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JEE Main 202331 JanuaryMorning Shift
Exam
JEE Main
Year
2023
Date
31 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2023 · 31 January · Morning Shift · Mathematics · Functions

If the domain of the function f(x) = [x] 1 + x ^ 2 , where [x] is...

If the domain of the function f\left(x\right) = \dfrac{\left[x\right]}{1 + {x}^{2}},where[x], where \left[x\right] is greatest integer ≤x\le x, is [2,6)[ 2, 6 ), then its range isIf the domain of the function f(x) = [x] 1 + x ^ 2 , where [x] is greatest integer ≤ x , is [ 2, 6 ) , then its range is

Options

  1. A(526,25]−{929,27109,1889,953}\left(\dfrac{5}{26}, \dfrac{2}{5}\right] - \left\{\dfrac{9}{29}, \dfrac{27}{109}, \dfrac{18}{89}, \dfrac{9}{53}\right\}
  2. B(526,25]\left(\dfrac{5}{26}, \dfrac{2}{5}\right]
  3. C(537,25]−{929,27109,1889,953}\left(\dfrac{5}{37}, \dfrac{2}{5}\right] - \left\{\dfrac{9}{29}, \dfrac{27}{109}, \dfrac{18}{89}, \dfrac{9}{53}\right\}
  4. D(537,25]\left(\dfrac{5}{37}, \dfrac{2}{5}\right]

Correct answer

D. (537,25]\left(\dfrac{5}{37}, \dfrac{2}{5}\right]

Step-by-step solution

Given:

f(x)=[x]1+x2f\left(x\right) = \dfrac{\left[x\right]}{1 + {x}^{2}}

So,

{f(x)=21+x2;x∈[2,3)f(x)=31+x2;x∈[3,4)f(x)=41+x2;x∈[4,5)f(x)=51+x2;x∈[5,6)\begin{cases}f\left(x\right) = \dfrac{2}{1 + {x}^{2}}; & & x \in [ 2, 3 ) \\ f\left(x\right) = \dfrac{3}{1 + {x}^{2}}; & & x \in [ 3, 4 ) \\ f\left(x\right) = \dfrac{4}{1 + {x}^{2}}; & & x \in [ 4, 5 ) \\ f\left(x\right) = \dfrac{5}{1 + {x}^{2}}; & & x \in [ 5, 6 )\end{cases}

If the domain of the function f(x) = [x] 1 + x ^ 2 , where [x] is greatest integer ≤ x , is [ 2, 6 ) , then its range is

Hence,

f(x)∈(537,25]f\left(x\right) \in \left(\dfrac{5}{37}, \dfrac{2}{5}\right]

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