Question

Let for $a\ne a_1\ne 0$,
$f\left(x\right)=a{x}^2+bx+c,g^{9}x)=a_1{x}^2+b_{1}x+c_1$ and $p\left(x\right)=f\left(x\right)-g\left(x\right)$.
If $p\left(x\right)=0$ only for $x=-1$ and $p\left(-2\right)=2$, then the value of $p\left(2\right)$ is :

• A. 3
• B. 9
• C. 6
• D. 18

Answer 1

Answer: (D)

$P(x)=0$
$\Rightarrow f\left(x\right)=g\left(x\right)$
$\Rightarrow a{x}^2+bx+c=a_1{x}^2+b_{1}x+C,$
$\Rightarrow \left(a-a_1\right)x^2+\left(b-b_1\right)x+\left(c-c_1\right)=0.$
It has only one solution $x=-1$
$\Rightarrow b-b_1=a-a_1+c-c_1....\left(1\right)$
vertex $\left(-1,0\right)\Rightarrow \frac{b-b_1}{2\left(a-a_1\right)}=-1\Rightarrow b-b_1=2\left(a-a_1\right)....\left(2\right)$
$\Rightarrow f\left(-2\right)-g\left(-2\right)=2$
$\Rightarrow 4a-2b+c-4a_1+2b_1-c_1=2$
$\Rightarrow 4\left(a-a_1\right)-2\left(b-b_1\right)+\left(c-c_1\right)=2....\left(3\right)$
by $\left(1\right),\left(2\right)$ and $\left(3\right)\left(a-a_1\right)=\left(c-c_1\right)=\frac{1}{2}\left(b-b_1\right)=2$
Now $P\left(2\right)=f\left(2\right)-g\left(2\right)$
$=4\left(a-a_1\right)+2\left(b-b_1\right)+\left(c-c_1\right)$
$=8+8+2=18$