Q55 / 160 · TS EAMCET 2005 · Mathematics

$E_1: a+b+c=0, \quad$ if 1 is a root of $a x^2+b x+c=0, \quad E_2: b^2-a^2=2 a c$, if $\sin \theta$, $\cos \theta$ are the roots of $a x^2+b x+c=0$ Which of the following is true?

  • Correct answer
    A
    $E_1$ is true, $E_2$ is true
  • B
    $E_1$ is true, $E_2$ is false
  • C
    $E_1$ is false, $E_2$ is true
  • D
    $E_1$ is false, $E_2$ is false

Solution

Given that, 1 is a root of $a x^2+b x+c=0$ $\begin{aligned} & \Rightarrow \quad a+b+c=0 \\ & \therefore \quad E_1: a+b+c=0 \text { is true. } \end{aligned}$ Since $\cos \theta, \sin \theta$ are the roots of $\begin{aligned} & a x^2+b x+c=0 \\ \therefore \quad & \sin \theta+\cos \theta=-\frac{b}{a} \\ \text { and } \sin \theta \cos \theta= & \frac{c}{a} \end{aligned}$ On squaring both sides of equation (i) $\begin{array}{rlrl} & (\sin \theta+\cos \theta)^2 & =\frac{b^2}{a^2} \\ \Rightarrow \sin ^2 \theta+\cos ^2 \theta+2 \sin \theta \cos \theta & =\frac{b^2}{a^2} \\ \Rightarrow & 1+2\left(\frac{c}{a}\right) & =\frac{b^2}{a^2} \\ \Rightarrow & 2 \cdot \frac{c}{a} & =\frac{b^2-a^2}{a^2} \\ \Rightarrow & -a^2+b^2 & =2 a c \\ \therefore & E_2: b^2-a^2 & =2 a c \text { is true } \end{array}$ $\Rightarrow E_1$ and $E_2$ both are true.

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

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