#### The distance between Madurai and Chennai is 400 km. An express train from Chennai travels towards Madurai at an average speed of 70 km/hr. The word equation to find the remaining distance is given by, Remaining distance = Total distance - distance travelled f (t) = 400 - 70t The value of the function for an input of 5.1 is given by ____ .

#### The function f: R → R such that f(x) = 2 + sin x is a _________________.

The function f(x) = 2 + sin x is well defined in R. Therefore, the domain is R itself. Since it is a trigonometric function, f(x) is a many to one function. The function f: R → R such that f(x) = 2 + sin x is a many to one function.

#### The modulus function f: R → R is given by f (x) = |x|. Then, f is _____.

Given f(x) = |x| f(2) = 2 and f(-2) = 2, as|± 2| = 2. |x| is always positive. Therefore, negative real numbers in R do not have their Pre-images. Hence, f is neither one one nor onto.

#### Given A = {1, 8, 27, 64}; B= {1, 2, 3, 4} and R is the relation from set A to B defined by "The cube of", then R is ______.

R is given by 'is the cube of'. A = {1, 8, 27, 64}; B= {1, 2, 3, 4} 1 = 1^{3}; 8 = 2^{3}; 27 = 3^{3}; 64 = 4^{3} 1, 8, 27 and 64 ε A and 1, 2, 3, 4 ε B Therefore, R = {(1, 1), (8, 2), (27, 3), (64, 4)}

#### If R be the relation on Z, such that (a, b) ∈ R ⇒ a - b is divisible by 5, then ____.

If R be the relation on Z, such that (a, b)∈R⇒ a - b is divisible by 5, then (a, b)∈R and a - b is divisible by 10. Option D: (a, b)∈ R ⇒ a - b = 5n, but 5n is not divisible by 10 for all odd values of n, e.g 5, 15, 25.... are not divisible by 10 but divisible by 5. Option A : (a, a)∈ R ⇒ a - a = 0 ⇒ 0 is divisible by 5 ⇒ (a, a)∈R for all a ∈ Z Option B: ( a, b) ∈ R ⇒ a - b = 5n ⇒ b - a = - (a - b) = - (5n) = (-5) n ⇒b - a is divisible by 5 and hence ∈ R Option C: (a, b)∈R (b, c)∈R, then a - b = 5n and b - c = 5m and n, m ∈N (a - b) + (b - c) = 5n + 5m = 5(n + m) ⇒ a - c is divisible by 5 ⇒ (a - c)∈ R ⇒ (a , b)∈ R and (b, c)∈R ⇒ (a - c) ∈ R

#### P and Q are two sets such that P ∩ Q = P. Given n(P) = 3, n(Q) = 2, (1, a), (2, b), (3, a) are in P × Q, then Q is ____ .

Given n(P) = 3; n(Q) = 2 Consider the distinct elements of P from P × Q. (The first elements belong to P) Therefore, P = {1, 2, 3} Q = {a, b} (Second elements in P × Q)

#### If A = {1, 2} and B = {4, 5}, then (A × B) U (B × A) = ____.

A = {1, 2} B= {4, 5} A × B = {(1, 4), (1, 5), (2, 4), (2, 5)} B × A = {(4, 1), (4, 2), (5, 1), (5, 2)} (A ×B) U (B ×A) = {(1, 4), (1, 5), (2, 4), (2, 5), (4, 1), (4, 2), (5, 1), (5, 2)}

#### If A is a non-empty set such that A × B = A × C, then ____.

Given A × B = A × C and A ≠ ϕ Case 1: Given A ≠ ϕ. If B = ϕ, then A × B = ϕ A × C = ϕ C = ϕ (since A ≠ ϕ) Therefore, B = C = ϕ Case 2: If B ≠ ϕ, let a ε B then, a ε A × B and a ε A × C Therefore, a ε C ⇒ B ⊆ C Similarly, C ⊆ B Hence, B = C (Since, B ⊆ C and C ⊆ B)

#### If A × A has 9 elements and (a, b), (c, a) ε A, then A is ____________

n(A × A) = 9 = 3^{2} n(A) = 3 Therefore, A has three distinct elements from the given set. ∴A = {a, b, c}

#### Let f(x) = x^{2} - 1 and g(x) = 9 - 3x. If f(x) and g(x) are equal, the domain is __________.

f(x) = g(x) x^{2} - 1 = 9 - 3x x^{2} + 3x - 10 = 0 = (x + 5)(x - 2) Therefore, f(x) and g(x) are equal on the set {-5, 2}.

#### Let f be a function on the set N of natural numbers defined by f = {(n, n^{2}): n ∈ N}. Then, the range of the function is ___________________________.

The range of the function is the set of all perfect squares in natural numbers. f(x) = x^{2}, but x is a natural number. Therefore, f(x) is also a natural number. But, every f(x) should be a perfect square. So, range is the set of all perfect squares in natural numbers.

