Question

If $f\left(x\right)=a{\log }_e\left|x\right|+b{x}^2+x$ has extremum 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

The correct option is A $a=-\frac{3}{4},b=-\frac{1}{8}$

Around $x=1,3$ we have $\left|x\right|=x$

$\therefore f\left(x\right)=a{\log }_e\left|x\right|+b{x}^2+x$

$\Rightarrow f^{\prime }\left(x\right)=\frac{a}{x}+2bx+1$

From the question, $f^{\prime }\left(1\right)=0,f^{\prime }\left(3\right)=0$

$\therefore a+2b+1=0,\frac{a}{3}+6b+1=0$

Solving we get $a=-\frac{3}{4}$ and $b=-\frac{1}{8}$