Q75 / 120 · VITEEE 2009 · Mathematics

The function $f(x)=x^{3}+a x^{2}+b x+c, a^{2} \leq 3 b$ has

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

Solution

Given, $f(x)=x^{3}+a x^{2}+b x+c, a^{2} \leq 3 b$ $$ \Rightarrow f^{\prime}(x)=3 x^{2}+2 a x+b $$ Put $f^{\prime}(x)=0$ $$ \begin{array}{l} \Rightarrow 3 x^{2}+2 a x+b=0 \\ \Rightarrow x=\frac{-2 a \pm \sqrt{4 a^{2}-12 b}}{2 \times 3} \\ =\frac{-2 a \pm 2 \sqrt{a^{2}-3 b}}{3} \end{array} $$ Since, $a^{2} \leq 3 b$, $\therefore x$ has an imaginary value. Hence, no extreme value of $x$ exists.

Hence, the correct answer is option: 'C'.

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