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Values of 'a' for the inequality \\\mathrm{\frac{x^2+ax -2}{x^2-x+1}> -3} holds true for all real values of x

Option: 1

(-1,0)


Option: 2

(-1,7)


Option: 3

(-7, 0)


Option: 4

(0, 7)


Answers (1)

best_answer

This equation can be written as 

\\\mathrm{\frac{x^2+ax -2 +3(x^2-x+1)}{x^2-x+1}> 0} \\\mathrm{\Rightarrow \frac{4x^2+(a-3)x+1}{x^2-x+1}>0} \\\mathrm{since\; denominator\;is \; always +ve,\; hence \; numerator\; must\; be \; +ve } \\\mathrm{\Rightarrow 4x^2+(a-3)x+1> 0 \Rightarrow D < 0 } \\\mathrm{\Rightarrow (a-3)^2-16 <0 \Rightarrow a \in (-1, 7)}

Posted by

Ritika Harsh

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