Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

Range of the function f (x) = [x / (1 + x^2)] is (1) (-∞,-∞) (2) [-1,1] (3) [(-1 / 2), (1 / 2)] (4) [-√2, √2]

Solution: (3)

f (x) = [x / (1 + x2)]

1 + x2 ≠ 0

x2 ≠ -1

x ∈ R

Domain: x ∈ R

1 + x2 ≥ 1

f (x) = y = [x / (1 + x2)]

yx2 – x + y = 0

D ≥ 0

(-1)2 – 4yy ≥ 0

1 – 4y2 ≥ 0

4y2 – 1 ≤ 0

(2y – 1) (2y + 1) ≤ 0

y ∈ [-1 / 2, 1 / 2]

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