Functions and its properties MCQ Quiz - Objective Question with Answer for Functions and its properties - Download Free PDF

Last updated on Apr 23, 2025

Latest Functions and its properties MCQ Objective Questions

Functions and its properties Question 1:

If a continuous function f:R → R is defined as f(x + y) = f(x) + f(y), ∀x, y ∈ R and f(1) = 10, then, r=1n(f(r))2 =

  1. 72n(n + 1)
  2. 5n(n + 1)
  3. 503n(n + 1)(2n + 1)
  4. 1004n2(n + 1)2

Answer (Detailed Solution Below)

Option 3 : 503n(n + 1)(2n + 1)

Functions and its properties Question 1 Detailed Solution

Given, f(x + y) = f(x) + f(y), 

⇒ f(x) = kx

and f(1) = 10 

⇒ f(x) = 10x

So,  r=1n(f(r))2 =  r=1n(10r)2 =  r=1n100(r)2 

⇒  r=1n(f(r))2 = 100  r=1n(r)2 

 ⇒  r=1n(f(r))2 = 100(12 + 22 + ....+ n2)

⇒  r=1n(f(r))2 = 100(n(n+1)(2n+1)6)

⇒  r=1n(f(r))2 =  503n(n + 1)(2n + 1)

∴ The correct answer is option (3).

Functions and its properties Question 2:

If 2f(x)+f(1x)=logx, for all x > 0, then f(ex) is 

  1. x2
  2. x
  3. 2x
  4. x2

Answer (Detailed Solution Below)

Option 2 : x

Functions and its properties Question 2 Detailed Solution

Given:

2f(x)+f(1x)=logx

Concept:

In such questions, we generally try to reduce the expression to a single function f(x) by elimination method.

Formula:

log(ex) = x

log(1/ex) = log(e-x) = -x

Calculation:

2f(x)+f(1x)=logx - (i)

Replacing x by 1/x in (i),

⇒ 2f(1x)+f(x)=log(1x) - (ii)

Now, 2 × (i) - (ii) :

⇒ 3f(x) = 2log(x) - log(1/x)

⇒ f(x) = 13(2log(x) - log(1/x)) - (iii)

Replacing x by ex in (iii) -

⇒ f(ex) =  13(2log(ex) - log(1/ex)) 

⇒ f(ex) =  13(2x + x)

⇒ f(ex) =  x

Functions and its properties Question 3:

The value of b and c for which the identity f(x + 1) - f(x) = 8x + 3 is satisfied, where f(x) = bx2 + cx + d, are

  1. b = 2, c = 1
  2. b = 4, c = -1
  3. b = -1, c = 4
  4. None of these

Answer (Detailed Solution Below)

Option 2 : b = 4, c = -1

Functions and its properties Question 3 Detailed Solution

Given:

f(x) = bx2 + cx + d

f(x + 1) - f(x) = 8x + 3

Calculation:

f(x) = bx2 + cx + d    ___(i)

⇒ f(x + 1) = b(x + 1)2 + c(x + 1) + d      __(ii)

Putting value of (i) and (ii) in the given equation f(x + 1) - f(x) = 8x + 3, we get

⇒ b(x + 1)2 + c(x + 1) + d - (bx2 + cx + d) = 8x + 3

⇒ b(x2 + 2x + 1) + c - bx2 = 8x + 3

⇒ 2bx + b + c = 8x + 3

Equating the coefficients of x and constant terms, we get

⇒ 2b = 8 and b + c = 3

⇒ b = 4 and c = -1

∴ The correct answer is option (2).

Functions and its properties Question 4:

If f(x)=1+x1x, then f(x).f(x2)1+[f(x)]2 is equal to

  1. 14
  2. 16
  3. 18
  4. 12

Answer (Detailed Solution Below)

Option 4 : 12

Functions and its properties Question 4 Detailed Solution

Formula used:

  • (a - b)2 = a2 - 2ab + b2
  • (a + b)2 = a2 + 2ab + b2
  • (a + b)(a - b) = a2 - b2

 

Calculation:

Given that,

f(x)=1+x1x    ----(1)

We have to find the value of

F(x) = f(x).f(x2)1+[f(x)]2 -----(2)

⇒  F(x) = (1+x1x).(1+x21x2)1+(1+x1x)2

⇒  F(x) =  (1+x)(1+x2)(1x)(1x2)(1x)2+(1+x)2(1x)2

By using above formula, 

⇒  F(x) =  (1+x)(1+x2)(1x)(1x)(1+x)1+x22x+1+x2+2x(1x)2

⇒  F(x) =  (1+x)(1+x2)(1x)2(1+x)2(1+x2)(1x)2

⇒  F(x) =  12

∴ The value of the given function F(x) = 12.

