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प्रश्न
If f A→ A and A=R - `{8/5}` , show that the function `f (x) = (8x + 3)/(5x - 8)` is one-one onto. Hence,find `f^-1`.
उत्तर
`f(x)= (8x + 3)/(5x - 8)`
`f(x_1)= (8x_1 + 3)/(5x_1- 8) and f(x_2)= (8x_2 + 3)/(5x_2- 8)`
`f(x_1) = f(x_2)`
`(8x_1 + 3)/(5x_1 - 8)= (8x_2 + 3)/(5x_2-8)`
`(8x_1 + 3)(5x_2-8) =(8x_2 + 3)(5x_1 - 8)`
`40x_1 x_2 - 64x_1 + 15x_2 - 24 = 40x_1x_2 - 64x_2 + 15x_1 -24`
`40x_1 x_2 - 40x_1 x_2 - 64x_1+64x_2 + 15x_2- 15x_1 = 0`
`64(x_2 - x_1) + 15(x_2 - x_1) = 0`
`79(x_2 - x_2) = 0`
∴ `x_1= x_2`
∴ f(x) is one-one function
Let f(x)= y
`y=(8x + 3)/ (5x - 8) `
5yx- 8y = 8x + 3
5yx - 8x = 3 + 8y
x(5y - 8) = 8y + 3
`x = (8y + 3)/(5y - 8) ; y ∈ R - {8/5}`
∴ co - domain = Range
∴ f (x) is onto function.
∴ `x = (8y + 3)/(5y - 8)`
Replacing x by y and y by x.
`y = (8x + 3)/(5x-8)`
`f^-1 (x) = (8x + 3)/(5x - 8)`
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