If $x=-2$ and $x=4$ are the extreme points of $y=x^3-\alpha x^2-\beta x+5$, then

If $x=-2$ and $x=4$ are the extreme points of $y=x^3-\alpha x^2-\beta x+5$, then
  1. $\alpha=3, \beta=24$
  2. $\alpha=-24m \beta=-3$
  3. $\alpha=-3, \beta=-24$
  4. $\alpha=24 \beta, \beta=3$

Solution

$\begin{aligned} & y=x^3-\alpha x^2-\beta x+5 \\ & \therefore \frac{d y}{d x}=3 x^2-2 \alpha x-\beta \end{aligned}$ We have $\mathrm{x}=-2$ and $\mathrm{x}=4$ as extreme points. $\begin{aligned} & \therefore\left(\frac{d y}{d x}\right)_{x=-2}=3(-2)^2-2 \alpha(-2)-\beta=0 \\ & \therefore 12+4 \alpha-\beta=0 \Rightarrow 4 \alpha-\beta=-12 \\ & \left(\frac{d y}{d x}\right)_{x=4}=3(4)^2-2 \alpha(4)-\beta=0 \\ & \therefore 48-8 \alpha-\beta=0 \Rightarrow 8 \alpha+\beta=48 \end{aligned}$ Solving eq. (1) and (2), we get $\alpha=3, \beta=24$

Asked in: MHT CET 2021 (23 Sep Shift 2)

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