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NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce PDF Download

Q.1. Let A = {–1, 2, 3} and B = {1, 3}. Determine
(i) A × B
(ii) B × A
(iii) B × B
(iv) A × A
Ans.
Given that: A = {- 1, 2, 3} and B = {1, 3}
(i) A × B {- 1, 2, 3} × {1, 3} = {(- 1, 1), (- 1, 3),(2, 1), (2, 3), (3, 1), (3, 3)}
(ii) B × A = {1, 3} ×{- 1, 2, 3} = {(1, - 1), (3,- 1), (1, 2), (3, 2),(1, 3), (3, 3)}
(iii) B × B = {1, 3} × {1, 3} = {(1, 1), (1, 3), (3, 1), (3, 3)}
(iv) A = { -1, 2, 3}× {-1, 2, 3}
= {{-1, -1), (-1,2), (-1,3),(2,-1),(2, 2),(2, 3),(3,-1),(3,2),(3,3)}

Q.2. If  P = {x : x < 3, x ∈ N}, Q = {x : x ≤ 2, x ∈ W}. Find (P ∪  Q) × (P ∩ Q), where W is the set of whole numbers.
Ans.
Given that:
P = {x : x < 3, x  N}
 P = {1, 2}
Q = {x : x  2, x ∈W}
 Q = {0, 1, 2}
Now (P ∪ Q) = {0, 1, 2} and (P ∩ Q) = {1, 2}
 (P ∪ Q) × (P ∩ Q) = {0, 1, 2} × {1, 2} = {(0, 1), (0, 2), (1, 1), (1, 2), (2, 1),(2, 2)}

Q.3. If A = {x : x ∈ W, x < 2} B = {x : x ∈ N, 1 < x < 5} C = {3, 5} find
(i) A × (B ∩ C)
(ii) A × (B ∪ C)
Ans.
Given that:
A = {x : x  W, x < 2}
 A = {0, 1}
B = {x : x  N, 1 < x < 5}
 B = {2, 3, 4}
⇒ C = {3, 5}
Now (B ∩ C) = {3} and (B ∪ C) = {2, 3, 4, 5}
(i) A × (B ∩ C) = {0, 1} × {3} = {(0, 3), (1, 3)}
(ii) (B ∪ C) = {0, 1} × {2, 3, 4, 5}
= {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)}

Q.4. In each of the following cases, find a and b.
(i) (2a + b, a – b) = (8, 3) 
(ii) (a/4, a- 2b) = (0, 6 + b)
Ans.
(i) Given that: (2a + b, a – b) = (8, 3)
Comparing the domains and ranges, we get
2a + b = 8 ...(i)
a – b = 3 ...(ii)
Solving (i) and (ii) we get a = 11/3 and b = 2/3
(ii) Given that: (a/4, a- 2b) = (0, 6 + b)
Comparing the domains and ranges, we get
 a/4 = 0 ⇒ a = 0, a – 2b = 6 + b
6  0 - 3b = 6 ⇒ b = - 2.
∴a = 0, b = -2.

Q.5. Given A = {1, 2, 3, 4, 5}, S = {(x, y) : x ∈ A,  y ∈ A}. Find the ordered pairs which satisfy the conditions given below:
(i) x + y = 5
(ii) x + y < 5
(iii) x + y > 8
Ans.
Given that: A = {1, 2, 3, 4, 5} and
S = {(x, y) : x ∈ A, y ∈ A}
(i) x + y = 5, so, the ordered pairs satisfying the given conditions are (1, 4), (4, 1), (2, 3), (3, 2).
(ii) x + y < 5, so, the ordered pairs satisfying the given conditions are (1, 1), (1, 2), (2, 1), (1, 3), (2, 2), (3, 1).
(iii) x + y > 8, so the ordered pairs satisfying the given conditions are (4, 5), (5, 4), (5, 5).

