$y=x^3-a x^2+48 x+7$ is an increasing function for all real values of $x$, then $a$ lies in the interval

Question

Aicharya · Accepted Answer

Given, equation $y=x^3-a x^2+48 x+7$ Differentiating the above equation with respect to the variable $x$, we have $\frac{d y}{d x}=3 x^2-2 a x+48$ Thus, above function is a quadratic function. So, $D < 0$. In $y^{\prime}=3 x^2-2 a x+48, A=3, B=-2 a$ and $c=48$ $	herefore \quad D < 0$ $\begin{aligned} & \Rightarrow \quad(-2 a)^2-4 \cdot 3 \cdot 48 < 0 \Rightarrow 4 a^2-4 \cdot 3 \cdot 48 < 0 \ & \Rightarrow \quad 4\left(a^2-144ight) < 0 \Rightarrow a^2-(12)^2 < 0 \ & \Rightarrow \quad(a-12)(a+12) < 0 \ & \Rightarrow \quad a \in(-12,12)\end{aligned}$

$y=x^3-a x^2+48 x+7$ is an increasing function for all real values of $x$, then $a$ lies in the interval

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