Find the area of the region bounded by t | Class 12 Mathematics Chapter Application of Integrals, Application of Integrals NCERT Solutions

(i)  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.

(ii) R = {(x, y): y = x + 5 and x < 4} = {(1, 6), (2, 7), (3, 8)}

It is seen that (1, 1) ∉ R.

∴ R is not reflexive.

(1, 6) ∈R

But,

(1, 6) ∉ R.

∴ R is not symmetric.

Now, since there is no pair in R such that (x, y) and (y, z) ∈R, then (x, z) cannot belong to R.

∴ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

(iii)  A = {1, 2, 3, 4, 5, 6}

R = {(x, y): y is divisible by x}

We know that any number (x) is divisible by itself.

 (x, x) ∈R

∴ R is reflexive.

Now,

(2, 4) ∈R [as 4 is divisible by 2]

But,

(4, 2) ∉ R. [as 2 is not divisible by 4]

∴ R is not symmetric.

Let (x, y), (y, z) ∈ R. Then, y is divisible by x and z is divisible by y.

∴ z is divisible by x.

⇒ (x, z) ∈R

∴ R is transitive.

Hence, R is reflexive and transitive but not symmetric.

(iv) R = {(x, y): x − y is an integer}

Now, for every x ∈ Z, (x, x) ∈R as x − x = 0 is an integer.

∴ R is reflexive.

Now, for every x, y ∈ Z if (x, y) ∈ R, then x − y is an integer.

⇒ −(x − y) is also an integer.

⇒ (y − x) is an integer.

∴ (y, x) ∈ R

∴ R is symmetric.

Now,

Let (x, y) and (y, z) ∈R, where x, y, z ∈ Z.

⇒ (x − y) and (y − z) are integers.

⇒ x − z = (x − y) + (y − z) is an integer.

∴ (x, z) ∈R

∴ R is transitive.

Hence, R is reflexive, symmetric, and transitive.

(v)  (a) R = {(x, y): x and y work at the same place}

(x, x) ∈ R

∴ R is reflexive.

If (x, y) ∈ R, then x and y work at the same place.

⇒ y and x work at the same place.

⇒ (y, x) ∈ R.

∴ R is symmetric.

Now, let (x, y), (y, z) ∈ R

⇒ x and y work at the same place and y and z work at the same place.

⇒ x and z work at the same place.

⇒ (x, z) ∈R

∴ R is transitive.

Hence, R is reflexive, symmetric, and transitive.

(b) R = {(x, y): x and y live in the same locality}

Clearly (x, x) ∈ R as x and x is the same human being.

∴ R is reflexive.

If (x, y) ∈R, then x and y live in the same locality.

⇒ y and x live in the same locality.

⇒ (y, x) ∈ R

∴ R is symmetric.

Now, let (x, y) ∈ R and (y, z) ∈ R.

⇒ x and y live in the same locality and y and z live in the same locality.

⇒ x and z live in the same locality.

⇒ (x, z) ∈ R

∴ R is transitive.

Hence, R is reflexive, symmetric, and transitive.

(c) R = {(x, y): x is exactly 7 cm taller than y}

Now, (x, x) ∉ R

Since human being x cannot be taller than himself.

∴ R is not reflexive.

Now, let (x, y) ∈R.

⇒ x is exactly 7 cm taller than y.

Then, y is not taller than x.

∴ (y, x) ∉R

Indeed if x is exactly 7 cm taller than y, then y is exactly 7 cm shorter than x.

∴R is not symmetric.

Now,

Let (x, y), (y, z) ∈ R.

⇒ x is exactly 7 cm taller than y and y is exactly 7 cm taller than z.

⇒ x is exactly 14 cm taller than z .

∴ (x, z) ∉R

∴ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

(d) R = {(x, y): x is the wife of y}

Now, (x, x) ∉ R

Since x cannot be the wife of herself.

∴R is not reflexive.

Now, let (x, y) ∈ R

⇒ x is the wife of y.

Clearly y is not the wife of x.

∴ (y, x) ∉ R

Indeed if x is the wife of y, then y is the husband of x.

∴ R is not transitive.

Let (x, y), (y, z) ∈ R

⇒ x is the wife of y and y is the wife of z.

This case is not possible. Also, this does not imply that x is the wife of z.

∴ (x, z) ∉ R

∴ R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.

(e) R = {(x, y): x is the father of y}

Now  (x, x) ∉ R

As x cannot be the father of himself.

∴R is not reflexive.

Now, let (x, y) ∈R.

⇒ x is the father of y.

⇒ y cannot be the father of y.

Indeed, y is the son or the daughter of y.

∴ (y, x) ∉ R

∴ R is not symmetric.

Now, let (x, y) ∈ R and (y, z) ∈ R.

⇒ x is the father of y and y is the father of z.

⇒ x is not the father of z.

Indeed x is the grandfather of z.

∴ (x, z) ∉ R

∴R is not transitive.

Hence, R is neither reflexive, nor symmetric, nor transitive.