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Class 11 Maths MCQ Questions of Relations and Functions with Answers?

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Click below for Class 11 Maths MCQ Questions of Relations and Functions with Answers, easy to learn concepts and study notes, do free online tests and other study material and check where you stand in this chapter. So, you can How much you need to spend on this chapter.

Students can solve the  MCQ Questions for Class 11 Maths with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Relations and Functions Objective types Questions.
 

Practice MCQ Questions for class 11 Maths Chapter-Wise

1. If f(x) = (a – x)1/n, a > 0 and n ∈ N, then the value of f(f(x)) is

(a) 1/x
(b) x
(c) x2
(d) x1/2

2. The domain of the definition of the real function f(x) = \(\sqrt{(log_{12}x^2)}\) of the real variable x is

(a) x > 0
(b) |x| ≥ 1
(c) |x| > 4
(d) x ≥ 4

3. Assertion: If f(x) is an odd function, then f′(x) is an even function

Reason: If f' (x) is an even function, then f(x) is an odd function.

(a) Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
(b) Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
(c) Assertion is correct but Reason is incorrect
(d) Both Assertion and Reason are incorrect

4. If f(x) is an odd differentiable function on R, then df(x)/dx is a/an

(a) Even function
(b) Odd function
(c) Either even or odd function
(d) Neither even nor odd function

5. The domain of the function \(f(x)= \frac{1}{\sqrt{x^2-3x+2}}\) is

(a) (−∞,1)
(b) (−∞,1)∪(2,∞) 
(c) (−∞,1]∪[2,∞)
(d) (2,∞)

6. The domain of tan-1 (2x + 1) is

(a) R
(b) R -{1/2}
(c) R -{-1/2}
(d) None of these

7. The function f(x) = x – [x] has period of

(a) 0
(b) 1
(c) 2
(d) 3

8. A relation R is defined from the set of integers to the set of real numbers as (x, y) = R if x2 + y2 = 16 then the domain of R is

(a) (0, 4, 4)
(b) (0, -4, 4)
(c) (0, -4, -4)
(d) None of these

9. Find the number of binary operations on the set {a,b} 

(a) 10
(b) 16
(c) 20
(d) 8

10. Let n(A)=n. Then the number of all relations on A is

(a) 2n
(b) 2(n)!
(c) 2n2
(d) none

11. If f (x) = ax + b and g (x) = cx + d, then f {g (x)} = g {f (x)} is equivalent to

(a) f(a) = g(c)
(b) f(b) = g (b)
(c) f(d) = g (b)
(d) none

12. If f is an even function and g is an odd function the fog is a/an

(a) Even function
(b) Odd function
(c) Either even or odd function
(d) Neither even nor odd function

13. If n is the smallest natural number such that n + 2n + 3n + …. + 99n is a perfect square, then the number of digits in square of n is

(a) 1
(b) 2
(c) 3
(d) 4

14. The number of relations from A={1,2,3} to B={4,6,8,10} is

(a) 43
(b) 27
(c) 212
(d) 34

15. The number of relations from A={1,2,3,4,8} to B={4,6,8,} is

(a) 43
(b) 215
(c) 212
(d) 314

16. If f :Z→Z is defined by f(x)= \(f(x)=\begin{cases} \frac{x}{2}\;if\;x\;is\;even & \\ 0\;if\;x\;is\;odd \end{cases}\) then f is

​(a) Onto but not one to one
(b) One to one but not onto
(c) One to one and onto
(d) Neither one to one nor onto

17.  Express the function f: A—R. f(x) = x2 – 1. where A = { -4, 0, 1, 4) as a set of ordered pairs.

(a) {(-4, 15), (0, -1), (1, 0), (4, 15)}
(b) {(-4, -15), (0, -1), (1, 0), (4, 15)}
(c) {(4, 15), (0, -1), (1, 0), (4, 15)}
(d) {(-4, 15), (0, -1), (1, 0)

18. f : R→R is a function defined by f(x)=10x−7. If  g=f −1 then  g(x)=  

(a) \(\frac{1}{10x-7}\)
(b) \(\frac{1}{10x+7}\)
(c) \(\frac{x+7}{10}\)
(d) \(\frac{x-7}{10}\)

19. ​​Let R be the relation in the set N given by = {(a, b): a = b - 2, b > 6}. Choose the correct answer

(a) (2,4)ϵR
(b) (3,8)ϵR
(c) (6,8)ϵR
(d) (8,7)ϵR

20. If the graph of the function y=f(x) is symmetrical about the line x=2, then

(a) f(x+2)=f(x−2)
(b) f(x−2)=f(2−x)
(c) f(x)=f(−x)
(d) f(x)=−f(−x)

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Answer:

1. Answer: (b) x

Explanation: Given, f(x) = (a – x)1/n

Now, f(f(x)) = [(a – f(x))n]1/n

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

⇒ f(f(x)) = [a – (a – xn)]1/n

⇒ f(f(x)) = [a – a + xn)]1/n

⇒ f(f(x)) = (xn)1/n

⇒ f(f(x)) = x

2. Answer: (b) |x| ≥ 1

Explanation: f(x) = \(\sqrt{(log_{12}x^2)}\) 

Since, loga k ≥ 0 if a > 1, k ≥ 1

or 0 < a < 1 and 0 < k ≤ 1

So, the function f(x) exists if

log12 x2 ≥ 0

⇒ x2 ≥ 1

⇒ |x| ≥ 1

3. Answer: (c) Assertion is correct but Reason is incorrect

Explanation: Statement 1 is always true, but Statement 2 is not always true, as if f′(x) = cos x, then f(x) can be sin x which is odd function, but if f′(x)= sin x+2, then f(x) is neither odd not even. 

