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Question-1 :-  The relation f is defined by relation The relation g is defined by function Show that f is a function and g is not a function.

Solution :-
    It is observed that for 0 ≤ x < 3, 
    f(x) = x²
    3 < x ≤ 10, 
    f(x) = 3x
    Also, at x = 3, 
    f(x) = 3² = 9 or 
    f(x) = 3 × 3 = 9 i.e., 
    at x = 3, f(x) = 9.
    Therefore, for 0 ≤ x ≤ 10, the images of f(x) are unique. 
    Thus, the given relation is a function.
    It can be observed that for x = 2, g(x) = 2² = 4 and g(x) = 3 × 2 = 6
    Hence, element 2 of the domain of the relation g corresponds to two different images i.e., 4 and 6. 
    Hence, this relation is not a function.
   

Question-2 :-  If f(x) = x², find = [f(1.1)-f(1)]/(1.1 - 1).

Solution :-
   Given that f(x) = x².
   [f(1.1) - f(1)]/(1.1 - 1)
 = [(1.1)² - (1)²]/(1.1 - 1) 
 = (1.21 - 1)/(0.1)
 = 0.21/0.1
 = 2.1
   

Question-3 :-  Find the domain of the function f(x) = (x² + 2x + 1)/(x² - 8x + 12).

Solution :-
   Given that f(x) = (x² + 2x + 1)/(x² - 8x + 12). 
   (x² - 8x + 12) = x² - 6x - 2x + 12
 = x(x - 6) - 2(x - 6)
 = (x - 2)(x - 6)
   f(x) = (x² + 2x + 1)/(x - 2)(x - 6)
   It can be seen that function f is defined for all real numbers except at x = 6 and x = 2. 
   Hence, the domain of f is R – {2, 6}.
    

Question-4 :-  Find the domain and the range of the real function f defined by f (x) = √x - 1

Solution :-
   Given that f(x) = f (x) = √x - 1.
   It can be seen that  is defined for (x – 1) ≥ 0.
   i.e.,  is defined for x ≥ 1.
   Therefore, the domain of f is the set of all real numbers greater than or equal to 1 i.e., the domain of f = [1, ∞).
   As x ≥ 1 ⇒ (x – 1) ≥ 0 ⇒ √x - 1 ≥ 0
   Therefore, the range of f is the set of all real numbers greater than or equal to 0 i.e., the range of f = [0, ∞).
   

Question-5 :-  Find the domain and the range of the real function f defined by f (x) = |x - 1|.

Solution :-
   Given that f(x) = |x - 1|.
   It is clear that |x – 1| is defined for all real numbers. 
   Domain of f = R
   Also, for x ∈ R, |x – 1| assumes all real numbers.
   Hence, the Range of f is the set of all non-negative real numbers.
   

Question-6 :-  Let f = {(x, x²/(1 + x²)) : x ∈ R} be a function from R into R. Determine the range of f.

Solution :-
   Given that f(x) = {(x, x²/(1 + x²)) : x ∈ R}. 
   f(x) = {(0, 0), (±0.5, 1/5), (±1, 1/2), (±1.3, 9/13), (±2, 4/5), (3, 9/10),.....}
   The range of f is the set of all second elements. 
   It can be observed that all these elements are greater than or equal to 0 but less than 1.
   Range of f = [0, 1)
    

Question-7 :-  Let f, g : R → R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find f + g, f – g and f/g.

Solution :-
   Given that f(x) = x + 1, g(x) = 2x – 3.
   f + g = x + 1 + 2x - 3 = 3x - 2
   f - g = x + 1 - 2x + 3 = -x + 4
     f/g = (x + 1)/(2x - 3)
   

Question-8 :-  Let f = {(1,1), (2,3), (0,–1), (–1, –3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b. Determine a, b.

Solution :-
   Given that f = {(1,1), (2,3), (0,–1), (–1, –3)} and 
   f(x) = ax + b
   (1, 1) ∈ f
⇒ f(1) = 1 
⇒ a × 1 + b = 1 
⇒ a + b = 1 
  (0, –1) ∈ f
⇒ f(0) = –1 
⇒ a × 0 + b = –1 
⇒ b = –1 
  On substituting b = –1 in a + b = 1, we obtain a + (–1) = 1 ⇒ a = 1 + 1 = 2. 
  Thus, the respective values of a and b are 2 and –1.
   

Question-9 :-  Let R be a relation from N to N defined by R = {(a, b) : a, b ∈ N and a = b²}. Are the following true?
(i) (a,a) ∈ R, for all a ∈ N
(ii) (a,b) ∈ R, implies (b,a) ∈ R
(iii) (a,b) ∈ R, (b,c) ∈ R implies (a,c) ∈ R.
Justify your answer in each case.

Solution :-
   Given that R = {(a, b) : a, b ∈ N and a = b²}. 
(i) (a,a) ∈ R, for all a ∈ N 
   It can be seen that 2 ∈ N;
   However, 2 ≠ 2² = 4. 
   Therefore, the statement '(a, a) ∈ R, for all a ∈ N' is not true.

(ii) (a,b) ∈ R, implies (b, a) ∈ R 
   It can be seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 3².
   Now, 3 ≠ 9² = 81; 
   Therefore, (3, 9) ∉ N.
   Therefore, the statement '(a, b) ∈ R, implies (b, a) ∈ R' is not true.

(iii) (a,b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R. 
   It can be seen that (9, 3) ∈ R, (16, 4) ∈ R because 9, 3, 16, 4 ∈ N and 9 = 3² and 16 = 4². 
   Now, 9 ≠ 4² = 16; 
   Therefore, (9, 4) ∉ N
   Therefore, the statement '(a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R' is not true.
    

Question-10 :-  Let A ={1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)} Are the following true?
(i) f is a relation from A to B
(ii) f is a function from A to B.
Justify your answer in each case.

Solution :-
   Given that A ={1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)}.
   A × B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), 
           (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), 
           (4, 9), (4, 11), (4, 15), (4, 16)} 

(i) f is a relation from A to B 
   A relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B. 
   It is observed that f is a subset of A × B. 
   Thus, f is a relation from A to B.

(ii) f  is a function from A to B.
   Since the same first element i.e., 2 corresponds to two different images 
   i.e., 9 and 11, relation f is not a function. 
   

Question-11 :-  Let f be the subset of Z × Z defined by f = {(ab, a + b) : a, b ∈ Z}. Is f a function from Z to Z? Justify your answer.

Solution :-
   Given that f = {(ab, a + b) : a, b ∈ Z}.
   The relation f is defined as f = {(ab, a + b): a, b ∈ Z}
   We know that a relation f from a set A to a set B is said to be a function 
   if every element of set A has unique images in set B.
   Since 2, 6, –2, –6 ∈ Z, (2 × 6, 2 + 6), [(–2) × (–6), (–2) + (–6)) ∈ f i.e., 
   (12, 8), (12, –8) ∈ f
   It can be seen that the same first element i.e., 
   12 corresponds to two different images i.e., 8 and –8. 
   Thus, relation f is not a function.
   

Question-12 :-  Let A = {9, 10 ,11 ,12 ,13} and let f : A → N be defined by f (n) = the highest prime factor of n. Find the range of f.

Solution :-
   Given that A = {9,10,11,12,13} & f(n) = the highest prime factor of n. 
   Prime factor of 9 = 3
   Prime factors of 10 = 2, 5 
   Prime factor of 11 = 11 
   Prime factors of 12 = 2, 3 
   Prime factor of 13 = 13
   f(9) = The highest prime factor of 9 = 3
    
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