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Functions question

2023 · 13 Apr · Shift 2 · Q33
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Functions question

2023 · 13 Apr · Shift 2 · Q33

JEE MainMathematicsFunctionsMCQ+4 / −1
The range of f(x)=4sin⁡−1(x2x2+1)f(x)=4 \sin ^{-1}\left(\frac{x^{2}}{x^{2}+1}\right)f(x)=4sin−1(x2+1x2​) is
  1. A
    [0,2π][0,2 \pi][0,2π]
  2. B
    [0,2π)[0,2 \pi)[0,2π)
  3. C
    [0,π)[0, \pi)[0,π)
  4. D
    [0,π][0, \pi][0,π]
View written solutionFree

Correct answer: B

  1. We need the range of f(x)=4sin⁡−1(x2x2+1).f(x)=4\sin^{-1}\left(\frac{x^2}{x^2+1}\right).f(x)=4sin−1(x2+1x2​).

  2. First find the range of the inner expression t=x2x2+1.t=\frac{x^2}{x^2+1}.t=x2+1x2​. Since x2≥0x^2\ge 0x2≥0, let y=x2y=x^2y=x2. Then y∈[0,∞)y\in[0,\infty)y∈[0,∞) and t=yy+1=1−1y+1.t=\frac{y}{y+1}=1-\frac{1}{y+1}.t=y+1y​=1−y+11​. Now:

  • when y=0y=0y=0, t=0t=0t=0
  • as y→∞y\to\inftyy→∞, t→1t\to 1t→1 but never equals 111

So, x2x2+1∈[0,1).\frac{x^2}{x^2+1}\in[0,1).x2+1x2​∈[0,1).

  1. Now apply sin⁡−1\sin^{-1}sin−1. The principal range of sin⁡−1u\sin^{-1}usin−1u is [−π2,π2]\left[-\frac\pi2,\frac\pi2\right][−2π​,2π​], and on [0,1)[0,1)[0,1) it is increasing. Therefore, sin⁡−1(x2x2+1)∈[0,π2).\sin^{-1}\left(\frac{x^2}{x^2+1}\right)\in\left[0,\frac\pi2\right).sin−1(x2+1x2​)∈[0,2π​). Indeed:
  • at x=0x=0x=0, value is sin⁡−1(0)=0\sin^{-1}(0)=0sin−1(0)=0
  • value π2\frac\pi22π​ would require the argument to be 111, which is impossible

  1. Multiply by 444: f(x)=4sin⁡−1(x2x2+1)∈4[0,π2)=[0,2π).f(x)=4\sin^{-1}\left(\frac{x^2}{x^2+1}\right)\in 4\left[0,\frac\pi2\right)=[0,2\pi).f(x)=4sin−1(x2+1x2​)∈4[0,2π​)=[0,2π).

  2. Hence the range is [0,2π).\boxed{[0,2\pi)}.[0,2π)​.

  3. Check options:

  • A: [0,2π][0,2\pi][0,2π] ❌ because 2π2\pi2π is not attained
  • B: [0,2π)[0,2\pi)[0,2π) ✅
  • C: [0,π)[0,\pi)[0,π) ❌ too small
  • D: [0,π][0,\pi][0,π] ❌ incorrect upper bound

Therefore, the correct option is B.

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