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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Answer:

Correct Answer: 7. (a)

Solution:

  1. 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, $P=\alpha \beta$

alt text

$ \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 $-4<a<2$.



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