MHT CET · Maths · Application of Derivatives
If \(x=-1\) and \(\cdot x=2\) are extreme points of \(f(x)=\alpha \log |x|+\beta x^2+x\), then
- A \(\alpha=-6, \beta=\frac{1}{2}\)
- B \(\alpha=-6, \beta=-\frac{1}{2}\)
- C \(\alpha=2, \beta=-\frac{1}{2}\)
- D \(\alpha=2, \beta=\frac{1}{2}\)
Answer & Solution
Correct Answer
(C) \(\alpha=2, \beta=-\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
According to the given condition, \(\mathrm{f}^{\prime}(-1)=0\) and \(\mathrm{f}^{\prime}(2)=0\) \(\mathrm{f}(x)=\alpha \log |x|+\beta x^2+x\)
\(\begin{array}{ll}
\therefore & \mathrm{f}^{\prime}(x)=\frac{\alpha}{x}+2 \beta x+1 \\
\therefore & \mathrm{f}^{\prime}(-1)=0 \Rightarrow \alpha+2 \beta=1 ...(i)\\
& \text { and } \mathrm{f}^{\prime}(2)=0 \Rightarrow \alpha+8 \beta=-2...(ii)
\end{array}\)
From (i) and (ii), we get
\(\alpha=2 \text { and } \beta=\frac{-1}{2}\)
\(\begin{array}{ll}
\therefore & \mathrm{f}^{\prime}(x)=\frac{\alpha}{x}+2 \beta x+1 \\
\therefore & \mathrm{f}^{\prime}(-1)=0 \Rightarrow \alpha+2 \beta=1 ...(i)\\
& \text { and } \mathrm{f}^{\prime}(2)=0 \Rightarrow \alpha+8 \beta=-2...(ii)
\end{array}\)
From (i) and (ii), we get
\(\alpha=2 \text { and } \beta=\frac{-1}{2}\)
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