MHT CET · Maths · Application of Derivatives
If \(x=-1\) and \(x=2\) are extreme points of \(\mathrm{f}(x)=\alpha \log x+\beta x^2+x, \alpha\) and \(\beta\) are constants, then the value of \(\alpha^2+2 \beta\) is
- A \(-3\)
- B \(3\)
- C \(\frac{3}{2}\)
- D \(5\)
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
According to the given condition, \(\mathrm{f}^{\prime}(1)=0\) and \(\mathrm{f}^{\prime}(2)=0\)
\(
\mathrm{f}(x)=\alpha \log x+\beta x^2+x
\)
\(\therefore \mathrm{f}^{\prime}(x)=\frac{\alpha}{x}+2 \beta x+1 \)
\( \therefore \mathrm{f}^{\prime}(-1)=0 \Rightarrow \alpha+2 \beta=1 \)
\( \text { and } \mathrm{f}^{\prime}(2)=0 \Rightarrow \alpha+8 \beta=-2\)
\(\therefore \) From (i) and (ii), we get
\(\beta=\frac{-1}{2} \text { and } \alpha=2 \)
\( \therefore \alpha^2+2 \beta=4-1=3\)
\(
\mathrm{f}(x)=\alpha \log x+\beta x^2+x
\)
\(\therefore \mathrm{f}^{\prime}(x)=\frac{\alpha}{x}+2 \beta x+1 \)
\( \therefore \mathrm{f}^{\prime}(-1)=0 \Rightarrow \alpha+2 \beta=1 \)
\( \text { and } \mathrm{f}^{\prime}(2)=0 \Rightarrow \alpha+8 \beta=-2\)
\(\therefore \) From (i) and (ii), we get
\(\beta=\frac{-1}{2} \text { and } \alpha=2 \)
\( \therefore \alpha^2+2 \beta=4-1=3\)
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