Question

If $f\left(\frac{3x-4}{3x-4}\right)=x+2,x\ne -\frac{4}{3}$ and $\int f(x)dx=A\log |1-x|+Bx+C$, then the ordered pair $(A,B)$ is equal to : (where $C$ is a constant of integration)

• A. $\left(\frac{8}{3},\frac{2}{3}\right)$
• B. $\left(-\frac{8}{3},\frac{2}{3}\right)$
• C. $\left(-\frac{8}{3},-\frac{2}{3}\right)$
• D. $\left(\frac{8}{3},-\frac{2}{3}\right)$

Answer 1

Answer: (B)

$f\left(\frac{3x-4}{3x-4}\right)=x+2,x\ne =\frac{4}{3}$
Let $\frac{3x-4}{3x+4}=t$
$3x-4=3tx+4t$
$x=\frac{4t+4}{3-3t}+2$
$f(t)=\frac{10-2t}{3-3t}$
$f(t)=\frac{2x-10}{3x-3}$
$\int f(x)d=\int \frac{2x-10}{3x-3}dx$
$=\int \frac{2x}{3x-3}dx-10\int \frac{dx}{3x-3}$
$=\frac{2}{3}\int \frac{x-1}{x-1}dx+\frac{2}{3}\int \frac{dx}{x-1}-\frac{10}{3}\int \frac{dx}{x-1}$
$=\frac{2x}{3}-\frac{8}{3}\mathrm{In}(x-1)+C$