Question

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

$\therefore \int \frac{\frac{x^2-1}{x^2}}{\left(\frac{1}{x^4}+\frac{1}{x^2}\right) \sqrt{x^8+x^6+x^4}} d x=\sec ^{-1} f(x)+c$
$\Rightarrow \int \frac{\left(1-\frac{1}{x^2}\right)}{\frac{1}{x^3}\left(x+\frac{1}{x}\right) x^3 \sqrt{x^2+1+\frac{1}{x^2}}} d x=\sec ^{-1}(f(x))+C$
$ \Rightarrow \int \frac{\left(1-\frac{1}{x^2}\right) d x}{\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 $x=1$