Functions MCQ Quiz in मल्याळम - Objective Question with Answer for Functions - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

If A = m and B = n, then number of onto functions possible from A to B is

n^m – ⁿC₁(n - 1)^m + ⁿC₂(n-2)^m – ⁿC₃(n-3)^m+…………(-1)^{n-1 n}C_n-1(1)^m

Explanation:

Here m = 5 and n = 4

Required number of ways = 4⁵ – ⁴C₁(3)⁵ + ⁴C₂(2)⁵ – ⁴C₃(1)⁵ + 0

= 1024 – 243 × 4 + 6 × 32 – 4

= 1024 – 972 + 192 – 4

Let f(1) = a

Given:

f(n) = f(n/2) if n is even

f(n) = f(n + 5) if n is odd

Now, f (2) = f (2/2) = f(1) = a   [Because 2 is even]

For n = 3; 3 is odd

f(3) = f (3 + 5) = f (8) = f(8/2) = f(4) = f(4/2) = f (2) = f(2/2) = f(1) = a

For n = 4; 4 is even

f(4) = f (4/2) = f(2) = f(2/2) = f(1) = a;

For n = 5, 5 is odd

f(5) = f(5 + 5) = f(10) = f(10/2) =f(5) = f(5 + 5) = f(10) = f(10/2) = f(5) = b    [As f(5) is repeating ,let f(5) = b]

for n= 6

f(6) = f(6/2) = f(3) = x

So, there are two values only a and b. All multiples of 5 are b and others have value a.

So, the maximum possible size of R is 2.

The correct answer is (A), (B) and (C) Only

EXPLANATION:

The properties of the relation R defined as R ↔ xv = yu.

• (A) Reflexive:
  • For a relation to be reflexive, (a, a) must be in the relation for every element a in the set.
  • In this case, we have x = u and y = v. So, (u, u) and (v, v) must be in the relation.
  • Since N is the set of natural numbers, for any natural number u or v, (u, u) and (v, v) are in the relation. Therefore, the relation is reflexive.
• (B) Symmetric:
  • For a relation to be symmetric, if (a, b) is in the relation, then (b, a) must also be in the relation for every pair (a, b). In this case, (x, y) is in the relation if and only if (y, x) is in the relation, as xv = yu implies yu = xv. Therefore, the relation is symmetric.
• (C) Transitive:
  • For a relation to be transitive, if (a, b) and (b, c) are in the relation, then (a, c) must also be in the relation for every triplet (a, b, c). In this case, if xv = yu and yw = zv, then multiplying these equations gives xw = zu, which implies (x, w) is in the relation. Therefore, the relation is transitive.
• (D) Asymmetric:
  • For a relation to be asymmetric, if (a, b) is in the relation, then (b, a) must not be in the relation for any pair (a, b). In this case, since (x, y) is in the relation if and only if (y, x) is also in the relation, the relation cannot be asymmetric.

Based on the analysis: The correct answer is: 4) (A), (B) and (C) Only

Concept:

The function is continuous if it satisfies the following conditions.

\(\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{b}}{( 0-)}}} {\rm{f}}\left( {\rm{x}} \right) = \mathop {\lim }\limits_{{\rm{x}} \to {{\rm{b}}{ (0-)}}} {\rm{f}}\left( {\rm{x}} \right) = {\rm{f}}\left( {\rm{b}} \right){\rm{}}\)

Otherwise, the function is not continuous at x = b

Analysis:

To check the continuity at x = 0 and x = 1,

at x = 0,

\(\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{0}}}} f(x) = \mathop {\lim }\limits_{{\rm{x}} \to {{\rm{0}}}}( -x)\)

= 0

at x = 0^-,

f(x) = -x ; f(0^-) = 0

at x = 0⁺, 

f(x) = x ; f(0⁺) = 0

∵ Left-hand limit = Right-hand limit = Value of the function at the point. 

∴ The function is continuous at x = 0

at x = 1,

\(\mathop {\lim }\limits_{{\rm{x}} \to {{\rm{1}}}} f(x) = \mathop {\lim }\limits_{{\rm{x}} \to {{\rm{1}}}}(2 -x)\)

= 1

at x = 1-,

f(x) = x ; f(1-) = 1

at x = 1+, 

f(x) = 2 - x ; f(1+) = 1

∵ Left-hand limit = Right-hand limit = Value of the function at the point. 

∴ The function is continuous at x = 1

Important Points

The following properties are true in calculus:

• If a function is differentiable at any point, then it is necessarily continuous at the point.
• But the converse of this statement is not true i.e. continuity is a necessary but sufficient condition for the Existence of a finite derivative.
• Differentiability implies Continuity
• Continuity does not necessarily imply differentiability.

Concept:

Concept:

If root of the function are a, b and c

then Polynomial function f(x) = (x - a)(x - b)(x - c)

Explanation:

Three roots are 1, 2 and 3

f(x) = (x - 1)(x - 2)(x - 3)

f(0) = -1 × -2 × -3 = -6

f(4) = 3 × 2 × 1 = 6

∴ f(0)× (4) = -6 × 6 = -36

Hence f(0) f(4) < 0

Data:

g(x) = 1 – x

\(h\left( x \right) = \frac{x}{{x - 1}}\) 

Calculation:

\(g\left( {h\left( x \right)} \right) = 1 - \;\frac{x}{{x - 1}} = \; - \frac{1}{{x - 1}}\)

\(h\left( {g\left( x \right)} \right) = \;\frac{{1 - x}}{{1 - x - 1}} = \frac{{1 - x}}{{ - x}}\;\)

n(X) = 3

n(Y) = 8

Function from Y to X

Total number of function possible = n(X)^n(Y) = 3⁸ = 6561

Total number of onto function possible is

\({3^8}{ - ^3}{C_1}{\left( {3 - 1} \right)^8}{ + ^3}{C_2}{\left( {3 - 2} \right)^8} - {\;^3}{C_3}{\left( {3 - 3} \right)^8}\;\)

3⁸ – 3 × 2⁸ + 3 = 5796

the probability of g being one-to-one = \(\frac{{5796}}{{6561}} = 0.8834\;\)

Hence if \(gof\) is injective then \(f\) is injective but \(g\)  need not to be.

Hence if \(gof\)  is surjective then \(g\) has to be onto but \(f\)  need not be.

Given:

The conditions for a certain type of function, namely that it is both injective and surjective.

Concept used:

A bijective function is a function that is both injective (one-to-one) and surjective (onto).

Injective means Every element of domain has a preimage in Codomain and it should be unique.

Surjective means No element of Codomain left without preimage 

Calculation:

By the defination "f : A → B" is both injective (one-to-one) and surjective (onto)  This type of function is classified as a "Bijective function". 

Hence, Option D is correct

Given:

Our options represent various function types

Concept used:

A self-map, or endofunction, is a function whose domain is equal to its codomain.

Calculation:

Self Map having image equal to it pre image

Only option B aptly represents a function that maps to itself -- a Self-Map

Hence, Option B is Correct.