If (A +- 2).r +( + 2)x<1 V Xe R then de
Answers
Step-by-step explanation:
ANSWER
Given, function f:R→R such that f(x)=1+x
2
,
Let A and B be two sets of real numbers.
Let x
1
,x
2
∈A such that f(x
1
)=f(x
2
).
⇒1+x
1
2
=1+x
2
2
⇒x
1
2
−x
2
2
=0⇒(x
1
−x
2
)(x
1
+x
2
)=0
⇒x
1
=±x
2
. Thus f(x
1
)=f(x
2
) does not imply that x
1
=x
2
.
For instance, f(1)=f(−1)=2, i.e. , two elements (1, -1) of A have the same image in B. So, f is many-one function.
Now, y=1+x
2
⇒x=
y−1
⇒elements < y have no pre-image in A (for instance an element -2 in the codomain has no pre-image in the domain A). So, f is not onto.
Hence, f is neither one-one onto. So, it is not bijective.
Step-by-step explanation:
Option A is correct.
R is reflexive and symmetric but not transitive.
Reflexive Property
The Reflexive Property states that for every real number x, xRx.
For given, we have, {(1,1),(2,2),(3,3)}
Symmetric Property
The Symmetric Property states that for all real numbers x and y,
if xRy, then yRx.
For given, we have, {(1,2),(2,1),(2,3),(3,2)}
Transitive Property
The Transitive Property states that for all real numbers x ,y, and z,
if xRy and yRz , then xRz
we have 3R2 and 2R1 but 3
R1