Q.6. Given R = {(x, y) : x, y ∈ W, x2 + y2 = 25}. Find the domain and Range of R.
Ans.
Given that: R = {(x, y) : x, y  W, x2 + y2 = 25} So, the ordered pairs satisfying the given condition x2 + y2 = 25 are (0, 5), (3, 4), (5, 0), (4,3)   (∵ x, y ∈ W)
Hence, the domain = {0, 3, 4, 5} and the range = {0, 3, 4, 5}.

Q.7. If R1 = {(x, y) | y = 2x + 7, where x ∈ R and – 5  ≤ x ≤ 5} is a relation. Then find the domain and Range of R1.
Ans.
Given that: R1 = {(x, y)|y = 2x + 7 where x  R and -5  x  5}
Since R1 is defined for all real numbers that are greater than or equal to  5 and less than    or equal to 5 ,domain of R1 is {x:– 5 ≤ x  5} = [-5, 5] Here domain is  { -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5} and y = 2x + 7.
So, the values of y for the corresponding given values of x are {-3,-1,1, 3, 5,7, 9, 11, 13, 15,17}
Hence, the domain of R1 = [-5, 5] and range of R1 = [-3, 17]

Q.8. If R2 = {(x, y) | x and y are integers and x2 + y2 = 64} is a relation. Then find R2.
Ans.
Given that: x2 + y2 = 64, x, y  Z
The sum of the squares of two integers is 64
 For x = 0, y = ±8
For x = ±8, y = 0
Hence, R2 = {(0, 8), (0, -8), (8, 0), (-8, 0)}

Q.9. If R3 = {(x, x ) | x is a real number} is a relation. Then find domain and range of R3.
Ans.
Given that: R= {(x, |x|)|x is a real number}
Clearly, domain of R3 = R
and Range of R3 = (0, ∞) [∵ |x| =R+]

Q.10. Is the given relation a function? Give reasons for your answer.
(i) h = {(4, 6), (3, 9), (– 11, 6), (3, 11)}
(ii) f = {(x, x) | x is a real number}

(iii) g = n, 1/n, |n is a positive integer n
(iv) s = {(n, n2) | n is a positive integer}
(v) t = {(x, 3) | x is a real number.

Ans.
Given that: (i) h = {(4, 6), (3, 9), (– 11, 6), (3, 11)} Since in the given relation 3 has two images 9 and 11. So, h is not a function.
(ii) f = {(x, x)|x is a real number}. Here, we observe that for every element of domain has a unique image. So, f is a function.
(iii) Given that:
g = n, 1/n |n is a positive integer
Here, we observe that n is a positive integer so, for every element of domain, there is a unique 1/n image. Hence g is a function.
(iv) Given that: S = {(n, n2)|n is a positive integer} Here, we observe that the square of any integer is a unique number. So, for every element element in the domain there is unique image. Hence, S is a function.
(v) Given that: t = {(x, 3)|x is a real number} Here, we observe that for every real element in the domain, there is a constant number 3. Hence t is a constant function.

Q.11. If f and g are real functions defined by f (x) = x2 + 7 and g (x) = 3x + 5, find each of the following
(a) f (3) + g (– 5)
(b) f(1/2) × g(14)

(c) f (– 2) + g (– 1)
(d) f (t) – f (– 2)
(e) f(t) - f(5)/t-5, if t ≠ 5
Ans.
Given that: f(x) = x2 + 7 and g(x) = 3x + 5
(i) f(3) + g(– 5) = [(3)2 + 7] + [3(– 5) + 5]
= (9 + 7) + (– 15 + 5) = 16 – 10 = 6
Hence, f(3) + g(– 5) = 6
(ii)
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
Hence, NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
(iii) f(– 2) + g(– 1) = [(– 2)2 + 7] + [3(– 1) + 5] = (4 + 7) + (– 3 + 5) = 11 + 2 = 13
Hence, f(– 2) + g(– 1) = 13
(iv) f(t) – f(– 2) = (t2 + 7) – [(– 2)2 + 7] = t2 + 7 – 11 = t2 – 4
Hence, f(t) – f(– 2) = t2 – 4.
(v)
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
Hence, NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce

Q.12. Let f and g be real functions defined by f (x) = 2x + 1 and g (x) = 4x – 7. 
(a) For what real numbers x,  f (x) = g (x)?
(b) For what real numbers x,  f (x) < g (x)?
Ans.
Given that: f(x)= 2x + 1 and g(x) = 4x  7
(i) For f(x) = g(x), we get
2x + 1 = 4x  7
 2x  4x = -7  1 ⇒
- 2x = - 8
⇒ x = 4.
Hence,the required real number is 4.
(ii) For f(x) < g(x), we get
2x + 1 < 4x  7
⇒ 2x  4x < - 1  7
⇒-2x <- 8
 2x >8
 x >4
Hence, the required real number is x > 4.

