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All questions of Functions for EmSAT Achieve Exam

 The Shaded region in the following figure illustrates
  • a)
    A ∩ ( B ∪ C)
  • b)
    A ∩ B ∩ C
  • c)
    A ∪ B ∪ C
  • d)
    (A ∩ B) ∪ (A ∩ C)
Correct answer is option 'D'. Can you explain this answer?

New Words answered
First which region is over which region Then We will see that A is on the B so A intersection B and after C is on the A so, A intersection C after that we have to take all intersection part so A intersection B is Union with A intersection C.
The shaded region represents (A ∩ B) ∪ (A ∩ C).

If A = {a, b, c} then the number of proper subsets of A are:
  • a)
    3
  • b)
    10
  • c)
    7
  • d)
    8
Correct answer is option 'C'. Can you explain this answer?

Pooja Nair answered
  • Number of proper subsets of a given set = 2m - 1, where m is the number of elements.
  • Here the number of elements is 3. So the number of proper subsets of A = 23 - 1 = 7.

Can you explain the answer of this question below:

If A = {5, 10, 15}, B = ϕ, then B – A is

  • A:

    5

  • B:

    {5,10}

  • C:

    ϕ

  • D:

    {5,10, 15}

The answer is c.

Om Desai answered
If A = {5, 10, 15}, B = ϕ
B - A will have those elements which are in B but not in A.
B - A = ϕ

 Which of the following is not an empty set?
  • a)
    {x : x is a multiple of 7, x < 7, x ∈ N}
  • b)
    Set of common points of two parallel lines in a plane
  • c)
    {x : 6 + 2x > 5x + 3, x ∈ N}
  • d)
    Set of smallest whole number
Correct answer is option 'D'. Can you explain this answer?

Ayush Joshi answered
As, the set of Whole numbers is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...};
So, according to the set of whole numbers above, the smallest whole number should be "0" and therefore the set of smallest whole number is not empty.

For real number x and y, we writeis an irrational
number. Then the relation R is:​
a)Reflexive
b)Symmetric
c)Transitive
d)Equivalence
Correct answer is option 'A'. Can you explain this answer?

xRy => x - y + √2 is an irrational number.
Let R is a binary relation on real numbers x and y.
Now, R is transitive iff for all (x, y) ∈ R and (y, z) ∈ R implies (x, z) ∈ R
Given, xRy => x - y + √2 is irrational    ............1
and yRz => y - z + √2 is irrational       ............2
Add equation 1 and 2, we get
   (x - y + √2) + (y - z + √2) is irrational
= x - z + √2 is irrational
= xRz is irrational
So, the relation R is transitive.

If f(x) = x2 and g(x) = cosx, which of the following is true?
  • a)
    f + g is even function
  • b)
    f – g is an odd function
  • c)
    f + g is not defined
  • d)
    f + g is an odd function
Correct answer is option 'A'. Can you explain this answer?

Raghav Bansal answered
if f(x) is an odd function
So, f(−x)=−f(x)
F(−x)=cos(f(−x))
=cos(−f(x))
=cos(f(x))
=F(x)
So cos(f(x)) is an even function
So, f(x) and g(x) is an even function

The set A = {1,4,9,16,25—} in set builder form is written as
  • a)
    A = {x:x is a prime number}
  • b)
    A ={x:x is the cube of a natural number}
  • c)
    A = {x:x is the square of a natural number}
  • d)
    A = {x:x is an even natural number}
Correct answer is 'C'. Can you explain this answer?

Krishna Iyer answered
  • We know that, 12 = 1, 22 = 4, 32 = 9, 42 = 16, 52 = 25
  • Therefore the set A = {1, 4, 9, 16, 25...} can be written in set builder form as: 
    A = {x: x is the square of a natural number}

 If ordered pair (a + 2b, 9) = (7, 3a + 2b), then the values of a and b are
  • a)
    9, 7
  • b)
    1, 3
  • c)
    7, 9
  • d)
    3, 1
Correct answer is option 'B'. Can you explain this answer?

A+2B=7------(1)
3A+2B=9------(2)
Subtracting (2) from (1)
we get, 2A=2
A=1
Subsitute A=1 in (1)
1+2B=7
2B=6
B=3
therefore A = 1, B = 3
 

For the set of all natural numbers the universal set can be
  • a)
    Set of all odd numbers
  • b)
    Set of all even numbers
  • c)
    Set of all integers
  • d)
    Set of all prime numbers
Correct answer is option 'C'. Can you explain this answer?

Krishna Iyer answered
Integers contain all the natural numbers. So it can be a universal set for natural numbers. In other options, there are only some of the elements of natural numbers.

If A = { (1, 2, 3}, then the relation R = {(2, 3)} in A isa)symmetric and transitive 

Rajib Dhar answered
The relation is transitive for elements in the {a,b,c} if (a,b) belongs to r and again if (b,c) belongs to r and (a,c ) also belongs 

Number of binary sets on the set {p, q,r} is:​
  • a)
    39
  • b)
    16
  • c)
    18
  • d)
    36
Correct answer is option 'A'. Can you explain this answer?

Guru Randhawa answered
It os must to remember these basic results the answer is 1 at 8 corners 1/8 sphere is present :8*1/8=1 that means overall( effective) one atom is present in ssc

 Choose the incorrect statement
  • a)
    If a set has only one element, we call it a singleton set.
  • b)
    Set of all even prime numbers is a subset of set of all natural numbers.
  • c)
    Φ is not a subset of any set.
  • d)
    Every set is a subset of itself.
Correct answer is option 'C'. Can you explain this answer?

