सिद्ध कीजिए कि `f: R to {x in R:-1 lt x lt 1}` जहाँ `f(x) =(x)/(1+|x|), x in R ` द्वारा परिभाषित फलन एकैकी तथा आच्छादक है ।

Explanation

दिया गया हैं :
`f (x) = (x)/(1 + |x|) = {{:((x)/(1 + x) , "यदि" x ge 0), ((x)/(1 - x) , "यदि" x le 0):}`
[ `therefore |x| = {{:(x, "यदि" x ge 0), (- x , "यदि" x lt 0):}]`
f एकैकी हैं : माना `x_(1) "और" x_(2)` प्रांत R के दो भिन्न अवयव हैं ।
स्थिति I : यदि `x_(1) ge 0 , x_(2) ge 0 `
`therefore f (x_(1) = (x_(1))/(1 + x+(1)) "और" f (x _(2)) = (x_(2))/(1 + x_(2)))` .
अब , `x_(1) ne x_(2) rArr 1 1 + x_(1) ne 1 + x_(2) `
`rArr (1)/(1 + x_(1) ) ne (1)/(1 + x_(2))`
`rArr (-1)/(1 + x_(1)) ne (-1)/(1 + x_(2))`
`therefore 1 - (1)/(1 + x_(1)) ne 1 - (1)/(1 + x_(2))`
`rArr (x_(1))/(1 + x_(1)) ne f (x_(2)) `
स्थिति II : यदि `x_(1) lt 0 , x_(2) lt 0`
`therefore f (x_(1)) = (x_(1))/(1 - x_(1)) और f (x_(2)) = (x_(2))/(1 - x_(2))`
अब , `x_(1) ne x_(2)`
`rArr - x _(1) ne - x_(2)`
`rArr 1 - x_(1) ne 1 - x_(2)`
`rArr (1)/(1 - x_(1)) ne (1)/(1 - x_(2))`
`rArr (1)/(1 - x_(1)) -1 ne (1)/(1 - x_(2)) -1`
`rArr (x_(1) /(1- x_(1)) ne (x_(2))/(1 - x_(2)))`
`rArr f (x_(1)) ne f (x_(2))`
स्थिति III : यदि `x_(1) ge 0 , x_(2) lt 0`
`therefore x_(1) ne x_(2) rArr f (x_(1)) ne f ( x_(2))`,
यदि `x_(1) lt 0 , x_(2) ge 0`
`therefore f(x_(1) = (x_(1))/(1 - x_(1)) lt 0` और `f (x_(2)) = (x_(2))/(1 + x_(2)) gt 0)`
`therefore x_(1) ne x_(2) rArr f (x_(1) ne f (x_(2))`
`therefore f ` एकैकी फलन हैं ।
f आच्छादक हैं : माना y सहप्रांत R का स्वेच्छ अवयव हैं ।
`therefore y = f(x) = (x)/(1 + |x|) = {{:((x)/(1 + x) , "यदि" x ge 0), ((x)/(1 - x) , "यदि" x le 0):}`
`therefore y = (x)/(1 + x ) , "जब" y ge 0 `
`rArr y - xy = x rArr ( 1 + y ) = y rArr x = ( y )/( 1 + y ) lt 0`
`rArr f (x) = f ((y )/( 1+ y )) = (((y)/(1 + y)))/(1 - ((y)/(1 + y ))) = y`
`therefore f ` आच्छादक फलन हैं ।
अतः f एकैकी और आच्छादक फलन हैं । यही सिद्ध करना था ।