Question

Let $f$ be a real-valued function and satisfies $f\left(x\right)+f\left(y\right)=\frac{1}{x}+\frac{1}{y}\forall x,y\in R-\left\{0\right\}.$ If $\underset{2}{\overset{3}{\int }}\left(\frac{3(f\left(x\right){)}^5-f\left(x\right)}{1-(f\left(x\right){)}^4}\right)dx=\frac{1}{2}\ln \left(\frac{2^{\alpha }}{3^{\beta }}\right),$ then

• A. $\alpha \le \beta$
• B. $\alpha +\beta =17$
• C. ${\tan }^{-1}\sqrt{\alpha -\beta }={\sec }^{-1}\left(\frac{\alpha }{5}\right)$
• D. $(\alpha -\beta )$ is not a prime number

Answer 1

Answer: (B), (C)

The correct options are
B $\alpha +\beta =17$
C ${\tan }^{-1}\sqrt{\alpha -\beta }={\sec }^{-1}\left(\frac{\alpha }{5}\right)$

$f\left(x\right)+f\left(y\right)=\frac{1}{x}+\frac{1}{y}\forall x,y\in R-\left\{0\right\}$
Put $x=y,$ we get
$f\left(x\right)=\frac{1}{x}$

Now, $I=\underset{2}{\overset{3}{\int }}\left(\frac{\frac{3}{x^5}-\frac{1}{x}}{1-\frac{1}{x^4}}\right)dx$

On multiplying and dividing by $\frac{1}{x^2},$
$I=\underset{2}{\overset{3}{\int }}\left(\frac{\frac{3}{x^7}-\frac{1}{x^3}}{\frac{1}{x^2}-\frac{1}{x^6}}\right)dx$

Put $\frac{1}{x^2}-\frac{1}{x^6}=t$
$\Rightarrow dt=(\frac{-2}{x^3}+\frac{6}{x^7})dx$

$I=\frac{1}{2}\underset{\frac{15}{64}}{\overset{\frac{80}{729}}{\int }}\frac{dt}{t}$

$=\frac{1}{2}[\ln t{]}_{\frac{15}{64}}^{\frac{80}{729}}$

$=\frac{1}{2}\ln \left(\frac{2^{10}}{3^7}\right)$
On comparing with $\frac{1}{2}\ln \left(\frac{2^{\alpha }}{3^{\beta }}\right),$
$\alpha =10,\beta =7$