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

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(C) આપેલ છે $\frac{3x+5}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}$.બંને બાજુ $(x+1)(2x^2+3)$ વડે ગુણતા, $3x+5 = A(2x^2+3) + (Bx+C)(x+1)$ મ

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$4$

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(C) આપેલ છે $\frac{3x+5}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}$.બંને બાજુ $(x+1)(2x^2+3)$ વડે ગુણતા, $3x+5 = A(2x^2+3) + (Bx+C)(x+1)$ મળે.$x = -1$ લેતા: $3(-1)+5 = A(2(-1)^2+3) + 0 \implies 2 = 5A \implies A = \frac{2}{5}$.$x^2$ ના સહગુણકોની સરખામણી કરતા: $0 = 2A + B \implies B = -2A = -2(\frac{2}{5}) = -\frac{4}{5}$.અચળ પદોની સરખામણી કરતા: $5 = 3A + C \implies C = 5 - 3(\frac{2}{5}) = 5 - \frac{6}{5} = \frac{19}{5}$.આમ, $f(x) = \frac{2}{5}x^3 - \frac{4}{5}x^2 + 7x + \frac{19}{5}$.$f'(x) = \frac{6}{5}x^2 - \frac{8}{5}x + 7$.$f'(-2) = \frac{6}{5}(4) - \frac{8}{5}(-2) + 7 = \frac{24}{5} + \frac{16}{5} + 7 = \frac{40}{5} + 7 = 8 + 7 = 15$.અંતે, $5C - f'(-2) = 5(\frac{19}{5}) - 15 = 19 - 15 = 4$.

જો $\frac{3x+5}{(x+1)(2x^2+3)} = \frac{A}{x+1} + \frac{Bx+C}{2x^2+3}$ અને $f(x) = Ax^3 + Bx^2 + 7x + C$ હોય, તો $5C - f'(-2) = $

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