Question

If ${\log }_e(1+x+x^2+x^3)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\dots$, then values of $a_1$ and $a_2$ are

• A. $a_1=-1,a_2=\frac{1}{2}$
• B. $a_1=1,a_2=\frac{1}{2}$
• C. $a_0=1,a_1=-1$
• D. $a_0=1,a_1=1$

Answer 1

The correct option is B $a_1=1,a_2=\frac{1}{2}$

Given, $\log (1+x+x^2+x^3)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\dots$
$\Rightarrow \log \left[\right(1+x^2\left)\right(1+x\left)\right]=\log (1+x)+\log (1+x^2)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+..........\left(1\right)$
Putting $x=0$ we get $a_0=0$
Differentiating equation (1) both w.r.t $x$
$\frac{1}{1+x}+\frac{2x}{1+x^2}=a_1+2a_{2}x+3a_3{x}^2+..........\left(2\right)$
Putting $x=0$ in (2) we get $a_1=1$
Now differentiating (2)
$-\frac{1}{(1+x{)}^2}+\frac{2(1+x^2-x(2x\left)\right)}{(1+x^2{)}^2}=2a_2+6a_{3}x+..........\left(3\right)$
Putting $x=0$ in (3) we get $a_2=\frac{1}{2}$