__ Answer__:19 to 19

**Data**

f: R -> R

continuous in [-3.3]

differentiable in (-3,3)

f(-3) = 7

f'(x) < = 2

__ Calculation__:

=>f'(x) < = 2

Integrating both side from -3 to 3

\(\mathop \smallint \limits_{ - 3}^3 f'\left( x \right)dx <= 2\mathop \smallint \limits_{ - 3}^31dx\)

[f(3)-f(-3)] <= 2[3-(-3)]

f(3)<= 7 + 2(6)

f(3)<=19

Option 2 : \(\left( {\begin{array}{*{20}{c}} 0&-1\\ -1&0 \end{array}} \right)\)

__Concept:__

If x_{1} and x_{2} are the components of a vector X with respect to a standard basis, this means

X = [x_{1}, x_{2}]^{T} = x_{1} e_{1} + x_{2} e_{2} where {e1, e2} is standard ordered basis.

**Reflection theorem:**

Let T : R^{2} → R^{2} be a linear transformation given by reflecting vectors over the line x_{2} = m x_{1}. Then the matrix of T is given by

\(\frac {1}{1+{m^2}} \left[ {\begin{array}{*{20}{c}} {1-m^2}&2m\\ 2m&m^2-1 \end{array}} \right]\)

__Calculation:__

Given T is the linear transformation that reflects the points through the line x1 = -x2

⇒ m = -1

The line x1 = -x2 is the bisector of the second and fourth quadrant and the reflection through this line is represented by a matrix

\(T = \left( {\begin{array}{*{20}{c}} 0&-1\\ -1&0 \end{array}} \right)\)

Option 3 : Empty set

**General points: **

- sin(x) is the function having a range between -1 and +1.
- Log(x) is defined only when x is positive and greater than zero.

log (sin(x)) is defined only when 0 < sin(x) ≤ 1, and then range will be (−∞,0]

So, log [log (sin(x))] is undefined as the logarithm of non-positive numbers isn’t defined for real numbers.

Hence, Domain of log [ log (sin(x))]: ∅ (empty set)

Also, Range of log [log (sin(x))]: ∅ (empty set)

Option 4 : S_{1} is a basis for ℝ^{2} but S_{2} is not a basis for ℝ^{2}.

__Concept:__

Every basis for the vector space R^{n} consists of n vectors.

Let S be a subset of a vector space R^{n}, the **basis is a linearly independent spanning set** for R^{n}.

A set of vectors {v_{1}, v_{2},…, v_{p}} in a vector space V is said to be **linearly independent** if the vector equation c_{1}v_{1} + c_{2}v_{2 }+…+ c_{p}v_{p} = 0 has only one trivial solution **c _{1} = 0, c_{2} = 0,…, c_{p} = 0;**

The set is said to be **linearly dependent** if there exists weights **c _{1}, c_{2},…, c_{p} not all 0**, such that c

__Calculation:__

Given S1 = {[1, -2], [3, 5]} and S2 = {[1, 1], [0, 0]}

Let S_{1} = (v_{1}, v_{2}) and the vector equation be av_{1} + bv_{2} = 0

S1 = {[1, -2], [3, 5]} ⇒ v1 = [1, -2], v2 = [3, 5];

⇒ a + 3b = 0, -2a + 5b =0 ⇒ **a = 0, b = 0**

Since, both the constants are zero, **S1 will be linearly independent. **

Let S_{2} = (v1, v2) and the vector equation be av1 + bv2 = 0

S2 = {[1, 1], [0, 0]} ⇒ v1 = [1, 1], v2 = [0, 0];

⇒ a = 0, a = 0 ⇒ a = 0 and b can be any real number

Since many trivial solutions, **S2 will be linearly dependent.**

**∴ S1 is a basis for ℝ2 but S2 is not a basis for ℝ2.**

Option 4 : ABC + ABC' + AB'C

__Concept:__

Important Axioms and De Morgan's laws of Boolean Algebra:

- Double inversion \(\overline{\overline A} = A\)
- A . A = A
- A . \(\overline A \) = 0
- A + 1 = 1
- A + A = A
- A + \(\overline A \) = 1

De Morgan's laws:

Law 1: \(\overline {{\bf{A}} + {\bf{B}}} = \overline{A}\;.\overline B\)

Law 2: \(\overline {{\bf{A}}\;.{\bf{B}}} = \overline A +\overline B\)

**Calculation:**

Let the given function be Y

Y = AB + AC

Now expanding by using the important properties of boolean algebra:

Y = AB(C + C̅) + AC(B + B̅)

Y = ABC + ABC̅ + ACB + ACB̅

As ABC + ACB = ABC

Y = ABC + ABC̅ + ACB̅

Y can also be written as:

Y = ABC + ABC' + AB'C

Hence **option (4)** is the correct answer.