Functions and its properties Question 5:

If R is a relation on the set of integers Z such that R = {(a, b) : 2 divides a - b where a, b ∈ Z} then R is a/an ?

  1. Reflexive and symmetric but not transitive
  2. Reflexive and transitive but not symmetric
  3. Reflexive but neither symmetric nor transitive
  4. Equivalence relation

Answer (Detailed Solution Below)

Option 4 : Equivalence relation

Functions and its properties Question 5 Detailed Solution

Concept:

  • Reflexive:

Let R be a relation on a non-empty set A, if every element of A is related to itself then R is said to be a reflexive relation.

Thus, R is reflexive ⇔ (a, a) ∈ R, ∀ a ∈ A.

  • Symmetric:

Let R be a relation on a non-empty set A, then the relation R is said to be symmetric relation ⇔ (a, b) ∈ R ⇒ (b, a) ∀ a, b ∈ A.

  • Transitive:

Let R be a relation on a non-empty set A, then the relation R is said to be transitive relation ⇔ (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R, ∀ a, b, c ∈ A.

  • Equivalence:

Let R be a relation on a non-empty set A, then the relation R is said to be equivalence relation if R is reflexive, symmetric and transitive.

Calculation:

Given: R is a relation on the set of integers Z such that R = {(a, b) : 2 divides a - b where a, b ∈ Z}

Reflexive:

As we can see that a - a = 0 ∀ a ∈ Z ⇒ (a, a) ∈ R ∀ a ∈ Z

So, the given relation R is reflexive.

Symmetric:

Let (a, b) ∈ R ⇒ 2 divides a - b i.e a - b = 2k where k is some integer

So, if a - b = 2k ⇒ b - a = 2l where l = - k ∈ Z

So, 2 also divides b - a ⇒ (b, a) ∈ R

So, the given relation R is  also symmetric.

Transitive:

Let (a, b) and (b, c) ∈ R i.e 2 divides both a - b and b - c also.

⇒ a - b = 2k where k is some integer and similarly b - c = 2l where l is some integer.

⇒ (a - b) + (b - c) = 2k + 2l

⇒ a - c = 2m where m = l + k ∈ Z

So, 2 also divides a - c ⇒ (a, c) ∈ R

So, the given relation R is also transitive.

As we know that if a relation R is reflexive, symmetric and transitive then R is said to be an equivalence relation

Hence, option 4 is the correct answer.

Top Functions and its properties MCQ Objective Questions

The value of limx0[xcosxlog(1+x)x2] is:

  1. 1
  2. 12
  3. 3
  4. 0

Answer (Detailed Solution Below)

Option 2 : 12

Functions and its properties Question 6 Detailed Solution

Download Solution PDF

Calculation:

Given, limx0[xcosxlog(1+x)x2]

which is in the form (00)

using L-Hospital Rule:

on differentiating Nr and ωr

=limx0[cosxxsinx11+x2x]

again differentiating w.r.to x

=limx0[sinx(sinx+xcosx)+1(1+x)22]

put x → 0

=sin0(sin0+0.cos0)+11+02

= 1/2

Let f ∶ R → R be defined by f (x) = 1x ∀ x ∈ R. Then f is

  1. one-one
  2. onto
  3. bijective
  4. f is not defined

Answer (Detailed Solution Below)

Option 4 : f is not defined

Functions and its properties Question 7 Detailed Solution

Download Solution PDF

Explanation:

We have f(x) = 1x ∀ x ∈ R.

For x=0, f(x) is not defined.

Hence, the function f(x) is not defined function.

