Question

If $\int \frac{\left(\frac{1}{x}+\frac{1}{x^2}\right)(x-1)}{\left(\frac{1}{x^4}+\frac{1}{x^2}\right)\sqrt{\left(x^4-x^3+x^2\right)\left(x^4+x^3+x^2\right)}}dx={\sec }^{-1}(f(x))+C$ then

• A. Maximum value of $f(x)$ is $-2$ when $x<0$
• B. Minimum value of $f(x)$ is 2 when $x>0$
• C. $f(0)$ is not defined
• D. $f(e)<f(\pi )$

Answer 1

Answer: (A), (B), (C), (D)

$\therefore \int \frac{\frac{x^2-1}{x^2}}{\left(\frac{1}{x^4}+\frac{1}{x^2}\right)\sqrt{x^8+x^6+x^4}}dx={\sec }^{-1}f(x)+c$
$\Rightarrow \int \frac{\left(1-\frac{1}{x^2}\right)}{\frac{1}{x^3}\left(x+\frac{1}{x}\right)x^3\sqrt{x^2+1+\frac{1}{x^2}}}dx={\sec }^{-1}(f(x))+C$
$\Rightarrow \int \frac{\left(1-\frac{1}{x^2}\right)dx}{\left(x+\frac{1}{x}\right)\sqrt{{\left(x+\frac{1}{x}\right)}^2-1}}={\sec }^{-1}f(x)+C$
$\Rightarrow {\sec }^{-1}\left(x+\frac{1}{x}\right)+C={\sec }^{-1}(f(x))+C$
$\Rightarrow {\sec }^{-1}\left(x+\frac{1}{x}\right)+C={\sec }^{-1}(f(x))+C$
$\Rightarrow f(x)=\frac{(x+1)(x-1)}{x^2}=0\Rightarrow x=\pm 1$
Maximum occur at $x=-1$ and minimum occur at $x=1$