Question

If $\int \frac{xdx}{\sqrt{e^{2x}-(x+1{)}^2}}={\cos }^{-1}\left(f\right(x\left)\right)+C,$ where $C$ is arbitrary constant of integration, then

• A. $\int e^{2x}f\left(x\right)dx=e^x+C$
• B. $\int e^{2x}f\left(x\right)dx=x{e}^x+C$
• C. $\underset{x\to 0}{lim}\left(\frac{f\left(x\right)-1}{x^2}\right)=\frac{1}{2}$
• D. $\underset{x\to \infty }{lim}(1+6f(x){)}^{\frac{e^x}{3x}}=e^2$

Answer 1

Answer: (B), (D)

$\underset{x\to \infty }{lim}(1+6f(x){)}^{\frac{e^x}{3x}}=e^2$

$I=\int \frac{xdx}{\sqrt{e^{2x}-(x+1{)}^2}}=\int \frac{xdx}{e^x\sqrt{1-{\left(\frac{x+1}{e^x}\right)}^2}}$
Put $\frac{x+1}{e^x}=t$
$\Rightarrow \frac{-x}{e^x}dx=dt$
$\therefore I=\int \frac{-dt}{\sqrt{1-t^2}}$
$={\cos }^{-1}\left(\frac{x+1}{e^x}\right)+C$
$\therefore f\left(x\right)=\frac{x+1}{e^x}$


$\int e^{2x}f\left(x\right)dx$
$=\int e^{2x}\left(\frac{x+1}{e^x}\right)dx$
$=\int e^x(x+1)dx$
$=x{e}^x+C$


$\underset{x\to 0}{lim}\left(\frac{f\left(x\right)-1}{x^2}\right)$
$=\underset{x\to 0}{lim}\left(\frac{x+1-e^x}{x^2{e}^x}\right)$
$=\underset{x\to 0}{lim}\left(\frac{x+1-(1+x+\frac{x^2}{2!}+\cdots )}{x^2{e}^x}\right)$
$=-\frac{1}{2}$

$\underset{x\to \infty }{lim}{(1+\frac{6(x+1)}{e^x})}^{\frac{e^x}{3x}}$
$=\exp \left\{\underset{x\to \infty }{lim}\frac{e^x}{3x}\left(\frac{6(x+1)}{e^x}\right)\right\}$
$=\exp \left\{\underset{x\to \infty }{lim}\left(\frac{2(x+1)}{x}\right)\right\}=e^2$