Consider the following statements
Assertion (A) For $x \in R-\{1\}$;
$$ \frac{d}{d x}\left(\tan ^{-1}\left(\frac{1+x}{1-x}\right)\right)=\frac{d}{d x}\left(\tan ^{-1} x\right) $$
Reason (R) For $x<1, \tan ^{-1}\left(\frac{1+x}{1-x}\right)=\frac{\pi}{4}+\tan ^{-1} x$, for
$$ x>1, \tan ^{-1}\left(\frac{1+x}{1-x}\right)=-\frac{3 \pi}{4}+\tan ^{-1} x $$
The correct answer is
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If $y=\left(\sin ^{-1} x\right)^2$, then $\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}=$
The radius of a cone of height 9 units is changed from 2 units to 2.12 units. The exact change and approximate change in the volume of the cone are respectively
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