Q.13. If f and g are two real valued functions defined as f (x) = 2x + 1, g (x) = x2 + 1, then find.
(i) f  + g
(ii) f  –  g
(iii) f.g
(iv) f/g

Ans.
Given that: f(x) = 2x + 1 and g(x) = x2 + 1
(i) f + g = f(x) + g(x)
 2x + 1 +x2 + 1
 x2 + 2x + 2
(ii) f  g = f(x)  g(x)
 (2x + 1)  (x2 + 1)
= 2x +1  x2  1
 2x – x2
(iii) f.g = f(x).g(x)
 (2x + 1) (x2 + 1)
 2x3 + x2 + 2x + 1
(iv) f/g = f(x)/g(x) = 2x + 1/x2 + 1

Q.14. Express the following functions as set of ordered pairs and determine their range. f : X → R, f (x) = x3 + 1, where X = {–1, 0, 3, 9, 7}
Ans.
Given that: f : X → R, f(x) = x3 + 1, where X = {-1, 0, 3, 9, 7}
Here X = {-1, 0, 3, 9, 7}
For x = -1, f(-1) = (-1)3 + 1 = 0
For x = 0, f(0) = (0)3 + 1 = 1
For x = 3, f(3) = (3)3 + 1 = 28
For x = 9, f(9) = (9)3 + 1 = 730
For x = 7, f(7) = (7)3 + 1 = 344
∴ The ordered pairs are (-1, 0), (0, 1), (3, 28), (7, 344), (9, 730) and the range = {0, 1, 28, 344, 730}.

Q.15. Find the values of x for which the functions f (x) = 3x2 – 1 and g (x) = 3 + x are equal?
Ans.
Given that: f(x) = 3x2 – 1 and g(x) = 3 + x
Since f(x) = g(x) (given)
⇒  3x2 – 1 = 3 + x
⇒ 3x2 – x – 4 = 0
⇒  3x2 – 4x + 3x – 4 = 0
⇒  x(3x – 4) + 1(3x – 4) = 0
⇒ (3x – 4)(x + 1) = 0
⇒  3x – 4 = 0 or x + 1 = 0
⇒  3x = 4 or x = – 1
∴ x = 4/3
Hence, the value of x are – 1 and 4/3.

Long Answer Type
Q.16. Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function? Justify. If this is described by the relation, g (x) = αx + β, then what values should be assigned to α and β?
Ans.
Given that: g = {(1, 1), (2, 3), (3, 5), (4, 7)}
Since every element of the domain in this relations has unique image, so g is a function.
Now g(x) = αx + β
For (1, 1) g(1) = α (1) + β = 1 ⇒ α + β = 1 ...(i)
For (2, 3) g(2) = α(2) + β = 3⇒ 2α + β = 3 ...(ii)
Solving eqn. (i) and (ii) we have
α= 2 and β = -1
 [Note: We can take any other two ordered pairs]
Hence, the value of α= 2 and β = -1.

