Question

lf $\alpha ,\beta$ are the roots of $a{x}^2+bx+c=0$ observe the following lists.

List-I List-II

$A$: $a,b,c$ have same sign 1. $\alpha \beta =1$
$B$: lf $a=c$ 2. $-1$
$C$: lf $a+b+c=0$ then one root is 3. $\alpha$ and $\beta$ are negative
$D$: lf $a+c=b$ then one root is 4. $0$
5. $1$
The correct match for list-l from list-I is A B C D

• A. $1,3,2,5$
• B. $1,3,4,5$
• C. $3,1,5,4$
• D. $3,1,5,2$

Answer 1

Answer: (D)

$3,1,5,2$

If a, b and c have same sign
$\Rightarrow \alpha +\beta =\frac{-b}{a}<0$
$\alpha \beta =\frac{c}{a}>0$
$\therefore \alpha$ and $\beta$ are negative
If $a=c;\alpha \beta =\frac{c}{a}=1$
If $a+b+c=0;f\left(1\right)=0\Rightarrow a(1{)}^2+b(1)+c=0$
$\therefore$ 1 is a root.
If $a+c=b\Rightarrow a(1{)}^2+c=b(1)$
$a(1{)}^2-b(1)+c=0$
$-1$ is a root.