Q59 / 160 · AP EAMCET 2024 (18 May Shift 1) · Mathematics
The absolute value of the difference of the coefficients of $x^4$ and $x^6$ in the expansion of $\frac{2 x^2}{\left(x^2+1\right)\left(x^2+2\right)}$ is
- Correct answerA$\frac{13}{4}$
- B$\frac{1}{4}$
- C$\frac{9}{4}$
- D$1$
Solution
Since $\frac{2 x^2}{\left(x^2+1\right)\left(x^2+2\right)}=\frac{-2}{x^2+1}+\frac{4}{x^2+2}$
$=-2\left(1+x^2\right)^{-1}+\frac{4}{2}\left(1+\frac{x^2}{2}\right)^{-1}$
$=-2\left(1+x^2\right)^{-1}+2\left(1+\frac{x^2}{2}\right)^{-1}$
$=-2\left(1-x^2+x^4-x^6+\ldots.\right)+2\left(1-\frac{x^2}{2}+\frac{x^4}{4}-\frac{x^6}{8}+\ldots.\right)$
So, co-efficient of $x^4=-2+\frac{1}{2}=\frac{-3}{2}$
and co-efficient of $x^6=2-\frac{1}{4}=\frac{7}{4}$
Now, required value $=\frac{7}{4}+\frac{3}{2}=\frac{13}{4}$
Hence, the correct answer is option: 'A'.