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If (b^2-4a c)^2(1+4a^2)<64 a^2, a<0 , th...

If `(b^2-4a c)^2(1+4a^2)<64 a^2, a<0` , then maximum value of quadratic expression `a x^2+b x+c` is always less than a. 0 b. 2 c. -1 d. -2

A

0

B

2

C

`-1`

D

`-2`

Text Solution

Verified by Experts

The correct Answer is:
2
`((b^(2) - 4ac)^(2))/(16a^(2)) lt (4)/(1 + 4a^(2))`
Now , max `(ax^(2) + bx + c) = -((b^(2) - 4ac)/(4a)`
Also, `(-2)/(sqrt(1 + 4 a^(2))) lt - (b^(2) - 4ac)/(4a) lt (2)/(sqrt(1 + 4a^(2)))`
So, maximum value is alwasys less then 2 (when a `to` 0) .
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