11th Class Mathematics Complex Numbers and Quadratic Equations Question Bank Critical Thinking

  • question_answer Let \[a,b,c\] be real numbers \[a\ne 0\]. If \[\alpha \]is a root \[{{a}^{2}}{{x}^{2}}+bx+c=0\], \[\beta \] is a root of \[{{a}^{2}}{{x}^{2}}-bx-c=0\] and \[0<\alpha <\beta \], then the equation \[{{a}^{2}}{{x}^{2}}+2bx+2c=0\]has a root \[\gamma \]that always satisfies [IIT 1989]

    A) \[\gamma =\frac{\alpha +\beta }{2}\]

    B) \[\gamma =\alpha +\frac{\beta }{2}\]

    C) \[\gamma =\alpha \]

    D) \[\alpha <\gamma <\beta \]

    Correct Answer: D

    Solution :

    Since \[\alpha \]and \[\beta \] are the roots of given equations. So we have \[{{a}^{2}}{{\alpha }^{2}}+b\alpha +c=0\]and\[{{a}^{2}}{{\beta }^{2}}-b\beta -c=0\]. Let    \[f(x)={{a}^{2}}{{x}^{2}}+2bx+2c=0\] Then \[f(\alpha )={{a}^{2}}{{\alpha }^{2}}+2b\alpha +2c=0\] \[={{a}^{2}}{{\alpha }^{2}}+2(b\alpha +c)={{a}^{2}}{{\alpha }^{2}}-2{{a}^{2}}{{\alpha }^{2}}=-{{a}^{2}}{{\alpha }^{2}}=-ve\] and\[f(\beta )={{a}^{2}}{{\beta }^{2}}+2(b\beta +c)={{a}^{2}}{{\beta }^{2}}+2{{a}^{2}}{{\beta }^{2}}\]              \[=3{{a}^{2}}{{\beta }^{2}}=+ve\] Since \[f(\alpha )\] and \[f(\beta )\] are of opposite signs, therefore by theory of equations there lies a root \[\gamma \] of the equation \[f(x)=0\]between \[\alpha \]and \[\beta \]i.e.\[\alpha <\gamma <\beta \]

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