CBSE 2023 · Region 3 · Set 1 · Q24 · 2 marks
Evaluate : $\displaystyle 3 \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)+2 \cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)+\cos ^{-1}(0)$Draw the graph of $\displaystyle \mathrm{f}(\mathrm{x})=\sin ^{-1} \mathrm{x}, \mathrm{x} \in\left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]$. Also, write range of f(x).
Evaluate : $\displaystyle 3 \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)+2 \cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)+\cos ^{-1}(0)$
Draw the graph of $\displaystyle \mathrm{f}(\mathrm{x})=\sin ^{-1} \mathrm{x}, \mathrm{x} \in\left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]$. Also, write range of f(x).
Marking-scheme solution
(a)
Given expression $\displaystyle =\frac{3 \pi}{4}+\frac{2 \pi}{6}+\frac{\pi}{2}$
$$=\frac{19 \pi}{12}
$$(b)
Correct graph
Here, the points A, B, C and D are respectively $\displaystyle \left(0, \frac{\pi}{4}\right),\left(0,-\frac{\pi}{4}\right),\left(\frac{1}{\sqrt{2}}, 0\right),\left(-\frac{1}{\sqrt{2}}, 0\right)$.
Range $\displaystyle =\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$
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.