Question

Let $p\left(x\right)=a{x}^2+bx+c,q\left(x\right)=l{x}^2+mx+n.$ If $p\left(1\right)-q\left(1\right)=0$, $p\left(2\right)-q\left(2\right)=1$ and $p\left(3\right)-q\left(3\right)=4$, then $p\left(4\right)-q\left(4\right)$ equals

• A. $0$
• B. $5$
• C. $6$
• D. $9$

Answer 1

The correct option is D $9$

Let $p\left(x\right)=a{x}^2+bx+c$ and $q\left(x\right)=l{x}^2+mx+n$.

Now $p\left(x\right)-q\left(x\right)=(a-l)x^2+(b-m)x+(c-n)=A{x}^2+Bx+C$(let).

Given

$p\left(1\right)-q\left(1\right)=0$

$\Rightarrow A+B+C=0$......(1).

Also given

$p\left(2\right)-q\left(2\right)=1$

or, $4A+2B+C=1$........(2).

Again

$p\left(3\right)-q\left(3\right)=4$

or, $9A+3B+C=4$..........(3).

Solving from (1),(2) and (3) we get,

$A=1,B=-2,C=1$.

Now $p\left(4\right)-q\left(4\right)$

$=16A+4B+C=16\left(1\right)+4(-2)+1=9$