Let for $a \neq 0, a_{1} \neq 0, a \neq a_{1}, f (x) = {ax}^{2} + bx + c, g (x) = a_{1} x^{2} + b_{1} x + c_{1}$ and $p (x) = f (x) - g (x) . If p (x) = 0$ only for x = -1 and P(-2) = 2 then which of the following is not a value of p(2)?

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answer is A, B, C.

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Detailed Solution

Reason: $p(x)=f(x)-g(x)$ $p$ $($ $x$ $)$ $=$ $f$ $($ $x$ $)$ $-$ $g$ $($ $x$ $)$ is quadratic with a single zero at $x=-1$ $x$ $=$ $-1$ ⇒ double root.
So $p(x)=k(x+1)^2$ $p$ $($ $x$ $)$ $=$ $k$ $($ $x$ $+$ $1$ $)$ $2$ with $k\neq0$ $k$ $\neq$ $=$ $0$ .
Given $p(-2)=2\Rightarrow k\cdot 1=2\Rightarrow k=2$ $p$ $($ $-2$ $)$ $=$ $2$ $\Rightarrow$ $k$ $\cdot$ $1$ $=$ $2$ $\Rightarrow$ $k$ $=$ $2$ .
Thus $p(2)=2(2+1)^2=2\cdot9=18$ $p$ $($ $2$ $)$ $=$ $2$ $($ $2$ $+$ $1$ $)$ $2$ $=$ $2$ $\cdot$ $9$ $=$ $18$ .
So any choice not equal to 18 is “not a value of $p(2)$ $p$ $($ $2$ $)$ ”.