__Important Points__

Name |
AND Form |
OR Form |

Identity law |
1.A=A |
0+A=A |

Null Law |
0.A=0 |
1+A=1 |

Idempotent Law |
A.A=A |
A+A=A |

Inverse Law |
AA’=0 |
A+A’=1 |

Commutative Law |
AB=BA |
A+B=B+A |

Associative Law |
(AB)C |
(A+B)+C = A+(B+C) |

Distributive Law |
A+BC=(A+B)(A+C) |
A(B+C)=AB+AC |

Absorption Law |
A(A+B)=A |
A+AB=A |

De Morgan’s Law |
(AB)’=A’+B’ |
(A+B)’=A’B’ |

Consider two subsets of R^{3} given as

S_{1} = {[1, -1, 2], [3, 2, -1]} and S_{2} = {[2, 7, -3],[-6, -21, 9]}. Then:

Option 2 : S_{1} can be enlarged but S_{2} cannot be enlarged to a basis for R3

__Concept:__

Let S = {v_{1}, v_{2}, . . . , v_{n}} be a set of n vectors in a vector space V .

If **S is linearly independent and the dimension of V is n, then S is a basis of V.**

If the **dimension of S is less than Dimension of V, then S cannot be basis**, but **can be extended to a basis if S is linearly independent.**

A set of vectors {v1, v2,…, v_{n}} in a vector space V is said to be linearly independent if the vector equation c1v1 + c2v2 +…+ c_{n}v_{n} = 0 has only one trivial solution c1 = 0, c2 = 0,…, c_{n} = 0;

The set is said to be linearly dependent if there exists weights c1, c2,…, c_{n} not all 0, such that c1v1 + c2v2 +…+ c_{n}v_{n} = 0

__Calculation:__

Given S1 = {[1, -1, 2], [3, 2, -1]} and S2 = {[2, 7, -3],[-6, -21, 9]}

Let S1 = (v1, v2) and the vector equation be av1 + bv2 = 0

S1 = {[1, -1, 2], [3, 2, -1]} ⇒ v1 = [1, -1, 2], v2 = [3, 2, -1];

⇒ a + 3b = 0, -a + 2b =0, 2a - b = 0 ⇒ a = 0, b = 0

Since, both the constants are zero, S1 will be linearly independent.

Hence S1 can be extended to basis.

Let S2 = (v1, v2) and the vector equation be av1 + bv2 = 0

S2 = {[2, 7, -3],[-6, -21, 9]} ⇒ v1 = [2, 7, -3], v2 = [-6, -21, 9];

⇒ 2a - 6b = 0, 7a - 21b = 0, -3a + 9b = 0 ⇒ a = 3b, a and b can be any real number

Since many trivial solutions, S2 will be linearly dependent.

Hence S2 cannot be extended to basis.

Option 3 : (¬ p ⋀ q) ⋁ (p ⋀ ¬ q)

**Formula:**

p ↔ q ≡ (¬ p ∧ ¬ q) ∨ (p ∧ q) = p̅.q̅ + p.q

p → q = ¬ p ∨ q ≡ p̅ + q

**Derivation:**

__Option 1:__

(¬ p ⋁ q) ⋀ (p ⋁ ¬ q) ≡ (p̅ + q).(p + q̅ ) ≡ p̅.q̅ + pq

__Option 2:__

(¬ p ⋁ q) ⋀ (q → p) ≡ (¬ p ⋁ q) ⋀ (¬ q ∨ p) ≡ (p̅ + q).( q̅ + p) ≡ p̅.q̅ + pq

__Option 3:__

(¬ p ⋀ q) ⋁ (p ⋀ ¬ q) ≡ p̅.q + p.q̅ ≠ p̅.q̅ + pq

__Option 4:__

(¬ p ⋀ ¬ q) ⋁ (p ⋀ q) ≡ p̅.q̅ + p.q

Therefore, (¬ p ⋀ q) ⋁ (p ⋀ ¬ q) is NOT equivalent to p ↔q

The symmetric difference of sets A = {1, 2, 3, 4, 5, 6, 7, 8} and B = {1, 3, 5, 6, 7, 8, 9} is

Option 2 : {2, 4, 9}

Symmetric difference of two sets is set which contains elements which are in exactly one set.

