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प्रश्न
Show that f: [−1, 1] → R, given by f(x) = `x/(x + 2)` is one-one. Find the inverse of the function f: [−1, 1] → Range f.
(Hint: For y in Range f, y = `f(x) = x/(x +2)` for some x in [-1, 1] ie x = `2y/(1-y)`
उत्तर
`f: [−1, 1] → R is given as `f(x) = x/(x + 2)`
Let f(x) = f(y).
`=> x/(x + 2) = y/(y +2)`
=> xy + 2x = xy + 2y
=> 2x = 2y
=> x = y
∴ f is a one-one function.
It is clear that f: [−1, 1] → Range f is onto.
∴ f: [−1, 1] → Range f is one-one and onto and therefore, the inverse of the function:
f: [−1, 1] → Range f exists.
Let g: Range f → [−1, 1] be the inverse of f.
Let y be an arbitrary element of range f.
Since f: [−1, 1] → Range f is onto, we have:
`=> y = x/(x + 2)`
=> xy + 2y = x
=> x(1-y)= 2y
`=> x = (2y)/(1-y), y !=1`
`g(y) = (2y)/(1-y), y != 1`
Now `(gof) (x) = g(f(x)) = g(x/(x+2))= (2(x/(x+2)))/(1-x/(x+2)) = (2x)/(x + 2 - x) = 2x/2 = x`
`(fog)(y) = f(g(y)) = f("2y"/(1-y)) = ((2y)/((1-y)))/(((2y)/(1-y)) +2) = (2y)/(2y + 2 -2y) = (2y)/2 = y`
`:. gof = I_(-1,1) and fog - I_"Range f"`
`:. f^(-1) = g`
`=> f^(-1) (y) = "2y"/(1 - y), y != 1`
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