Theory Of Equations Ques 35
- Let $a, b, c, p, q$ be the real numbers. Suppose $\alpha, \beta$ are the roots of the equation $x^{2}+2 p x+q=0$. and $\alpha, \frac{1}{\beta}$ are the roots of the equation $a x^{2}+2 b x+c=0$, where $\beta^{2} \notin{-1,0,1}$.
Statement I $\left(p^{2}-q\right)\left(b^{2}-a c\right) \geq 0$
Statement II $b \notin p a$ or $c \notin q a$.
$(2008,3 M)$
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Answer:
Correct Answer: 35.(b)
Solution:
- Given, $x^{2}+2 p x+q=0$
$ \begin{aligned} & \therefore \quad \alpha+\beta=-2 p ……(i) \\ & \alpha \beta=q ……(ii) \\ & \text { And } \quad a x^{2}+2 b x+c=0 \\ & \therefore \quad \alpha+\frac{1}{\beta}=-\frac{2 b}{a} …… (iii)\\ & \text { and } \quad \frac{\alpha}{\beta}=\frac{c}{a}……(iv) \end{aligned} $
Now, $\left(p^{2}-q\right)\left(b^{2}-a c\right)$
$ \begin{aligned} & ={\frac{\alpha+\beta^{2}}{-2}-\alpha \beta}^{2} \frac{\alpha+\frac{1}{\beta}^{2}}{2}-\frac{\alpha}{\beta} a^{2} \\ & =\frac{(\alpha-\beta)^{2}}{16} (\alpha-\frac{1}{\beta})^{2} \cdot a^{2} \geq 0 \end{aligned} $
$\therefore$ Statement I is true.
Again, now $\quad p a=-(\frac{\alpha+\beta}{2}) \quad a=-\frac{a}{2}(\alpha+\beta)$
and $\quad b=-\frac{a}{2} (\alpha+\frac{1}{\beta})$
Since,
$ p a \neq b \Rightarrow \alpha+\frac{1}{\beta} \neq \alpha+\beta $
$\Rightarrow \quad \beta^{2} \neq 1, \beta \neq{-1,0,1}$, which is correct.
Similarly, if $c \neq q a$
$ \begin{array}{cc} \Rightarrow & a \frac{\alpha}{\beta} \neq a \alpha \beta \Rightarrow \alpha (\beta-\frac{1}{\beta}) \neq 0 \\ \Rightarrow & \alpha \neq 0 \text { and } \beta-\frac{1}{\beta} \neq 0 \\ \Rightarrow & \beta \neq{-1,0,1} \end{array} $
Statement II is true.
Both Statement I and Statement II are true. But Statement II does not explain Statement I.
BETA
