The condition for the function f(x)=x^3+p x^2 | vedclass

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(A) function $f(x)$ has no extreme value if its derivative $f^{\prime}(x)$ does not change sign,which means $f^{\prime}(x)$ is either always non-negat

Vedclass · Accepted Answer

$p^2 < 3 q$

Vedclass · Answer

(A) function $f(x)$ has no extreme value if its derivative $f^{\prime}(x)$ does not change sign,which means $f^{\prime}(x)$ is either always non-negative or always non-positive. Given $f(x) = x^3 + p x^2 + q x + r$. The derivative is $f^{\prime}(x) = 3x^2 + 2px + q$. For $f(x)$ to have no extreme value,$f^{\prime}(x)$ must not have distinct real roots. This implies the discriminant $D \le 0$. The discriminant of the quadratic $3x^2 + 2px + q$ is $D = (2p)^2 - 4(3)(q) = 4p^2 - 12q$. Setting $D \le 0$: $4p^2 - 12q \le 0$ $4p^2 \le 12q$ $p^2 \le 3q$. Since the options provided represent the strict inequality condition for the derivative to be strictly monotonic,the correct condition is $p^2 < 3q$.

The condition for the function $f(x)=x^3+p x^2+q x+r$ $(x \in R)$ to have no extreme value is

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