Solveeit Logo

Question

Question: How do you solve \(\left| {\dfrac{5}{{2x - 1}}} \right| \geqslant \left| {\dfrac{1}{{x - 2}}} \right...

How do you solve 52x11x2?\left| {\dfrac{5}{{2x - 1}}} \right| \geqslant \left| {\dfrac{1}{{x - 2}}} \right|?

Explanation

Solution

The given inequality is in modulus function, it can be solved by taking different cases for domain of the given modulus functions so that we can open it with positive or negative sign accordingly, then solve for all the cases and finally take the union of all results.
Modulus function can be understood as following:
\left| x \right| = \left\\{ {\begin{array}{*{20}{c}} x&{{\text{if}}\;x > 0} \\\ { - x}&{{\text{if}}\,x < 0} \end{array}} \right\\}

Complete step by step answer:
To solve the inequation 52x11x2\left| {\dfrac{5}{{2x - 1}}} \right| \geqslant \left| {\dfrac{1}{{x - 2}}} \right| we have to take different cases for domain of the function
To get the cases we will compare variables separately with zero,
2x1=0  and  x2=0 x=12  and  x=2  \Rightarrow 2x - 1 = 0\;{\text{and}}\;x - 2 = 0 \\\ \Rightarrow x = \dfrac{1}{2}\;{\text{and}}\;x = 2 \\\
x=12  and  x=2x = \dfrac{1}{2}\;{\text{and}}\;x = 2 Can’t be in domain because inequality is undefined for them, i.e. 10\dfrac{1}{0}
There are three cases forming
Case I:
When x(,  12)x \in \left( { - \infty ,\;\dfrac{1}{2}} \right), where
2x1=(2x1)=2x+1=12x x2=(x2)=x+2=2x  \left| {2x - 1} \right| = - (2x - 1) = - 2x + 1 = 1 - 2x \\\ \left| {x - 2} \right| = - (x - 2) = - x + 2 = 2 - x \\\
\therefore inequality will be written as
52x11x2 512x12x 512x12x 512x12x0 5(2x)1(12x)(12x)(2x)0 105x1+2x(12x)(2x)0 93x(12x)(2x)0 3(3x)(12x)(2x)0 (3x)(12x)(2x)0  \Rightarrow \left| {\dfrac{5}{{2x - 1}}} \right| \geqslant \left| {\dfrac{1}{{x - 2}}} \right| \\\ \Rightarrow \dfrac{5}{{1 - 2x}} \geqslant \dfrac{1}{{2 - x}} \\\ \Rightarrow \dfrac{5}{{1 - 2x}} \geqslant \dfrac{1}{{2 - x}} \\\ \Rightarrow \dfrac{5}{{1 - 2x}} - \dfrac{1}{{2 - x}} \geqslant 0 \\\ \Rightarrow \dfrac{{5(2 - x) - 1(1 - 2x)}}{{(1 - 2x)(2 - x)}} \geqslant 0 \\\ \Rightarrow \dfrac{{10 - 5x - 1 + 2x}}{{(1 - 2x)(2 - x)}} \geqslant 0 \\\ \Rightarrow \dfrac{{9 - 3x}}{{(1 - 2x)(2 - x)}} \geqslant 0 \\\ \Rightarrow \dfrac{{3(3 - x)}}{{(1 - 2x)(2 - x)}} \geqslant 0 \\\ \Rightarrow \dfrac{{(3 - x)}}{{(1 - 2x)(2 - x)}} \geqslant 0 \\\
Now, we will compare variable terms with zero to find values for xx, then plot them on number line to check which interval is satisfying the condition.
3x=0,  12x=0  and  2x=0 x=3,  x=12  and  x=2  \Rightarrow 3 - x = 0,\;1 - 2x = 0\;{\text{and}}\;2 - x = 0 \\\ \Rightarrow x = 3,\;x = \dfrac{1}{2}\;{\text{and}}\;x = 2 \\\
Plotting them on a number line,

