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The correct option is (a) True

The best I can explain: The given statement is true. If there is a binary OPERATION *:M×M → M with an IDENTITY element a∈ M is said to be invertible with respect to the binary operation * if there EXISTS an element b ∈ M such that a*b = e = b*a, b is called INVERSE of a.

The correct OPTION is (a) Transitive

The EXPLANATION is: Transitive is not a TYPE of binary operation. It is a type of relation. Distributive, associative, commutative are different types of binary operations.

Correct CHOICE is (d) 1541

To elaborate: It is GIVEN that a*b=3a^b+5.

Then, 8*3=3(8^3)+5=3(512)+5=1536+5=1541.

Right answer is (a) a*B=b*a

The explanation: 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.

Correct option is (a) f = {(7,7),(8,8),(9,9)}

For explanation: 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.

Right OPTION is (b) bijective

For explanation I WOULD say: The given function is bijective i.e. both one-one and onto.

one – one : Every element in the domain X has a distinct image in the codomain Y. THUS, the given function is one- one.

onto: Every element in the CO- domain Y has a PRE- image in the domain X. Thus, the given function is onto.

The CORRECT answer is (b) bijective

Explanation: 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.

Correct option is (b) False

The BEST explanation: The GIVEN statement is false. The BINARY operation ‘*’ is commutative if a*b=b*a

Here, a*b=6a^4-9b^4 and b*a=6b^4-9a^4

⇒a*b≠b*a

Hence, the ‘*’ is not commutative.

Right choice is (b) R = {(1, 2), (2, 1)}

Easy explanation: A RELATION in a set A is said to be symmetric if (A1, a2)∈R implies that (a1, a2)∈R,for every a1, a2∈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 (a1, a2)∈R and (a2, A3)∈R implies that (a1, a3)∈ R for every a1, a2, a3∈R.

Correct answer is (a) cos⁡(7x^3+6)

EASIEST explanation: GIVEN that, F:R→R, f(x)=cos⁡x and G(x)=7x^3+6

Then, fοg(x) = f(g(x))=cos⁡(g(x))=cos⁡(7x^3+6).

Correct option is (d) (10+x^4)^1/4

The EXPLANATION: 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.

Right OPTION is (d) f={(1,4),(2,5),(3,6)}

The explanation is: f={(1,4),(2,5),(3,6)} is a bijective function.

One-one: It is a one-one function because EVERY element in set A={1,2,3} has a DISTINCT image in set B={4,5,6}.

Onto: It is an onto function as every element in set B={4,5,6} is the image of some element in set A={1,2,3}.

f={(2,4),(2,5),(2,6)} and f={(1,4),(1,5),(1,6)} are many-one onto.

f={(1,5),(2,4),(3,4)} is NEITHER one – one nor onto.

Correct choice is (b) False

The explanation is: The above statement is false. f is neither one-one nor onto.

For one-one: CONSIDER f(x1)=f(x2)

∴ 5x1^4+2=5x2^4+2

⇒x1=± x2.

Hence, the function is not one – one.

For onto: Consider the real number 1 which lies in CO- domain R, and let \(x=(\FRAC{y-2}{5})^{\frac{1}{4}}\).

Clearly, there is no real value of x which lies in the domain R such that f(x)=y.

Therefore, f is not onto as every element LYING in the codomain must have a pre- image in the domain.

The correct option is (a) True

The best explanation: The given STATEMENT is true. f is both ONE-one and onto.

For one-one: Consider f(x1)=f(X2)

∴7x1+4=7x2+4

⇒ x1=x2.

Thus, f is one – one.

For onto: Now for any REAL number y which lies in the co- domain R, there EXISTS an element x=(y-4)/7

such that \(f(\frac{y-4}{7}) = 7*(\frac{y-4}{7}) + 4 = y\). Therefore, the function is onto.

Correct option is (a) Reflexive relation

For explanation I WOULD SAY: 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.

Correct option is (a) TRUE

Explanation: The given statement is true. A function is invertible if it is bijective.

For one – one: CONSIDER f(x1)=f(x2)

∴ 5x1+9=5x2+9

⇒x1=x2. Hence, the function is one – one.

For ONTO: For any real number y in the co-domain R, there exists an element x=\(\frac{y-9}{5}\) such that f(x)=\(f(\frac{y-9}{5})=5(\frac{y-9}{5})\)+9=y.

Therefore, the function is onto.

Right ANSWER is (b) f={(10,5),(20,10),(30,15)}

The best I can explain: The function f={(10,5),(20,10),(30,15)} is one -one and not onto. The function is one-one because element is set P={10,20,30} has a distinct image in set Q={5,10,15,20}. The function is not onto because every element in set Q={5,10,15,20} does not have a pre-image in set P={10,20,30} (20 does not have a pre-image in set P).

f={(10,5),(10,10),(10,15),(10,20)} and f={(10,5),(10,10),(20,15),(30,20)} are MANY – one onto.

f={(20,5),(20,10),(30,10)} is NEITHER one – one nor onto.

