X-x-ax-b-1-k-a-x-a-b-x-b- k-

The correct option is C a b x(x+a)(x+b)=LHS RHS=1k[(x+b)a (x+a)b(x+a)(x+b)] LHS=RHS k #160; x=(x+b)a (x+a)b since it is equality if ...

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Theorems for Differentiability

x/x+ax+b=1/k[...

Question

$\frac{x}{(x + a) (x + b)} = \frac{1}{k} [\frac{a}{x + a} - \frac{b}{x + b}] \Rightarrow k =$

a - b

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\frac{x^{3}}{(x - a) (x - b) (x - c)} = k + \frac{A}{x - a} + \frac{B}{x - b} + \frac{C}{x - c} \Rightarrow k =

\frac{x - a}{b c} + \frac{x - b}{c a} + \frac{x - c}{a b}

(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})

\frac{(x - a) (x - b)}{x - a - b} = \frac{(x - c) (x - d)}{x - c - d}

a x + \frac{y}{b} = a^{2} - b^{2}

\frac{x}{a} - b y = 1 + b^{4}

k = b^{2} x + \frac{a}{b} y

f (x) = \{\begin{cases} \frac{\log (1 + a x) - \log (1 - b x)}{x}, & x \neq 0 \\ k, & x = 0 \end{cases}

\frac{(x - a) (x - b)}{(x - c) (x - d)} = K + \frac{A}{x - c} + \frac{B}{x - d}

k

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