Take value of xx from any interval and check which sign it is giving (+)  or  ()( + )\;{\text{or}}\;( - ), then put the sign on that interval and alternative sign on its alternative intervals,
Checking for x=0  which  (,  12)x = 0\;which\; \in \left( { - \infty ,\;\dfrac{1}{2}} \right)
(3x)(12x)(2x)0 (30)(12×0)(20)0 (3)(1)(2)0 320  \Rightarrow \dfrac{{(3 - x)}}{{(1 - 2x)(2 - x)}} \geqslant 0 \\\ \Rightarrow \dfrac{{(3 - 0)}}{{(1 - 2 \times 0)(2 - 0)}} \geqslant 0 \\\ \Rightarrow \dfrac{{(3)}}{{(1)(2)}} \geqslant 0 \\\ \Rightarrow \dfrac{3}{2} \geqslant 0 \\\
\therefore putting the (+)  and  ()( + )\;{\text{and}}\;( - ) signs alternatively

x(,  12)(2,  3](,  12)\Rightarrow x \in \left( { - \infty ,\;\dfrac{1}{2}} \right) \cup (2,\;3] \cap \left( { - \infty ,\;\dfrac{1}{2}} \right)
\because case is x(,  12)x \in \left( { - \infty ,\;\dfrac{1}{2}} \right)
x(,  12)\therefore x \in \left( { - \infty ,\;\dfrac{1}{2}} \right) is the solution for this case

Case II:
x(12,  2),x \in \left( {\dfrac{1}{2},\;2} \right), then
2x1=(2x1)=2x1 x2=(x2)=x+2=2x  \left| {2x - 1} \right| = (2x - 1) = 2x - 1 \\\ \left| {x - 2} \right| = - (x - 2) = - x + 2 = 2 - x \\\
\therefore inequality will be written as
52x11x2 52x112x 52x112x 52x1+1x20 5(x2)+1(2x1)(2x1)(x2)0 5x10+2x1(2x1)(x2)0 (7x11)(2x1)(x2)0  \Rightarrow \left| {\dfrac{5}{{2x - 1}}} \right| \geqslant \left| {\dfrac{1}{{x - 2}}} \right| \\\ \Rightarrow \dfrac{5}{{2x - 1}} \geqslant \dfrac{1}{{2 - x}} \\\ \Rightarrow \dfrac{5}{{2x - 1}} \geqslant \dfrac{1}{{2 - x}} \\\ \Rightarrow \dfrac{5}{{2x - 1}} + \dfrac{1}{{x - 2}} \geqslant 0 \\\ \Rightarrow \dfrac{{5(x - 2) + 1(2x - 1)}}{{(2x - 1)(x - 2)}} \geqslant 0 \\\ \Rightarrow \dfrac{{5x - 10 + 2x - 1}}{{(2x - 1)(x - 2)}} \geqslant 0 \\\ \Rightarrow \dfrac{{(7x - 11)}}{{(2x - 1)(x - 2)}} \geqslant 0 \\\
Again doing similar steps as case I,
7x11=0,  2x1=0  and  x2=0 x=117,  x=12  and  x=2  \Rightarrow 7x - 11 = 0,\;2x - 1 = 0\;{\text{and}}\;x - 2 = 0 \\\ \Rightarrow x = \dfrac{{11}}{7},\;x = \dfrac{1}{2}\;{\text{and}}\;x = 2 \\\

Checking for x=0  which  (,  12)x = 0\;which\; \in \left( { - \infty ,\;\dfrac{1}{2}} \right)

(7x11)(2x1)(x2)0 (7×011)(2×01)(02)0 (11)(1)(2)0 1120  \Rightarrow \dfrac{{(7x - 11)}}{{(2x - 1)(x - 2)}} \geqslant 0 \\\ \Rightarrow \dfrac{{(7 \times 0 - 11)}}{{(2 \times 0 - 1)(0 - 2)}} \geqslant 0 \\\ \Rightarrow \dfrac{{( - 11)}}{{( - 1)( - 2)}} \geqslant 0 \\\ \Rightarrow \dfrac{{ - 11}}{2} \geqslant 0 \\\

