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Mathematics · 2026 · 5 marks
Inverse Trigonometric FunctionsBasic Concepts of Inverse Trigonometric Functions5 marksApplylong answer
CBSE 2026 · Region 2 · Set 1 · Q34
Find the domain of g (x)= cos^-1(x^2-1). Hence, find the value of x for which g (x)=(π)/3. Also, write the range of cos^-1 x other than its principal…
Find the domain of $\displaystyle \mathrm{g}(x)=\cos ^{-1}\left(x^{2}-1\right)$. Hence, find the value of $\displaystyle x$ for which $\displaystyle \mathrm{g}(x)=\frac{\pi}{3}$. Also, write the range of $\displaystyle \cos ^{-1} x$ other than its principal branch.
Marking-scheme solution
Since domain of $\displaystyle \cos^{-1} x$ is $\displaystyle [-1,1] \therefore -1 \leq x^{2}-1 \leq 1$
$\displaystyle \Rightarrow 0 \leq x^{2} \leq 2 \Rightarrow-\sqrt{2} \leq x \leq \sqrt{2}$
$\displaystyle \therefore$ Domain of $\displaystyle g(x)=\left[-\sqrt{2}, \sqrt{2}\right]$
When $\displaystyle g(x)=\dfrac{\pi}{3}$, we have $\displaystyle \cos^{-1}\left(x^{2}-1\right)=\dfrac{\pi}{3}$
$\displaystyle \Rightarrow x^{2}-1=\cos \dfrac{\pi}{3}=\dfrac{1}{2}$
$\displaystyle \Rightarrow x^{2}=\dfrac{3}{2} \Rightarrow x=\sqrt{\dfrac{3}{2}}, -\sqrt{\dfrac{3}{2}}$
Range of $\displaystyle \cos^{-1} x$ other than its principal branch is $\displaystyle [-\pi, 0]$ or $\displaystyle [\pi, 2 \pi]$.
Note: Any valid branch like $\displaystyle [2 \pi, 3 \pi],[-2 \pi,-\pi]$, ... etc can be considered.All Inverse Trigonometric Functions questions2026 question paperNCERT solutions: Inverse Trigonometric FunctionsPractice Inverse Trigonometric Functions →
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CBSE Class 12 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.