Let y = frac{x^{2}}{(x+1)^{2}(x+2)}. Then fra | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $y = \frac{x^{2}}{(x+1)^{2}(x+2)}$.Using partial fractions, we write:$\frac{x^{2}}{(x+1)^{2}(x+2)} = \frac{A}{x+2} + \frac{B}{x+1} + \frac{C

Vedclass · Accepted Answer

$2\left[\frac{3}{(x+1)^{4}}-\frac{3}{(x+1)^{3}}+\frac{4}{(x+2)^{3}}\right]$

Vedclass · Answer

(A) Given $y = \frac{x^{2}}{(x+1)^{2}(x+2)}$.Using partial fractions, we write:$\frac{x^{2}}{(x+1)^{2}(x+2)} = \frac{A}{x+2} + \frac{B}{x+1} + \frac{C}{(x+1)^{2}}$.Solving for coefficients, we get $A=4, B=-3, C=1$.So, $y = 4(x+2)^{-1} - 3(x+1)^{-1} + (x+1)^{-2}$.Differentiating with respect to $x$ once:$y' = -4(x+2)^{-2} + 3(x+1)^{-2} - 2(x+1)^{-3}$.Differentiating again:$y'' = 8(x+2)^{-3} - 6(x+1)^{-3} + 6(x+1)^{-4}$.$y'' = 2\left[\frac{4}{(x+2)^{3}} - \frac{3}{(x+1)^{3}} + \frac{3}{(x+1)^{4}}\right]$.This matches option $A$.

Let $y = \frac{x^{2}}{(x+1)^{2}(x+2)}$. Then $\frac{d^{2} y}{dx^{2}}$ is

Similar Questions

For $y=\sin ^{-1}\left\{\frac{5 x+12 \sqrt{1-x^{2}}}{13}\right\} ;|x| \leq 1$, if $a\left(1-x^{2}\right) y_{2}+b x y_{1}=0$ then $(a, b)=$

If $\cos ^{-1}\left(\frac{y}{b}\right)=2 \log \left(\frac{x}{2}\right)$,where $x>0$,then $x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}$ is equal to

$y=e^{a \sin ^{-1} x} \Rightarrow (1-x^2) y_{n+2}-(2 n+1) x y_{n+1}$ is equal to

If $f(x)=b \cdot e^{a x}+a \cdot e^{b x}$,then $f^{\prime \prime}(0)=$

If $y=5 \cos x-3 \sin x,$ prove that $\frac{d^{2} y}{d x^{2}}+y=0.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module