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Let for a!=a(1)!=0f(x)=ax^(2)+bx+c, g(x)...

Let for `a!=a_(1)!=0f(x)=ax^(2)+bx+c, g(x)=a_(1)x^(2)+b_(1)x+c_(1)` and `p(x)=f(x)-g(x)`. If `p(x)=0` only for `x=(-1)` and `p(-2)=2`, the value of `p(2)` is

A

9

B

6

C

18

D

3

Text Solution

Verified by Experts

The correct Answer is:
C
Let `p(x) = ax^(2) + bx + c`. Then, `a = a- a_(1), b = b-b_(1) and c = c - c_(1)`.
It is given that p(x) = 0 only for x = - 1. Therefore, p(x) has equal roots both equal to -1.
`therefore" "-2 = -(b)/(a) and -1 xx -1=(c)/(a)`
`rArr" "b = 2a and c = a`
Now, `p(-2) = 2 rArr 4a - 2b + c = 2 rArr c = 2`
`therefore" "c = a = 2 and b = 4`
Hence, `p(2) = 4a + 2b + c = 8 + 8 + 2 = 18`.
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