Question

If the curve $y=a{x}^2+bx+c,x\in R,$ passes through the point $(1,2)$ and the tangent line to this curve at origin is $y=x,$ then the possible values of $a,b,c$ are

• A. $a=1,b=1,c=0$
• B. $a=-1,b=1,c=1$
• C. $a=1,b=0,c=1$
• D. $a=\frac{1}{2},b=\frac{1}{2},c=1$

Answer 1

The correct option is A $a=1,b=1,c=0$

$y=a{x}^2+bx+c$ passes through $(1,2)$
$\Rightarrow 2=a+b+c$
$\frac{dy}{dx}=2ax+b,\frac{dy}{dx}{|}_{(0,0)}=1$
$\Rightarrow b=1$ and $a+c=1$
Since $(0,0)$ lies on the curve,
$\therefore c=0,a=1$

$TRICK:$ $(0,0)$ lies on the curve. Only option $a=1,b=1,c=0$ has $c=0.$