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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 y = 2 ^ - 1 [ x ^ 2 + (1)/(2)] + ^ - 1 [...

The range of the function y=2sin⁡−1[x2+12]+cos⁡−1y = 2\sin^{- 1} \left[{x}^{2} + \dfrac{1}{2}\right] + \cos^{- 1}\left[{x}^{2} - \dfrac{1}{2}\right]is(where,[⋅]is (where, \left[\cdot\right] denotes the greatest integer function)

Options

  1. A(0,π)\left(0, \pi\right)
  2. B[π,3π2]\left[\pi, \dfrac{3\pi}{2}\right]
  3. C{π}\left\{\pi\right\}
  4. D{π,3π2}\left\{\pi, \dfrac{3\pi}{2}\right\}

Correct answer

D. {π,3π2}\left\{\pi, \dfrac{3\pi}{2}\right\}

Step-by-step solution

Let, [x2−12]=[x2+12−1]=α∈I\left[{x}^{2} - \dfrac{1}{2}\right] = \left[{x}^{2} + \dfrac{1}{2} - 1\right] = \alpha \in I
⇒[x2+12]−1=α⇒ \left[{x}^{2} + \dfrac{1}{2}\right] - 1 = \alpha
∴y=2sin⁡−1(α+1)+cos⁡−1α,α={−1,0}∴ y = 2\sin^{- 1} \left(\alpha + 1\right) + \cos^{- 1} \alpha, \alpha = \left\{- 1,0\right\}
At α=−1\alpha = - 1
y=2sin⁡−1(0)+cos⁡−1(−1)=πy = 2\sin^{- 1} \left(0\right) + \cos^{- 1} \left(- 1\right) = \pi
At α=0\alpha = 0
y=2sin⁡−1(1)+cos⁡−1(0)=3π2y = 2\sin^{- 1} \left(1\right) + \cos^{- 1} \left(0\right) = \dfrac{3\pi}{2}
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