Let for `a != a_1 != 0` , `f(x)=ax^2+bx+c` ,`g(x)=a_1x^2+b_1x+c_1` and `p(x) = f(x) - g(x)`. If `p(x) = 0` only for `x = -1` and `p(-2) = 2` then the value of `p(2)`.

Options

A. 18
B. 3
C. 9
D. 6

Answer

B

Explanation

Given `p(x)=f(x)-g(x)`
`impliesp(x)=(a-a_(1))x^(2)+(b-b_(1))x+(c-c_(1))`
It is clear that `p(x)=0` has both equal roots `-1` then
`-1-1=-((b-b_(1)))/((a-a_(1)))`
and `-1xx-1=(c-c_(1))/(a-a_(1))`
`impliesb-b_(1)=2(a-a_(1))` and `c-c_(1)=(a-a_(1))`............i
Also given `p(-2)=2`
`=4(a-a_(1))-2(b-b_(1))+(c-c_(1))=2`..........ii
From eqs i and ii we get
`4(a-a_(1))-4(a-a_(1))+(a-a_(1))=2`
`:.(a-a_(1))=2`........iii
`impliesb-b_(1)=4` and `c-c_(1)=2`[from Eq. i ] ...iv
Now `p(2)=4(a-a_(1)+2(b-b_(1))+(c-c_(1))`
`=8+8+2=18` [from Eqs. iii and iv]