If a,b∈R and ax2+bx+6=0 (a≠0) does not have two distinct real roots, then

Roots are equal or complex b 2 − 4 ac ≤ 0 ⇒ f ( x ) ≥ 0 ⇒ f ( 3 ) ≥ 0 ⇒ 9 a + 3 b + 6 ≥ 0 ⇒ 3 a + b ≥ − 2 f ( 6 ) ≥ 0 ⇒ 36 a + 6 b + 6 ≥ 0 ⇒ 6 a + 6 ≥ − 1

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If $a, b \in R$ and ${ax}^{2} + bx + 6 = 0$ $(a \neq 0)$ does not have two distinct real roots, then

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minimum value of $3 a + b$ is $- 2$

minimum value of $3 a + b$ is $2$

minimum value of $6 a + b$ is $1$

minimum value of $6 a + b$ is $- 1$

answer is A, C.

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Detailed Solution

Roots are equal or complex $b^{2} - 4 ac \leq 0$
$\Rightarrow f (x) \geq 0 \Rightarrow f (3) \geq 0 \Rightarrow 9 a + 3 b + 6 \geq 0 \Rightarrow 3 a + b \geq - 2$
$f (6) \geq 0 \Rightarrow 36 a + 6 b + 6 \geq 0 \Rightarrow 6 a + 6 \geq - 1$