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Question

If roots of the equation f(x+2)=ax2+bx+c=0 and α,β are such that α<2<β, then for the equation f(x)=Ax2+Bx+C.

A
AC<0
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B
AC>0
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C
B24AC>0
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D
B24AC<0
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Solution

The correct option is C B24AC>0
We are given
f(x+2)=ax2+bx+c=0(1)
have roots α & β & α<2<β it show thta α & β are real and distinct roots So, b24ac>0
Now, we also given,
f(x)=Ax2+Bx+C(2)
In the equation (1) let put x+2
So in the left hand side we have to put xx2
f(x)=a(x2)2+b(x2)+c
=a(x24x+4)+b(x2)+c
=ax24ax+4a+bx2b+c
f(x)=ax2+(4a+b)+x+4a2b+1(3)
Now, compare equation (2) and (3) we get
A=a;B=4a+b & C=4a2b+c
If both (2) & (3) equation are equal then the roots of the equation (3) must be real & distinct
So, B24AC>0
(4a+b)24(a)(4a2b+c)>0
+16a28ab+b216a2+8ab4ac>0
(1616)a2+(8+8)ab+(b24ac)>0
b24ac>0
Hence it proof that for equation f(x)=Ax2+Bx+C
B24AC>0

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