JEE MAIN - Mathematics (2019 - 11th January Morning Slot - No. 10)
Let f : R $$ \to $$ R be defined by f(x) = $${x \over {1 + {x^2}}},x \in R$$. Then the range of f is :
$$\left[ { - {1 \over 2},{1 \over 2}} \right]$$
$$R - \left[ { - {1 \over 2},{1 \over 2}} \right]$$
($$-$$ 1, 1) $$-$$ {0}
R $$-$$ [$$-$$1, 1]
Explanation
f(0) = 0 & f(x) is odd
Further, if x > 0 then
f(x) = $$f(x) = {1 \over {x + {1 \over x}}} \in \left( {0,{1 \over 2}} \right]$$
Hence, $$f(x) \in \left[ { - {1 \over 2},{1 \over 2}} \right]$$
Further, if x > 0 then
f(x) = $$f(x) = {1 \over {x + {1 \over x}}} \in \left( {0,{1 \over 2}} \right]$$
Hence, $$f(x) \in \left[ { - {1 \over 2},{1 \over 2}} \right]$$
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