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CBSE Class 12-science Answered

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Asked by alanpeter9611 | 12 Feb, 2019, 06:08: PM
Expert Answer
Bijective space is space one space minus space one space and space onto
for space one minus one
straight f left parenthesis straight x right parenthesis equals left parenthesis straight y right parenthesis
fraction numerator straight x minus 2 over denominator straight x minus 3 end fraction equals fraction numerator straight y minus 2 over denominator straight y minus 3 end fraction
cross space multiply space and space simplify
3 straight x minus 2 straight x equals 3 straight y minus 2 straight y
straight x equals straight y
so space one space minus space one
now space for space onto comma space for space every space straight x element of straight A space there space should space be
straight f open parentheses straight x close parentheses equals straight y
fraction numerator straight x minus 2 over denominator straight x minus 3 end fraction equals straight y
straight x minus 2 equals xy minus 3 straight y
straight x equals fraction numerator 2 minus 3 straight y over denominator 1 minus straight y end fraction element of straight A...... straight y not equal to 1
thus space for space any space straight y element of straight B comma space there space exists space fraction numerator 2 minus 3 straight y over denominator 1 minus straight y end fraction element of straight A.
straight f open parentheses fraction numerator 2 minus 3 straight y over denominator 1 minus straight y end fraction close parentheses equals fraction numerator space fraction numerator 2 minus 3 straight y over denominator 1 minus straight y end fraction minus 2 over denominator space fraction numerator 2 minus 3 straight y over denominator 1 minus straight y end fraction minus 3 end fraction equals straight y..... so space this space is space true
straight f space is space onto
so space straight f space is space bijective
straight f to the power of negative 1 end exponent open parentheses straight x close parentheses equals fraction numerator 2 minus 3 straight x over denominator 1 minus straight x end fraction equals 4
straight x equals 2
straight f to the power of negative 1 end exponent open parentheses 7 close parentheses equals fraction numerator 2 minus 3 open parentheses 7 close parentheses over denominator 1 minus 7 end fraction equals 19 over 6
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