Inequalities
3.0 Wavy curve method
3.0 Wavy curve method
This method is used to find the solution set for a given inequality without actually solving the inequalities. In order to solve the inequalities of the form $$f(x) = \frac{{{{(x - {a_1})}^{{n_1}}}{{(x - {a_2})}^{{n_2}}}......{{(x - {a_k})}^{{n_k}}}}}{{{{(x - {b_1})}^{{m_1}}}{{(x - {b_2})}^{{m_2}}}......{{(x - {b_p})}^{{m_p}}}}} \geqslant 0\quad ( > 0, < 0, \leqslant 0)$$
where ${n_1},{n_2},...,{n_k}{\text{ and }}{m_1},{m_2},...,{m_p}$ are real numbers and ${a_1},{a_2},...,{a_k}{\text{ and }}{b_1},{b_2},...,{b_p}$ are any real number such that ${a_i} \ne {b_j}$ where $i = 1,2,3,...k{\text{ and }}j = 1,2,3,...p$
Step-1: Factorize the numerator and denominator of the expression into linear factor and make the coefficients of $x$ positive in all the linear factors.
Step-2: Equate each linear factor to zero and find the value of $x$ in each case. These values of $x$ are called critical points.
Step-3: Value of $x$ at which numerator becomes zero should be marked with dark circles and all points of discontinuities i.e., value of $x$ at which denominator becomes zero should be marked on number line with empty circles.
Step-4: Check the value of the expression for any real number greater than the right most marked number on the number line.
Step-5: From right to left, beginning above the number line (if the value of expression is positive in step-4, otherwise from below the number line if the value of expression is negative in step-4) a wavy curve is drawn to pass through all the marked points so that when it passes through a critical point, the curve intersects the number line. But if the critical point is from a factor which is repeated even number of times, the curve will not intersect the number line but remain on the same side after touching it.
Step-6: Continue drawing the curve as mentioned in step-5 to cover all the critical points and reach the extreme left on the number line.
Now the expression is positive whenever the curve is situated above the number line and negative whenever the curve is situated below the number line.
If the question said expression $>0$, positive parts of the wavy curve are our solution. For expression $ \geqslant 0$, the roots of the equation (critical points marked with dark circles) are also part of the final solution.
If the question said expression $<0$, negative parts of the wavy curve are our solution. For expression $ \leqslant 0$, the roots of the equation (critical points marked with dark circles) are also part of the final solution.
The critical points marked with empty circles are never part of the solution as they make the expression undefined.
The union of different regions (either positive or negative) represent our final solution set.
Question 2. Solve the following inequalities:
(a) ${x^2} - 5x + 6 \leqslant 0$
(b) $ - {x^2} + 8x - 12 \geqslant 0$
(c) $(x - 1)(x - 2)(x - 3) \geqslant 0$
(d) $\frac{{2x + 4}}{{x - 1}} \geqslant 5$
Solution: (a) $$\begin{equation} \begin{aligned} {x^2} - 5x + 6 \leqslant 0 \\ {x^2} - 3x - 2x + 6 \leqslant 0 \\ x(x - 3) - 2(x - 3) \leqslant 0 \\ (x - 3)(x - 2) \leqslant 0 \\\end{aligned} \end{equation} $$ Now we get two critical points i.e., $x=3$ and $x=2$. Using wavy curve method, a curve is shown in figure.
As there is sign of $ \leqslant 0$, we take negative values of $x$ i.e., $$x \in \left[ {2,3} \right]$$
(b) $$ - {x^2} + 8x - 120$$ Multiply both sides by $-1$ to make the coefficient of highest power of $x$ i.e., ${x^2}$ positive. $$\begin{equation} \begin{aligned} {x^2} - 8x + 12 \leqslant 0 \\ {x^2} - 6x - 2x + 12 \leqslant 0 \\ x(x - 6) - 2(x - 6) \leqslant 0 \\ (x - 6)(x - 2) \leqslant 0 \\\end{aligned} \end{equation} $$
Now, we get two critical points i.e., $x=6$ and $x=2$. Using wavy curve method, a curve is shown in figure.
As there is sign of $ \leqslant 0$, we take negative values of $x$ i.e., $$x \in \left[ {2,6} \right]$$
(c) We get three critical points i.e., $x=1$, $x=2$ and $x=3$. Using wavy curve method, a curve is shown in figure.
As there is sign of $ \geqslant 0$, we take positive values of $x$ i.e., $$\left[ {1,2} \right] \cup \left[ {3,\infty } \right)$$
(d) $$\begin{equation} \begin{aligned} \frac{{2x + 4}}{{x - 1}} \geqslant 5 \\ \frac{{2x + 4}}{{x - 1}} - 5 \geqslant 0 \\ \frac{{2x + 4 - 5x + 5}}{{x - 1}} \geqslant 0 \\ \frac{{ - 3x + 9}}{{x - 1}} \geqslant 0 \\ \frac{{3x - 9}}{{x - 1}} \leqslant 0 \\ \frac{{(3x - 9)(x - 1)}}{{{{(x - 1)}^2}}} \leqslant 0 \\ (3x - 9)(x - 1) \leqslant 0 \\ 3(x - 3)(x - 1) \leqslant 0 \\ (x - 3)(x - 1) \leqslant 0 \\\end{aligned} \end{equation} $$
Now, we get two critical points i.e., $x=1$ and $x=3$. Using wavy curve method, a curve is shown in figure.
As there is sign of $ \leqslant 0$, we take negative values of $x$ i.e., $$x \in \left[ {1,3} \right]$$ But from the given inequality, it is clearly seen that $x=1$ is a point of discontinuity and denominator becomes zero at $x=1$. So, excluding the value of $x=1$ and the answer becomes $$x \in \left( {1,3} \right]$$
Question 3. Solve the inequality $$2x + y \geqslant 6{\text{ and }}3x + 4y \leqslant 12$$ to find the common region which satisfy both the inequalities using graphical method.
Solution: Take both equations $2x+y=6$ and $3x+4y=12$ by neglecting the greater than sign to draw a graph as shown in figure. Now, satisfy the origin $(0,0)$ in each of the inequalities. If it satisfies then the region in the graph which satisfy the inequality will be towards the origin and if does not satisfy then the region in the graph which satisfy the inequality will be away from the origin and find the common region as shown in figure.
