Question

$f\left(x\right)=2x^2+bx+c$ and $f\left(0\right)=3$ and $f\left(2\right)=1$, then $f\left(1\right)$ is equal to

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

The correct option is A. $0$

From question,
$f\left(x\right)=2x^2+bx+c$

As, $f\left(0\right)=3$

$\therefore f\left(0\right)=2(0{)}^2+b(0)+c=3$

$\Rightarrow 0+0+c=3$

$\Rightarrow c=3$

Also, $f\left(2\right)=1$ -------------------- (1)

$\therefore f\left(2\right)=2\times (2{)}^2+b(2)+3=1$ [As $c=3$]

$\Rightarrow 8+3+2b=1$

$\Rightarrow 11+2b=-10$

$\Rightarrow 2b=-10$

$\Rightarrow b=-5$ --------------------------- (2)

From (1) and (2),
$f\left(x\right)=2x^2+(-5)x+3$
$f\left(x\right)=2x^2-5x+3$

Hence, $f\left(1\right)=2(1{)}^2-5(1)+3=2-5+3=-3+3=0$