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**Objective Type Questions****Q.28. Let T be the set of all triangles in the Euclidean plane, and let a relation R on T be defined as aRb if a is congruent to b ∀ a, b ∈ T. Then R is****(a) Reflexive but not transitive ****(b) Transitive but not symmetric ****(c) Equivalence ****(d) None of these****Ans. **(c)**Solution.**

If a ≌ b ∀ a, b ∈ T

then a R a ⇒ a ≌ a which is true for all a ∈ T

So, R is reflexive.

Now, aRb and bRa.

i.e., a ≌ b and b ≌ a which is true for all a, b ∈ T

So, R is symmetric.

Let aRb and bRc.

⇒ a ≌ b and b ≌ a ⇒ a ≌ c ∀ a, b, c ∈ T

So, R is transitive.

Hence, R is equivalence relation.**Q.29. Consider the non-empty set consisting of children in a family and a relation R defined as aRb if a is brother of b. Then R is****(a) Symmetric but not transitive ****(b) Transitive but not symmetric ****(c) Neither symmetric nor transitive ****(d) Both symmetric and transitive****Ans. **(b)**Solution.**

Here, a R b ⇒ a is a brother of b.

a R a ⇒ a is a brother of a which is not true.

So, R is not reflexive.

a R b ⇒ a is a brother of b.

b R a ⇒ which is not true because b may be sister of a.

⇒ a R b ≠ b R a

So, R is not symmetric.

Now, a R b, b R c ⇒ a R c

⇒ a is the brother of b and b is the brother of c.

∴ a is also the brother of c.

So, R is transitive.**Q.30. The maximum number of equivalence relations on the set A = {1, 2, 3} are****(a) 1 ****(b) 2 ****(c) 3 ****(d) 5****Ans. **(d)**Solution.**

Here, A = {1, 2, 3} The number of equivalence relations are as follows:

R_{1} = {(1, 1), (1, 2), (2, 1), (2, 3), (1, 3)}

R_{2} = {(2, 2), (1, 3), (3, 1), (3, 2), (1, 2)}

R_{3} = {(3, 3), (1, 2), (2, 3), (1, 3), (3, 2)}**Q.31. If a relation R on the set {1, 2, 3} be defined by R = {(1, 2)}, then R is****(a) Reflexive ****(b) Transitive ****(c) Symmetric ****(d) None of these****Ans. **(b)

Solution:

Given that: R = {(1, 2)}

a R b and b R c ⇒ a R c which is true.

So, R is transitive.**Q.32. Let us define a relation R in R as aRb if a ≥ b. Then R is****(a) An equivalence relation ****(b) Reflexive, transitive but not symmetric ****(c) Symmetric, transitive but not reflexive ****(d) Neither transitive nor reflexive but symmetric.****Ans. **(b)

Solution:

Here, aRb if a ≥ b

⇒ aRa ⇒ a ≥ a which is true, so it is reflexive.

R is not symmetric.

Now, a ≥ b, b ≥ c ⇒ a ≥ c which is true.

So, R is transitive.**Q.33.**** Let A = {1, 2, 3} and consider the relation R = {1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1,3)}.**

(a) Reflexive but not symmetric

Solution:

Given that: R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}

Here, 1 R 1, 2 R 2 and 3 R 3, so R is reflexive.

1 R 1 and 1 R 2 ⇒ 1 R 3, so, R is transitive.

Solution:

Given that: a * b =

Let e be the identity element

∴ a * e = ae/2 = a ⇒ e = 2

Solution:

If A and B sets have m and n elements respectively, then the number of one-one and onto mapping from A to B is

n! if m = n

and 0 if m ≠ n

Here, m = 5 and n = 6

5 ≠ 6

So, number of mapping = 0

Solution:

Here, A = {1, 2, 3, ..., n} and B = {a, b}

Let m be the number of elements of set A

and n be the number of elements of set B

∴ Number of surjections from A to B is

Here, m = 2 (given)

∴ Number of surjections from A to B =

**Q.37. Let f: R → R be defined by f (x) = 1/x ∀ x∈ R. Then f is****(a) one-one ****(b) onto****(c) bijective ****(d) f is not defined****Ans. **(d)

Solution:

Given that f(x) = 1/x** **

Put x = 0 ∴ f(x) = 1/0 = ∞

So, f(x) is not defined. **Q.38. Let f : R → R be defined by f (x) = 3x ^{2} – 5 and g : R → R by g (x) = x/(x^{2} + 1). **

Solution:

Here, f(x) = 3x

∴ gof = gof (x) = g[3x

Solution.