Q.17. Find the domain of each of the following functions given by
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
 (iii) f(x) = x |x|
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
Ans.
(i) Given that: f(x)
= NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
We know that – 1  cos ≤ x ≤ 1
⇒1  - cos x  - 1
⇒ 1 + 1  1  cos x  -1 + 1
⇒ 2  1  cos x ≥ 0
⇒ 0  1  cos x  2
For real value of domain
1  cos x  0  cos x  1
⇒ x  2nπ ∀ n  Z
Hence, the domain of f = R  {2nπ, n  Z}
(ii) Given that:
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
 x + |x| = x + x = 2x if  x  0
and x + |x|= x  x = 0 if x < 0
So far x < 0, f is not defined.
Hence, the domain f = R+.
(iii) Given that: f(x) = x|x|
It is clear that f(x) is defined for all x  R.
Hence, the domain of f = R.
(iv) Given that:
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
Here, f(x) is only defined if  x2 -1   0
(x–1)(x+1)  0
 x  1, x  - 1
Hence, the domain of f = R  {-1, 1}
(v) Given that:
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
Here, f(x) is only defined if  28– x  0  x  28
Hence, the domain = R  {28}.

Q.18. Find the range of the following functions given by
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
(ii) f(x) = 1 - |x-2|
(iii) f(x) = |x  3|
(iv) f(x)= 1 + 3 cos 2x
Ans.
(i) Given that:
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
Let y = f(x)
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
⇒ y(2– x2) = 3
 2y   yx2 = 3
⇒ yx2 = 2y  3
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
Here, x is real if 2y – 3 ≥ 0 and y  0
⇒  NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
Hence, the range of f =
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
(ii) Given that: f(x) = 1 - |x  2|
We know that |x  2|= - (x  2) if x < 2
and |x– 2|= (x–2), x  2
 -|x– 2|  0
 1 - |x  2|≤ 1
Hence, the range of f = (-∞, 1].
(iii) Given that: f(x) = |x  3|
We know that |x  3|  0
 f(x)  0
Hence, the range of  f  = [0, ∞)
(iv) Given that: f(x) = 1 + 3 cos 2x
We know that  1  cos 2x  1
⇒ -3  3 cos 2x  3
⇒- 3 + 1  1 + 3 cos 2x  3 + 1
⇒ -2  1 + 3 cos 2x  4
⇒- 2  f(x)  4
Hence, the range of f = [- 2, 4].

Q.19.  Redefine the function f (x) =  x − 2 + 2 + x ,   – 3   ≤ x ≤ 3.
Ans.
Given that: f(x) = |x  2| + |2 + x|, - 3  x  3
Since |x  2| = - (x  2), x < 2
and |x– 2| = (x  2), x  2
|2+ x| =- (2 + x), x < - 2
|2+x| = (2+ x), x  - 2
Now f(x) = |x–2|+ |2 +x|, - 3 ≤x  3.
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
f(x) = NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce

Q.20. If f (x) = x - 1/x + 1, then show that
Ans.
Given that:
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
Hence,
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce

Q.21. Let f (x) = √x and g (x) = x be two functions defined in the domain R+ ∪ {0}.Find
(i) (f + g) (x)
(ii) (f – g) (x)
(iii) (fg) (x)
(iv) (f/g) (x) 
Ans.
Given that: f(x) = √x and g(x) = x be two functions defined in the domain R+ ∪ {0}
(i) (f + g)(x) = f(x) + g(x) = √x + x
(ii) (f – g)(x) = f(x) – g(x) = √x - x
(iii) (fg)(x) = f(x).g(x) = √x . x = x 3/2
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce

Q.22. Find the domain and Range of the function f (x), 
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
Ans.
Given that:
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
Here, it is clear that f(x) is real when x   5  > 0  x > 5
Hence, the domain = (5, ∞)
Now to find the range put
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
For x  (5, ∞), y  R+.
Hence, the range of f = R+.

Q.23. If  f (x) = y = ax - b/cx - a, then prove that f (y) = x.
Ans.
Given that:f(x) = y
= NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
putting x = y in f(x) we get
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
Hence, f(y) = x.

