Question

If $f\left(x\right)=a{\log }_e\left|x\right|+b{x}^2+x$ has extremums at $x=1$ and $x=3$ then

• A. $a=-\frac{3}{4},b=-\frac{1}{8}$
• B. $a=\frac{3}{4},b=-\frac{1}{8}$
• C. $a=-\frac{3}{4},b=\frac{1}{8}$
• D. none of these

Answer 1

Answer: (A)

$a=-\frac{3}{4},b=-\frac{1}{8}$

$f\left(x\right)=a{\log }_e\mid x\mid +b{x}^2+x$
$f^{\prime }\left(x\right)=a/x+2bx+1$
Given $x=1,3$ are extremum of $f\left(x\right)$
$\Rightarrow f^{\prime }\left(1\right)=0=f^{\prime }\left(3\right)$
$\Rightarrow a+2b+1=0.....\left(1\right)$ and $a/3+6b+1=0.....\left(2\right)$
solving (1) and (2) we get, $a=-\frac{3}{4},b=-\frac{1}{8}$