Question

The function $f\left(x\right)=x^3+a{x}^2+bx+c,a^2\le 3b$ 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, $f(x)=x^3+a{x}^2+bx+c,a^2\le 3b$
On differentiating w.r.t. $x$, we get
$f^{{}^{\prime }}(x)=3x^2+2ax+b$
Put $f^{{}^{\prime }}(x)=0$
$\Rightarrow 3x^2+2ax+b=0$
$\Rightarrow x=\frac{-2a\pm \sqrt{4a^2-12b}}{2\times 3}$
$=\frac{-2a\pm 2\sqrt{a^2-3b}}{3}$
Since, $a^2\le 3b$,
$\therefore x$ has an imaginary value.
Hence, no extreme value of $x$ exist.