If f(a) = 2, f'(a) = 1, g(a) = -1, g'(a) = 2 then limxa.g(x)f(a)g(a)f(x)xa is

  1. -5
  2. 15
  3. 5
  4. 0

Answer (Detailed Solution Below)

Option 3 : 5

Functions and its properties Question 8 Detailed Solution

Download Solution PDF

Given:

f(a) = 2, f'(a) = 1, g(a) = -1, g'(a) = 2

Concept:

L'Hospital's Rule: It tells us that if we have an indeterminate form '0/0' or '∞/∞' all we need to do is differentiate the numerator and the denominator separately and then take the limit.

Calculation: 

Let  y = limxa.g(x)f(a)g(a)f(x)xa

when x = a, it gives 0/0 form. Hence, we can solve the limit by simply using L'Hospital's rule. 

Hence, differentiate the numerator and the denominator separately with respect to x, we will get

 y = limxa.g(x)f(a)g(a)f(x)1

Put x = a

  y = g(a)f(a)g(a)f(a)

But according to question,

f(a) = 2, f'(a) = 1, g(a) = -1, g'(a) = 2

⇒ y = (2)(2) - (-1)(1) 

⇒ y = 5

Hence, the value of above limit is 5.

If R is a relation on the set of integers Z such that R = {(a, b) : 2 divides a - b where a, b ∈ Z} then R is a/an ?

  1. Reflexive and symmetric but not transitive
  2. Reflexive and transitive but not symmetric
  3. Reflexive but neither symmetric nor transitive
  4. Equivalence relation

Answer (Detailed Solution Below)

Option 4 : Equivalence relation

Functions and its properties Question 9 Detailed Solution

Download Solution PDF

Concept:

  • Reflexive:

Let R be a relation on a non-empty set A, if every element of A is related to itself then R is said to be a reflexive relation.

Thus, R is reflexive ⇔ (a, a) ∈ R, ∀ a ∈ A.

  • Symmetric:

Let R be a relation on a non-empty set A, then the relation R is said to be symmetric relation ⇔ (a, b) ∈ R ⇒ (b, a) ∀ a, b ∈ A.

  • Transitive:

Let R be a relation on a non-empty set A, then the relation R is said to be transitive relation ⇔ (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R, ∀ a, b, c ∈ A.

  • Equivalence:

Let R be a relation on a non-empty set A, then the relation R is said to be equivalence relation if R is reflexive, symmetric and transitive.

Calculation:

Given: R is a relation on the set of integers Z such that R = {(a, b) : 2 divides a - b where a, b ∈ Z}

Reflexive:

As we can see that a - a = 0 ∀ a ∈ Z ⇒ (a, a) ∈ R ∀ a ∈ Z

So, the given relation R is reflexive.

Symmetric:

Let (a, b) ∈ R ⇒ 2 divides a - b i.e a - b = 2k where k is some integer

So, if a - b = 2k ⇒ b - a = 2l where l = - k ∈ Z

So, 2 also divides b - a ⇒ (b, a) ∈ R

So, the given relation R is  also symmetric.

Transitive:

Let (a, b) and (b, c) ∈ R i.e 2 divides both a - b and b - c also.

⇒ a - b = 2k where k is some integer and similarly b - c = 2l where l is some integer.

⇒ (a - b) + (b - c) = 2k + 2l

⇒ a - c = 2m where m = l + k ∈ Z

So, 2 also divides a - c ⇒ (a, c) ∈ R

So, the given relation R is also transitive.

As we know that if a relation R is reflexive, symmetric and transitive then R is said to be an equivalence relation

Hence, option 4 is the correct answer.

Functions and its properties Question 10:

If a continuous function f:R → R is defined as f(x + y) = f(x) + f(y), ∀x, y ∈ R and f(1) = 10, then, r=1n(f(r))2 =

  1. 72n(n + 1)
  2. 5n(n + 1)
  3. 503n(n + 1)(2n + 1)
  4. 1004n2(n + 1)2

Answer (Detailed Solution Below)

Option 3 : 503n(n + 1)(2n + 1)

Functions and its properties Question 10 Detailed Solution

Given, f(x + y) = f(x) + f(y), 

⇒ f(x) = kx

and f(1) = 10 

⇒ f(x) = 10x

So,  r=1n(f(r))2 =  r=1n(10r)2 =  r=1n100(r)2 

⇒  r=1n(f(r))2 = 100  r=1n(r)2 

 ⇒  r=1n(f(r))2 = 100(12 + 22 + ....+ n2)

⇒  r=1n(f(r))2 = 100(n(n+1)(2n+1)6)

⇒  r=1n(f(r))2 =  503n(n + 1)(2n + 1)

∴ The correct answer is option (3).

