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Question

The functionf:XYdefined by f(x)=sinx is one-one but not onto ifXandY are respectively equal to.


A

and

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B

[0,π]and[0,1]

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C

0,π2and-1,1

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D

-π2,π2and [-1,1]

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Solution

The correct option is C

0,π2and-1,1


Explanation for the correct option:

Option C: 0,π2and-1,1

f:XY

X: domain and Y: Co-domain

Domain =X= 0,π2 and co-domain=Y=-1,1

By graph we can clearly observe function is one -one but range for given domian is 0,1.

For a unique value of x there exist a unique value of y, but for y in the interval [-1,0) does not have any pre-image.

So, the function is one-one but not onto.

Explanation for the in-correct option:

Option B [0,π]and[0,1]

In the given domain sinπ6=sin5π6=12,so function is not one one

and range of function in given domain is [0,1]

So function is not one -one but onto.

Option A : and

Sinx has range only -1,1.So, function is onto

and Sinxis periodic function,so one -one is not possible.

Therefore function is neither one-one nor onto.

Option D : -π2,π2and [-1,1]

sinxis one-one function in the given domain and range of function also is [-1,1]as shown in figure.

Therefore function is one-one and onto in the given domain and co-domain.

Hence, the correct option is OptionC.


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