Question

A differentiable function $f$ satisfies the relation $f\left(\frac{x+1}{x-1}\right)=2f\left(x\right)+\frac{3}{x-1}\forall x\in R-\left\{1\right\}.$ If
$\left(i\right)$ the range of the function excludes the interval $(a_1,a_2)$ on the real number line,
$\left(ii\right)$ relative maximum and minimum values of $f\left(x\right)$ occur at $x=b_1$ and $x=b_2,$
then the value of $a_1^2+a_2^2+b_1^2+b_2^2$ is

Answer 1

$f\left(\frac{x+1}{x-1}\right)=2f\left(x\right)+\frac{3}{x-1}\cdots \left(1\right)$
Let $t=\frac{x+1}{x-1}$
Then, $x=\frac{t+1}{t-1}$ and $x-1=\frac{2}{t-1}$$\therefore f\left(t\right)=2f\left(\frac{t+1}{t-1}\right)+\frac{3(t-1)}{2}$
$\Rightarrow f\left(x\right)=2f\left(\frac{x+1}{x-1}\right)+\frac{3(x-1)}{2}$
$\Rightarrow f\left(\frac{x+1}{x-1}\right)=\frac{1}{2}f\left(x\right)-\frac{3(x-1)}{4}\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),$ we have
$2f\left(x\right)+\frac{3}{x-1}=\frac{1}{2}f\left(x\right)-\frac{3(x-1)}{4}$
$\Rightarrow f\left(x\right)=-\frac{2}{x-1}-\frac{x-1}{2}$
$\Rightarrow f\left(x\right)=-(\frac{2}{x-1}+\frac{x-1}{2})$

For maximum and minimum, $f^{\prime }\left(x\right)=0$
$f^{\prime }\left(x\right)=\frac{2}{(x-1{)}^2}-\frac{1}{2}=0$
$\Rightarrow x=1\pm 2$
$\Rightarrow x=-1,3$
and $f^{\prime \prime }\left(x\right)=-\frac{4}{(x-1{)}^3}$
For $x=-1,f^{\prime \prime }\left(x\right)=\frac{1}{2}>0$
$\therefore x=-1=b_2$ is point of local minimum.
$f(-1)=2$
For $x=3,f^{\prime \prime }\left(x\right)=-\frac{1}{2}<0$
$\therefore x=3=b_1$ is point of local maximum.
$f\left(3\right)=-2$

Now,
$f\left(x\right)\to -\infty$ as $x\to \infty$
$f\left(x\right)\to \infty$ as $x\to -\infty$
$f\left(x\right)\to \infty$ as $x\to 1^{-}$
$f\left(x\right)\to -\infty$ as $x\to 1^{+}$
Hence, range does not include the interval $(–2,2).$
$\Rightarrow a_1=-2$ and $a_2=2$
$\therefore a_1^2+a_2^2+b_1^2+b_2^2=4+4+9+1=18$