If frac{3x+5}{(x+1)(2x^2+3)} = frac{A}{x+1} + | vedclass

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

Question

(C) Given $\frac{3x+5}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}$.Multiplying both sides by $(x+1)(2x^2+3)$, we get $3x+5 = A(2x^2+3) + (Bx

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given $\frac{3x+5}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}$.Multiplying both sides by $(x+1)(2x^2+3)$, we get $3x+5 = A(2x^2+3) + (Bx+C)(x+1)$.Setting $x = -1$: $3(-1)+5 = A(2(-1)^2+3) + 0 \implies 2 = 5A \implies A = \frac{2}{5}$.Comparing coefficients of $x^2$: $0 = 2A + B \implies B = -2A = -2(\frac{2}{5}) = -\frac{4}{5}$.Comparing constant terms: $5 = 3A + C \implies C = 5 - 3(\frac{2}{5}) = 5 - \frac{6}{5} = \frac{19}{5}$.Thus, $f(x) = \frac{2}{5}x^3 - \frac{4}{5}x^2 + 7x + \frac{19}{5}$.$f'(x) = \frac{6}{5}x^2 - \frac{8}{5}x + 7$.$f'(-2) = \frac{6}{5}(4) - \frac{8}{5}(-2) + 7 = \frac{24}{5} + \frac{16}{5} + 7 = \frac{40}{5} + 7 = 8 + 7 = 15$.Finally, $5C - f'(-2) = 5(\frac{19}{5}) - 15 = 19 - 15 = 4$.

If $\frac{3x+5}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}$ and $f(x) = Ax^3 + Bx^2 + 7x + C$, then $5C - f'(-2) = $

Similar Questions

If $\frac{1}{(3-5 x)(2+3 x)}=\frac{A}{3-5 x}+\frac{B}{2+3 x}$,then $A+B$ is equal to

If $\frac{2x^2-3x+5}{(x-7)^3}=\frac{A}{x-7}+\frac{B}{(x-7)^2}+\frac{C}{(x-7)^3}$, then $2A-3B+C=$

$\frac{1}{x(x+1)(x+2) \ldots(x+n)} = \sum_{r=0}^{n} \frac{A_r}{x+r}$. Then $A_r$ is equal to:

If $\frac{3x+2}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}$,then $A-B+C=$

If $\frac{x^2}{(x^2 + a^2)(x^2 + b^2)} = k \left( \frac{a^2}{x^2 + a^2} - \frac{b^2}{x^2 + b^2} \right)$,then $k =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module