Question

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

Answer 1

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

$\because$ At $x=1$ and $x=3,\frac{\delta f\left(x\right)}{\delta x}=0$
Now we find $\frac{\delta f\left(x\right)}{\delta x}$
$\frac{\delta f\left(x\right)}{\delta x}=\frac{9}{\left|x\right|}+b\left(2x\right)+1$
So at $x=1$ $0=\frac{a}{1}+b\times 2+1$
$\Rightarrow a+2b+1=0⟶\left(1\right)$
At $x=3$
$0=\frac{a}{3}+2\cdot 3\cdot b+1$
$\Rightarrow a+3\cdot 6b+3=00ra+18b+3=0⟶\left(2\right)$
So using $\left(1\right)\&\left(2\right)$
$a+2b+1=a+18b+3$
$1-3=18b-2b$
$b=\frac{-2}{16}or\frac{-1}{8}$
Also putting it in equation $\left(1\right)$ we get
$a+2\left(\frac{-1}{8}\right)+1=0$
$a+\left(\frac{-1}{4}\right)+1=0$
$\Rightarrow a+\frac{3}{4}=0$
$a=\frac{-3}{4}$