Question

Let \(y=\frac{x^{2}}{(x+1)^{2}(x+2)}\). Then \(\frac{d^{2} y}{d x^{2}}\) is
(A) \(2\left[\frac{3}{(x+1)^{4}}-\frac{3}{(x+1)^{3}}+\frac{4}{(x+2)^{3}}\right]\)
(B) \(3\left[\frac{2}{(x+1)^{3}}+\frac{4}{(x+1)^{2}}-\frac{5}{(x+2)^{3}}\right]\)
(C) \(\frac{6}{(x+1)^{3}}-\frac{4}{(x+1)^{2}}+\frac{3}{(x+1)^{3}}\)
(D) \(\frac{7}{(x+1)^{3}}-\frac{3}{(x+1)^{2}}+\frac{2}{(x+1)^{3}}\)

Answer 1

Ans: (A)
Hint : By partial fraction technique
$$
\begin{aligned}
&y=\frac{x^{2}}{(x+1)^{2}(x+2)}=\frac{4}{(x+2)}-\frac{3}{(x+1)}+\frac{1}{(x+1)^{2}} \\
&\Rightarrow y^{\prime \prime}=\frac{6}{(x+1)^{4}}-\frac{6}{(x+1)^{3}}+\frac{8}{(x+2)^{3}}
\end{aligned}
$$