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  • Find all the value of a for the inequality, <img alt="\\\mathrm{\frac{x^2+ax -2}{x^2-x+1}<2}" src="http://entrancecorner.oncodecogs.com/gif.latex?%5C%5C%5Cmathrm%7B%5Cfrac%7Bx%5E2

Find all the value of a for the inequality, \\\mathrm{\frac{x^2+ax -2}{x^2-x+1}<2} holds true for all values of x

Option: 1

a \in (-\infty,-6]\cup (2,\infty)


Option: 2

a \in [-6,2]


Option: 3

a \in (-\infty,-6)\cup (2,\infty)


Option: 4

a \in (-6,2)


Answers (1)

best_answer

This equation can be written as 

\\\mathrm{\frac{x^2+ax -2-2(x^2-x+1)}{x^2-x+1}<0} \\\Rightarrow \frac{-x^2+(a+2)x-4}{x^2-x+1}<0 \\\\\mathrm{x^2-x+1, D = 1-4 = -3\,\, also \,\,a>0,\,it \,\,is\,\,always\,\,positive, so\,it\,can\,be\,cross\,multiplied } \\\\\Rightarrow -x^2+(a+2)x-4<0 \\\Rightarrow x^2-(a+2)x+4>0 \\\mathrm{ \Rightarrow \text{for this to be always positive}\,discriminant \;must\; be\; -ve, hence} \\\mathrm{(a+2)^2-16 < 0} \\ (a+2+4)(a+2-4)<0\\ (a+6)(a-2)<0 \\\Rightarrow a \in (-6,2)

correct option is (d)

Posted by

Rishi

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