Question

The coefficient of the fourth term in Taylor series of $x^4+x^2-2$ centered at $a=1$.

• A. $4$
• B. $1$
• C. $3$
• D. $6$

Answer 1

The correct option is B $1$

The Taylor series is given by $f\left(x\right)={\sum }_{k=0}^n\frac{f^{\left(k\right)}\left(a\right)}{k!}(x-a{)}^k$
We have $f\left(x\right)=x^4+x^2-2,a=1$
Since we have to find the coefficient of the fourth term, let us take $n=5$.
$\therefore f\left(x\right)\approx {\sum }_{k=0}^5\frac{f^{\left(k\right)}\left(1\right)}{k!}(x-1{)}^k$
$f^{\left(0\right)}\left(x\right)=x^4+x^2-2,\Rightarrow f^{\left(0\right)}\left(1\right)=1+1-2=0$
$f^{\left(1\right)}\left(x\right)=4x^3+2x,\Rightarrow f^{\left(1\right)}\left(1\right)=4+2=6$
$f^{\left(2\right)}\left(x\right)=12x^2+2,\Rightarrow f^{\left(2\right)}\left(1\right)=12+2=14$
$f^{\left(3\right)}\left(x\right)=24x,\Rightarrow f^{\left(3\right)}\left(1\right)=24$
$f^{\left(4\right)}\left(x\right)=24,\Rightarrow f^{\left(4\right)}\left(1\right)=24$
$f^{\left(5\right)}\left(x\right)=0,\Rightarrow f^{\left(5\right)}\left(1\right)=0$
$\therefore f\left(x\right)\approx 0+\frac{6}{1!}(x-1)+\frac{14}{2!}(x-1{)}^2+\frac{24}{3!}(x-1{)}^3+\frac{24}{4!}(x-1{)}^4+0$
$\Rightarrow f\left(x\right)\approx 6(x-1)+7(x-1{)}^2+4(x-1{)}^3+(x-1{)}^4$
Thus the coefficient of fourth term is $1$.