If $f(x) = \cot^{-1} \left( \frac{3x - x^3}{1 - 3x^2} \right)$ and $g(x) = \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)$, then $\lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)}$ for $0 < a < \frac{1}{2}$ is
- A$\frac{3}{2}$
- B$\frac{1}{2}$
- C$-\frac{3}{2}$
- D$-\frac{1}{2}$
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Let $Z$ denote the set of integers. Then match the items in List-$I$ with those of the items in List-$II$.
List-$I$ List-$II$ $A$. $\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)+\sin ^{-1} \frac{1}{3}$ $I$. $k \pi \pm(-1)^k \frac{\pi}{6}, k \in Z$ $B$. $\sin ^{-1}\left(\frac{(-1)^n}{2}\right), n \in Z$ $II$. $k \pi \pm 1, k \in Z$ $C$. $\tan ^{-1}\left(\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right)$ $III$. $\frac{3}{2}$ $D$. $\sin ^{-1}|\sin x|=\sqrt{\sin ^{-1}|\sin x|} \Rightarrow x \in$ $IV$. $\frac{3 \pi}{8}$ $V$. $\frac{\pi}{2}$
The correct match is:
| List-$I$ | List-$II$ |
| $A$. $\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)+\sin ^{-1} \frac{1}{3}$ | $I$. $k \pi \pm(-1)^k \frac{\pi}{6}, k \in Z$ |
| $B$. $\sin ^{-1}\left(\frac{(-1)^n}{2}\right), n \in Z$ | $II$. $k \pi \pm 1, k \in Z$ |
| $C$. $\tan ^{-1}\left(\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right)$ | $III$. $\frac{3}{2}$ |
| $D$. $\sin ^{-1}|\sin x|=\sqrt{\sin ^{-1}|\sin x|} \Rightarrow x \in$ | $IV$. $\frac{3 \pi}{8}$ |
| $V$. $\frac{\pi}{2}$ |
The correct match is:
If $\sin^{-1}(\tan \frac{\pi}{4}) - \sin^{-1}(\sqrt{\frac{3}{x}}) = \frac{\pi}{6}$, then $x$ is a root of the equation:
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