Question

Let for $a\ne a_1\ne 0$, $f\left(x\right)=a{x}^2+bx+c$, $g\left(x\right)=a_1{x}^2+b_{1}x+c_1$ and $p\left(x\right)=f\left(x\right)-g\left(x\right)$. If $p\left(x\right)=0$ only for $x=-1$ and $p(-2)=2$, then the value of $p\left(2\right)$ is :

• A. $3$
• B. $9$
• C. $6$
• D. $18$

Answer 1

The correct option is D $18$

$p\left(x\right)=(a-a_1)x^2+(b-b_1)x+(c-c_1)$

$p(-1)=0$

$p(-2)=2$

$p\left(0\right)=2$ (one to symmetry around x = -1 )

$p\left(2\right)=?$

$p\left(0\right)=2\Rightarrow c=c_1+2$

$p(-1)=0\Rightarrow (a-a_1)-(b-b_1)+2=0$ __ (1)

$p(-2)=2\Rightarrow 4(a-a_1)-2(b-b^1)+2=2$

$e{q}^n\left(1\right)\times 2$

$\Rightarrow 2(a-a_1)-2(b-b_1)+4=0$

$2(a-a_1)=2+2$

$\Rightarrow a-a_1=2$

$\Rightarrow b-b^1=4$

$p\left(2\right)=4(a-a_1)+2(b-b_1)+(c-c_1)$

$=4\times 2+2\times 4+2$

$=8+8+2$

$p\left(2\right)=18$