4. Answer: (a) Even function

Explanation: Given, f(x) is an odd differentiable function on R

⇒ f(-x) = -f(x) for all x ∈ R

differentiate on both side, we get

⇒ -df(-x)/dx = -df(x)/dx for all x ∈ R

⇒ df(-x)/dx = df(x)/dx for all x ∈ R

⇒ df(x)/dx is an even function on R.

5. Answer: (b) (−∞,1)∪(2,∞)

Explanation: For f (x) to be defined, we must have

x2−3x+2 = (x−1)(x−2)>0

⇒ x<1 or >2

Domain of f = (−∞,1)∪(2,∞).

6. Answer: (a) R

Explanation: Since tan-1 x exists if x ∈ (-∞, ∞)

So, tan-1 (2x + 1) is defined if

-∞ < 2x + 1 < ∞

⇒ -∞ < x < ∞

⇒ x ∈ (-∞, ∞)

⇒ x ∈ R

So, domain of tan-1 (2x + 1) is R.

7. Answer: (b) 1

Explanation: Let T is a positive real number.

Let f(x) is periodic with period T.

Now, f(x + T) = f(x), for all x ∈ R

⇒ x + T – [x + T] = x – [x], for all x ∈ R

⇒ [x + T] – [x] = T, for all x ∈ R

Thus, there exist T > 0 such that f(x + T) = f(x) for all x ∈ R

Now, the smallest value of T satisfying f(x + T) = f(x) for all x ∈ R is 1

So, f(x) = x – [x] has period 1

8. Answer: (b) (0, -4, 4)

Explanation: Given that: (x, y) ∈ R 

⇔ x2 + y2 = 16

⇔ y = \(\pm\sqrt{(16-x^2)}\)

when x = 0 ⇒ y = ±4

(0, 4) ∈ R and (0, -4) ∈ R

when x = ±4 ⇒ y = 0

(4, 0) ∈ R and (-4, 0) ∈ R

Now for other integral values of x, y is not an integer.

Hence R = {(0, 4), (0, -4), (4, 0), (-4, 0)}

So, Domain(R) = {0, -4, 4}

9. Answer: (b) 16

Explanation: Considering 'n' elements in a set, the no. of binary operations on that set will be \(=n^{n^2}\)

Applying this concept to the above set gives us

\(=2^{2^2}.....\) ∵ n = 2

=24

=16

10. Answer: (c) 2n2

Explanation: For any set A such that n(A)=n then number of all relations on A is 2n2.

As the total number of Relations that can be defined from a set A to B is the number of possible subsets of A×B. 

If n(A) =p and n(B)=q then n(A×B)=pq and the number of subsets of A×B = 2pq

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11. Answer: (c) f(d) = g (b)  

Explanation: f (x) = ax + b and g (x) = cx + d

f {g (x)}= a (cx + d) + b

= acx + ad + b

= g {f (x)}= c (ax + b) + d

= acx + bc + d.

since f {g (x)} = g {f (x)}

⇒ acx + ad + b = acx + bc + d

⇒ ad + b = c . b + d

⇒ f (d) = g (b)

[∵ ad + b = f (d) and bc + d = g (b)]

12. Answer: (a) Even function

Explanation: Fog function is an even function

Let f(x) is a even function and g(−x) is odd function.

So, f(g(−x))=f(−g(x)) = even

13. Answer: (c) 3

Explanation: n+2n+3n+......+99n

n(1+2+3+..........+99)

\(n\frac{(99)(100)}{2}=9\times25\times22\times n\) is a perfect square when n=22

Number of digits in n = 3

14. Answer: (c) 212

Explanation: n(B) =4 

n(A)=3

Therefore there exists 23×4 relations from A to B i.e 212 relations.

15. Answer: (b) 215

Explanation: n(B) =3 

n(A)=5

Therefore there exists 23×5 relations from A to B i.e 215 relations.

16. Answer: ​(a) Onto but not one to one

Explanation: f is onto and but not one to one as all odd values x has a 0 assigned in f(x).
Function is onto as every element in f(x) is mapped to some element in x.

17. Answer: (a) {(-4, 15), (0, -1), (1, 0), (4, 15)}

Explanation: A = {-4, 0, 1, 4}

f(x) = x2 – 1

f(-4) = (-4)2 – 1 

= 16 – 1

=15

f(0) = (0)2 – 1 = -1

f(1) = (1)2 – 1 = 0

f(4) = (4)2 – 1 

= 16 – 1 

= 15

Therefore, the set of ordered pairs = {(-4, 15), (0, -1), (1, 0), (4, 15)}

18. Answer: (c) \(\frac{x+7}{10}\)

Explanation: Given f(x)=10x −7 and g=f −1

Let f(x) =10x−7 =y

\(x=\frac{y+7}{10}=\) f−1 (y) = g(y)

∴g(x) \(=\frac{x+7}{10}\)

19. Answer: (c) (6,8)ϵR

Explanation: Firstly b should be greater than 6 & also difference between b& a should be 2 , option C is satisfing both the conditions hence the correct answer is C.

20. Answer: (b) f(x−2)=f(2−x)

Explanation: When a graph is symmetric about x=2, the values of f(x) at equal distance from x=2 on either side are equal.If we translate the origin to (2,0), in the new coordinate system, for the function we can write f(X)=f(−X). The relation between new and old coordinates are as follows:

x =X+2 and y=Y+0 (as shown in figure)

Hence, in terms of x,y we can write  f(x−2)=f(2−x)

Click here to practice MCQ Questions for Relations and Functions class 11

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