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Q1. The function f(x) = 3x + 1 is

• injective
• not injective
• Insufficient information
• It is injective in some intervals of the domain

Solution

 A function  is said to be injective (or one-one), if for every     Consider  f(x) = 3x + 1   Therefore f is an injective function

Q2. Let f and g be given by f = {(5,2) , (6 , 3)} and g = { (2,5), (3,6)}. Find gof.

• {(5,5) , (6,6)}
• {(5,5)(5,2) , (6,6)}
• {(2,5), (3,6)}.
• {(5,2) , (6 , 3)}

Solution

  Domain of gof = Domain of f = {5,6} Now, gof(5) = g {f(5)} = g (2) = 5 ; gof(6) = g {f(6)} = g (3) = 6 gof = { (5,5) , (6,6)}  

Q3. Given below is the table corresponding to some binary operation a * b on a set {0,1,2,3,4,5}. How many  elements of this operation have an inverse?. *012345 0 0 1 2 3 4 5 1 1 2 3 4 5 0 2 2 3 4 5 0 1 3 3 4 5 0 1 2 4 4 5 0 1 2 3 5 5 0 1 2 3 4

• 0
• 1
• 6
• 4

Solution

  From the table, the second row and second column are the same as the original set. 0*0 = 0, 1*0 = 0*1 = 1 , 2*0= 0*2 = 2 , 3*0= 0*3 = 3 , 4*0 = 0*4 =4, 0*5 = 5*0 = 5 ⇒ ’0’is the identity element of the operation ‘*’ Now , the identity element ‘0’ appears in the table 6 times .Therefore, the number of  elements of this operation that  have an inverse is 6

Q4. Given below is the table corresponding to some binary operation a * b on a set {0, 1,2,3,4,5}. Identify the identity element of this operation. * 0 1 2 3 4 5 0 0 1 2 3 4 5 1 1 2 3 4 5 0 2 2 3 4 5 0 1 3 3 4 5 0 1 2 4 4 5 0 1 2 3 5 5 0 1 2 3 4

• 0
• 1
• 6
• 4

Solution

From the table, the second row and second column are the same as the original set. 0*0 = 0, 1*0 = 0*1 = 1 , 2*0= 0*2 = 2 , 3*0= 0*3 = 3 , 4*0 = 0*4 =4, 0*5 = 5*0 = 5  ’0’is the identity element of the operation ‘*’

Q5. If f(x) = ax + b and g(x) = cx + d, then f[g(x)] – g[f(x)] is equivalent to

• f(a) – g(c)
• f(c) + g(a)
• f(b) – g(b)
• f(d) – g(b)

Solution

f(g(x)–g(f(x)) = a(cx + d) + b – {c(ax + b) + d} = ad + b – bc – d = (ad + b) – (bc + d) = f(d) – g(b)

Q6. What type of function is the sine function in R ?

•    one-one
•   many one
•   one-one and onto
•   neither onto  nor one-one

Solution

  We know that a sine function is a periodic function with period 2, so after every 2 the same value repeats sin2 = sin 4 = sin6and so on. So f(x) = sinx is not a one-one function Also  , i.e Range of sin x = [-1, 1]  which is a subset of R So f(x)  = sin x is not an onto function

Q7. If a relation f:X→Y is a function , then for g:Y→X to be a function ,function f need to be

•  one-one
• many one
• one-one and onto
• onto but not one-one

Solution

By definitiong:Y→X will be a function if every element of Y has a unique correspondence in X.This is possible only if for f  the codomain = range , i.e it is an onto function . Also because of requirement of uniqueness of correspondence ,  f needs to be a one-one function.  

Q8. Let A = {1, 2, 3, 4} and let R = {(2, 2), (3, 3), (4, 4), (1, 2)} be a relation on A. Then, R is

• Reflexive
• Symmetric
• Transitive
• Equivalence relation

Solution

R is not reflexive as (1, 1) R. Moreover, (1, 2)  R but (2, 1) R. Thus, R is not symmetric. Now, for (a, b)R and (b, c) R, (a, c) R . So, R is transitive.

