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2

Q1.

Solution

Q2.

Solution

Q3. Let A = {2, 4, 6, 8, 10} and B = {(a, b) : a divides b; a , b  A}. Write the elements of B.

Solution

B = {(2, 2), (2, 4), (2, 6), (2, 8), (2, 10), (4, 8)}
Q4. If the set A has 4 elements and the set B = {4, 5, 6}, then find the number of elements in (A  B).

Solution

Since, n(A) = 4 and n(B) = 3 Therefore n((A  B) = 12
Q5. If A = {1, 2, 3}, B = {4}, C = {5}, verify that A  (B  C) = (A  B)  (A  C)

Solution

(B  C) = {4, 5} So, A  (B  C) = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)} Now    (A  B) = {(1, 4), (2, 4), (3, 4)} And    (A  C) = {(1, 5), (2, 5), (3, 5)} Therefore, (A  B)  (A  C) =  {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)} Hence, A  (B  C) = (A  B)  (A  C)
Q6. The function f is defined by f open parentheses x close parentheses equals open curly brackets table row cell 2 minus x comma space i f space x less than 0 end cell row cell 2 comma space i f space x equals 0 end cell row cell x plus 2 comma space i f space x greater than 0 end cell end table close Draw the graph of f(x).

Solution

Here, f(x) = 2 – x, x < 0, this gives f(-3) = 5, f(-2) = 4, f(-1) = 3,… f(3) = 5, f(2) = 4, f(1) = 3,… for f(x) = x + 2, x > 0. Thus the graph of f is as follows-
Q7. If R is the relation ‘is less than’ from A = {2, 4, 6, 8, 10} and B = {2, 6, 10}, write down the Cartesian product corresponding to R.

Solution

Clearly R = {(a, b)  A  B; a < b}  R = {(2, 6), (2, 10), (4, 6), (4, 10), (6, 10), (8, 10)}
Q8. Let X = {2, 4, 6}, Y = {3, 5}and Z= {2, 3}, then find X  (Y  Z).

Solution

(Y  Z) = {2, 3, 5} Therefore, X  (Y  Z) = {(2, 2), (2, 3), (2, 5), (4, 2), (4, 3), (4, 5), (6, 2), (6, 3), (6, 5)}
Q9.

Solution

Q10. Let R = {(3,1),(3, 5), (5, 1), (5, 5)} is a relation from A = {2,3,4,5} to B = {1,3, 5}  Find the Codomain of R.

Solution

Codomain of R = {1, 3, 5}
Q11.

Solution

Q12. If A = {x, y}, then find A  A  A.

Solution

A  A  A = {(x, x, x ), (x, x, y), (x, y, x), (x, y, y), (y, x, x), (y, x, y), (y, y, x) (y, y, y)}
Q13.

Solution

Q14. Let f: R  R be defined as                                                                   Show that f is not a function.

Solution

W h e n space x less than 5 comma space w e space h a v e space f open parentheses x close parentheses equals 3 x plus 4 W h e n space x greater than 5 comma space w e space h a v e space f open parentheses x close parentheses equals x plus 5 B u t space w h e n space x equals 5 comma space w e space h a v e comma space f open parentheses 5 close parentheses equals 3 open parentheses 5 close parentheses plus 4 equals 19 A l s o comma space f open parentheses 5 close parentheses equals 5 plus 5 equals 10 I t space i s space c l e a r space t h a t space f open parentheses 5 close parentheses space i s space n o t space u n i q u e l y space d e t e r m i n e d space a n d h e n c e space f space i s space n o t space a space f u n c t i o n.
Q15.

Solution

Q16. Determine the domain and range of the relation R defined by R = {(x + 2, x + 4): x  {0, 1, 2, 3, 4}

Solution

Domain = {2, 3, 4, 5, 6}, Range = {4, 5, 6, 7, 8}
Q17. Draw the graph of the function y = x2 + 4x + 6.

Solution

The given function is y = x2 + 4x + 6    = (x + 2)2 + 2 The graph of the given function will be a parabola. The parabola will open upward. The least value of (x + 2)2 is zero and will be so when x = -2. When x = -2, y = 2 The vertex is (-2, 2). So, the graph of the given function is as follows.  
Q18. Determine the domain and range of the following relation on R: R equals open curly brackets open parentheses x comma 2 over x close parentheses : 0 less than x less than 8 comma space x element of N close curly brackets

Solution

R equals open curly brackets open parentheses x comma 2 over x close parentheses : 0 less than x less than 8 comma space x element of N close curly brackets Domain of R = {1, 2, 3, 3, 5, 6, 7} R a n g e space o f space R equals open curly brackets 2 comma 1 comma 2 over 3 comma 1 half comma 2 over 5 comma 1 third comma 2 over 7 close curly brackets
Q19. Let A = {2, 4, 6} and B = {6, 8, 12, 18}. Define a relation R from A to B by R= {(x, y): x divides y; x A, y B}. Express R as a set of ordered pairs.

Solution

R = {(2, 6), (2, 8), (2, 12), (2, 18), (4, 8), (4, 12), (6, 6), (6, 12), (6, 18)}
Q20. Find the domain of the function given by: .

Solution

f(x) is defined only when or when x -1, -2 So, Domain of f(x) = R – {-1, -2}
Q21. Let A = {1, 2, 3}, B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} f: A B be defined as f(x) = x + 5. Find the domain and range of the function.

Solution

We have, f(x) = x + 5 f(1) = 6, f(2) = 7, f(3) = 8 So, f = {(1, 6), (2, 7), (3, 8)} Domain of f = {1, 2, 3} Range of f  = {6, 7, 8}


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