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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NTA Abhyas 2020 · Mathematics · Inverse Trigonometric Functions

The range of the function f (x) = sin^ - 1 [x^ 2 - (1)/(3)] - cos^...

The range of the function f(x)=sin−1f \left(x\right) = sin^{- 1}\left[x^{2} - \dfrac{1}{3}\right] -cos−1cos^{- 1}\left[x^{2} + \dfrac{2}{3}\right]is(where,[x]is (where, \left[x\right] represents the greatest integer value of xx)

Options

  1. A[−π,0]\left[- \pi, 0\right]
  2. B{−π,0}\left\{- \pi, 0\right\}
  3. C{0,π}\left\{0, \pi\right\}
  4. D{0,π,−π}\left\{0, \pi, - \pi\right\}

Correct answer

B. {−π,0}\left\{- \pi, 0\right\}

Step-by-step solution

\left[x^{2} - \dfrac{1}{3}\right] =\left[x^{2} + \dfrac{2}{3} -1\right] = \left[x^{2} + \dfrac{2}{3}\right]−-1 = k(let)k\left(let\right)
Now, the function is sin−1k−cos−1(k+1)sin^{- 1}k - cos^{- 1} \left(k + 1\right)

For the above function to be defined,
- 1 ≤ k ≤ 1 & - 1 \le \left(k + 1\right)≤ 1<br/><br />∴ k =\left\{- 1,0\right\}<br/><br />∴RangeRange= \left\{sin^{- 1}\left(-1\right)−cos−1(0)- cos^{- 1}\left(0\right), sin^{- 1}\left(0\right)- cos^{- 1}\left(1\right)\right\}= \left\{- \pi,0\right\}

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