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PS MATHS - SSLC - TWO MARKS QUESTION AND ANSWER

TENTH MATHS - TWO MARK QUESTIONS 06. Let A = {1, 2, 3},
AND ANSWERS B ={x / x is a prime number less than 10} find A  B
01. If A  B = {(3, 2), (3, 4), (5, 2), (5, 4)}then find and B  A
A and B A = {1, 2, 3}, B = {2, 3, 5, 7}
A = {3, 5} A B = {1, 2, 3} {2, 3, 5, 7}
B = {2, 4} = {(1, 2), (1, 3), (1, 5), (1, 7), (2, 2), (2, 3),
02. If B  A= {(-2, 3),(-2, 4), (0, 3), (0, 4),(3, 3),(3, 4)} (2, 5), (2, 7), (3, 2), (3, 3), (3, 5), (3, 7)}
then find A and B
B A = {2, 3, 5, 7} {1, 2, 3}
A = {3, 4}
= {(2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3),
B = {-2, 0, 3}
(5, 1), (5, 2), (5, 3), (7, 1), (7, 2), (7, 3)}
03. If A = {2, -2, 3}, B = {1, -4} then find A  B,
A  A and B  A 07. If A = {1, 3, 5}and B = {2, 3} then (i) find A  B
A  B = {2, -2, 3} {1, -4} and B  A(ii) Is A  B = B  A?
= {(2, 1), (2, -4), (-2, 1), (-2, -4), (3, 1), (3, -4)} (iii) show that n(A  B) = n(B  A) = n(A)  n(B)
A  A = {2, -2, 3} {2, -2, 3} A = {1, 3, 5},
= {(2, 2), (2, -2), (2, 3), (-2, 2), (-2, -2), (-2, 3), B = {2, 3}
(3, 2), (3, -2), (3, 3)} A  B = {(1, 2), (1, 3), (3, 2), (3, 3), (5, 2), (5, 3)}
B  A = {1, -4}  {2, -2, 3}
B  A = {(2, 1), (2, 3), (2, 5), (3, 1), (3, 3), (3, 5)}
= {(1, 2), (1, -2), (1, 3), (-4, 2), (-4, -2), (-4, 3)}
n(A  B)  n(B  A)
04. If A = B = {p, q} then find A  B, A  A and B  A n(A  B) = 6
A = {p, q} n(B  A) = 6
B = {p, q}
n(A)  n(B) = 3  2 = 6
A  B = {p, q} {p, q}
n(A  B) = n(B  A) = n(A)  n(B) Proved.
= {(p, p), (p, q), (q, p), (q, q)}
08. A Relation R is given by the set
A  A = {p, q} {p, q}
 {(x, y) / y = x + 3, x {0,1,2,3, 4,5}}.
= {(p, p), (p, q), (q, p), (q, q)}
B  A = {p, q} {p, q} Determine its domain and range.
= {(p, p), (p, q), (q, p), (q, q)} R = {(0, 3), (1, 4), (2, 5), (3, 6), (4, 7), (5, 8)}

05. If A = {m, n}, B =  then find A  B, A  A and Domain= {0, 1, 2, 3, 4, 5}
BA Range= {3, 4, 5, 6, 7, 8}
A = {m, n},
09. Let A={1, 2, 3, 4, 5,... 45}and be the relation
B=
defined as “square is of a number” on A. Write R as a
A  B = {m, n}  { }
={} subset of AA. Also find the domain and range of R.
A  A = {m, n}  {m, n} R = {(1, 1), (2, 4), (3, 9), (4, 16), (5, 25), (6, 36)}
= {(m, m), (m, n), (n, m), (n, n)} Domain= {1, 2, 3, 4, 5, 6}
B  A = { } {m, n} Range= {1, 4, 9, 16, 25, 36}
= { }
PV THEVAR GOVT. HR. SEC. SCHOOL, KARUPPAMPULAM, NAGAI DT. ..1..

