A) one maximum value
B) one minimum value
C) no extreme value
D) one maximum and one minimum value
Correct Answer: C
Solution :
Given, \[f(x)={{x}^{3}}+a{{x}^{2}}+bx+c,\] \[{{a}^{2}}\le 3b.\] On differentiating w.r.t. x, we get \[f(x)=3{{x}^{2}}+2ax+b\] Put \[f(x)=0\] \[\Rightarrow \] \[3{{x}^{2}}+2ax+b=0\] \[\Rightarrow \] \[x=\frac{-\,2a\pm \sqrt{4{{a}^{2}}-12b}}{2\times 3}\] \[=\frac{-\,2a\pm 2\sqrt{{{a}^{2}}-12b}}{3}\] Since, \[{{a}^{2}}\le 3b.\] \[\therefore \]\[x\] has an imaginary value. Hence, no extreme value of \[x\] exist.
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