MCQ
$\sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}\left( {\frac{3}{{{r^2} - r + 9}}} \right)} $ is-
- A$\frac{\pi }{3}$
- B$\frac{\pi }{6}$
- ✓$\frac{\pi }{2}$
- D$\frac{\pi }{12}$
$\mathrm{T}_{\mathrm{r}}=\tan ^{-1}\left(\frac{\mathrm{r}}{3}\right)-\tan ^{-1}\left(\frac{(\mathrm{r}-1)}{3}\right)$
$\mathrm{T}_{1}=\tan ^{-1}\left(\frac{1}{3}\right)-\tan ^{-1}(0)$
$\mathrm{T}_{2}=\tan ^{-1}\left(\frac{2}{3}\right)-\tan ^{-1}\left(\frac{1}{3}\right)$
$\mathrm{T}_{\mathrm{n}}=\tan ^{-1}\left(\frac{\mathrm{n}}{3}\right)-\tan ^{-1}\left(\frac{(\mathrm{n}-1)}{3}\right)$
$S_{n}=\tan ^{-1}\left(\frac{n}{3}\right)=\frac{\pi}{2}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\frac{d y}{d x}+\alpha y=x e^{\beta x}, y(1)=1$
Let $S=\left\{y_{\alpha \beta}(x): \alpha, \beta \in R \right\}$. Then which of the following functions belong(s) to the set $S$ ?
$(A)$ $f( x )=\frac{ x ^2}{2} e ^{- x }+\left( e -\frac{1}{2}\right) e ^{- x }$
$(B)$ $f( x )=-\frac{ x ^2}{2} e ^{- x }+\left( e +\frac{1}{2}\right) e ^{- x }$
$(C)$ $f( x )=\frac{ e ^{ x }}{2}\left( x -\frac{1}{2}\right)+\left( e -\frac{ e ^2}{4}\right) e ^{- x }$
$(D)$ $f( x )=\frac{ e ^{ x }}{2}\left(\frac{1}{2}- x \right)+\left( e +\frac{ e ^2}{4}\right) e ^{- x }$
$x \tan \left(\frac{y}{x}\right) d y=\left(y \tan \left(\frac{y}{x}\right)-x\right) d x,-1 \leq x \leq 1, y\left(\frac{1}{2}\right)=\frac{\pi}{6}$
Then the area of the region bounded by the curves $x=0, x=\frac{1}{\sqrt{2}}$ and $y=y(x)$ in the upper half plane is :