If frac{2 x^3+x^2-5}{x^4-25}=frac{A x+B}{x^2- | vedclass

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

Question

(C) Given,$\frac{2 x^3+x^2-5}{x^4-25}=\frac{A x+B}{x^2-5}+\frac{C x+1}{x^2+5}$ Since $x^4-25 = (x^2-5)(x^2+5)$,we have: $2 x^3+x^2-5 = (A x+B)(x^2+5)

Vedclass · Accepted Answer

$(1, 0, 1)$

Vedclass · Answer

(C) Given,$\frac{2 x^3+x^2-5}{x^4-25}=\frac{A x+B}{x^2-5}+\frac{C x+1}{x^2+5}$ Since $x^4-25 = (x^2-5)(x^2+5)$,we have: $2 x^3+x^2-5 = (A x+B)(x^2+5) + (C x+1)(x^2-5)$ $2 x^3+x^2-5 = A x^3 + 5Ax + Bx^2 + 5B + Cx^3 - 5Cx + x^2 - 5$ $2 x^3+x^2-5 = x^3(A+C) + x^2(B+1) + x(5A-5C) + (5B-5)$ Equating the coefficients of $x^3, x^2, x$ and the constant term: $A+C = 2$ $B+1 = 1 \Rightarrow B = 0$ $5A-5C = 0 \Rightarrow A = C$ $5B-5 = -5 \Rightarrow 5(0)-5 = -5$ (which is consistent) Substituting $A=C$ into $A+C=2$,we get $2C=2$,so $C=1$ and $A=1$. Thus,$(A, B, C) = (1, 0, 1)$.

If $\frac{2 x^3+x^2-5}{x^4-25}=\frac{A x+B}{x^2-5}+\frac{C x+1}{x^2+5}$,then $(A, B, C)$ equals to

Similar Questions

If the vertex of the conic $y^{2}-4y=4x-4a$ always lies between the straight lines $x+y=3$ and $2x+2y-1=0$, then

If the angle between the circles $x^2+y^2-12x-6y+41=0$ and $x^2+y^2+kx+6y-59=0$ is $45^{\circ}$,then a value of $k$ is

Which of the following gases has the highest heat of combustion?

$\frac{d}{d x}\left[a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right)\right]=\frac{1}{x^4-1}$
$\Rightarrow a-2 b$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module