Question

The function $f \left(x\right)=x^{3}+ax^{2}+bx+c, a^{2} \le 3b$ has

• A. one maximum value
• B. one minimum value
• C. no extreme value
• D. one maximum and one minimum value

Answer 1

Answer: (C)

Given, $f(x)=x^{3}+a x^{2}+b x+c, a^{2} \leq 3 b$
On differentiating w.r.t. $x$, we get
$f^{'}(x)=3 x^{2}+2 a x+b$
Put $f^{'}(x)=0$
$\Rightarrow 3 x^{2}+2 a x+b=0$
$\Rightarrow x=\frac{-2 a \pm \sqrt{4 a^{2}-12 b}}{2 \times 3}$
$=\frac{-2 a \pm 2 \sqrt{a^{2}-3 b}}{3}$
Since, $a^{2} \leq 3 b$,
$\therefore x$ has an imaginary value.
Hence, no extreme value of $x$ exist.