Given that f: Z → Z

Let x

f(x

So, f(x) is one-one function.

Now, let y = x + 2 ∴ x = y – 2 ∈ Z ∀ y ∈ Z

So, f(x) is onto function.

∴ f(x) is bijective function.

Solution:

Given that f(x) = x

Let y = x

∴ x = (y – 5)

Solution.

Here, f: A → B and g: B → C

∴ (gof)

Solution.

Given that

⇒ y(5x – 3) = 3x + 2

⇒ 5xy – 3y = 3x + 2

⇒ 5xy – 3x = 3y + 2

⇒ x(5y – 3) = 3y + 2

⇒ f – 1(x) = f (x)

Solution.

Given that f: [0, 1] → [0, 1]

∴ f = f

So, (fof)x = x (identity element)

(b) [1, ∞)

(d) [5, ∞)

Solution.

Given that f(x) = x

Let y = x

⇒ x

∴ For real value of x, y – 1 ≥ 0 ⇒ y ≥ 1.

So, the range is [1, ∞).

(b) – 1

(c) 7/2

Solution.

Here, f(x) = 2x-1/2 and g(x) = x + 2

∴ gof(x) = g[(f(x)]

= f(x) + 2

Solution.

Given that:

∴ f(–1) + f(2) + f(4) = 3(– 1) + (2)

Q.47. Let f: R → R be given by f (x) = tan x. Then f

(d) None of these

Solution.

Given that f(x) = tan x

Let f(x) = y = tan x

⇒ x = tan

⇒ f

⇒ f

⇒ f

[tan (π/4)] = π/4

⇒ 3b = 30 – 2a

⇒b = 30-2a/3

for a = 3, b = 8

a = 6, b = 6

a = 9, b = 4

a = 12, b = 2

Hence, R = {(3, 8), (6, 6), (9, 4), (12, 2)}

So, clearly, R = {(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (4, 3) (3, 4), (4, 4), (5, 5)}

gof(1) = g[f(1)] = g(2) = 3

gof(3) = g[f(3)] = g(5) = 1

gof(4) = g[f(4)] = g(1) = 3

∴ gof = {(1, 3), (3, 1), (4, 3)} fog(2) = f[g(2)] = f(3) = 5

fog(5) = f[g(5)] = f(1) = 2

fog(1) = f[g(1)] = f(3) = 5

∴ fog = {(2, 5), (5, 2), (1, 5)}

fofof (x) = fof [f(x)] = f[f{f(x)}]

Let y = [4 – (x – 7)

⇒ (x – 7)

⇒ x – 7 = (4 – y)

Hence, f

(3, 3) ∈ R, so R is reflexive.

(3, 1) ∈ R and (1, 3) ∈ R, so R is symmetric.

Now, (3, 1) ∈ R and (1, 3) ∈ R but (1, 1) ∉ R

So, R is not transitive.

Hence, the statement is ‘False’.

f(x) is not one-one.

Hence, the statement is ‘False’.

R = {(1, 2), (2, 1), (2, 3), (1, 3)}

Here, (1, 2) ∈ R and (2, 1) ∈ R, so R is symmetric.

(1, 2) ∈ R, (2, 3) ∈ R ⇒ (1, 3) ∈ R, so R is transitive.

But (1, 1) ∉ R, (2, 2) ∉ R and (3, 3) ∉ R.

Hence, the statement is ‘False’.

If k = 1 m = n, so z is reflexive.

Clearly z is not symmetric but z is transitive.

Hence, the statement is ‘False’.

f(2n – 1) = 0 and f(2n) = 1 ∀ n ∈ N

So, f: N → A is a onto function.

Hence, the statement is ‘True’.

Here, (1, 1) ∈ R, so R is Reflexive.

(1, 2) ∈ R and (2, 1) ∈ R, so R is Symmetric.

(1, 2) ∈ R but (2, 3) ∉ R

So, R is not transitive.

Hence, the statement is ‘False’.

fog(x) = f[g(x)] = (2x + 3)

gof(x) = g[f(x)] = 2x

So, fog(x) ≠ g o f(x)

Hence, the statement is ‘False’.

fo{goh(x)} = fo{g(2x + 3)}

= f(2x + 3 – 1) = f(2x + 2) = 2(2x + 2) = 4x + 4.

and (fog)oh(x) = (fog) {h(x)}

= fog(2x + 3)

= f(2x + 3 – 1) = f(2x + 2) = 2(2x + 2) = 4x + 4

So, fo{goh(x)} = {(fog)oh(x)} = 4x + 4

Hence, the statement is ‘True’.

Hence, the statement is ‘False’.

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