#### If A = {a, b}, B = {c} and C = {d}, then A × (B U C) = _______________.

A = { a, b}, B= {c}, C = {d} B U C = {c} U {d} = {c, d} A × (B U C) = {a, b} × {c, d} = {(a, c), (a, d), (b, c), (b, d)}

#### Given P = {a, b}, Q = {1, 2}, (P × P) × Q = _______________.

P = {a, b} Q = {1, 2} P × P = {(a, a), (a, b), (b, a), (b, b)} (P × P) × Q = {(a, a, 1), (a, a, 2), (a, b, 1), (a, b, 2), (b, a, 1), (b, a, 2), (b, b, 1), (b, b, 2)}

#### If A = {a, b}, B = {c, d}, C= {d}, then ______________.

A = {a, b}, B = {c, d}, C = {d} B ∩ C = {d} A × (B ∩ C) = {(a, d), (b, d)} A × B = {(a, c), (a, d), (b, c), (b, d)} A × C = {(a, d), (b, d)} (A × B) ∩ (A × C) = {(a, d), (b, d)} Therefore, A × (B ∩ C) = (A × B) ∩ (A × C)

#### If R be a relation from set A to B, then ___________________.

Assume values for A and B and check each of the given options to find the true statement.

#### Given n(A) = 3; n(B) = 2, the number of relation from A to B is ____ .

Given n(A) = 3; n(B) = 2 n(A × B) = n (A) × n (B) = 3 × 2 = 6 Number of relations = number of subsets of A × B = 2^{6 } = 64

#### Consider a function f: z → defined by f(x) = x (x + 1) for all x ∈ z. Then f is ______.

Let, x, y ∈ z, then, f(x) = f(y) ⇒ x(x + 1) = y(y + 1) ⇒ x^{2} + x = y^{2} + y ⇒ x^{2 }- y^{2} = y – x ⇒ (x^{2} - y^{2}) + (x - y) = 0 ⇒ (x - y) (x + y + 1) = 0 ⇒ x = y or y = - (x + 1) As f(x) = f(y), x = y does not have a unique solution. As y = - (x + 1) ⇒ x ≠ y but f(x) = f(y), f is many one.

#### If f: R → R is given by f (x) = 3x, then f: R → R is ____.

Let x_{1}, x_{2 }∈ R such that f(x_{1}) = f(x_{2}). ⇒ 3x_{1} = 3x_{2 } ⇒ x_{1} = x_{2} ⇒ f : R → R is one one Let y be real in R. f(x) = y 3x = y ⇒x =y/3 y/3∈R such that f(x) = y ⇒ there is a pre-image in the domain. Therefore, f: R → R is onto. Hence, R is one one and onto.

#### The Domain and Range of a relation is given by Domain = {x/0 ≤ x ≤ 8} and Range = {y/2 ≤ y ≤ 7}. Then, the domain of its inverse relation is given by ____.

We reflect each point across y = x and connect them, to graph the inverse relation. So, the Domain of the inverse is the range of the given relation.

#### If R be the relation from N ---> N defined by R = {(x, y)/ x, y ε N and x = y^{2}}, then _______________

If R be the relation from N---> N defined by R = {(x, y)/ x, y ε N and x = y^{2}}, then (x, y)∈ R⇒(y, x)∈ R^{-1}. Option A is incorrect, as (2, 2), (3, 3).....∉R Option B is incorrect, as (4, 2) ∈ R but ( 2, 4)∉R, (9,3)∈R but (3, a)∉R .... Option C is incorrect, as (16, 4) ∈ R, (4, 2) ∈R but (16, 2)∉R

#### The values of x and y, if (3x - 4, y + 5) = (2x + 4, -3) are ____.

Given (3x - 4, y + 5) = (2x + 4, - 3) (Two ordered pairs are equal, if and only if, x1 = x2 and y1 = y2) (i.e) 3x - 4 = 2x + 4 ⇒ 3x - 2x = 4 + 4 ⇒ x = 8 y + 5 = -3 ⇒ y = -3 -5 ⇒ y = -8

#### The range of the relation R defined by, R = {(x, x + 1) / x ε {0, 1, 2, 3, 4}} is ____.

Given R = {(x, x + 1) / x ε {0, 1, 2, 3, 4}} When x = 0, x + 1 = 1 x = 1, x + 1 = 2 x = 2, x + 1 = 3 x = 3, x + 1 = 4 x = 4, x + 1 = 5 Therefore, R = {(0, 1), (1, 2), (2, 3), (3, 4), (4, 5)} Range = {y/ (x, y) ε R} = {1, 2, 3, 4, 5}

#### Let f(x) = 2[x], where [x] is the greatest integer function. Then, f(-2.13) is ____________.

f(x) = 2[x] f(-2.13) = 2 [-2.13] The greatest integer is -3. Therefore, f(-2.13) = -6

#### If R is a relation defined on N by, a R b such that 2a + b = 5 and a, b ε N, then R-1 is ____.

If a = 1, b = 5 - 2 = 3 ε N a = 2, b = 5 - 4 = 1 ε N a = 3, b = 5 - 6 = - 1 ∉N Only a = 1 and a = 2 satisfy the given condition. Therefore, R = {(1, 3), (2, 1)} Hence, R^{-1}= {(3, 1), (1, 2)}

#### Let f be a function from Z into Z and and f = {(1, -1), (0, -3), (2, 1). If f is defined by f(x) = ax + b, then the values of a and b are___________ respectively.

The values of a and b are 2 and -3 respectively. f(1) = a(1) + b = -1 f(0) = a(0) + b = -3 implies, b = -3 Substituting this in f(1), a = 2. Let's verify with f(2) = 2(2) - 3 = 1 Hence, a = 2 and b = -3

Share your Results: