Theory of Equations 4 Question 6
7. If the roots of the equation $x^{2}-2 a x+a^{2}+a-3=0$ are real and less than 3 , then
(1999, 2M)
(a) $a<2$
(b) $2 \leq a \leq 3$
(c) $3<a \leq 4$
(d) $a>4$
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Solution:
- Let $f(x)=x^{2}-2 a x+a^{2}+a-3$
Since, both root are less than 3 .
$$ \begin{array}{lrl} \Rightarrow & \alpha<3, \beta & <3 \\ \Rightarrow & \text { Sum, } S=\alpha+\beta & <6 \\ \Rightarrow & \frac{\alpha+\beta}{2}<3 \\ \Rightarrow & \frac{2 a}{2}<3 \\ \Rightarrow & a<3 \end{array} $$
Again, product, $P=\alpha \beta$
$$ \begin{array}{lc} \Rightarrow & P<9 \quad \Rightarrow \quad \alpha \beta<9 \\ \Rightarrow & a^{2}+a-3<9 \\ \Rightarrow & a^{2}+a-12<0 \\ \Rightarrow & (a-3)(a+4)<0 \\ \Rightarrow & -4<a<3 \quad \ldots \text { (ii) } \quad \alpha \\ \text { Again, } D=B^{2}-4 A C \geq 0 \\ \Rightarrow & (-2 a)^{2}-4 \cdot 1\left(a^{2}+a-3\right) \geq 0 \\ \Rightarrow & 4 a^{2}-4 a^{2}-4 a+12 \geq 0 \\ \Rightarrow & -4 a+12 \geq 0 \quad \Rightarrow \quad a \leq 3 \\ \text { Again, } & a f(3)>0 \\ \Rightarrow & 1\left[(3)^{2}-2 a(3)+a^{2}+a-3\right]>0 \\ \Rightarrow & 9-6 a+a^{2}+a-3>0 \\ \Rightarrow & a^{2}-5 a+6>0 \\ \Rightarrow & (a-2)(a-3)>0 \\ \therefore & a \in(-\infty, 2) \cup(3, \infty) \end{array} $$
From Eqs. (i), (ii), (iii) and (iv), we get
$$ a \in(-4,2) \text {. } $$
NOTE There is correction in answer $a<2$ should be $-4<a<2$.
