If the roots of the equation 1x+p+1x+q=1rare equal in magnitude but opposite in sign, then the product of the roots will be

If the roots of the equation $\frac{1}{\mathrm{x}+\mathrm{p}}+\frac{1}{\mathrm{x}+\mathrm{q}}=\frac{1}{\mathrm{r}}$are equal in magnitude but opposite in sign, then the product of the roots will be

1. A

$\frac{{\mathrm{p}}^{2}+{\mathrm{q}}^{2}}{2}$

2. B

$-\frac{\left({\mathrm{p}}^{2}+{\mathrm{q}}^{2}\right)}{2}$

3. C

$\frac{{\mathrm{p}}^{2}-{\mathrm{q}}^{2}}{2}$

4. D

$-\frac{\left({\mathrm{p}}^{2}-{\mathrm{q}}^{2}\right)}{2}$

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Solution:

Given equation $\frac{1}{\mathrm{x}+\mathrm{p}}+\frac{1}{\mathrm{x}+\mathrm{q}}=\frac{1}{\mathrm{r}}$can be rewritten as
${\mathrm{x}}^{2}+\mathrm{x}\left(\mathrm{p}+\mathrm{q}-2\mathrm{r}\right)+\mathrm{pq}-\mathrm{pr}-\mathrm{qr}=0$……………………….(i)
Let roots are $\alpha$ and - $\alpha$, then the product of roots
$-{\mathrm{\alpha }}^{2}=\mathrm{pq}-\mathrm{pr}-\mathrm{qr}=\mathrm{pq}-\mathrm{r}\left(\mathrm{p}+\mathrm{q}\right)$………………………(ii)
and            sum of roots, $0=-\left(\mathrm{p}+\mathrm{q}-2\mathrm{r}\right)$
……………………….(iii)
On solving Eqs. (ii) and (iii), we get

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