Question

Let for $a\ne a_1\ne o$, $f\left(x\right)=a{x}^2+bx+C$, $g\left(x\right)=a_1{x}^2+b_{1}x+c_1$ 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 value of $P\left(2\right)$ is

• A. $9$
• B. $6$
• C. $18$
• D. $3$

Answer 1

Answer: (C)

$18$

$f\left(x\right)=a{x}^2+bx+c$

$g\left(x\right)=a_1{x}^2+b_{1}x+c_{1}a\ne a_1\ne 0$

$P\left(x\right)=f\left(x\right)-g\left(x\right)$

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

$P\left(x\right)=0$ at only $x=-1$

$\therefore {(b-b_1)}^2=4(a-a_1)(c-c_1)$ ($\because$ only one solution)

$P(-1)=0$

$(a-a_1)-(b-b_1)+(c-c_1)=0\Rightarrow (b-b_1)=(a-a_1)+(c-c_1)\to 2$

$1\Rightarrow {((a-a_1)+(c-c_1))}^2-4(a-a_1)(c-c_1)=0$

$\therefore (a-a_1)=(c-c_1)\to 2$

$P(-2)=2$ (given)

$(a-a_1){(-2)}^2+(b-b_1)(-2)+(c-c_1)=2$

$2\Rightarrow (b-b_1)=2(a-a_1)$ (From $3$)

$4(a-a_1)-2(b-b_1)+(c-c_1)=2$

$4(a-a_1)-4(a-a_1)+(a-a_1)=2$

$\therefore (a-a_1)=2=(c-c_1)$

$\therefore (b-b_1)=4$

$\therefore P\left(x\right)=2x^2+4x+2=2{(x+1)}^2$

$P\left(2\right)=2{(2+1)}^2=18$