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Surjective is not a type of relation. It is a type of function. Reflexive, Symmetric and Transitive are type of relations.
Let a binary operation ‘*’ be defined on a set A. The operation will be commutative if ________
A binary operation ‘*’ defined on a set A is said to be commutative only if a * b = b *a, ∀ a, b ∈ A.
If (a * b) * c = a * (b * c), then the operation is said to associative ∀ a, b∈ A.
If (b ο c) * a = (b * a) ο (c * a), then the operation is said to be distributive ∀ a, b, c ∈ A.
tan^{−1}√3+sec^{−1}2–cos−^{1}1 is equal to ________
tan^{−1 }√3 = π/3, sec^{−1}2 = π/3, cos^{−1}1 = 0
tan^{−1}√3 + sec^{−1}2 – cos−^{1}1 = π/3 + π/3
= 2π/3
Let sin^{1}x = y
⇒ x = siny
⇒ x = √1  cos^{2}y
⇒ x^{2} = 1  cos^{2}y
⇒ cos^{2}y = 1  x^{2}
∴ y = cos^{1 }√1  x^{2} = sin^{1}x
Which of the following relations is symmetric but neither reflexive nor transitive for a set A = {1, 2, 3}.
A relation in a set A is said to be symmetric if (a_{1}, a_{2})∈R implies that (a_{1}, a_{2})∈R,for every a_{1}, a_{2}∈R.
Hence, for the given set A={1, 2, 3}, R={(1, 2), (2, 1)} is symmetric. It is not reflexive since every element is not related to itself and neither transitive as it does not satisfy the condition that for a given relation R in a set A if (a_{1}, a_{2})∈R and (a_{2}, a_{3})∈R implies that (a_{1}, a_{3})∈ R for every a_{1}, a_{2}, a_{3}∈R.
If f : R→R, g(x) = 3 x 2 + 7 and f(x) = √x, then gοf(x) is equal to _______
Given that, g(x) = 3 x 2 + 7 and f(x) = √x
∴ gοf(x) = g(f(x)) = g(√x) = 3(√x)2 + 7 = 3x + 7.
Hence, gοf(x) = 3x + 7.
Let I be a set of all lines in a XY plane and R be a relation in I defined as R = {(I_{1}, I_{2}):I_{1} is parallel to I_{2}}. What is the type of given relation?
This is an equivalence relation. A relation R is said to be an equivalence relation when it is reflexive, transitive and symmetric.
Reflexive: We know that a line is always parallel to itself. This implies that I_{1} is parallel to I_{1} i.e. (I_{1}, I_{2})∈R. Hence, it is a reflexive relation.
Symmetric: Now if a line I_{1}  I_{2} then the line I_{2}  I_{1}. Therefore, (I_{1}, I_{2})∈R implies that (I_{2}, I_{1})∈R. Hence, it is a symmetric relation.
Transitive: If two lines (I_{1}, I_{3}) are parallel to a third line (I_{2}) then they will be parallel to each other i.e. if (I_{1}, I_{2}) ∈R and (I_{2}, I_{3}) ∈R implies that (I_{1}, I_{3}) ∈R.
Let = y
sec y = 2/√3
sec y = secπ/6
⇒ y = π/6
Let sec^{1}x = y
⇒ x = secy
⇒ x = √ 1 + tan^{2}y
⇒ x^{2}  1 = tan^{2}y
∴ y = tan^{1}√x^{2}  1 = sec^{1}x
Given that, A = and B =
Then A + B =
=
If f : R → R is given by f(x) = (5 + x^{4})^{1/4}, then fοf(x) is _______
Given that f(x) = (5 + x^{4})^{1/4}
∴ fοf(x) = f(f(x)) = (5 + {(5 + x^{4})^{1/4}}^{4})^{1/4}
= (5 + (5 + x^{4}))^{1/4 }= (10+x^{4})^{1/4}
Let ‘*’ be a binary operation on N defined by a * b =a  b + ab^{2}, then find 4 * 5.
The binary operation is defined by a * b = a  b + ab^{2}.
∴ 4 * 5 = 4  5 + 4(5^{2}) = 1 + 100 = 99.
[1, 1] is the domain for which of the following inverse trigonometric functions?
[1, 1] is the domain for sin^{1}x.
The domain for cot^{1}x is (∞,∞).
The domain for tan^{1}x is (∞,∞).
The domain for sec^{1}x is (∞,1] ∪ [1,∞).
Given that,
A + B = and A =
⇒ B = (A + B)  A =
B =
A function is invertible if and only if it is bijective i.e. the function is both injective and surjective. If a function f: A → B is bijective, then there exists a function g: B → A such that f(x) = y ⇔ g(y) = x, then g is called the inverse of the function.
Let M={7,8,9}. Determine which of the following functions is invertible for f:M→M.
The function f = {(7,7),(8,8),(9,9)} is invertible as it is both one – one and onto. The function is one – one as every element in the domain has a distinct image in the co – domain. The function is onto because every element in the codomain M = {7,8,9} has a pre – image in the domain.
Let R be a relation in the set N given by R={(a,b): a+b=5, b>1}. Which of the following will satisfy the given relation?
(2,3) ∈ R as 2+3 = 5, 3>1, thus satisfying the given condition.
(4,2) doesn’t belong to R as 4+2 ≠ 5.
(2,1) doesn’t belong to R as 2+1 ≠ 5.
