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

જેને માટે \(\sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta) > 0, \theta \in(0,2 \pi)\) અથાર્થ થાય તેવો મોટામાં મોટો અંતરાલ \(( a , b ) \subset(0,2 \pi)\) છે.જો \(\alpha x^2+\beta x+\sin ^{-1}\left(x^2-6 x+10\right)+\cos ^{-1}\left(x^2-6 x+10=0\right)\) અને \(\alpha-\beta= b - a\) હોય,તો \(\alpha=...........\).

  1. A \(\frac{\pi}{48}\)
  2. B \(\frac{\pi}{16}\)
  3. C \(\frac{\pi}{8}\)
  4. D \(\frac{\pi}{12}\)
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Answer & Solution

Correct Answer

(D) \(\frac{\pi}{12}\)

Step-by-step Solution

Detailed explanation

\(\sin ^{-1} \sin \theta-\left(\frac{\pi}{2}-\sin ^{-1} \sin \theta\right) > 0\) \(\Rightarrow \sin ^{-1} \sin \theta > \frac{\pi}{4}\) \(\Rightarrow \sin \theta > \frac{1}{\sqrt{2}}\) \(\text { So, } \theta \in\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)\)…
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