MHT CET · Maths · Application of Derivatives
If \(x=-2\) and \(x=4\) are the extreme points of \(y=x^3-\alpha x^2-\beta x+5\), then
- A \(\alpha=3, \beta=24\)
- B \(\alpha=-24m \beta=-3\)
- C \(\alpha=-3, \beta=-24\)
- D \(\alpha=24 \beta, \beta=3\)
Answer & Solution
Correct Answer
(A) \(\alpha=3, \beta=24\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& y=x^3-\alpha x^2-\beta x+5 \\
& \therefore \frac{d y}{d x}=3 x^2-2 \alpha x-\beta
\end{aligned}
\)
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 \(\alpha=3, \beta=24\)
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