JEE Main & Advanced JEE Main Solved Paper-2014

  • question_answer
    Let \[\alpha \] and \[\beta \] be the roots of equation \[p{{x}^{2}}+qx+r=0,p\ne 0.\]If p, q, r are in A.P. and \[\frac{1}{\alpha }+\frac{1}{\beta }=4,\]then the value of \[|\alpha -\beta |\]is :   JEE Main  Solved  Paper-2014

    A) \[\frac{\sqrt{61}}{9}\]                   

    B) \[\frac{2\sqrt{17}}{9}\]

    C) \[\frac{\sqrt{34}}{9}\]                   

    D) \[\frac{2\sqrt{13}}{9}\]

    Correct Answer: D

    Solution :

    \[\alpha ,\beta \]are roots of \[p{{x}^{2}}+qx+r=0\] \[\alpha +\beta =-\frac{q}{p},\alpha \beta =\frac{r}{p}\text{ }\]p, q, r are in A.P. \[2q=p+r\]                                          ? (1) \[\frac{1}{\alpha }+\frac{1}{\beta }=4\] \[\alpha +\beta \]            \[=4\alpha \beta \] \[-\frac{q}{p}\] \[=\frac{4r}{p}\] \[-q=4r\] \[\frac{q}{r}=-4\]                                             ?(2) From (1) and (2)\[\frac{p}{r}=-9\]             ?.(3) \[|\alpha -\beta |=\sqrt{{{(\alpha +\beta )}^{2}}-4\alpha \beta }\]\[=\sqrt{\frac{{{q}^{2}}}{{{p}^{2}}}+\frac{4r}{p}}\] \[=\sqrt{\frac{16}{81}+\frac{4}{9}}=\frac{2\sqrt{13}}{9}\]

You need to login to perform this action.
You will be redirected in 3 sec spinner