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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

If the range of f (x) = tan^ - 1 x + 2sin^ - 1 x + cos^ - 1 x is...

If the range of f(x)=tan−1x+2sin−1x+cos−1xf \left(x\right) = tan^{- 1}x + 2sin^{- 1}x + cos^{- 1}x is \left[a, b\right]$, then

Options

  1. Aa=π4a = \dfrac{\pi}{4}
  2. Ba=−π2a = - \dfrac{\pi}{2}
  3. Cb=5π4b = \dfrac{5\pi}{4}
  4. Db=3π2b = \dfrac{3\pi}{2}

Correct answer

C. b=5π4b = \dfrac{5\pi}{4}

Step-by-step solution

f(x)=tan−1x+sin−1x+(sin−1x+cos−1x)f \left(x\right) = tan^{- 1}x + sin^{- 1}x + \left(sin^{- 1}x + cos^{- 1}x\right)
f(x)=tan−1x+sin−1x+π2;x∈[−1,1]{∵sin−1x+cos−1x=π2}f \left(x\right) = tan^{- 1}x + sin^{- 1}x + \dfrac{\pi}{2} ; x \in \left[-1,1\right] \left\{∵ sin^{- 1}x + cos^{- 1}x = \dfrac{\pi}{2}\right\}
∵∵f \left(x\right)isanincreasingfunction<br/>is an increasing function<br />∴RangeofRange off \left(x\right) \in \left[f \left(-1\right), f\left(1\right)\right]
f\left(-1\right) =tan−1(−1)+tan^{- 1}\left(-1\right) + sin^{- 1}\left(-1\right) + \dfrac{\pi}{2} =−-\dfrac{\pi}{4}<br/>f(1)=<br />f\left(1\right) = tan^{- 1}\left(1\right) +sin−1(1)+π2==5π4<br/>sin^{- 1}\left(1\right) + \dfrac{\pi}{2} = =5\dfrac{\pi}{4}<br />∴RangeofRange off \left(x\right)\in \left[- \dfrac{\pi}{4}, \dfrac{5\pi}{4}\right]
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