Search any question & find its solution
Question:
Answered & Verified by Expert
If $x=-2$ and $x=4$ are the extreme points of $y=x^3-\alpha x^2-\beta x+5$, then
Options:
Solution:
1306 Upvotes
Verified Answer
The correct answer is:
$\alpha=3, \beta=24$
$$
\begin{aligned}
& y=x^3-\alpha x^2-\beta x+5 \\
& \therefore \frac{d y}{d x}=3 x^2-2 \alpha x-\beta
\end{aligned}
$$
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 $\alpha=3, \beta=24$
\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 $\alpha=3, \beta=24$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.