Relation & Functions
Question
Q1.
If f(1) = 7 , f(2) = 11 , f(3) = 15 , f(4) = 19 and fn is described by f(x) = Ax + B , find A&B .
(a) (4,3)
(b) (4,1)
(c) (8,4)
(d) (7,3)
Solution. a
f(1) = 7
A + B = 7 ----------(1)
f(2) = 11
2A + B = 11 ----------(2)
On solving equation (1) and
(2)
A = 4 & B = 3
Solution
Q2.
Find the domain of f(x) =
1/x
(a) D = R - {1}
(b) D = R - {0}
(c) D = R
(d) D = R - {2}
Solution. b
f(x) =
1/x
is defined for each
value except 0 .
Solution
Q3.
Find the domain of f(x) =
(x + 3)2/x2 - 8x + 12
(a) D = R - {6}
(b) D = R - {1}
(c) D = R - {2,6}
(d) D = R
Solution. c
(x + 3)2/(x - 6)(x - 2)
,
denominator will be zero if x = 6 or 2 (or function will not be defined for x = 6 or 2)
Solution
Q4.
Find the domain of a given real valued function , f(x) =
4x + 6/x2 - 4
(a) D = R - {1}
(b) D = R
(c) D = R - {-2,2}
(d) D = R - {6}
Solution. c
4x + 6/(x + 2) (x - 2)
if x = -2 , or 2 then denominator will be zero and function will not be defined and is the value for which a function is defined. So
D = R - {-2,2}
Solution
Q5.
Is f = {(1,1),(2,3),(3,5),(4,7)} a function ? If this is described by the formula , f(x) = Ax + B , then what values should be
assigned to A and B ?
(a) (1,2)
(b) (1,-1)
(c) (2,-1)
(d) (2,3)
Solution. c
Since no two ordered pair in f have the same first component . So , f is a function
Such that f(1) = 1 , f(2) = 3 , f(3)
= 5 and f(4) = 7
It is given that f(x) = Ax + B
∴ f(1) = 1 , so A + B = 1 ----------(1)
f(2) = 3 , so 2A + B = 3 ----------(2)
On solving equation 1 and 2 we get,
A = 2 , B = -1
Solution