Objective Type Questions
Q.24. Let n (A) = m, and n (B) = n. Then the total number of non-empty relations that can be defined from A to B is
(a) mn
(b) nm – 1
(c) mn – 1
(d) 2mn – 1
Ans. (d)
Solution.
Given that: n(A) = m and n(B) = n
 n(A × B) = n(A) . n(B) = mn
So, the total number of relations from A to B 2mn  1.
Hence, the correct option is (d)

Q.25. If [x]2 – 5 [x] + 6 = 0, where [ . ] denote the greatest integer function, then
(a) x ∈ [3, 4]
(b) x ∈ (2, 3]
(c) x ∈ [2, 3]
(d) x ∈ [2, 4)
Ans. (c)
Solution.
we have [x]2  5[x] + 6 = 0
⇒ [x]2  3[x] 2[x] + 6 = 0
⇒ [x]([x]– 3)–2([x]–3)= 0
⇒ ([x]– 3)([x]–2) =0 
 [x] = 2, 3
So, x  [2, 3].
Hence, the correct option is (c).

Q.26. Range of f (x) = 1/1-2 cos x is
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
Ans. (b)
Solution.
Given that:
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
We know that  1  cos x  1
⇒ 1  cos x ≥ - 1
 - 1  - cos x ≤1
⇒ -2  - 2 cos x  2
⇒- 2+ 1  1– 2 cos x ≤ 2 +1
⇒ -1  1  2 cos x  3
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
So the range of f(x) = [-1, 1/3]
Hence, the correct option is (b).

Q.27. Let   NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce , then
(a) f (xy) = f (x) . f (y)
(b) f (xy) ≥ f (x) . f (y)
(c) f (xy) ≤ f (x) . f (y)
(d) None of these
[Hint : find f (xy) = NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce f (x) . f (y) = 1+ NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce]
Ans. (c)
Solution.
Given that:
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
and f(x) . f(y)
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
Hence, the correct option is (c).

Q.28. Domain of √a2 - x(a>0) is 
(a) (– a, a)
(b) [– a, a]
(c) [0, a]
(d) (– a, 0]
Ans. (b)
Solution.
Let
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
f(x) is defined if a2   x2  0
⇒ x2  a2  0
 x2  a2 
⇒ x  ±a
 -a≤x  a
 Domain of f(x) = [-a, a]
Hence, the correct option is (b).

Q.29. If f (x) = ax + b, where a and b are integers, f (–1) = – 5 and f (3) = 3, then a and b are equal to
(a) a = – 3, b = –1
(b) a = 2, b = – 3
(c) a = 0, b = 2
(d) a = 2, b = 3
Ans. (b)
Solution.
Given that: f(x) = ax+ b
⇒ f(-1) = a(-1) +b
⇒ - 5 = - a+b
 a  b = 5 ….(i)
f(3) =3a + b
⇒ 3 = 3a + b
 3a + b = 3…..(ii)
On solving eqn. (i) and (ii), we get a = 2, b = - 3
Hence, the correct option is (b).

Q.30. The domain and range of the real function f  defined by f (x) =  NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
is given by
(A) Domain = R, Range = {–1, 1}
(B) Domain = R – {1}, Range = R
(C) Domain = R – {4}, Range = {– 1}
(D) Domain = R – {– 4}, Range = {–1, 1}
Ans. (a)
Solution.
Given that:
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
f(x) is defined if
4– x ≥ 0 or x2  1 > 0
⇒ - x  - 4 or  (x  1)(x + 1) >0
⇒ x  4 or x < - 1 and x > 1
 Domain of f(x)  is (- ∞, - 1) ∪ (1, 4]
Hence, the correct option is (a).

Q.31. The domain and range of the real function f defined by f (x) = NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
is given by
(a) Domain = R, Range = {–1, 1}
(b) Domain = R – {1}, Range = R
(c) Domain = R – {4}, Range = {– 1}
(d) Domain = R – {– 4}, Range = {–1, 1}
Ans. (c)
Solution.
Given that:
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
We know that f(x) is defined if x  4≠ 0⇒x≠4
So,the domain of f(x) is = R – {4}
Let,
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
⇒ yx  4y = 4  x
 yx + x = 4y + 4
⇒ x(y + 1) = 4y + 4
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
If x is real number, then 1 + y  0 ⇒ x ≠1
 Range of f(x) = R–{ -1)
Hence, the correct option is (c).

