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Let $\mathrm{a,b}$ be two non-zero real numbers. If $\mathrm{p \;and\;r}$ are the roots of the equation $\mathrm{x^{2}-8 \mathrm{a} x+2 \mathrm{a}=0}$ and $\mathrm{q\; and \; s}$ are the roots of the equation $\mathrm{x^{2}+12 \mathrm{~b} x+6 \mathrm{~b}=0}$ , such that $\mathrm{\frac{1}{\mathrm{p}}, \frac{1}{\mathrm{q}}, \frac{1}{\mathrm{r}}, \frac{1}{\mathrm{~s}}}$ are in A.P, then $\mathrm{a^{-1}-b^{-1}}$ is equal to _______.
Option: 1
38

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

$\mathrm{x^{2}-8 a x+2 a=0 \qquad x^{2}+12 b x+6 b=0} \\$

$\mathrm{p+r=8 a \qquad } \\$ $\mathrm{q+s=+2 b } \\$

$\mathrm{p r=2 a \qquad }\\$ $\mathrm{q s=6 b }\\$

$\mathrm{\frac{1}{p}+\frac{1}{r}=4 \qquad } \\$ $\mathrm{\frac{1}{q}+\frac{1}{s}=-2}$

$\mathrm{\frac{2}{q}=4 \qquad} \\$ $\mathrm{\frac{2}{r}=-2}$

$\mathrm{q=\frac{1}{2}}$ $\mathrm{r=-1}$

$\mathrm{p=\frac{1}{5}\qquad } \\$ $\mathrm{s=\frac{-1}{4}} \\$

$\mathrm{\text { Now, } \frac{1}{a}-\frac{1}{b}=\frac{2}{p r}-\frac{6}{q s}=38 }$

Hence answer is 38

Posted by

himanshu.meshram

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