from the origin is the same as the distance of point P from the origin.
In other words, if O (0, 0) is the origin and OP = k, then the set of all points related to P is at a distance of k from the origin.
Hence, this set of points forms a circle with the centre as the origin and this circle passes through point P.

R = {(P1, P2): P1 and P2 have same the number of sides}
R is reflexive, Since (P1, P1) ∈ R, as the same polygon has the same number of sides with itself.
Let (P1, P2) ∈ R.
⇒ P1 and P2 have the same number of sides.
⇒ P2 and P1 have the same number of sides.
⇒ (P2, P1) ∈ R,
∴ R is symmetric.
Now, Let (P1, P2), (P2, P3) ∈ R.
⇒ P1 and P2 have the same number of sides.
Also, P2 and P3 have the same number of sides.
⇒ P1 and P3 have the same number of sides.
⇒ (P1, P3) ∈ R ∴ R is transitive.
Hence, R is an equivalence relation.
The elements in A related to the right-angled triangle (T) with sides 3, 4, and 5 are those polygons which have 3 sides (Since T is a polygon with 3 sides).
Hence, the set of all elements in A related to triangle T is the set of all triangles.

R = {(L1, L2): L1 is parallel to L2}
R is reflexive as any line L1 is parallel to itself
i.e., (L1, L1) ∈ R.
Now, let (L1, L2) ∈ R.
⇒ L1 is parallel to L2
⇒ L2 is parallel to L1.
⇒ (L2, L1) ∈ R
∴ R is symmetric.
Now, let (L1, L2), (L2, L3) ∈ R.
⇒ L1 is parallel to L2. Also, L2 is parallel to L3.
⇒ L1 is parallel to L3.
∴ R is transitive.
Hence, R is an equivalence relation.
The set of all lines related to the line y = 2x + 4 is the set of all lines that are parallel to the line y = 2x + 4.
Slope of line y = 2x + 4 is m = 2
It is known that parallel lines have the same slopes.
The line parallel to the given line is of the form y = 2x + c, where c ∈ R. Hence, the set of all lines related to the given line is given by y = 2x + c, where c ∈ R.

The functions f: {1, 3, 4} → {1, 2, 5} and
g: {1, 2, 5} → {1, 3} are defined as
f = {(1, 2), (3, 5), (4, 1)} and
g = {(1, 3), (2, 3), (5, 1)}.
gof(1) = g[f(1)] = g(2) = 3 [as f(1) = 2 and g(2) = 3]
gof(3) = g[f(3)] = g(5) = 1 [as f(3) = 5 and g(5) = 1]
gof(4) = g[f(4)] = g(1) = 3 [as f(4) = 1 and g(1) = 3]
∴ gof = {(1, 3), (3, 1), (4, 3)}

f: R → R is given as f(x) = x³.
For one – one Suppose f(x) = f(y), where x, y ∈ R.
⇒ x³ = y³ … (1)
Now, we need to show that x = y.
Suppose x ≠ y, their cubes will also not be equal.
⇒ x³ ≠ y³
However, this will be a contradiction to (1).
∴ x = y Hence, f is injective.