Q.32. The domain and range of real function f defined by f (x) = √x - 1 
is given by
(a) Domain = (1, ∞), Range = (0, ∞)
(b) Domain = [1, ∞), Range = (0, ∞)
(c) Domain = [1, ∞), Range = [0, ∞)
(d) Domain = [1, ∞), Range = [0, ∞)
Ans. (d)
Solution.
Given that: 
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
f(x) is defined if  x  1  0  x ≥ 1
 Domain of f(x) = [0, ∞)
Let f(x) = y = √x - 1  
⇒ y2 = x  1
⇒ x = y2 + 1
If x is real then y  R
 Range of f(x) = [0, ∞)
Hence, the correct option is (d).

Q.33. The domain of the function f given by f(x) = x2 + 2x + 1/x2 - x - 6 is
(a) R – {3, – 2}
(b) R – {–3, 2}

(c) R – [3, – 2]
(d) R – (3, – 2)

Ans. (a)
Solution.
Given that:
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
F(x) is defined if x2  x  6  0
⇒ x2  3x + 2x– 6≠0
⇒ (x– 3)(x + 2) ≠0
 x ≠ - 2, x  3
So, the domain of f(x) = R  {-2, 3}
Hence, the correct option is (a).

Q.34. The domain and range of the function f given by f (x) = 2 – x − 5 is
(a) Domain = R+, Range =  ( – ∞, 1]
(b) Domain = R, Range =  ( – ∞, 2]
(c) Domain = R, Range =  (– ∞, 2)
(d) Domain = R+, Range =  (– ∞, 2]
Ans. (b)
Solution.
Given that: f(x) = 2  |x– 5|
Here, f(x) is defined for x  R
 Domain of f(x) = R
Now,|x– 5|≥0
⇒ -|x  5|≤0
⇒ 2-|x– 5|≤2
⇒ f(x)  2
 Range of f(x) = (- ∞, 2]
Hence, the correct option is (b).

Q.35. The domain for which the functions defined by  f (x) = 3x2 – 1 and g (x) = 3 + x are equal is
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
Ans.
Given that: f(x) = 3x2 – 1 and g(x) = 3 + x
f(x) = g(x)
⇒ 3x2  1 = 3 + x
⇒ 3x2 – x – 4 = 0
 3x2  4x + 3x  4 = 0
⇒ x(3x – 4) + 1(3x  4)= 0
 (x+1)(3x–4)= 0
⇒ x + 1 = 0 or 3x  4 = 0
⇒ x = - 1, or x = 4/3
∴ Domain = {-1, 4/3}
Hence, the correct option is (a).

Fill in the blanks:
Q.36. Let f and g be two real functions given by

f = {(0, 1), (2, 0), (3, – 4), (4, 2), (5, 1)}
g = {(1, 0), (2, 2), (3, – 1), (4, 4), (5, 3)}
then the domain of f . g is given by _______.
Ans.
Given that: f(x) = {(0, 1), (2, 0), (3, – 4), (4, 2), (5, 1)}
and g(x) = {(1, 0), (2, 2), (3, – 1), (4, 4), (5, 3)}
∴ Domain of f = {0, 2, 3, 4, 5}
and domain of g = {1, 2, 3, 4, 5}
So, domain of f.g = Domain of f ∩ Domain of g = {2, 3, 4, 5}
Hence, the filler is {2, 3, 4, 5}.

Q.37. Let f = {(2, 4), (5, 6), (8, – 1), (10, – 3)}
g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, 5)}
be two real functions. Then Match the following:
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce

Ans.
Given that: f = {(2, 4), (5, 6), (8, – 1), (10, – 3)}
and g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, 5)}
f – g, f + g, f.g, f/g are defined in the domain
(domain of f ∩ domain of g)
(i) (f – g)2= f(2)  g(2) = 4  5 = - 1
(f – g)8 = f(8)  g(8) = - 1  4 = - 5
(f  g)10 = f(10)  g(10) = - 3  13 = - 16
 (f – g) = {(2, - 1), (8, - 5), (10, - 16)}
(ii) (f + g)2 = f(2) + g(2) = 4 + 5 = 9
(f + g)8 = f(8) + g(8) =  1 + 4 = 3
(f + g)10 = f(10) + g(10) =  3 + 13 = 10
 (f + g) = {(2, 9), (8, 3), (10, 10)}
(iii) (f . g)2 = f(2). g(2) = 4. 5 = 20
(f . g)8 = f(8) . g(8) = (-    1). (4)= - 4
(f . g)10 = f(10) . g(10)= - 3. 13 = - 39
 (f . g) = {(2, 20), (8,- 4), (10, - 39)}
(iv) 
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
∴  NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
Hence,    the    correct    option    is
(a) ↔ (iii)
(b) ↔ (iv)
(c) ↔ (ii)
(d) ↔ (i)

State True or False for the following statements
Q.38. The ordered pair (5, 2) belongs to the relation R = {(x, y) : y = x – 5, x, y ∈ Z}
Ans.
Given that: R = {(x, y) : y = x  5, x, y  Z}
For (5, 2), y = x  5
Put x = 5, y = 5  5 = 0  2
So (5, 2) is not the ordered pair of R.
Hence, the statement is ‘False’.

Q.39. If P = {1, 2}, then P × P × P = {(1, 1, 1), (2, 2, 2), (1, 2, 2), (2, 1, 1)}
Ans. 
Given that P = {1, 2}
∴ P × P = {1, 2} ´ {1, 2} = {(1, 1), (1, 2), (2, 1), (2, 2)}
P × P × P = {(1, 1), (1, 2), (2, 1), (2, 2)} × {1, 2}
= {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}
So, given statement is ‘False’.

Q.40. If A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}, then (A × B) ∪ (A × C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}.
Ans.
Given that: A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}
A × B = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
And A × C = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(A × B) ∪ (A × C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5),
 (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}
Hence, the given statement is ‘True’.

Q.41. If (x – 2, y + 5) = (-2, 1/3) are two equal ordered pairs, then x = 4, y = -14/3 
Ans.
Given that: (x – 2, y + 5) = (-2, 1/3)
⇒ x – 2 = – 2
⇒ x = 0
and
NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce
Hence, the given statement is ‘False’.

Q.42. If A × B = {(a, x), (a, y), (b, x), (b, y)}, then A = {a, b}, B = {x, y}
Ans.
Given that: A = {a, b} and B = {x, y}
∴ A × B = {(a, x), (a, y), (b, x), (b, y)}
Hence, the statement is ‘True’.

The document NCERT Exemplar - Relations and Functions | Mathematics (Maths) Class 11 - Commerce is a part of the Commerce Course Mathematics (Maths) Class 11.
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FAQs on NCERT Exemplar - Relations and Functions - Mathematics (Maths) Class 11 - Commerce

1. What are relations and functions in mathematics?
Ans. In mathematics, relations and functions are two fundamental concepts. A relation is a set of ordered pairs, where each ordered pair consists of two elements taken from two different sets. On the other hand, a function is a special type of relation in which each element from the first set is related to exactly one element of the second set.
2. How are relations and functions represented graphically?
Ans. Relations can be represented graphically using a Cartesian plane by plotting the ordered pairs. Each point on the graph represents an ordered pair. Functions, being a special type of relation, can also be represented graphically. In a function, each element from the first set is related to exactly one element of the second set, which can be depicted by a single-valued curve on the graph.
3. What is the domain and range of a function?
Ans. The domain of a function is the set of all possible input values or independent variables for which the function is defined. It represents the x-values of the function. The range of a function is the set of all possible output values or dependent variables that the function can produce. It represents the y-values of the function.
4. How do you determine if a relation is a function?
Ans. To determine if a relation is a function, we use the vertical line test. If any vertical line intersects the graph of the relation at more than one point, then the relation is not a function. If every vertical line intersects the graph at most once, then the relation is a function.
5. What are the different types of functions?
Ans. There are several types of functions in mathematics, such as linear functions, quadratic functions, exponential functions, logarithmic functions, trigonometric functions, etc. Each type of function has its own unique characteristics and properties, which are used to solve various mathematical problems and real-life applications.
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