Q9. If f is the greatest integer function and g is the modulus function . Write the value of g o f(-1/3) - f o g ( -1/3 )

• -1
• 1
• 2
• 0

Solution

 f(x) = [x] , the greatest integer £ x and g(x ) = |x | , xR . g o f(-1/3) = g{f(-1/3)} =  g(-1) = | -1 |= 1 f o g (-1/3) = f{| -1/3 |= f{1/3} = [1/3]= 0 g o f(-1/3) - f o g (-1/3 )= 1 - 0 = 1  

Q10. A function f: A→B is said to be invertible, if there exists a function g: B→A such that

• g o f = I_A and f o g = I_B
• g o f = I_B and f o g = I_A
• g o f = I_A and f o g = I_A
• g o f = I_B and f o g = I_B

Solution

A function f: A→B is said to be invertible, if there exists a function g: B→A such that : g o f = I_A and f o g = I_B  

Q11. What type of function is the exponential function e^x function on R ?

•   one-one and into
•    Many one
•   One-one and onto
•    Neither onto  nor one-one

Solution

  As the horizontal line test would tell us. Every value of e^x has a unique pre image in R So e^x is a one-one function on R Range of f = (0,∞)R So e^x is not an onto function, it is an into function.

Q12. The law a + b = b + a is called

• Closure law
• Commutative law
• Associative law
• Distributive law

Solution

For a binary operation ‘+’, a + b = b + a, is a commutative law on a and b.

Q13. If A = {1, 2, 3, 4} and B = {5, 6, 7, 8}, then which function is one-one and onto?

• f₁= {(1, 5), (2, 7), (3, 8), (4, 6)}
• f₂ = {(1, 6), (2, 8), (3, 8), (4, 5)}
• f₃ = {(1, 5), (2, 7), (3, 8), (4, 5)}
• f₄ = {(1, 8), (2, 7), (3, 6), (4, 7)}

Solution

The function  is one-one and onto.   The second entry in each ordered pair is unique and , the set of second entries of the orderd pairs is the set B  

Q14. The function, f(x) = 2x + 1 is

• It is surjective
• It is not surjective
• Insufficent information
• It is surjective in some intervals of the domain

Solution

A function  is said to be surjective (or onto), if for every  there exists an element  such that   For example, f(x) = 2x + 1 is a onto function, because

Q15. If f(x) = (a – xⁿ) ^1/n, find f{f(x)}

• (a – xⁿ) ^1/n
• (a – xⁿ)
• (a – x) ^1/n
• x

Solution

If f(x) = (a – xⁿ) ^1/n⇒f{f(x)}= (a – f(x) ⁿ) ^1/n =  (a – [(a – xⁿ) ^1/n] ⁿ) ^1/n = (a-[a-xⁿ])^1/n = (xⁿ)^1/n = x

Q16. Number of binary sets on the set {p, q,r} is:

• 18
• 16
• 3⁹
• 3⁶

Solution

If a finite set containing n elements, then total number of binary operations on this set is . So, total number of binary sets on the set {p, q,r} = = 3⁹ .

Q17. Let A = {1, 2, 3} and B = {5, 6, 7, 8, 9} and let f(x) = {(1, 8), (2, 7), (3, 6)} then f is

• Injective
• bijective
• surjective
• not a function

Solution

Here, f(1) = 8, f(2) = 7, f(3) = 6. Since, different points of domain have the different f-image in the range, therefore f is a  one-one function.

Q18. If A = N x N and * be any binary operation on A defined by (a, b) * (c, d) = (a + c, b + d), then the binary operation is

• Commutative
• Associative
• Commutative and associative
• Commutative but not associative

Solution

(a, b) * (c, d) = (a + c, b + d) and (c, d) * (a, b) = (c + a, d + b) = (a + c, b + d) = (a, b) * (c, d) So, * is commutative.   (a, b) * [(c, d) * (e, f)] = (a, b) * (c + e, d + f) = (a + c + e, b + d + f) and [(a, b) * (c, d)] * (e, f) = (a + c, b + d) * (e, f) = (a + c + e, b + d + f). So, * is associative also.

