logo

Ask Questions, Get Answers

X
 

If $x=-1$ and $x = 2$ extreme points of $f(x)= \alpha \log |x| +\beta x^2+x$ then

$\begin{array}{1 1}(A)\;\alpha =-6, \beta=\large\frac{1}{2} \\ (B)\;\alpha =-6, \beta=\large\frac{-1}{2} \\(C)\;\alpha =2, \beta=\large\frac{-1}{2} \\(D)\;\alpha =2, \beta=\large\frac{1}{2} \end{array} $

1 Answer

Given : $\begin{align*}x& = -1 \\ x &= 2 \end{align*}$
$f(x) = \alpha\; log\; |x| + \beta x^2 + x$
$f'(x) = \alpha . \frac{1}{x} + 2 \beta x + 1 $
for extreme values $f'(x) = 0$
when $x = -1$
$0 = -\alpha -2 \beta + 1 $
$\alpha + 2 \beta = 1 ----(1)$
when $x = 2$
$0 = \frac{\alpha}{2} + 4 \beta + 1$
$\implies \alpha + 8 \beta = -2 ----(2) $
on solving (1) and (2)
$\beta = \frac{-1}{2}$ and $\alpha = 2$
answered Apr 7, 2014 by meena.p
edited Nov 6 by priyanka.c
 

Related questions

...