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JEE Mains · Maths · STD 12 - 2. inverse trigonometric function

વિધેય \(f(x)=\sin ^{-1}\left(\frac{3 x^{2}+x-1}{(x-1)^{2}}\right)+\cos ^{-1}\left(\frac{x-1}{x+1}\right)\) નો પ્રદેશ મેળવો.

  1. A \(\left[0, \frac{1}{4}\right]\)
  2. B \([-2,0] \cup\left[\frac{1}{4}, \frac{1}{2}\right]\)
  3. C \(\left[\frac{1}{4}, \frac{1}{2}\right] \cup\{0\}\)
  4. D \(\left[0, \frac{1}{2}\right]\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\left[\frac{1}{4}, \frac{1}{2}\right] \cup\{0\}\)

Step-by-step Solution

Detailed explanation

\(f(x)=\sin ^{-1}\left(\frac{3 x^{2}+x-1}{(x-1)^{2}}\right)+\cos ^{-1}\left(\frac{x-1}{x+1}\right)\) \(-1 \leq \frac{x-1}{x+1} \leq 1 \Rightarrow 0 \leq x<\infty...(1)\)…
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