(5,0) doesn’tbelong to R as 0⊁1
If f: N→N, g: N→N and h: N→R is defined f(x) = 3x  5, g(y) = 6y2 and h(z) = tanz, find ho(gof).
Given that, f(x) = 3x  5, g(y) = 6y^{2} and h(z) = tanz,
Then, ho(gof) = hο(g(f(x)) = h(6(3x5)^{2}) = tan(6(3x  5)^{2})
∴ ho(gof) = tan(6(3x  5)^{2})
Let ‘*’ be defined on the set N. Which of the following are both commutative and associative?
The binary operation ‘*’ is both commutative and associative for a * b = a + b.
The operation is commutative on a * b = a + b because a + b = b + a.
The operation is associative on a * b = a + b because (a + b) + c = a + (b + c).
Let ‘*’ be a binary operation defined by a * b = 4ab. Find (a * b) * a.
Given that, a * b = 4ab.
Then, (a * b) * a = (4ab) * a
= 4(4ab)(a) = 16a^{2} b.
(a,a) ∈ R, for every a ∈ A. This condition is for which of the following relations?
The above is the condition for a reflexive relation. A relation is said to be reflexive if every element in the set is related to itself.
Let ‘*’ be a binary operation defined by a * b = 3a^{b} + 5. Find 8 * 3.
It is given that a * b=3a^{b} + 5.
Then, 8 * 3 = 3(8^{3}) + 5 = 3(512) + 5 = 1536 + 5 = 1541.
Which of the following is not a type of binary operation?
Transitive is not a type of binary operation. It is a type of relation. Distributive, associative, commutative are different types of binary operations.
Which of the following relations is reflexive but not transitive for the set T = {7, 8, 9}?
The relation R= {(7, 7), (8, 8), (9, 9)} is reflexive as every element is related to itself i.e. (a,a) ∈ R, for every a∈A. and it is not transitive as it does not satisfy the condition that for a relation R in a set A if (a_{1}, a_{2})∈R and (a_{2}, a_{3})∈R implies that (a_{1}, a_{3}) ∈ R for every a_{1}, a_{2}, a_{3} ∈ R.
Which of the following condition is incorrect for matrix multiplication?
Matrix multiplication is never commutative i.e. AB ≠ BA. Therefore, the condition AB = BA is incorrect.
Which of the following is not the property of transpose of a matrix?
(AB)’ = (BA)’is incorrect. We know that matrix multiplication is not commutative i.e. AB ≠ BA. Hence, its transpose will also not be commutative.
(AB)’=B’A’
To find the transpose of the matrix of the given matrix, interchange the rows with columns and columns with rows.
Hence, we get A’ =
Which of the following is the reversal law of transposes?
According to the reverse law of transposes the transpose of the product is the product of the transposes taken in the reverse order i.e. (AB)’ = B’ A’.
Given that, A =
⇒ A’ =
i.e.A=A’. Hence, it is a symmetric matrix.
Which of the following conditions holds true for a skewsymmetric matrix?
A matrix is said to be skewsymmetric if it is equal to the negative of its transpose i.e. A = A’.
The given matrix A = is skew symmetric.
⇒ A’ = = A
∴ A = A’. Hence, it is a skewsymmetric matrix.
If A = , then which of the following statement is incorrect?
Given that, A =
∴ A’ =
⇒ A ’= ≠ A. Hence, it is not a skew symmetric matrix.
Which of the following conditions holds true for a symmetric matrix?
A matrix is A said to be a symmetric matrix if it is equal to its transpose i.e. A = A’.
How many elementary operations are possible on Matrices?
There are a total of 6 elementary operations that are possible on matrices, three on rows and three on columns.
Which of the following matrices will remain same if the elementary operation R_{1} → 2R_{1} + 3R_{2} is applied on the matrix?
Consider matrix A = , applying the elementary operation R_{1} → 2R_{1} + 3R_{2}.
=
herefore, the matrix A = , remains same after applying the elementary operation.
Which of the following is not a valid elementary operation?
The elementary operation R_{i} →1 + kR_{i} is incorrect, the valid elementary operations on matrices are as follows.
Which of the following column operation is incorrect for the matrix A = ?
The column operation C_{2}→2+2C_{2} is incorrect. A nonzero number cannot be directly added to any column or row in a matrix.
Which of the following relations is transitive but not reflexive for the set S = {3, 4, 6}?
For the above given set S = {3, 4, 6}, R = {(3, 4), (4, 6), (3, 6)} is transitive as (3,4)∈R and (4,6) ∈R and (3,6) also belongs to R . It is not a reflexive relation as it does not satisfy the condition (a,a)∈R, for every a∈A for a relation R in the set A.
Which of the following relations is symmetric and transitive but not reflexive for the set I = {4, 5}?
R= {(4, 5), (5, 4), (4, 4)} is symmetric since (4, 5) and (5, 4) are converse of each other thus satisfying the condition for a symmetric relation and it is transitive as (4, 5)∈R and (5, 4)∈R implies that (4, 4) ∈R. It is not reflexive as every element in the set I is not related to itself.
Which of the following is not a property of invertible matrices if A and B are matrices of the same order?
(AB)^{1 }= A^{1} B^{1} is incorrect. The correct formula is (AB)^{1 }= B^{1} A^{1}. B^{1} A^{1 }≠ A^{1} B^{1 }as matrix multiplication is not commutative.
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