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Mathematics · 2026 · 5 marks
Inverse Trigonometric FunctionsBasic Concepts of Inverse Trigonometric Functions5 marksApplylong answer
CBSE 2026 · Region 2 · Set 3 · Q33
Find the domain of q (x)= cos^-1(4 x^2-3). Hence, find the value of x for which q (x)=0. Also, write the range of 3 q (x)-π.
Find the domain of $\displaystyle \mathrm{q}(x)=\cos ^{-1}\left(4 x^{2}-3\right)$. Hence, find the value of $\displaystyle x$ for which $\displaystyle \mathrm{q}(x)=0$. Also, write the range of $\displaystyle 3 \mathrm{q}(x)-\pi$.
Marking-scheme solution
Since domain of $\displaystyle \cos^{-1} x$ is $\displaystyle [-1,1] \therefore -1 \leq 4x^{2}-3 \leq 1$
$\displaystyle \Rightarrow 2 \leq 4x^{2} \leq 4$
$\displaystyle \Rightarrow \dfrac{1}{2} \leq x^{2} \leq 1$
$\displaystyle \Rightarrow-1 \leq x \leq-\dfrac{1}{\sqrt{2}}$ or $\displaystyle \dfrac{1}{\sqrt{2}} \leq x \leq 1$
$\displaystyle \therefore$ Domain of $\displaystyle \mathrm{q}(x)=\left[-1,-\dfrac{1}{\sqrt{2}}\right] \cup\left[\dfrac{1}{\sqrt{2}}, 1\right]$
When $\displaystyle \mathrm{q}(x)=0$, we have $\displaystyle \cos^{-1}\left(4x^{2}-3\right)=0$
$\displaystyle \Rightarrow 4x^{2}-3=\cos 0=1$
$\displaystyle \Rightarrow x^{2}=1 \Rightarrow x=1, -1$
Since range of $\displaystyle \mathrm{q}(x)=\cos^{-1}\left(4x^{2}-3\right)$ is $\displaystyle [0, \pi]$
i.e. $\displaystyle 0 \leq \mathrm{q}(x) \leq \pi \Rightarrow 0 \leq 3\mathrm{q}(x) \leq 3\pi$
$\displaystyle \Rightarrow-\pi \leq 3\mathrm{q}(x)-\pi \leq 2\pi$
Hence the required range is $\displaystyle [-\pi, 2\pi]$.
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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.