How to solve $ \frac{x-1}{x-c} < \frac{1}{2}$?

Question

I have this inequality:

$$\frac{x-1}{x-c} < \frac{1}{2} \quad c \in \mathbb{R}$$

can someone show me how to do it?

EDIT: Your answers are what I did, I studied the two cases and got:

$$x < 2 - c \quad \quad \quad for\quad x > c$$

and

$$x > 2 - c \quad \quad \quad for \quad x< c$$

But my book solution says:

$$[S = {\quad x < c ∨ x > 1 \quad for\quad c ≤ 1;\quad x < 1 ∨ x > c\quad for\quad c > 1}]$$

Answer 1

Updated answer: Here's a simpler way.

Let $$ f(x)=\frac{x-1}{x-c}-\frac12 = \frac{x-(2-c)}{2(x-c)} . $$ We would like to know when $\color{blue}{f(x)<0}$.

Make a table of the signs of the factors $x-(2-c)$ and $2(x-c)$; they change their signs when $x=2-c$ and when $x=c$, and when we write out the table we need to know which one of the numbers $c$ and $2-c$ to put on the left and which one to put on the right. So we have to split into cases. Obviously, $c<2-c$ if and only if $c<1$.

First case, $c<1$: $$ \begin{array}{c|ccccc} x & &c& &2-c& \\[0.3ex] \hline x-(2-c) & - && - &0& + \\ 2(x-c) & - &0& + && + \\ \hline f(x) & + &\text{undef.}& - &0& + \end{array} $$ Looking where the minus sign in the last line is, we find that $f(x)<0 \iff \color{blue}{c < x < 2-c}$ in this case.

Second case, $c>1$: $$ \begin{array}{c|ccccc} x & &2-c& &c& \\[0.3ex] \hline x-(2-c) & - &0& + && + \\ 2(x-c) & - && - &0& + \\ \hline f(x) & + &0& - &\text{undef.}& + \end{array} $$ Hence $f(x)<0 \iff \color{blue}{2-c < x < c}$ in this case.

Third case, $c=1$: Here $f(x)=1/2$ identically, hence we never have $f(x)<0$, so there are no solutions in this case.

Original answer: To simplify, let $y=x-1$ and $d=c-1$: $$ \frac{x-1}{x-c} < \frac{1}{2} \iff \frac{y}{y-d} < \frac{1}{2} \iff 1 + \frac{d}{y-d} < \frac{1}{2} \iff \frac{d}{y-d} < -\frac{1}{2} . $$ Now if $d=0$ (i.e., if $c=1$), this is false for all $y$. So we can assume $d \neq 0$ and write $$ \dots \iff \frac{1}{\frac{y}{d}-1} < -\frac{1}{2} \iff \frac{1}{z-1} < -\frac{1}{2} , $$ where $z = \frac{y}{d}$. Solve this inequality in what ever way you prefer (most easily by just looking at the graph of $1/(z-1)$); you get $-1<z<1$, i.e., $$ \left| \frac{y}{d} \right|<1 \iff \frac{|y|}{|d|}<1 \iff |y|<|d| $$ (using in the last step that $|d|$ is positive).

Now split into cases: if $d>0$ you get $|y|<d$, and if $d<0$ you get $|y|<-d$.

In terms of the original quantities:

• If $c>1$ then $|x-1|<c-1$, which gives $-(c-1)<x-1<c-1$, i.e., $$ 2-c<x<c . $$

• If $c<1$, then $|x-1|<1-c$, which gives $-(1-c)<x-1<1-c$, i.e., $$ c < x < 2-c . $$

• If $c=1$, there are no solutions.

Answer 2

HINT: $2(x-1)<x-c$ for $x-c>0$. Then the other case.

Answer 3

You want to know when $\dfrac{x-1}{x-c} < \dfrac{1}{2} \quad c \in \mathbb{R}$.

If $x=c$, the left side is not defined, so we can assume that $x \ne c$.

If $c=1$, this is always false.

Since $\dfrac{x-1}{x-c} =\dfrac{x-c+c-1}{x-c} =1+\dfrac{c-1}{x-c} $, this is equivalent to $1+\dfrac{c-1}{x-c} \lt \dfrac12 $ or $\dfrac{1-c}{x-c} \gt \dfrac12 $.

If $c <1$, we must have $x > c$, so we can clear the fraction and get $2-2c > x-c$ or $x < 2-c$.

If $c > 1$ we must have $x < c$ so we can rewrite this as $\dfrac{c-1}{c-x} \gt \dfrac12$. Clearing the fraction, we get $2c-2 > c-x$ or $x > 2-c$.