STD 12 - 6. Application of derivatives — Mathematics STD 12 Science — Question

$y=\frac{2 x-1}{x-2}$

$x y-2 y=2 x-1$

$x(y-2)=2 y-1$

$\therefore x=\frac{2 y-1}{y-2}$ and $y \neq 2$

Hence the function is its own inverse.

Differentiating $y$ with respect to $x$

$y^{\prime}=\frac{(x-2)(2)-(2 x-1)}{(x-2)^{2}}=\frac{-3}{(x-2)^{2}}$

$(x-2)^{2}$ is positive for all values of $x .(x \neq 2$ given. $)$

hence $y^{\prime}$ is negative. Therefore the function $y$ is decreasing at all values of $x$

For the graph to be entirely above $x$ axis, $y>0$ $\frac{2 x-1}{x-2}>0 \Longrightarrow x>\frac{1}{2}$ and $x>2$ or $x<\frac{1}{2}$ and $x<2$

For $(2, \infty) \cup\left(-\infty, \frac{1}{2}\right)$ graph is above $\times$ axis.

and for $\left(\frac{1}{2}, 2\right)$ graph is below $x$ axis.

At $x=\frac{1}{2}, y=0$

At $x=2, y=\infty$ Hence unbounded.