माना कि फलन `f : R rarr R` जो `f(x) = (x^(2) + 4x + 30)/(x^(2) - 8x + 18)` से परिभाषित है। क्या f एकैकी फलन है?

Explanation

दिया है, `f(x) = (x^(2) + 4x + 30)/(x^(2) - 8x + 18)`
माना कि f(x) = f(0)
`rArr (x^(2) + 4x + 30)/(x^(2) - 8x + 18) = (3)/(15) = (5)/(3)`
`rArr 3x^(2) + 12x + 90 = 5x^(2) - 40x + 90`
`rArr 2x^(2) - 52 x = 0`
`rArr x = 0.26`
अतः `f(0) = f(26) rArr f` एकैकी फलन नहीं है।
दूसरी विधि : माना कि `x_(1), x_(2), in` प्रान्त R.
माना कि `f(x_(1)) = f(x_(2))`
`rArr (x_(1)^(2) + 4x_(1) + 30)/(x_(1)^(2) - 8x_(1) + 18) = (x_(2)^(2) + 4x_(2) + 30)/(x_(2)^(2) - 8x_(2) + 18)`
`rArr (x_(1)^(2) + 4x_(1) + 30)/(x_(1)^(2) - 8x_(1) + 18) -1 = (x_(2)^(2) + 4x_(2) + 30)/(x_(2)^(2) - 8x_(2) + 18) - 1`
`rArr (12(x_(1) + 1))/(x_(1)^(2) - 8x_(1) + 18) = (12(x_(2) + 1))/(x_(2)^(2) - 8x_(2) + 18)`
`rArr x_(1) x_(2)^(2) - 8x_(1)x_(2) + 18 x_(1) + x_(2)^(2) - 8x_(2) + 18`
`= x_(1)^(2) x_(2) - 8x_(1)x_(2) + x_(1)^(2) - 8x_(1) + 18`
`rArr x_(1)x_(2) (x_(2)-x_(1)) + (x_(2)^(2) - x_(1)^(2)) - 26 (x_(2) - x_(1)) = 0`
`rArr (x_(2) - x_(1)) (x_(1)x_(2) + x_(2) + x_(1) - 26) = 0`
`rArr x_(2) = x_(1), x_(2) = (26- x_(1))/(x_(1) + 1)`
यदि `x_(1) = 0`, तो `x_(2) = 26`, अतः f(0) = f(26)
इस प्रकार f एकैकी नहीं।