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Mathematics · 2026 · 5 marks
Inverse Trigonometric FunctionsBasic Concepts of Inverse Trigonometric Functions5 marksApplylong answer
CBSE 2026 · Region 2 · Set 2 · Q32
Find the domain of p (x)= sin^-1(1-2 x^2). Hence, find the value of x for which p (x)=(π)/6. Also, write the range of 2 p (x)+(π)/2.
Find the domain of $\displaystyle \mathrm{p}(x)=\sin ^{-1}\left(1-2 x^{2}\right)$. Hence, find the value of $\displaystyle x$ for which $\displaystyle \mathrm{p}(x)=\frac{\pi}{6}$. Also, write the range of $\displaystyle 2 \mathrm{p}(x)+\frac{\pi}{2}$.
Marking-scheme solution
Since domain of $\displaystyle \sin^{-1} x$ is $\displaystyle [-1,1] \therefore -1 \leq 1-2x^{2} \leq 1$
$\displaystyle \Rightarrow-2 \leq-2x^{2} \leq 0 \Rightarrow 0 \leq x^{2} \leq 1$
$\displaystyle \Rightarrow-1 \leq x \leq 1$
$\displaystyle \therefore$ Domain of $\displaystyle p(x)=[-1,1]$
When $\displaystyle p(x)=\dfrac{\pi}{6}$, we have $\displaystyle \sin^{-1}\left(1-2x^{2}\right)=\dfrac{\pi}{6}$
$\displaystyle \Rightarrow 1-2x^{2}=\sin \dfrac{\pi}{6}=\dfrac{1}{2}$
$\displaystyle \Rightarrow x^{2}=\dfrac{1}{4} \Rightarrow x=\dfrac{1}{2}, -\dfrac{1}{2}$
Since range of $\displaystyle p(x)=\sin^{-1}\left(1-2x^{2}\right)$ is $\displaystyle \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$
i.e. $\displaystyle -\dfrac{\pi}{2} \leq p(x) \leq \dfrac{\pi}{2} \Rightarrow-\pi \leq 2p(x) \leq \pi$
$\displaystyle \Rightarrow-\dfrac{\pi}{2} \leq 2p(x)+\dfrac{\pi}{2} \leq \dfrac{3\pi}{2}$
Hence the required range is $\displaystyle \left[-\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right]$
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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.