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Differential Equations

Question
CBSEENMA12035734

Show that the function f in straight A space equals space straight R to the power of minus open curly brackets 2 over 3 close curly brackets space defined space as space straight f left parenthesis straight x right parenthesis space equals space fraction numerator 4 straight x plus 3 over denominator 6 straight x minus 4 end fraction space is space one minus one space and space onto. space Hence space find space straight f to the power of negative 1 end exponent.

Solution
straight f left parenthesis straight x right parenthesis space equals space fraction numerator 4 straight x plus 3 over denominator 6 straight x minus 4 end fraction
Let space straight f left parenthesis straight x subscript 1 right parenthesis space equals space straight f left parenthesis straight x subscript 2 right parenthesis
rightwards double arrow fraction numerator 4 straight x subscript 1 plus 3 over denominator 6 straight x subscript 1 minus 4 end fraction equals fraction numerator 4 straight x subscript 2 plus 3 over denominator 6 straight x subscript 2 minus 4 end fraction
rightwards double arrow space space 24 straight x subscript 1 straight x subscript 2 minus 16 straight x subscript 1 plus 18 straight x subscript 2 minus 12 space equals space 24 straight x subscript 1 straight x subscript 2 plus 18 straight x subscript 1 minus 16 straight x subscript 2 minus 12
rightwards double arrow space space 18 straight x subscript 2 plus 16 straight x subscript 2 space equals space 18 straight x subscript 1 plus 16 straight x subscript 1
rightwards double arrow 34 straight x subscript 2 space equals 34 straight x subscript 1
rightwards double arrow straight x subscript 1 space equals space straight x subscript 2
Since, fraction numerator 4 straight x plus 3 over denominator 6 straight x minus 4 end fraction is a real number, therefore, for every y in the co-domain of f, there exists a number x in straight R to the power of minus open curly brackets 2 over 3 close curly brackets space such space that space straight f left parenthesis straight x right parenthesis space equals space straight y space equals space fraction numerator 4 straight x plus 3 over denominator 6 straight x minus 4 end fraction
Therefore, f(x) is onto.
Hence, straight f to the power of negative 1 end exponent exists.
Now, let space straight y space equals space fraction numerator 4 straight x plus 3 over denominator 6 straight x minus 4 end fraction
rightwards double arrow 6 xy minus 4 straight y space equals space 4 straight x plus 3
rightwards double arrow space 6 xy minus 4 straight x space equals space 4 straight y plus 3
rightwards double arrow straight x left parenthesis 6 straight y minus 4 right parenthesis space equals space 4 straight y plus 3
rightwards double arrow space space space straight x space equals space fraction numerator 4 straight y plus 3 over denominator 6 straight y minus 4 end fraction
rightwards double arrow space space straight y equals space fraction numerator 4 straight x plus 3 over denominator 6 straight x minus 4 end fraction space space left square bracket interchanging space the space variables space straight x space and space straight y right square bracket
rightwards double arrow straight f to the power of negative 1 end exponent left parenthesis straight x right parenthesis space equals space fraction numerator 4 straight x plus 3 over denominator 6 straight x minus 4 end fraction left square bracket put space straight y space equals space straight f to the power of negative 1 end exponent left parenthesis straight x right parenthesis right square bracket

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