Q80 / 160 · WBJEE 2012 · Mathematics

Let $f(x)=a x^{2}+b x+c, g(x)=p x^{2}+q x+r$ $\begin{array}{lll}\text { such that } & f(1)=g(1), f(2)=g(2) & \text { and }\end{array}$ $f(3)-g(3)=2 .$ Then, $f(4)-g(4)$ is

  • A
    4
  • B
    5
  • Correct answer
    C
    6
  • D
    7

Solution

Given, $f(x)=e x^{2}+b x+c, g(x)=p x^{2}+q x+r$ since, $f(1)=g(1)$ $\Rightarrow \quad a+b+c=p+q+r$ $f(2)=8(2)$ $\Rightarrow \quad 4 a+2 b+c=4 p+2 q+r$ Subtracting Eq. (tii) from Eq. (i), we get $3 a+b=3 p+q$ $f(3)-g(3)=2$ $\Rightarrow(9 a+3 b+c)-(9 p+3 q+r)=2$ $\Rightarrow \quad 3(3 a+b)+c-3(3 p+q)-r=2$ $\Rightarrow$ $c-r=2 \quad \ldots \text { (iv) }$ $(3 a+b=3 p+q)$ From Eq. (i), $(a-p)+(b-q)+(c-r)=0$ $\Rightarrow \quad(a-p)+(b-q)+2=0$ From Eq. (ii), $4(a-p)+2(b-q)+c-r=0$ $\Rightarrow \quad 2(a-p)+(b-q)+1=0$ Subtracting Eq.(v) from Eq. (vi), we get $(a-p)-1=0$ $a-p=1$ $\therefore$ From Eq. $(v), \quad b-q=-3$ Now, $f(4)-g(4)=(16 a+4 b+c)-(16 p+4 q+r)$ $=16(a-p)+4(b-q)+(c-r) \ldots(v$ Substituting the values of $(a-p),(b-q),(c-r)$ from above in Eq. (vii). we get $\begin{aligned} f(4)-g(4) &=16 \times 1+4(-3)+2 \\ &=16-12+2=6 \end{aligned}$

Hence, the correct answer is option: 'C'.