Question

If $f(x)=4x^2+ax+(a-3)$ is negative for atleast one negative $x$, then the possible value(s) of $a$, is/are

• A. 3
• B. 4
• C. 10
• D. 14

Answer 1

Answer: (A), (D)

$\text{ Case-I: if }f(0)<0$
$a-3<0\Rightarrow a<3$
$a\in (-\infty ,+3)$

Case-II: $\text{ if }f(0)\ge 0\text{ and }$
$D>0$ and $-\frac{b}{2a}<0$
$f(0)>0$ gives $a-3\ge 0\Rightarrow a\ge 3$ ....(1)

$D>0$ gives $a^2-16(a-3)>0$
$a^2-16a-48>0$
$(a-12)(a-4)>0\Rightarrow a>12\text{ or }a<4$ ....(2)
$-\frac{b}{2a}<0$ gives $-\frac{a}{8}<0\Rightarrow a>0$ ....(3)
$\text{ from (1), (2) and (3) }$
$a\in [3,4)\cup (12,\infty )$
$\text{ finally }a\in (-\infty ,4)\cup (12,\infty )$