\(A\Delta B = (A-B) \cup (B-A)\)

= {2, 4, 9}

Option 4 : 𝑅 is symmetric but not reflexive and not transitive

R be the relation on the set of positive integers such that aRb if and only if a and b are distinct and have a common divisor other than 1.

Let A = {1, 2, 3, 4, 5, 6…}

aRb ≡ (a, b}

- R cannot be reflexive because aRa, a and a are not distinct.

Example: R = {(4,6)}

- R is symmetric because if aRb is there then bRa is also possible, in both a and b are distinct.

Example: R = {(6, 4), (4,6)}

- R is not transitive as aRb and bRc doesn’t mean aRc.

Example: R = {(6, 4), (4,6)} since (6,6) cannot be included and hence it cannot be transitive

So, given relation is symmetric but not transitive and not reflexive.

**Data:**

R = {1, 2, 3}

Number of elements in a set = n = 3

**Formula: **

Total number of binary relations on a set with n elements = \({2^{{n^2}}}\)

Total number of reflexive relations on a set with n elements = \({2^{n\left( {n - 1} \right)}}\)

**Calculation:**

Number of binary relations possible = \({2^{{3^2}}} = {2^9}\) = 512

Number of reflexive relations = \({2^{3*2}} = {2^6} = 64\)

Required probability: \(\frac{{64}}{{512}} = \frac{1}{8} = 0.125\)

If A = {x, y, z} and B = {u, v, w, x} and the universe is {s, t, u, v, w, x, y, z}

Then (A ∪ B̅) ∩ (A ∩ B) is equal toOption 2 : {x}

**Data:**

A = {x, y, z}

B = {u, v, w, x}

Universe set = {s, t, u, v, w, x, y, z}

**To find:** (A ∪ B̅) ∩ (A ∩ B)

**Explanation:**

B̅ = {s, t, y, z}

(A ∪ B̅) = {x, y, z} ∪ {s, t, y, z} = {s, t, x, y, z}

(A ∩ B) = {x, y, z} ∩ {u, v, w, x} = {x}

∴ (A ∪ B̅) ∩ (A ∩ B) = {s, t, x, y, z} ∩ {x} = {x}

**Note:**

This question was excluded for evaluation in ISRO 2020 CSE.

Option 2 is changed to get correct answer.Option 3 : the primal objective value at x is less than the dual objective value at y

__Concept:__

Duality in Linear programming problem (LPP): It means a linear programming problem has another LPP which is derived from it.

**Original LPP is known as primal and derived LPP is known as Dual.**

Dual:

Dual can be found by using the below formula,

If primal is given as

Maximize CTx, subject to Ax ≤ b

Then dual will be

Minimize bTy, subject to A^{T}y ≥ c

If x is feasible for the primal, and y is feasible for the dual, then

**C ^{T}x ≤ b^{T}y**

**That is primal objective is less than or equal to Dual objective.**

**At an optimal feasible solution, the primal objective is equal to the dual objective.**

**At a non-optimal feasible solution, the primal objective is less than the dual objective.**

If either the primal or the dual problem has a finite optimal solution, then the other problem also has a finite optimal solution.

If either problem has an unbounded optimum solution, then the other problem has no feasible solution at all

Option 3 : unbounded above in **R**

__Concept:__

A natural number is a number that occurs commonly and obviously in nature. As such, it is a non-negative number. The set of natural numbers can be denoted by

N = {1, 2, 3, 4,....}

The set of natural numbers is bounded below and **not bounded above in R.**

We can prove not bounded above in R using contradiction.

**Proof:**

Assume by way of contradiction that N is a bounded above. Then, since N is not empty, it follows from the **completeness axiom that sup(N) exists.** Thus there must be m ∈ N such that

sup(N) - 1 < m (**sup(N) means supremum of N or least upper bound**)

⇒ sup(N) < m + 1

As m ∈ N , also **m + 1 ∈ N, Which is a contradiction.**

∴ **N is not bounded above**

Option 2 : f(x) = -x, x ∈ R

__Explanation:__

Function:

A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. Let A & B be any two non-empty sets, mapping from A to B will be a function only when every element in set A has one end only one image in set B.

