Question

Let for $a \ne a_{1} \ne 0$,
$f\left(x\right) = ax^{2} + bx + c, g^{9}x) = a_{1}x^{2} + b_{1}x + c_{1}$ 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)

$P(x) = 0$
$\Rightarrow \quad f\left(x\right) = g\left(x\right)$
$\Rightarrow \quad ax^{2} + bx + c = a_{1}x^{2} + b_{1}x + C,$
$\Rightarrow \quad\left(a - a_{1}\right) x^{2} + \left(b - b_{1}\right) x + \left(c - c_{1}\right) = 0.$
It has only one solution $x = - 1$
$\Rightarrow \quad b - b_{1} = a - a_{1} + c - c_{1} \quad\quad.... \left(1\right)$
vertex $\left(-1, 0\right) \quad\Rightarrow \quad \frac{b-b_{1}}{2\left(a-a_{1}\right)} = -1\quad\Rightarrow \quad b - b_{1} = 2\left(a - a_{1}\right) \quad\quad.... \left(2\right)$
$\Rightarrow \quad f\left(-2\right) - g\left(-2\right) = 2$
$\Rightarrow \quad4a - 2b + c - 4a_{1} + 2b_{1} - c_{1} = 2$
$\Rightarrow \quad4\left(a - a_{1}\right) - 2\left(b - b_{1}\right) + \left(c - c_{1}\right) = 2\quad\quad .... \left(3\right)$
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 $\quad P\left(2\right) = f\left(2\right) - g\left(2\right)$
$= 4 \left(a - a_{1}\right) + 2 \left(b - b_{1}\right) + \left(c - c_{1}\right)$
$= 8 + 8 + 2 = 18$