Let&nbsp;α,&nbsp;β&nbsp;be the roots of&nbsp;ax2+bx+c=0, where&nbsp;1&lt;α&lt;βThen&nbsp;limx→x0|ax2+bx+c|ax2+bx+c=1&nbsp;then which of the following statements is incorrecta)&nbsp;a&gt;0&nbsp;and&nbsp;x0&lt;1b)&nbsp;a&gt;0&nbsp;and&nbsp;x0&lt;βc)&nbsp;a&lt;0&nbsp;and&nbsp;α&lt;x0&lt;βd)&nbsp;a&lt;0&nbsp;and&nbsp;x0&lt;1

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Let&amp;nbsp;α,&amp;nbsp;β&amp;nbsp;be the roots of&amp;nbsp;ax2+bx+c=0, where&amp;nbsp;1&amp;lt;α&amp;lt;βThen&amp;nbsp;limx→x0|ax2+bx+c|ax2+bx+c=1&amp;nbsp;then which of the following statements is incorrecta)&amp;nbsp;a&amp;gt;0&amp;nbsp;and&amp;nbsp;x0&amp;lt;1b)&amp;nbsp;a&amp;gt;0&amp;nbsp;and&amp;nbsp;x0&amp;lt;βc)&amp;nbsp;a&amp;lt;0&amp;nbsp;and&amp;nbsp;α&amp;lt;x0&amp;lt;βd)&amp;nbsp;a&amp;lt;0&amp;nbsp;and&amp;nbsp;x0&amp;lt;1

louran20z47 · Accepted Answer

Step 1See this graph for&amp;nbsp;a&amp;gt;0&amp;nbsp;(The labeling of the graph is incorrect, and I can&#039;t find another one) Those points are the roots. first one is&amp;nbsp;α&amp;nbsp;and second one is&amp;nbsp;β. Both are greater than 1limx→x0|ax2+bx+c|ax2+bx+c=1Forget the modulus, without the modulus, the limit is always 1, provided&amp;nbsp;x0≠α,β. But since modulus is there, the limit can be -1 too.Hint: If&amp;nbsp;a&amp;gt;0&amp;nbsp;and&amp;nbsp;x0&amp;gt;β, both numerator and denominator will be positive. now you can check all the options, I suppose.

Let \alpha,\ \beta be the roots of ax^{2}+bx+c=0, where 1<\alpha<\beta Then

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