CBSE 2025 · Region 1 · Set 3 · Q24 · 2 marks
Solve for $\displaystyle \mathrm{x}$, \[2 \tan ^{-1} \mathrm{x}+\sin ^{-1}\left(\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}\right)=4 \sqrt{3} \]
Marking-scheme solution
\[2 \tan ^{-1} \mathrm{x}+\sin ^{-1}\left(\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}\right)=4 \sqrt{3} \Rightarrow 2 \tan ^{-1} \mathrm{x}+2 \tan ^{-1} \mathrm{x}=4 \sqrt{3}
\]
$\displaystyle \Rightarrow \boldsymbol{\operatorname { t a n }}^{-\mathbf{1}} \mathrm{x}=\sqrt{3} \notin\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
$\displaystyle \therefore \mathrm{x}=\boldsymbol{\operatorname { t a n }}(\sqrt{3})$ which has no solution.
Inverse Trigonometric FunctionsProperties of Inverse Trigonometric FunctionsApplyvery_short_answerhard
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CBSE Class 12 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.