Right ANSWER is (b) ONTO

The best explanation: The above FUNCTION is onto or surjective. A function f:X→Y is SAID to be surjective or onto if, every element of Y is the IMAGE of some elements in X.

The condition for a surjective function is for every y∈Y, there is an element in X such that f(x)=y.

The correct option is (c) Symmetric relation

The best I can explain: The above is a condition for a symmetric relation.

For EXAMPLE, a relation R on set A = {1,2,3,4} is GIVEN by R={(a,b):a+b=3, a>0, b>0}

1+2 = 3, 1>0 and 2>0 which implies (1,2) ∈ R.

Similarly, 2+1 = 3, 2>0 and 1>0 which implies (2,1)∈R. Therefore both (1, 2) and (2, 1) are converse of each other and is a part of the relation. HENCE, they are symmetric.

Right answer is (d) EQUIVALENCE relation

For explanation: 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 I1 is parallel to I1 i.e. (I1, I2)∈R. Hence, it is a reflexive relation.

Symmetric: Now if a line I1 || I2 then the line I2 || I1. Therefore, (I1, I2)∈R implies that (I2, I1)∈R. Hence, it is a symmetric relation.

Transitive: If two lines (I1, I3) are parallel to a third line (I2) then they will be parallel to each other i.e. if (I1, I2) ∈R and (I2, I3) ∈R implies that (I1, I3) ∈R.

Correct OPTION is (C) 3x+7

For explanation I would SAY: Given that, G(x)=3x^2+7 and F(x)=√x

∴ gοf(x)=g(f(x))=g(√x)=3(√x)^2+7=3x+7.

Hence, gοf(x)=3x+7.

Right choice is (a) F={(5,3),(5,4),(6,4),(8,9)}

The explanation: The function f={(5,3),(5,4),(6,4),(8,9)} is neither one -one nor onto.

The function is not one – one 8 does not have an image in the codomain N and we KNOW that a function can only be one – one if every element in the set M has an image in the codomain N.

A function can be onto only if each element in the co-DOMAIN has a pre-image in the domain X. In the function f={(5,3),(5,4),(6,4),(8,9)}, 10 in the co-domain N does not have a pre- image in the domain X.

f={(5,3),(6,4),(7,9),(8,10)} is both one-one and onto.

f={(5,4),(5,9),(6,3),(7,10),(8,10)}and f={(6,4),(7,3),(7,9),(8,10)} are many – one onto.

The correct answer is (C) injective

The EXPLANATION: The above FUNCTION is an injective or one-one function.

Consider F(X1)=f(x2)

∴ x1^2+12=x2^2+12

⇒x1=x2

Hence, it is an injective function.

The correct answer is (a) True

For explanation: The given STATEMENT is true. A relation R in a SET A is said to be an EQUIVALENCE relation if R is reflexive, SYMMETRIC and transitive. Hence, an equivalence relation is always symmetric.

The CORRECT answer is (b) tan⁡(6(3x-5)^2)

The best I can explain: Given that, F(x)=3x-5, g(y)=6y^2 and h(z)=tan⁡z,

Then, ho(GOF)=hο(g(f(x))=h(6(3x-5)^2)=tan⁡(6(3x-5)^2)

∴ ho(gof)=tan⁡(6(3x-5)^2)

Correct answer is (a) \(\sqrt{X-9}\)

For EXPLANATION: The FUNCTION F(x)=x^2+9 is bijective.

Therefore, f(x)=x^2+9

i.e.y=x^2+9

x=\(\sqrt{y-9}\)

⇒f^-1 (x)=\(\sqrt{x-9}\).

Right OPTION is (a) (2,3) ∈ R

Explanation: (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

Right answer is (C) MANY-one

Easy EXPLANATION: The above is a many -one function.

Consider F(X1)=f(x2)

∴5x1^3-8=5x2^3-8

5x1^3=5x2^3

⇒x1 = ±x2. Hence, the function is many – one.

Correct ANSWER is (d) R = {(4, 5), (5, 4), (4, 4)}

For explanation: 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.

Right option is (a) R = {(3, 4), (4, 6), (3, 6)}

For explanation: 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.

Right choice is (B) False

To EXPLAIN: The given statement is false. The composition of functions is associative i.e. fο(g ο H)=(f ο g)οh. The composition of functions is not COMMUTATIVE i.e. g ο f ≠ f ο g.

The CORRECT answer is (B) Surjective

The explanation is: Surjective is not a TYPE of relation. It is a type of function. Reflexive, Symmetric and TRANSITIVE are type of relations.