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If a, b in R and ax^2 + bx +6 = 0,a!= 0 ...

If `a, b in R` and `ax^2 + bx +6 = 0,a!= 0` does not have two distinct real roots, then :

A

(a) minimum possible value of `3a+b` is `-2`

B

(b) minimum possible value of `3a+b` is 2

C

(c) minimum possible value of `6a+b` is `-1`

D

(d) minimum possible value of `6a+b` is 1

Text Solution

Verified by Experts

The correct Answer is:
A, C
Here `Dle0`
and `f(x)ge0,AA x epsilonR`
`:.(f)ge0`
`implies9a+3b+6ge0`
or `3a+bge-2`
`implies`Minimum value of `3a+b` is `-2`.
and `f(6)ge0`
`implies36a+6b+6ge0`
`implies6a=bge-1`
`implies` Minimum value of `6a+b` is `-1`.
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