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 $p\left(-2\right)=2$, then the value of $p\left(2\right)$ is :

• A. $3$
• B. $9$
• C. $6$
• D. $18$

Answer 1

The correct option is D $18$

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

$g\left(x\right)=a_1{x}^2+b_{1}x+c_1$

and $p\left(x\right)=f\left(x\right)-g\left(x\right)$

$=a{x}^2+bx+c-(a_1{x}^2+b_{1}x+c_1)$

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

Now $p(-1)=0$

i.e., $(a-a_1)+(b-b_1)(-1)+(c-c_1)=0$

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

Also $p(-2)=2$

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

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

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

$=6(a-a_1)+3(c-c_1)$

$=3\left(2\right(a-a_1)+(c-c_1\left)\right)$

$=3(2+c-c_1+c-c_1)$

$=6(1-c-c_1)$

But $c-c_1$ is not equal to zero so $1+c-c_1>1$ and $p\left(2\right)$ is a multiple of $6$, so, $p\left(2\right)=18$ is possible.