Let is not a positive integer. Define a function as , then is
- A.injective but not surjective
- B.not injective
- C.surjective but not injective
- D.neither injective nor surjective
Show answer and solution
Correct option: A
Solution
f is injective: suppose f(a) = f(b), then 2a/(a-1) = 2b/(b-1), so a(b-1) = b(a-1), giving ab - a = ab - b, hence a = b. So f is one-one.
For surjectivity, note f(x) = 2x/(x-1) = 2 + 2/(x-1). As x varies over A (all reals except positive integers), we check if every real y is achieved. Setting y = 2 + 2/(x-1) gives x = 1 + 2/(y-2), which requires y ≠ 2. When y = 2, there's no solution. Also, x = 1 + 2/(y-2) must not be a positive integer — for example if y = 4, x = 2, which IS a positive integer and is excluded from A, so y = 4 is not in the range either. Thus f is not surjective.
For surjectivity, note f(x) = 2x/(x-1) = 2 + 2/(x-1). As x varies over A (all reals except positive integers), we check if every real y is achieved. Setting y = 2 + 2/(x-1) gives x = 1 + 2/(y-2), which requires y ≠ 2. When y = 2, there's no solution. Also, x = 1 + 2/(y-2) must not be a positive integer — for example if y = 4, x = 2, which IS a positive integer and is excluded from A, so y = 4 is not in the range either. Thus f is not surjective.
[JEE Main 2019 · 9 Jan · Shift II]