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JEE MAIN 2025
04-04-2025 SHIFT-1
Question
Considering the principal values of the inverse trigonometric functions, ${\sin ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}x + \frac{1}{2}\sqrt {1 - {x^2}} } \right), - \frac{1}{2} < x < \frac{1}{{\sqrt 2 }}$, is equal to
Select the correct option:
A
$\frac{{5\pi }}{6} - {\sin ^{ - 1}}x$
B
$\frac{\pi }{4} + {\sin ^{ - 1}}x$
C
$\frac{{ - 5\pi }}{6} - {\sin ^{ - 1}}x$
D
$\frac{\pi }{6} + {\sin ^{ - 1}}x$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
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JEE Main
Mathematics
Medium
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