MHT CET · Maths · Inverse Trigonometric Functions
If \(\sin ^{-1}(4 x)+\sin ^{-1}(4 \sqrt{3} x)=-\frac{\pi}{2}\), then the absolute value of \(x\) is
- A \(\frac{1}{8}\)
- B \(\frac{1}{6}\)
- C \(\frac{2}{3}\)
- D \(\frac{1}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{8}\)
Step-by-step Solution
Detailed explanation
\(\sin ^{-1}(4 x)+\sin ^{-1}(4 \\sqrt{3} x)=-\frac{\\pi}{2}\) \(\implies \sin ^{-1}(-4 x)+\sin ^{-1}(-4 \\sqrt{3} x)=\frac{\\pi}{2}\) (since \(x \le 0\))
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