Rohit Joshi answered
set A is a proper subset of a set B if A is a subset of B and there is at least one element of B that's not an element of A. Thus, the void set is a subset of all sets, and it's a proper subset of every set except itself

If A = {1, 2, 3}, and B = {3, 6} then the number of relations from A to B is
  • a)
    32
  • b)
    23
  • c)
    23
  • d)
    26
Correct answer is option 'D'. Can you explain this answer?

Preeti Iyer answered
The number of relations between sets can be calculated using 2mn where m and n represent the number of members in each set.
So, number of relations from A to B is 26.

 If U= set of all whole numbers less than 12, A=set of all whole numbers less than 10, B= Set of all odd natural numbers less than 10, then what is (A∩B)’?
  • a)
    {3,5,7,9}
  • b)
    {0,1,3,5,7,9}
  • c)
    {0,2,4,6,8,10,11}
  • d)
    {1,3,5,7}
Correct answer is option 'A'. Can you explain this answer?

Hansa Sharma answered
U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
B = {1, 3, 5, 7, 9}
A ∩ B = {1, 3, 5, 7, 9}
(A ∩ B)’ = U - (A ∩ B)
(A ∩ B)’ = {0, 2, 4, 6, 8, 10, 11}

Let R be a relation on N (set of natural numbers) such that (m, n) R (p, q)mq(n + p) = np(m + q). Then, R is​
  • a)
    An Equivalence Relation
  • b)
    Only Reflexive
  • c)
    Symmetric and reflexive.
  • d)
    Only Transitive
Correct answer is option 'C'. Can you explain this answer?

Anaya Patel answered
(m, n) R (p, q) <=> mq(n + p) = np(m + q)
For all m,n,p,q € N
Reflexive:
(m, n) R (m, n) <=> mn(n + m) = nm(m + n)
⇒ mn2 + m2n = nm2 + n2m
⇒ mn2 + m2n = mn2 + m2n
⇒ LHS = RHS
So, (m, n) R (m, n) exists.
Hence, it is Reflexive
Symmetric:
Let (m, n) R (p, q) exists
mq(n + p) = np(m + q) --- (eqn1)
(p, q) R (m, n) <=> pn(q + m) = qm (p + n)
⇒ np(m + q) = mq(n + p)
⇒ mq(n + p) = np(m + q)
This equation is true by (eqn1).
So, (p, q) R (m, n) exists
Hence, it is  not symmetric.
Transitive:
Let (m, n) R (p, q) and (p, q) R (r, s) exists.
Therefore,
mq(n + p) = np(m + q) --- (eqn1)
ps(q + r) = qr (p + s) --- (eqn2)
We cannot obtain ms(n+r) = nr(m+s) using eqn1 and eqn2.
So, ms(n + r) ≠ nr(m + s)
Therefore, (m, n) R (r, s) doesn’t exist.
Hence, it is transitive.

From the sets given below, select equal sets :
A = { 2, 4, 8, 12}, B = { 1, 2, 3, 4},
C = { 4, 8, 12, 14}, D = { 3, 1, 4, 2}
  • a)
    A and C
  • b)
    A and B
  • c)
    B and D
  • d)
    B and C
Correct answer is option 'C'. Can you explain this answer?

Mansi Chopra answered
 The sets are equal, if they have the exact same elements in them. Since option B & D have exactly same number of elements in them So, B & D are equal sets. 

If f(x) = ax + b and g(x) = cx + d, then f[g(x)] – g[f(x)] is equivalent to​
  • a)
    f(c) + g(a)
  • b)
    f(a) – g(c)
  • c)
    f(d) – g(b)
  • d)
    f(b) – g(b)
Correct answer is option 'C'. Can you explain this answer?

Leelu Bhai answered
F(g(x)) - g(f(x)) = f(cx + d) - g(ax + b)= a(cx + d) + b - c(ax + b) - d= acx + ad + b - acx - bc - d= ad + b - bc - dnow by putting each 

Consider the set A of all divisors of 30. How many subsets of A contains even divisors only?
  • a)
    2
  • b)
    16
  • c)
    28
  • d)
    4
Correct answer is 'B'. Can you explain this answer?

Himaja Ammu answered
Set of divisors of 30={1,2,3,6,10,15,30} in these the even divisors r={2,6,10,30} we know no.of subsets to any set=2^n so answer is 2^4=16

 The number of elements in the Power set P(S) of the set S = [ [ Φ] , 1, [ 2, 3 ]] is
  • a)
    2
  • b)
    4
  • c)
    8
  • d)
    None of these
Correct answer is option 'C'. Can you explain this answer?

Preeti Iyer answered
There’s a result in mathematics used for this. It says that a power set B of any set A is a set of all the subsets of A and the number of elements of B will be 2^n where n is the number of elements of A.
So taking your question as an example;
A = {1,2,3}
B : set of all subsets of A
List out all the subsets of A - {1},{2},{3},{1,2},{2,3},{1,3},{1,2,3},{empty set}
Number of elements in A (n) = 3 
so 23 = 8
So, B = {{1},{2},{3},{1,2},{2,3},{1,3},{1,2,3},{empty set}} 
and the number of elements are 8.

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