MCQ
$\lim_{n \rightarrow \infty} \sum\limits_{r=1}^{\infty }{{{\tan }^{-1}}\left( \frac{1}{2{{r}^{2}}} \right)=...........}$
- A$\frac{\pi }{6}$
- ✓$\frac{\pi }{4}$
- C$\frac{\pi }{2}$
- D$\frac{\pi }{3}$
$\lim_{n \rightarrow \infty}\sum_{r=1}^\infty \tan^{-1}\left(\frac{1}{2r^2}\right)$
$=\sum_{r=1}^\infty \tan^{-1}\left(\frac{2}{4r^2}\right) $
$=\sum_{r=1}^\infty \tan^{-1}\left(\frac{2}{1+4r^2-1}\right)$
$=\sum_{r=1}^\infty \tan^{-1}\left(\frac{(2r+1)-(2r-1)}{1+(2r-1)(2r+1)}\right) $
$=\sum_{r=1}^\infty [\tan^{-1}(2r+1)-\tan^{-1}(2r-1)]$
$=\lim_{n \rightarrow \infty}\sum_{r=1}^n[\tan^{-1}(2r+1)-\tan^{-1}(2r-1)]$
$=\lim_{n \rightarrow \infty}\tan^{-1}(2n+1)-\frac{\pi}{4} \\ =\frac{\pi}{2}-\frac{\pi}{4}$
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$(I)$ $y=f(x)$ એ $x$-અક્ષને બરાબર એક બિંદુએ છેદ છે.
$(II)$ $y=f(x)$ એ $x$-અક્ષને $x=\cos \frac{\pi}{12}$ આગળ છેદ છે. તો.......