Functions and its properties Question 11:

If ϕ(x) = ax, then [ϕ(p)]3 is equal to

  1. ϕ(3p)
  2. 3ϕ(p)
  3. 6ϕ(p)
  4. 2ϕ(p)

Answer (Detailed Solution Below)

Option 1 : ϕ(3p)

Functions and its properties Question 11 Detailed Solution

Calculation:

ϕ(x) = ax          -----(1)

⇒ [ϕ(p)]3 = (ap)3 

⇒ [ϕ(p)]3 = a3p 

⇒ [ϕ(p)]3  ϕ(3p)    [Using the equation (1)]

∴  Required value is  ϕ(3p). 

Functions and its properties Question 12:

The value of b and c for which the identity f(x + 1) - f(x) = 8x + 3 is satisfied, where f(x) = bx2 + cx + d, are

  1. b = 2, c = 1
  2. b = 4, c = -1
  3. b = -1, c = 4
  4. None of these

Answer (Detailed Solution Below)

Option 2 : b = 4, c = -1

Functions and its properties Question 12 Detailed Solution

Given:

f(x) = bx2 + cx + d

f(x + 1) - f(x) = 8x + 3

Calculation:

f(x) = bx2 + cx + d    ___(i)

⇒ f(x + 1) = b(x + 1)2 + c(x + 1) + d      __(ii)

Putting value of (i) and (ii) in the given equation f(x + 1) - f(x) = 8x + 3, we get

⇒ b(x + 1)2 + c(x + 1) + d - (bx2 + cx + d) = 8x + 3

⇒ b(x2 + 2x + 1) + c - bx2 = 8x + 3

⇒ 2bx + b + c = 8x + 3

Equating the coefficients of x and constant terms, we get

⇒ 2b = 8 and b + c = 3

⇒ b = 4 and c = -1

∴ The correct answer is option (2).

Functions and its properties Question 13:

The value of limx0[xcosxlog(1+x)x2] is:

  1. 1
  2. 12
  3. 3
  4. 0

Answer (Detailed Solution Below)

Option 2 : 12

Functions and its properties Question 13 Detailed Solution

Calculation:

Given, limx0[xcosxlog(1+x)x2]

which is in the form (00)

using L-Hospital Rule:

on differentiating Nr and ωr

=limx0[cosxxsinx11+x2x]

again differentiating w.r.to x

=limx0[sinx(sinx+xcosx)+1(1+x)22]

put x → 0

=sin0(sin0+0.cos0)+11+02

= 1/2

Functions and its properties Question 14:

Let f ∶ R → R be defined by f (x) = 1x ∀ x ∈ R. Then f is

  1. one-one
  2. onto
  3. bijective
  4. f is not defined

Answer (Detailed Solution Below)

Option 4 : f is not defined

Functions and its properties Question 14 Detailed Solution

Explanation:

We have f(x) = 1x ∀ x ∈ R.

For x=0, f(x) is not defined.

Hence, the function f(x) is not defined function.

Functions and its properties Question 15:

If 2f(x)+f(1x)=logx, for all x > 0, then f(ex) is 

  1. x2
  2. x
  3. 2x
  4. x2

Answer (Detailed Solution Below)

Option 2 : x

Functions and its properties Question 15 Detailed Solution

Given:

2f(x)+f(1x)=logx

Concept:

In such questions, we generally try to reduce the expression to a single function f(x) by elimination method.

Formula:

log(ex) = x

log(1/ex) = log(e-x) = -x

Calculation:

2f(x)+f(1x)=logx - (i)

Replacing x by 1/x in (i),

⇒ 2f(1x)+f(x)=log(1x) - (ii)

Now, 2 × (i) - (ii) :

⇒ 3f(x) = 2log(x) - log(1/x)

⇒ f(x) = 13(2log(x) - log(1/x)) - (iii)

Replacing x by ex in (iii) -

⇒ f(ex) =  13(2log(ex) - log(1/ex)) 

⇒ f(ex) =  13(2x + x)

⇒ f(ex) =  x

Get Free Access Now
Hot Links: teen patti dhani teen patti real cash game teen patti bodhi