JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
Let \(S=\left\{x \in R: 0 < x < 1\right.\) and \(\left.2 \tan ^{-1}\left(\frac{1-x}{1+x}\right)=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right\}\). If \(n ( S )\) denotes the number of elements in \(S\) then:
- A \(n(S)=2\) and only one element in \(S\) is less then \(\frac{1}{2}\).
- B \(n ( S )=1\) and the element in \(S\) is more than \(\frac{1}{2}\).
- C \(n(S)=1\) and the element in \(S\) is less than \(\frac{1}{2}\).
- D \(n(S)=0\)
Answer & Solution
Correct Answer
(C) \(n(S)=1\) and the element in \(S\) is less than \(\frac{1}{2}\).
Step-by-step Solution
Detailed explanation
\(0 < x < 1\) \(2 \tan ^{-1}\left(\frac{1- x }{1+ x }\right)=\cos ^{-1}\left(\frac{1- x ^2}{1+ x ^2}\right)\) \(\tan ^{-1} x =\theta \in\left(0, \frac{\pi}{4}\right) \therefore x =\tan \theta\)…
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