Mark the correct alternative in each of the following:
Let f : R –{n} → R be a function defined by
where m ≠ n. Then,
Given that f : R –{n} → R where
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Let f(x) = f(y)
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⇒ (x-m)(y-n)=(x-n)(y-m)
⇒ xy – xn – my + mn = xy – xm – ny + mn
⇒ x = y
So, f is one-one.
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⇒ y(x-n)=(x-m)
⇒ xy – ny = x – m
⇒ x(y-1) = ny – m
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For y = 1 , no x is defined.
So, f is into.
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