Question

Let $x_1=\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\cdots$ upto $10$ terms,
$y_1=\text{Sum of roots of equation of}{x}^2-7x+10=0$. If $(x_1,y_1)$ lies inside the hyperbola $\frac{(21x{)}^2}{100}-\frac{y^2}{(7b{)}^2}=-1$, then number of integeral values for $b$ is

Answer 1

$x_1=\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\cdots$ upto $10$ terms
Let $t_r$ be the $r^{\text{th}}$ term of the given series.
Then $t_r=\frac{1}{(2r-1)(2r+1)},r=1,2,3\cdots n$
$t_r=\frac{1}{2}[\frac{1}{2r-1}-\frac{1}{2r+1}]$
$\therefore S_n=\sum _{r=1}^n{t}_r$
$\Rightarrow S_n=\frac{1}{2}[(1-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+\cdots (\frac{1}{2n-1}-\frac{1}{2n+1})]$
$S_n=\frac{1}{2}[1-\frac{1}{2n+1}]$
$\therefore S_{10}=\frac{10}{21}=x_1$
and $y_1=\frac{7}{1}=7$

For point to be internal to hyperbola $S_1>0$ ,
where $S::-\frac{(21x{)}^2}{100}+\frac{y^2}{(7b{)}^2}-1$
$\Rightarrow -\frac{{(21\times \frac{10}{21})}^2}{100}+\frac{49}{49b^2}-1>0$
$\Rightarrow 2-\frac{1}{b^2}<0$
$\Rightarrow 2b^2-1<0$
$\Rightarrow b\in (-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$
But for $b=0$ equation will not valid
$\therefore b\in (-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})-\left\{0\right\}$
So, number of integers lying in the interval $=0$