JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
The number of solutions of \(\tan^{-1}4x+\tan^{-1}6x=\frac{\pi}{6}\) where \(-\frac{1}{2\sqrt{6}}< x <\frac{1}{2\sqrt{6}}\) is equal to
- A 3
- B 0
- C 1
- D 2
Answer & Solution
Correct Answer
(C) 1
Step-by-step Solution
Detailed explanation
\(\tan ^{-1} 4 x+\tan ^{-1} 6 x=\frac{\pi}{6}\) \(\Rightarrow \tan ^{-1}\left(\frac{4 x+6 x}{1+24 x^2}\right)=\frac{\pi}{6}\) \(\Rightarrow \frac{10 x}{1-24 x^2}=\frac{1}{\sqrt{3}}\) \(\Rightarrow 24 x^2+10 \sqrt{3} x-1=0\) \(x=\frac{-10 \sqrt{3} \pm \sqrt{300+96}}{48}\)…
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