Question

$f\left(x\right)=a{x}^2+bx+c$ and $g\left(x\right)=p{x}^2+qx$ with $g\left(1\right)=f\left(1\right),g\left(2\right)-f\left(2\right)=1,g\left(3\right)-f\left(3\right)=4,$ then $g\left(4\right)-f\left(4\right)$ is.

• A. $0$
• B. $5$
• C. $6$
• D. None of these

Answer 1

Answer: (D)

None of these

Given, $f\left(x\right)=a{x}^2+bx+c$ and $g\left(x\right)=p{x}^2+qx$

$g\left(1\right)=f\left(1\right)\Rightarrow p+q=a+b+c\Rightarrow p-a+q-b-c=0$

$g\left(2\right)-f\left(2\right)=1\Rightarrow 4p+2q=4a+2b+c+1$

$g\left(3\right)-f\left(3\right)=4\Rightarrow 9p+3q=9a+3b+c+4$

$g\left(4\right)-f\left(4\right)=16p+4q-16a-4b-c$

Let p-a be x and q-b be y

Then

$x+y-c=0$ ..........(1)

$4x+2y-c=1$ ......(2)

$9x+3y-c=4$ ........(3)

From 1 and 2

$3x+y=1$ .......(4)

From 2 and 3

$5x+y=3$ .......(5)

From 4 and 5

$2x=2\Rightarrow x=1$

from (4)

$y=1-3=-2$

from (1)

$c=1-2=-1$

$g\left(4\right)-f\left(4\right)=16x+4y-c=16\left(1\right)+4(-2)-(-1)=16-8+1=9$