Determine for what value(s) of m that the equation $ |2x-1| = mx+1$ has exactly one solution only.

Question

Determine for what value(s) of m that the equation $|2x-1| = mx+1$ has exactly one solution only.

I don't really understand the steps I should undertake in solving this problem!

Answer 1

CASE $\bf1$:$\ \ x>1/2$
$2x-1=mx+1$
$x=2/(2-m)$
So, $x=2/(2-m)$ AND $x>1/2$
$\Rightarrow 2/(2-m)>1/2$
$\Rightarrow m>-2, m\neq2$
$\Rightarrow m\in(-2,\infty)\sim2$

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CASE $\bf2$: $\ \ x\leq1/2$
$1-2x=mx+1$
$(m+2)x=0$
$\Rightarrow m\neq-2,x=0 \cap x\leq1/2\qquad$ OR $\qquad m=-2,x\in R \cap x\leq1/2$
$\Rightarrow m\neq-2,x=0$ OR $m=-2,x\leq1/2$
Since we are interested in only $1$ solution so, $m\neq-2,x=0$

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Now, our final answer is CASE $1$ $\cup$ CASE $2$
Since we are interested in only $1$ solution so we need $m$ s.t. it satisfies CASE $1$ or CASE $2$ but not both.
This can be done by excluding all the values of $m$ s.t. CASE $1$ is satisfied.
So, $(m\neq-2)-(m\in(-2,\infty)\sim2)$
$\Rightarrow m\in(-\infty,-2)$

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Thus, for $m\in(-\infty,-2)$, we have only $1$ solution, $x=0$.