Question

The function $y=a\log \left|\begin{array}{c}x\end{array}\right|+b{x}^2+x$ has its extremum values at $x=-1$ and $x=2$ then

• A. $a=2,b=-1$
• B. $a=2,b=-1/2$
• C. $a=-2,b=1/2$
• D. None of these

Answer 1

Answer: (B)

$a=2,b=-1/2$

$f\left(x\right)=a{\log }_e\mid x\mid +b{x}^2+x$
$f^{\prime }\left(x\right)=a/x+2bx+1$
Given $x=-1,2$ are extremum of $f\left(x\right)$
$\Rightarrow f^{\prime }(-1)=0=f^{\prime }\left(2\right)$
$\Rightarrow -a-2b+1=0.....\left(1\right)$ and $a/2+4b+1=0.....\left(2\right)$
solving (1) and (2) we get, $a=2,b=-\frac{1}{2}$