Q113 / 160 · TS EAMCET 2017 · Mathematics

If $\alpha$ and $\beta$ are the roots of the equation $a x^2+b x+c=0$ and the equation having roots $\frac{1-\alpha}{\alpha}$ and $\frac{1-\beta}{\beta}$ is $p x^2+q x+r=0$, then $r=$

  • A
    $a+2 b$
  • B
    $a b+b c+c a$
  • Correct answer
    C
    $a+b+c$
  • D
    $a b c$

Solution

We have, $\alpha, \beta$ are the roots of the equations $a x^2+b x+c=0$ $$ \therefore \quad \alpha+\beta=-\frac{b}{a}, \alpha \beta=\frac{c}{a} $$ Equations whose roots are $\frac{1-\alpha}{\alpha}$ and $\frac{1-\beta}{\beta}$ are $$ \begin{aligned} & x^2-\left(\frac{1-\alpha}{\alpha}+\frac{1-\beta}{\beta}\right) x+\left(\frac{1-\alpha}{\alpha}\right)\left(\frac{1-\beta}{\beta}\right)=0 \\ & \Rightarrow \quad x^2-\left(\frac{1}{\alpha}+\frac{1}{\beta}-2\right) x+\left(\frac{1-(\alpha+\beta)+\alpha \beta}{\alpha \beta}\right)=0 \\ & \Rightarrow x^2-\left(\frac{\alpha+\beta-2 \alpha \beta}{\alpha \beta}\right) x+\left(\frac{1-(\alpha+\beta)+\alpha \beta}{\alpha \beta}\right)=0 \\ & \Rightarrow \alpha \beta x^2-(\alpha+\beta-2 \alpha \beta) x+(1-\langle\alpha+\beta\rangle+\alpha \beta)=0 \\ & \Rightarrow \quad \frac{c}{a} x^2-\left(-\frac{b}{a}-\frac{2 c}{a}\right) x+\left(1+\frac{b}{a}+\frac{c}{a}\right)=0 \\ & \Rightarrow \quad a x^2+(b+2 c) x+(a+b+c)=0 \\ & \end{aligned} $$ Comparing with $p x^2+q x+r=0$, we get $$ r=a+b+c $$

Hence, the correct answer is option: 'C'.

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