, PS MATHS - SSLC - TWO MARKS QUESTION AND ANSWER
10. Represent the relation
(ii)Graph
{(x, y) / x = 2y, x {2, 3, 4, 5}, y {1, 2, 3, 4}}by 
Y  (6, 9)
(i) an arrow diagram (ii) a graph (iii) a set in roster
form.  (5, 8)
(i) Arrow diagram x y  (4, 7)
2 >1  (3, 6)
3 >2  (2, 5)
4 3 
(1, 4)
5 4

(ii)Graph 
Y
(4, 2)
(2, 1) 
  O X
O X
(iii)Set in roster form.
R = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}

(iii)Set in roster form.
12.The arrow diagram shows a relationship between the
R = {(2, 1), (4, 2)}
sets P and Q. Write the relation in (i) set builder form
11. Represent the relation (ii) Roster form (iii) What is the domain and range of
{(x, y)/y = x+3, x, y are natural numbers< 10}by Relation?
(i) an arrow diagram (ii) a graph (iii) a set in roster
form
(i) Arrow diagram
x y
1 1 (i) Set builder form
2 2 R = {(x, y)/y = x - 2, x P, y Q}
3 3 (ii) Roster form
4 4
>




{(5, 3), (6, 4), (7, 5)}
>




5 5
(iii) Domain = {5, 6, 7}
6 6
>




(iv) Range= {3, 4, 5}
7 7
>




8 8
>




13. Let X = {1, 2, 3, 4} and Y = {2, 4, 6, 8, 10}and
9 9
>




R = {(1, 2), (2, 4), (3, 6), (4, 8)}. Show that R is a
function and find its domain, Co-domain and range.


PV THEVAR GOVT. HR. SEC. SCHOOL, KARUPPAMPULAM, NAGAI DT. ..2..

, PS MATHS - SSLC - TWO MARKS QUESTION AND ANSWER

Arrow diagram 16. Let f(x) = 2x + 5. If  x  0, then find
f ( x  2)  f ( 2)
X Y 
x
1 >2 f(x) = 2x + 5
2 >4 f(x + 2) = 2(x+2) + 5
= 2x + 4 + 5
3 >6
= 2x + 9
4 >8
f(2) = 2(2) + 5
10
=4+5
All elements in X have only one image in Y. =9
Therefore R is function. f ( x  2)  f ( 2) 2x  9  9
=
x x
Domain X = {1, 2, 3, 4}
2x
Co-domain Y = {2, 4, 6, 8, 10} =
x
Range = {2, 4, 6, 8}
=2
14. A relation f : x  y is defined by f(x) = x2 - 2 17. A Plane is flying at a speed of 500 km per hour.
where, X = {-2, -1, 0, 3} andY = R. (i) List the ele- Express the distance ‘d’ travelled by the plane
ments of f (ii) Is f a function? as function of time ‘t’ in hours.
f(x) = x2 - 2 Distance
Speed 
2 Time
f(-2) = (-2) - 2 = 4 - 2 = 2
f(-1) = (-1)2 - 2 = 1 - 2 = -1 d
500 =
f(0) = 02 - 2 = 0 - 2 = -2 t
f(3) = 32 - 2 = 9 - 2 = 7 d = 500t
(i) f = {(-2, 2), (-1, -1), (0, -2), (3, 7)}
18. If A = {-2, -1, 0, 1, 2} and f : A  B is an onto
(ii) All elements in X have only one image in Y.
function defined by f(x) = x2 + x + 1, then find B.
Therefore it is a function.
f(x) = x2 + x + 1
15.Let f ={(x, y) / x, y N and  y = 2x} be a
f(-2) = (-2)2 + (- 2) + 1 = 4 - 2 + 1 = 3
relation on N. Find the domain, co-domain and
f(-1) = (-1)2 + (- 1) + 1 = 1 - 1 + 1 = 1
range. Is this relation a function?
f = {(1, 2), (2, 4), (3, 6), (4, 8), . . . . . . .} f(0) = (0)2 + 0 + 1 = 0 + 0 + 1 = 1
Domain= {1, 2, 3, 4, .....} f(1) = (1)2 + 1 + 1 = 1 +1 + 1 = 3
Co-domain = {1, 2, 3, 4, .....} f(2) = (2)2 + 2+ 1 = 4 +2 + 1 = 7
Range= {2, 4, 6, 8, .......}
B = {1, 3, 7}
All elements in X have only one image in Y.
Therefore it is a function.

PV THEVAR GOVT. HR. SEC. SCHOOL, KARUPPAMPULAM, NAGAI DT. ..3..