Question

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.​

Answer 1

Answer:

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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