यदि f : R - `{(7)/(5)} to R - {(3)/(5)}` जहाँ f (x) = `(3x + 4)/(5x - 7 )` और g : R `- {(3)/(5)} to R - {(7)/(5)}` जहाँ g(x) = `(7x + 4 )/(5x + 3)` परिभाषित हैं । दर्शाइये कि gof = `I_(A) "और" fog = I_(B)` जहाँ B = R - `{(3)/(5)} ` और `A = R - {(7)/(5)}` .

Explanation

यहाँ f : `A to B` , f (x) = (3x + 4 )/(5x - 7 )
और g : B `to` A , g (x) = `(7x + 4)/(5x - 7 )`
और g : B `to` A , g (x) = `(7x + 4)/(5x - 3)` .
`therefore gof : A to A` और fog : `B to B`

अब `x in A` के लिए ,
(gof )(x) = g [ f (x) ]
`= g ((3x + 4 )/(5x - 7 ) = (7 ((3x + 4)/(5x - 7)) + 4 )/(5 ((3x + 4)/(5x - 7)) - 3))`
= `(21x + 28 = 20x - 28)/(15x + 20 - 15x + 21) = (41x)/(41) = x.`
`rArr (gof) (x) = x AA x in A.`
`rArr (gof)(x) = I_(A) (x) `
`rArr gof = I_(A)`

पुनः `x in B` के लिए ,
(fog)(x) = f [ g (x)]
= `f ((7x + 4) /(5x - 3)) = (3((7x + 4)/(5x - 3)) + 4)/(5 ((7x + 4)/(5x - 3)) - 7 )`
` = (21x + 12 + 20x - 12)/(35 x + 20 - 35x + 21) = (41x)/(41) = x`
`rArr (fog)(x) = xAAxin B`
`rArr (fog)(x) = I_(B) (x) `
`rArr fog = I_(B)` . यही सिद्ध करना था ।