Question

Statement-I : If $f\left(x\right)$ is a quadratic expression such that $f\left(0\right)+f\left(1\right)=0$. If $-2$ is a root of $f\left(x\right)=0$, then the other root is $\frac{3}{5}$.
Statement-II : If $f\left(x\right)=a{x}^2+bx+c$, then sum of roots $=-\frac{b}{a}$ and product of roots $=\frac{c}{a}.$

• A. Statement-I is true, Statement-II is true; Statement-II is correct explanation for Statement-I.
• B. Statement-I is true, Statement-II is true; Statement-II is NOT a correct explanation for statement-I.
• C. Statement-I is true, Statement-II is false.
• D. Statement-I is false, Statement-II is true.

Answer 1

Answer: (A)

Statement-I is true, Statement-II is true; Statement-II is correct explanation for Statement-I.

Let us assume that $f\left(x\right)=a{x}^2+bx+c$
Then
$f\left(0\right)+f\left(1\right)=0$ implies
$c+a+b+c=0$
Or
$a+b+2c=0$
Or
$c=-\left(\frac{a+b}{2}\right)$
Hence $f\left(x\right)=a{x}^2+bx-\frac{(a+b)}{2}$
It is given that
$f(-2)=0$
Or
$4a-2b-\left(\frac{a+b}{2}\right)=0$
$8a-4c-a-b=0$
$7a-5b=0$
$b=\frac{7a}{5}$.
Hence
$f\left(x\right)=a{x}^2+\frac{7a}{5}x-\frac{12a}{10}$
Now sum of roots is
$=\frac{-B}{A}$
$=\frac{-7}{5}$
Since one root is -2, let the other root be $y$
Hence
$y-2=\frac{-7}{5}$
$y=2-\frac{7}{5}$
$=\frac{3}{5}$.
Hence the other root is $\frac{3}{5}$.

Hence, statement I is correct and statement II is the correct explanation for it.