Q19. Given below is the table corresponding to some binary operation a * b on a set S={0,1,2,3,4,5}. What is 5 ^-1 from this operation? * 0 1 2 3 4 5 0 0 1 2 3 4  5 1 1 2 3 4 5 0 2 2 3 4 5 0 1 3 3 4 5 0 1 2 4 4 5 0 1 2 3 5 5 0 1 2 3 4

• 1
• 2
• 5
• 4

Solution

    0 is the identity element of this table or this operation. as, 0*a=a*0 = 0   a  S={0,1,2,3,4,5} From the 6 ^th  row and 6 ^th column ,we observe 5*1 = 1*5= 0,the identity . So 5 ^-1 = 1  

Q20. f: R→R and g:R→R . f and g are defined as f(x) =2x - 3 and g(x) = x² + 3x + 1. Then, fog(x) is

•     4x²-6x -1
•     2x²+6x -1
•     x⁴+6x³+14x²+15x +5
•      4x -9

Solution

  fog(x) = f[g(x)] =2(x² + 3x + 1) -3 = 2x²+6x +2 -3 = 2x²+6x -1

Q21. Let R = { (P,Q) : OP  = OQ , O being the origin} be an equivalence relation on A . The equivalence class [( 1,2)] is

•   {(x,y) : x²+y² = 5}
•   {(x,y) : x²+y² = 4}
•   {(x,y) : x²+y² = 1}
•   {(x,y) : x² = y² }

Solution

  [( 1,2)] is the set of all points whose distance from the origin is same as the distance of the point (1,2) from (0,0) So it is  a circle with radius and centre at the origin

Q22. Let f : R → R and f(x) = x² and g: R → R such that g(x) = sinx. Then, g o f = ?

• sin x³
• sin x²
• x² + sin x
• (sin x)²

Solution

(g o f)(x) = g(f(x)) = g(x²) = sin x².

Q23. The  number of binary operations which can be defined  on the set P= { p, q}is

• 0
• 1
• 6
• 16

Solution

If a set has n elements, then there are binary operations, which can be defined on that set. Here n =2 so Number of binary operations on A  =  =16

Q24. If  * and ^O are two binary operations defined by a * b = a + b and a ^O b = ab, then

• ^O is distributive on *.
• * is distributive on ^O.
• (a * b) ^O c = ab + ac
• a ^O (b * c) = ac + bc

Solution

a ^O (b * c) = a ^O (b + c) = a(b + c) = ab + ac   (a ^Ob )* (a ^Oc) = ab * ac  = ab + ac Thus we note that  a ^O (b * c) =(a ^Ob )* (a ^Oc) Therefore , o distributes over *    

Q25. * is a binary operation on Z such that : a * b =  - a + b + ab.Then the value of 2 *(-3)*5  is

•   -11
•   11
•     39
• -39

Solution

2*(-3) = -2-3+2(-3) = -5-6 =-11 2*(-3)* 5 =(-11)* 5 = -(-11) +5 + (-11) x 5  = 11+5 - 55 = -39

Q26. Let f = { (1,3), (2,1) , (3,2)} and g= {(1,2), (2,3) , (3,1)} . What is gof(2)?

•     1
•     2
•     3
•      Does not exist

Solution

  gof(2) = g {f(2)} = g(1) = 2

Q27. Let A = {1,2,3,4,5,6,7}. P={1,2}, Q = {3, 7}. Write the elements of the set R so that P, Q and R form a partition that results in equivalence relation.

•   {4,5,6}
• {1,2,3,4,5,6,7}
•   { }
•   {0}

Solution

  With the option (i) the sets P , Q and R will be pair wise disjoint and their Union would result in A

Q28. If f(x) = (3 - x²)^1/2, then fof is:

• x
• x^1/2
• ½ x
• -x^1/2

Solution

Q29. Which domain and co-domain makes the function one-one if it is onto and onto if it is one - one ?

•   A→A, where A is any set
•     R→N
•     N→R
•     None of these 

Solution

  When a mapping is from a set to itself then one-one and onto implies each other.

