MCQ
The function $y =\frac{{2x\,\, - \,\,1}}{{x\,\, - \,\,2}} (x \ne 2)$
- Ais its own inverse
- Bdecreases for all values of x
- Chas a graph entirely above $x-$ axis
- ✓Both $(A)$ and $(B)$
✓
Answer
Correct option: D.
Both $(A)$ and $(B)$
d
Given $y=\frac{2 x-1}{x-2}$ where $x \neq 2$
Given $y=\frac{2 x-1}{x-2}$ where $x \neq 2$
$y=\frac{2 x-1}{x-2}$
$x y-2 y=2 x-1$
$x(y-2)=2 y-1$
$\therefore x=\frac{2 y-1}{y-2}$ 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.
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