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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View SolutionMatch List $I$ with List $II$ and select the correct answer using the code given below the lists:
List $I$ List $II$ $P$. $\left(\frac{1}{y^2}\left(\frac{\cos (\tan ^{-1} y)+y \sin (\tan ^{-1} y)}{\cot (\sin ^{-1} y)+\tan (\sin ^{-1} y)}\right)^2+y^4\right)^{1 / 2}$ takes value $1$. $\frac{1}{2} \sqrt{\frac{5}{3}}$ $Q$. If $\cos x+\cos y+\cos z=0=\sin x+\sin y+\sin z$ then possible value of $\cos \frac{x-y}{2}$ is $2$. $\sqrt{2}$ $R$. If $\cos (\frac{\pi}{4}-x) \cos 2 x+\sin x \sin 2 x \sec x=\cos x \sin 2 x \sec x+\cos (\frac{\pi}{4}+x) \cos 2 x$ then possible value of $\sec x$ is $3$. $\frac{1}{2}$ $S$. If $\cot (\sin ^{-1} \sqrt{1-x^2})=\sin (\tan ^{-1}(x \sqrt{6})), x \neq 0$,then possible value of $x$ is $4$. $1$
Codes: $P \quad Q \quad R \quad S$
| List $I$ | List $II$ |
|---|---|
| $P$. $\left(\frac{1}{y^2}\left(\frac{\cos (\tan ^{-1} y)+y \sin (\tan ^{-1} y)}{\cot (\sin ^{-1} y)+\tan (\sin ^{-1} y)}\right)^2+y^4\right)^{1 / 2}$ takes value | $1$. $\frac{1}{2} \sqrt{\frac{5}{3}}$ |
| $Q$. If $\cos x+\cos y+\cos z=0=\sin x+\sin y+\sin z$ then possible value of $\cos \frac{x-y}{2}$ is | $2$. $\sqrt{2}$ |
| $R$. If $\cos (\frac{\pi}{4}-x) \cos 2 x+\sin x \sin 2 x \sec x=\cos x \sin 2 x \sec x+\cos (\frac{\pi}{4}+x) \cos 2 x$ then possible value of $\sec x$ is | $3$. $\frac{1}{2}$ |
| $S$. If $\cot (\sin ^{-1} \sqrt{1-x^2})=\sin (\tan ^{-1}(x \sqrt{6})), x \neq 0$,then possible value of $x$ is | $4$. $1$ |
Codes: $P \quad Q \quad R \quad S$
DifficultIIT 2013
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