Q30. If f: A→B and g:B→C are onto , then gof:A→C is:

• an onto function
•  an into function
•  a many-one and onto function
• a bijective function

Solution

Let c∈C g:B→C is onto ⇒c∈Chas a pre image ⇒There exists a b ∈B  such that g(b) = c b ∈B   and f: A→B is onto ⇒ b∈ B  has a pre image ⇒There exists an a ∈A  such that f(a) = b c = g(b) = g[f(a)] = gof (a) So,  c∈Chas a pre image a∈A , such that gof(a) = c →gof is an onto function

Q31. Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)} be a relation on the set A = {3, 6, 9, 12}. Then, R is

• An equivalence relation
• Reflexive and symmetric only
• Symmetric only
• Reflexive and transitive only

Solution

For all a R, (a, a) R. So, R is reflexive. Now, (6, 12)  R but (12, 6)R. So, R is not symmetric. If (a, b), (b, c) R then (a, c) R for all a, b, c in A.  So, R is transitive.

Q32. A binary operation _* on R is such that a_*b=a-b. Then a_*(b_*c)=

• a-b+c
• a-b-c
• a+b-c
• b-a-c

Solution

a*(b*c)= a*(b- c)=a-(b-c) =a-b+c

Q33. * is a binary operation on Z such that: a * b = a + b + ab. The  solution  of  (3* 4) *x = - 1 is

•   -1
•   1
• 3
• 4

Solution

  (3* 4) *x = - 1 (3* 4) *x = (3 + 4 +3 .4)*x = (7 +12)*x =19*x= -1  ( given a*b = a + b +ab ) 19 +x + 19(x) = 19 + 20x = -1 20 x = -20 x =-1

Q34. For  which one of these mappings the function  f(x) = x ² will be one-one?

•   N→N
•    Z→Z
•   R→R
•   None of these

Solution

  Let x₁ and x₂ ∈ Domain  such that f(x₁) = f(x₂) i.e x₁ ² = x₂ ² ⇒ x₁ =± x₂ If the domain is N then x₁ = x₂ (which is true)

Q35. If f(x) = |x| and g(x) = | 5x – 2 |. Then, fog =

• f
• |g|
• | f |
• f²

Solution

(fog)(x) = f(g(x)) = f(| 5x – 2 |) == |g|

Q36. Binary operation 'subtraction' on integers is

• Commutative and associative
• Commutative, but not associative
• Associative, but not commutative
• Neither commutative nor associative

Solution

A binary operation on  is said to be  a)ASSOCIATIVE: If (a*b)*c=a*(b*c)   a,b,cZ b)COMMUTATIVE:If a*b=b*a  a,b Z. But,subtraction is not associative on integers as (8-5)-38-(5-3) Let , then . So, subtraction is not commutative on integers.

Q37. Let * be any binary operation on the set R defined by a * b = a + b – ab, then the binary operation * is

• Commutative
• Commutative and associative
• Commutative but not associative
• Associative

Solution

For commutativity: a * b = a + b – ab and b * a = b + a – ba = a + b – ab So, a * b = b * a. So, * is commutative.   For associativity: a * (b * c) = a * (b + c – bc) = a + b + c – bc – ab – ac +abc, and (a * b) * c = (a + b – ab) * c = a + b – ab + c – ac – bc + abc. So, a * (b * c) = (a * b) * c.

Q38. Which one of these co domains would make the function bijective. f : N → N, given by f (x) = 2x?

•   A = {a: a = 2n , n ∈N }
•   A = {a: a = n , n ∈N }
•   A = {a: a = n/2 , n ∈N }
•    A = {a: a = 2n , n ∈Z }

Solution

  The function f is one-one, for f (x₁) = f (x₂) ⇒2x₁= 2x₂ ⇒x₁= x₂. Further, f is not onto, as for 1 ∈N, there does not exist any x in N such that f (x) = 2x = 1.  So, fis one-one but not onto. If , the Codomain is A = {a: a = 2n , n ∈N }, then any y in A, there will be an x ∈N  such that f (y) = y

Q39. Let A = {1,2,3,4} and B = { x,y,z}. Then R = {(1,x) , ( 2,z), (1,y), (3,x)} is

•    relation from A to B
•    relation from B to A
•    relation from B to B
•   Is  not  a relation

Solution

  The given set R= {(1,x) , ( 2,z), (1,y), (3,x)}  is a subset of the cross product A x B  and any subset of the cross product A x B  is a relation from A to B , so  R= {(1,x) , ( 2,z), (1,y), (3,x)} is a relation from A to B.  

Q40. How many onto functions from set A to set A can be formed for the set A = { 1,2,3,4,5,......n} ?

• n
• 2n
• n²
• n!