• # 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) $a\le -2$

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 ${{x}^{2}}-(a+1)\,x+{{a}^{2}}+a-8=0$                 ${{(a+1)}^{2}}-4({{a}^{2}}+a-8)\ge 0$ $\Rightarrow$               ${{a}^{2}}+1+2a-4{{a}^{2}}-4a+32\ge 0$ $\Rightarrow$               $-3{{a}^{2}}-2a+33\ge 0$ $\Rightarrow$               $3{{a}^{2}}+2a+33\ge 0$ $\Rightarrow$               $3{{a}^{2}}+11a-9a-33\le 0$ $\Rightarrow$               $a(3a+11)-3(3a+11)\le 0$ $\Rightarrow$               $(3a+11)\,(a-3)\le 0$                 $\Rightarrow$               $a\in \,\left[ \frac{-11}{3},\,3 \right]$                 $f(2)<0$                 $4-(a+1)\,.\,2+{{a}^{2}}+a-8<0$                 $4-2a-2+{{a}^{2}}+a-8<0$                 ${{a}^{2}}-a-6<0$                                 ${{a}^{2}}-3a+2a-6<0$                 $a(a-3)+2(a-3)<0$ $\Rightarrow$               $(a-3)(a+2)<0$ $\Rightarrow$               $a\in (-2,\,3)$