Question

In the curve $y=p{x}^2+qx+r$ passes through the point (1,2) and the line $y=x$ touches it at the origin, then the values of p,q and r are

• A. $p=1,q=-1,r=0$
• B. $p=1,q=1,r=0$
• C. $p=-1,q=1,r=0$
• D. $p=1,q=2,r=3$

Answer 1

The correct option is B $p=1,q=1,r=0$

Given,

$y=p{x}^2+qx+r$

at $(0,0)$ we get,

$0=p(0{)}^2+q(0)+r$

$\therefore r=0$

$\frac{dy}{dx}=2px+q$ $\to y=x,m=1$

${\frac{dy}{dx}}_{(0,0)}=2p\left(0\right)+q=1$

$\Rightarrow q=1$

$y=p{x}^2+q$ at $(1,2)$

$2=p(1{)}^2+1$

$\Rightarrow p=1$

$\therefore p=1,q=1,r=0$