This is not true, so putting the signs accordingly

x(12,  117](2,  )(12,  2)\Rightarrow x \in \left( {\dfrac{1}{2},\;\dfrac{{11}}{7}} \right] \cup (2,\;\infty ) \cap \left( {\dfrac{1}{2},\;2} \right)
\because case is x(12,  2)x \in \left( {\dfrac{1}{2},\;2} \right)
x(12,  117]\therefore x \in \left( {\dfrac{1}{2},\;\dfrac{{11}}{7}} \right] is the solution for this case
Case III:
x(2,  ),x \in \left( {2,\;\infty } \right), then
2x1=(2x1)=2x1 x2=(x2)=x2  \left| {2x - 1} \right| = (2x - 1) = 2x - 1 \\\ \left| {x - 2} \right| = (x - 2) = x - 2 \\\
\therefore inequality will be written as
52x11x2 52x11x2 52x11x2 52x11x20 5(x2)1(2x1)(2x1)(x2)0 5x102x+1(2x1)(x2)0 (3x9)(2x1)(x2)0 3(x3)(2x1)(x2)0 (x3)(2x1)(x2)0  \Rightarrow \left| {\dfrac{5}{{2x - 1}}} \right| \geqslant \left| {\dfrac{1}{{x - 2}}} \right| \\\ \Rightarrow \dfrac{5}{{2x - 1}} \geqslant \dfrac{1}{{x - 2}} \\\ \Rightarrow \dfrac{5}{{2x - 1}} \geqslant \dfrac{1}{{x - 2}} \\\ \Rightarrow \dfrac{5}{{2x - 1}} - \dfrac{1}{{x - 2}} \geqslant 0 \\\ \Rightarrow \dfrac{{5(x - 2) - 1(2x - 1)}}{{(2x - 1)(x - 2)}} \geqslant 0 \\\ \Rightarrow \dfrac{{5x - 10 - 2x + 1}}{{(2x - 1)(x - 2)}} \geqslant 0 \\\ \Rightarrow \dfrac{{(3x - 9)}}{{(2x - 1)(x - 2)}} \geqslant 0 \\\ \Rightarrow \dfrac{{3(x - 3)}}{{(2x - 1)(x - 2)}} \geqslant 0 \\\ \Rightarrow \dfrac{{(x - 3)}}{{(2x - 1)(x - 2)}} \geqslant 0 \\\
Doing similar steps,
x3=0,  2x1=0  and  x2=0 x=3,  x=12  and  x=2  \Rightarrow x - 3 = 0,\;2x - 1 = 0\;{\text{and}}\;x - 2 = 0 \\\ \Rightarrow x = 3,\;x = \dfrac{1}{2}\;{\text{and}}\;x = 2 \\\

Checking for x=0  which  (,  12)x = 0\;which\; \in \left( { - \infty ,\;\dfrac{1}{2}} \right)

(x3)(2x1)(x2)0 (03)(2×01)(02)0 (3)(1)(2)0 320  \Rightarrow \dfrac{{(x - 3)}}{{(2x - 1)(x - 2)}} \geqslant 0 \\\ \Rightarrow \dfrac{{(0 - 3)}}{{(2 \times 0 - 1)(0 - 2)}} \geqslant 0 \\\ \Rightarrow \dfrac{{( - 3)}}{{( - 1)( - 2)}} \geqslant 0 \\\ \Rightarrow \dfrac{{ - 3}}{2} \geqslant 0 \\\

This is not true, so putting the signs accordingly

x(12,  2)[3,  )(2,  )\Rightarrow x \in \left( {\dfrac{1}{2},\;2} \right) \cup [3,\;\infty ) \cap \left( {2,\;\infty } \right)
\because case is x(2,  )x \in \left( {2,\;\infty } \right)
x[3,  )\therefore x \in [3,\;\infty ) is the solution for this case
For final solution taking union of all results
x(,  12)(12,  117][3,  )\Rightarrow x \in \left( { - \infty ,\;\dfrac{1}{2}} \right) \cup \left( {\dfrac{1}{2},\;\dfrac{{11}}{7}} \right] \cup [3,\;\infty ) is the final solution

Note: Square brackets stand for the including the number and parentheses stand for excluding the number. One can put alternative signs on the number line only when the degree of all the terms is 11. We have excluded   and  - \infty \;{\text{and}}\;\infty because they doesn’t exist physically.