Application of Derivatives 4 Question 16
####17. If $x=-1$ and $x=2$ 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) $\alpha=2, \beta=\frac{1}{2}$
(2014 Main)
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Answer:
Correct Answer: 17. (c)
Solution:
- Here, $x=-1$ and $x=2$ are extreme points of $f(x)=\alpha \log |x|+\beta x^{2}+x$, then
$ \begin{aligned} f^{\prime}(x) & =\frac{\alpha}{x}+2 \beta x+1 \\ f^{\prime}(-1) & =-\alpha-2 \beta+1=0 \end{aligned} $
[at extreme point, $f^{\prime}(x)=0$ ]
$ f^{\prime}(2)=\frac{\alpha}{2}+4 \beta+1=0 $
On solving Eqs. (i) and (ii), we get
$ \alpha=2, \beta=-\frac{1}{2} $