An injective function means one-one.

Consider f(x) = -x

Let f(x) = f(y) ∀ x, y ∈ R

⇒−x = −y ⇒ x = y

For every value of x, we get a different value of f. Hence, it is injective.

__Additional Information__

**One – One function (Injective function)**

If each element in the domain of a function has a distinct image in the co-domain, the function is said to be one-one function.

**For examples** f; R R given by f(x) = 3x + 5 is one – one.

**Many – one function**

On the other hand, if there are at least two elements in the domain whose images are same, the function is known as many to one.

**For example** f : R R given by f(x) = x^{2} + 1 is many one.

If (G, ⋅) is a group such that (ab)-1 = a-1 b-1, ∀ a, b ∈ G, then G is a/an

Option 2 : Abelian group

A group is said to be abelian if (a*b) = (b*a) ∀a,b ∈G

Since (G, ⋅) is a group

∴ (ab)^{-1 }= (b^{-1}a^{-1}) (1)

Given: (ab)^{–1} = a^{–1}b^{–1 }(2)

From (1) and (2)

(a*b) = (b*a)

Therefore it is abelian

Option 2 : x - 2

**Given:**

Factor of the x3 + ax + b is,

(x - 1), and (x + 3)

**Calculation:**

**Let,**

(x - β) is the remaining factor of the given equation then,

⇒ (x - β)(x - 1)(x + 3) = x3 + ax + b

⇒ (x - β)[x^{2} + 2x - 3] = x3 + ax + b

⇒ x^{3} + 2x^{2} - 3x - βx^{2} - 2βx + 3β = x3 + ax + b

⇒ x^{3} + x^{2}(2 - β) - x(3 + 2β) + 3β = x3 + ax + b

**On comparing both side,**

⇒ 2 - β = 0

**β = 2**

Hence,** the remaining factor is (x - 2)**

Option 2 : a ∈ H ⇒ a^{-1} ∈ H

**Concept:**

A non-empty subset H of a group (G, ∗) is a group of G iff,

⇒ **a, b ∈ H ⇒ a ∗ b ∈ H**

⇒ a ∈ H ⇒ 1/a ∈ H ⇒ a^{-1} = 1/a

Which of the following is a functionally complete set of gates?

(i) NAND (ii) NOT

Option 1 : I but not II

The Correct Answer is __ I but not II__.

**NAND gate**is a functionally**complete set of gates.**- In the logic gate, a functionally complete collection of logical connectives or Boolean operators is one which can be used to express all possible truth tables by combining members of the set into a Boolean expression.
- A well-known complete set of connectors is {AND, NOT} and each of the singleton sets {NAND} is functionally complete, consisting of binary conjunction and negation.
- A NAND gate is a logic gate that generates a false output only if all its inputs are valid, so its output is complementary to that of an AND gate.
- A low output only results if all the inputs to the gate are high; a high output results if any input is low.

__Key Points__

Input A |
Input B |
Output |

0 | 0 | 1 |

0 | 1 | 1 |

1 | 0 | 1 |

1 | 1 | 0 |

Suppose A is a finite set with n elements. The number of elements and the rank of the largest equivalence relation on A are

Option 3 : {n^{2}, 1}

**Concept:**

A relation ‘R’ on a set A is said to be equivalent relation of ‘A’ if A is

1) Reflexive 2) Symmetric 3) Transitive

**Example**

A = {p, q} ∴ n = 4

R = {(p, p), (p, q), (q, p), (q, q)}

∴ Reflexive

The largest equivalence relation is n^{2} = 2^{2} = 4

Diagraph of this relation will have only one connected component, hence this relation has rank 1.

Let *G *be a finite group on 84 elements. The size of a largest possible proper subgroup of *G *is ________.

**Lagrange’s theorem specifies: **Order of subgroup divides the order of a group

order of a group= number of elements in a group = 84

Let X be its any subgroup

X ⊆ G

Divisor of 84 = 1, 2, 3, 4, 6, 7, 12, 14, 21 and 42

∴ size of X must be 1, 2, 3, 4, 6, 7, 12, 14, 21 and 42

Since in the question, it said that size largest possible proper subgroup(X ≠ 84) of *G*

Hence size of X is 42