JEE MAIN - Mathematics (2020 - 8th January Evening Slot - No. 13)
Let ƒ : (1, 3) $$ \to $$ R be a function defined by
$$f(x) = {{x\left[ x \right]} \over {1 + {x^2}}}$$ , where [x] denotes the greatest integer $$ \le $$ x. Then the range of ƒ is
$$f(x) = {{x\left[ x \right]} \over {1 + {x^2}}}$$ , where [x] denotes the greatest integer $$ \le $$ x. Then the range of ƒ is
$$\left( {{2 \over 5},{1 \over 2}} \right) \cup \left( {{3 \over 4},{4 \over 5}} \right]$$
$$\left( {{3 \over 5},{4 \over 5}} \right)$$
$$\left( {{2 \over 5},{4 \over 5}} \right]$$
$$\left( {{2 \over 5},{3 \over 5}} \right] \cup \left( {{3 \over 4},{4 \over 5}} \right)$$
Explanation
f(x) = $$\left\{ {\matrix{
{{x \over {{x^2} + 1}},} & {1 < x < 2} \cr
{{{2x} \over {{x^2} + 1}},} & {2 \le x < 3} \cr
} } \right.$$
$$ \therefore $$ f(x) is decreasing function
$$ \therefore $$ Range is $$\left( {{2 \over 5},{1 \over 2}} \right) \cup \left( {{3 \over 4},{4 \over 5}} \right]$$.
$$ \therefore $$ f(x) is decreasing function
$$ \therefore $$ Range is $$\left( {{2 \over 5},{1 \over 2}} \right) \cup \left( {{3 \over 4},{4 \over 5}} \right]$$.
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