• # question_answer 36) 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) $\alpha <\gamma <\beta$

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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