CBSE 2023 · Region 4 · Set 1 · Q21 · 2 marks
Find the domain of $\displaystyle y=\sin ^{-1}\left(\mathrm{x}^{2}-4\right)$.Evaluate : \[\cos ^{-1}\left[\cos \left(-\frac{7 \pi}{3}\right)\right] \]
Find the domain of $\displaystyle y=\sin ^{-1}\left(\mathrm{x}^{2}-4\right)$.
Evaluate : \[\cos ^{-1}\left[\cos \left(-\frac{7 \pi}{3}\right)\right] \]
Marking-scheme solution
Domain of $\displaystyle \sin ^{-1} \mathrm{x}$ is $\displaystyle -1 \leq \mathrm{x} \leq 1$
$$\therefore-$\displaystyle 1$ \leq \mathrm{x}^{$\displaystyle 2$}-$\displaystyle 4$ \leq $\displaystyle 1$
$$or $\displaystyle \mathrm{x}^{2} \geq 3, \mathrm{x}^{2} \leq 5$
∴ Domain is $\displaystyle [-\sqrt{5},-\sqrt{3}] \cup[\sqrt{3}, \sqrt{5}]$
$\displaystyle \cos ^{-1}\left[\cos \left(-\frac{7 \pi}{3}\right)\right]=\cos ^{-1}\left[\cos \left(\frac{7 \pi}{3}\right)\right]$
$$=\cos ^{-$\displaystyle 1$}\left[\cos \left(2 \pi+\frac{\pi}{3}\right)\right]
Inverse Trigonometric FunctionsBasic Concepts of Inverse Trigonometric FunctionsApplyvery_short_answereasy
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CBSE Class 12 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.