question_answer

                The Values of ?a? for which one root of the equation \[{{x}^{2}}-(a+1)x+{{a}^{2}}+a-8=0\]exceeds 2 and the other is lesser than 2, are given by:           JEE Main Online Paper (Held On 09 April 2013)          

A)                 \[3<a<10\]                

B)                 \[a\ge 10\]                

C)                 \[-2<a<3\]                

D)                 (d) \[a\le -2\]                

Correct Answer: C

Solution :

                If one root is less than \[\alpha \] other root is greater than \[\beta \], then \[D\ge 0\] and \[f(\alpha )<0,\,\,f(\beta )<0\]                 Here, equation is \[{{x}^{2}}-(a+1)\,x+{{a}^{2}}+a-8=0\]                 \[{{(a+1)}^{2}}-4({{a}^{2}}+a-8)\ge 0\] \[\Rightarrow \]               \[{{a}^{2}}+1+2a-4{{a}^{2}}-4a+32\ge 0\] \[\Rightarrow \]               \[-3{{a}^{2}}-2a+33\ge 0\] \[\Rightarrow \]               \[3{{a}^{2}}+2a+33\ge 0\] \[\Rightarrow \]               \[3{{a}^{2}}+11a-9a-33\le 0\] \[\Rightarrow \]               \[a(3a+11)-3(3a+11)\le 0\] \[\Rightarrow \]               \[(3a+11)\,(a-3)\le 0\]                 \[\Rightarrow \]               \[a\in \,\left[ \frac{-11}{3},\,3 \right]\]                 \[f(2)<0\]                 \[4-(a+1)\,.\,2+{{a}^{2}}+a-8<0\]                 \[4-2a-2+{{a}^{2}}+a-8<0\]                 \[{{a}^{2}}-a-6<0\]                                 \[{{a}^{2}}-3a+2a-6<0\]                 \[a(a-3)+2(a-3)<0\] \[\Rightarrow \]               \[(a-3)(a+2)<0\] \[\Rightarrow \]               \[a\in (-2,\,3)\]