E_1: a+b+c=0,if 1 is a root of ax^2+bx+c=0. E | vedclass

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(A) Given that $1$ is a root of $ax^2+bx+c=0$.Substituting $x=1$,we get $a(1)^2+b(1)+c=0$,which implies $a+b+c=0$.Thus,$E_1$ is true.Given that $\sin

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$E_1$ is true,$E_2$ is true

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(A) Given that $1$ is a root of $ax^2+bx+c=0$.Substituting $x=1$,we get $a(1)^2+b(1)+c=0$,which implies $a+b+c=0$.Thus,$E_1$ is true.Given that $\sin 	heta$ and $\cos 	heta$ are roots of $ax^2+bx+c=0$.From the sum and product of roots:$\sin 	heta + \cos 	heta = -\frac{b}{a}$ and $\sin 	heta \cos 	heta = \frac{c}{a}$.Squaring the sum of roots:$(\sin 	heta + \cos 	heta)^2 = (-\frac{b}{a})^2$$\sin^2 	heta + \cos^2 	heta + 2 \sin 	heta \cos 	heta = \frac{b^2}{a^2}$$1 + 2(\frac{c}{a}) = \frac{b^2}{a^2}$Multiplying by $a^2$ (assuming $a 
eq 0$):$a^2 + 2ac = b^2$$b^2 - a^2 = 2ac$.Thus,$E_2$ is true.Therefore,both $E_1$ and $E_2$ are true.

List-$I$	List-$II$
$(A)$ The minimum value of $2x^2 + 4x + 5$	$(I)$ $-1$
$(B)$ The maximum value of $\frac{x^2 + 4x + 1}{x^2 + x + 1}$	$(II)$ $1$
$(C)$ If $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$, $\forall x \in [a, b]$ then $b =$	$(III)$ $2$
$(D)$ If $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$, $\forall x \in [a, b]$ then $a =$	$(IV)$ $3$
	$(V)$ $4$

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

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