JEE Mains · Maths · STD 12 - 1. relation and function
Tho damnin of tho finction \(\cos ^{-1}\left(\frac{2 \sin ^{-1}\left(\frac{1}{4 x^{2}-1}\right)}{\pi}\right)\) is
- A \(R-\left\{-\frac{1}{2}, \frac{1}{2}\right\}\)
- B \((-\infty,-1] \cup[1, \infty) \cup\{0\}\)
- C \(\left(-\infty, \frac{-1}{2}\right) \cup\left(\frac{1}{2}, \infty\right) \cup\{0\}\)
- D \(\left(-\infty, \frac{-1}{\sqrt{2}}\right] \cup\left[\frac{1}{\sqrt{2}}, \infty\right) \cup\{0\}\)
Answer & Solution
Correct Answer
(D) \(\left(-\infty, \frac{-1}{\sqrt{2}}\right] \cup\left[\frac{1}{\sqrt{2}}, \infty\right) \cup\{0\}\)
Step-by-step Solution
Detailed explanation
\(-1 \leq \frac{2 \sin ^{-1}\left(\frac{1}{4 x^{2}-1}\right)}{\pi} \leq 1\) \(-\pi / 2 \leq \sin ^{-1} \frac{1}{4 x^{2}-1} \leq \pi / 2\) Always \(-1 \leq \frac{1}{4 x^{2}-1} \leq 1\) \(x \in\left(\infty, \frac{1}{\sqrt{2}}\right) \cup\left[\frac{1}{\sqrt{2}}, \infty\right)\)
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