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

समीकरण \(\sin ^{-1}\left[x^{2}+\frac{1}{3}\right]+\cos ^{-1}\left[x^{2}-\frac{2}{3}\right]=x^{2}\), \(x \in[-1,1]\), जहाँ \([ x ]\) महत्तम पूर्णांक \(\leq x\) है, के हलों की संख्या है

  1. A \(2\)
  2. B \(0\)
  3. C \(4\)
  4. D \(Infinite\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(0\)

Step-by-step Solution

Detailed explanation

Given equation \(\sin ^{-1}\left[x^{2}+\frac{1}{3}\right]+\cos ^{-1}\left[x^{2}-\frac{2}{3}\right]=x^{2}\) Now, \(\sin ^{-1}\left[x^{2}+\frac{1}{3}\right]\) is defined if \(-1 \leq x^{2}+\frac{1}{3}<2 \Rightarrow \frac{-4}{3} \leq x^{2}<\frac{5}{3}\)…
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