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प्रश्न
Classify the following function as injection, surjection or bijection :
f : R → R, defined by f(x) = `x/(x^2 +1)`
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उत्तर
f : R → R, defined by f(x) = `x/(x^2 +1)`
Injection test:
Let x and y be any two elements in the domain (R), such that f(x) = f(y).
f(x) = f(y)
`x/(x^2+1) = y/(y^2 + 1)`
xy2+ x = x2y + y
xy2−x2y + x −y = 0
−xy (−y+x)+ 1 (x−y) = 0
(x−y) (1−xy) = 0
x = y or x = `1/y`
So, f is not an injection.
Surjection test:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).
f(x) = y
`x/(x^2 +1)= y`
yx2− x+y = 0
`x ((-1) ± sqrt(1-4x^2))/(2y)` if y ≠ 0
`= (1±sqrt(1-4y^2))/(2y) ,`which may not be in R
For example, if y=1, then
`(1±sqrt(1-4))/(2y) = (1± isqrt3)/2`
which is not in R
So, f is not surjection and f is not bijection.
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