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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2008

  • question_answer
        The function\[f:X\to Y\]defined   by \[f(x)=\sin x\]is one-one but not onto, if\[X\]and Y are respectively equal to

    A)  R and R

    B)  \[[0,\pi ]\]and [0,1]

    C)  \[\left[ 0,\frac{\pi }{2} \right]\]and \[[-1,1]\]

    D)  \[\left[ \frac{-\pi }{2},\frac{\pi }{2} \right]\]and \[[-1,1]\]

    Correct Answer: C

    Solution :

                    Since,\[f:X\to Y,\]then\[f(x)=sin\text{ }x\] Now, take option (c). Domain\[=\left[ 0,\frac{\pi }{2} \right],\]Range\[=[-1,1]\] For every value of\[x,\]we get unique value of y. But the value of y in\[[-1,0)\]does not have any pre-image. \[\therefore \]Function is one-one but not onto.


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