CBSE 2024 · Region 2 · Set 1 · Q21 · 2 marks
Find the value of $\displaystyle \tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)+\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)+\tan ^{-1}\left[\sin \left(-\frac{\pi}{2}\right)\right]$.Find the domain of the function $\displaystyle f(x)=\sin ^{-1}\left(x^{2}-4\right)$. Also, find its range.
Find the value of $\displaystyle \tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)+\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)+\tan ^{-1}\left[\sin \left(-\frac{\pi}{2}\right)\right]$.
Find the domain of the function $\displaystyle f(x)=\sin ^{-1}\left(x^{2}-4\right)$. Also, find its range.
Marking-scheme solution
The given expression $\displaystyle =\frac{-\pi}{6}+\frac{\pi}{3}-\frac{\pi}{4} =\frac{-\pi}{12}$\begin{aligned}
& -1 \leqslant x^{2}-4 \leqslant 1
& \Rightarrow 3 \leqslant x^{2} \leqslant 5
& \text { Domain }=[-\sqrt{5},-\sqrt{3}] \cup[\sqrt{3}, \sqrt{5}]
\end{aligned}
$$
Inverse Trigonometric FunctionsBasic Concepts of Inverse Trigonometric FunctionsApplyvery_short_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.