Question

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

• A. $4$
• B. $5$
• C. $6$
• D. $7$

Answer 1

The correct option is C $6$

Given,

$f\left(x\right)=a{x}^2+bx+c,g\left(x\right)=p{x}^2+qx+r$
Since, $f\left(1\right)=g\left(1\right)$
$\Rightarrow a+b+c=p+q+r$ ...... $\left(i\right)$

$f\left(2\right)=g\left(2\right)$
$\Rightarrow 4a+2b+c=4p+2q+r$ ....... $\left(ii\right)$

Subtracting Eq. (ii) from Eq. (i), we get
$3a+b=3p+q$ ..... $\left(iii\right)$

$f\left(3\right)-g\left(3\right)=2$
$\Rightarrow (9a+3b+c)-(9p+3q+r)=2$
$\Rightarrow 3(3a+b)+c-3(3p+q)-r=2$
$\Rightarrow c-r=2$ .... $(\because 3a+b=3p+q)$ $\left(iv\right)$

From Eq. $\left(i\right),$
$(a-p)+(b-q)+(c-r)=0$
$\Rightarrow (a-p)+(b-q)+2=0$ ..... $\left(v\right)$

From Eq. $\left(ii\right)$
$4(a-p)+2(b-q)+c-r=0$
$\Rightarrow 2(a-p)+(b-q)+1=0$ ..... $\left(vi\right)$

Subtracting Eq. $\left(v\right)$ from Eq. $\left(vi\right),$ we get
$(a-p)-1=0$
$a-p=1$
$\therefore$ From Eq. $\left(v\right),$ $b-q=-3$

Now,
$f\left(4\right)-g\left(4\right)=(16a+4b+c)-(16p+4q+r)$
$=16(a-p)+4(b-q)+(c-r)$ ..... $\left(vii\right)$

Substituting the values of $(a-p),(b-q)$ and $(c-r)$ from above in Eq. $\left(vii\right),$ we get
$f\left(4\right)-g\left(4\right)=16\times 1+4(-3)+2$
$=16-12+2=6$