Q1. (a) Given the function f(x) = x2 + 3x + 5 such that f:[1,2] → [5,15).
Check whether f(x) is
(i) One-one,
(ii) Onto.
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Answer:
- Hope this helps you
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.
solution
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