Equations and inequalities in one real variable

Exercises · Numbers and logic · with worked solutions · PDF

Exercise 1

Solve the following inequality in the real variable \(x\):

\[ \frac{x^{2}-2x}{x^{2}-4x+3}>0 \]

Solution

The numerator \(x^{2}-2x\) is positive for \(x<0\) or for \(x>2\), zero for \(x=0\), \(x=2\), and negative for \(0<x<2\). The denominator \(x^{2}-4x+3\) is positive for \(x<1\) or for \(x>3\), and negative for \(1<x<3\). The quotient is defined and has the required positive sign for

\[ x\in(-\infty,0)\cup(1,2)\cup(3,+\infty). \]

Exercise 2

Solve the following inequality in the real variable \(x\):

\[ \log(x-1)^{2}-\log(x-2)>0 \]

Solution

The function \(f(x)=\log(x-1)^{2}-\log(x-2)\) is defined for \(x>2\), since for such \(x\), and only for such \(x\), we have \(x-2>0\), \((x-1)^{2}>0\). In its domain, we have, as required, \(f(x)>0\) if and only if

\[ \log(x-1)^{2}>\log(x-2), \ \ \ x>2 \]

hence if and only if

\[ \left\{\begin{array}{l}(x-1)^{2}>x-2\\ \\ x>2\end{array} \right. \]

We obtain

\[ \left\{\begin{array}{l}x^{2}-3x+3>0\\ \\ x>2\end{array}\right. \]

which is equivalent to

\[ x>2 \]

since \(x^{2}-3x+3>0\) is satisfied for every \(x\). On its whole domain \((2,+\infty)\) the function \(f(x)\) takes positive values.

Exercise 3

Solve the following inequality in the real variable \(x\):

\[ e^{x}+e^{-x}<\frac{10}{3} \]

Solution

For every \(x\in\R\)

\[ e^{x}+e^{-x}<\frac{10}{3} \]

is equivalent to

\[ 3e^{2x}-10e^{x}+3<0 \]

hence to

\[ \frac{1}{3}<e^{x}<3. \]

It follows that the inequality is solved by

\[ -\log3<x<\log3. \]

Exercise 4

Solve the following inequality in the real variable \(x\):

\[ 8^{x+1}\geq2^{x^{2}} \]

Solution

For every \(x\in\R\)

\[ 8^{x+1}\geq2^{x^{2}} \]

is equivalent to

\[ 2^{3x+3}\geq2^{x^{2}} \]

hence to

\[ 3x+3\geq x^{2} \]

since \(2^{x}\) is a strictly increasing function on \(\R\). It follows that the inequality is solved by

\[ \frac{3-\sqrt{21}}{2}\leq x\leq \frac{3+\sqrt{21}}{2}. \]

Exercise 5

Solve the following inequality in the real variable \(x\):

\[ \log(1-\sin x)\geq0 \]

Solution

The function \(f(x)=\log(1-\sin x)\) is defined for \(\sin x<1\), hence for

\[ x\neq\frac{\pi}{2}+2k\pi,\ \ \ k\in{\bf Z}. \]

For such \(x\) we have \(f(x)\geq0\) if and only if

\[ 1-\sin x\geq1 \]

hence for

\[ \sin x\leq 0. \]

The given inequality is solved for

\[ \pi+2k\pi\leq x\leq 2\pi+2k\pi,\ \ \ k\in{\bf Z} \]

i.e., for

\[ x\in\bigcup_{k\in{\bf Z}}[\pi+2k\pi, 2\pi+2k\pi]. \]

Exercise 6

Solve the following inequality in the real variable \(x\):

\[ |x|\sqrt{1-2x^{2}}>2x^{2}-1 \]

Solution

\(\sqrt{1-2x^{2}}\) is defined for \(1-2x^{2}\geq0\). When \(1-2x^{2}=0\) the inequality is not satisfied, since it reduces to \(0>0\). When \(1-2x^{2}>0\) the inequality is satisfied, since the left-hand side \(|x|\sqrt{1-2x^{2}}\) is the product of two positive numbers while the right-hand side \(2x^{2}-1=-(1-2x^{2})\) is negative. Therefore the inequality is solved for

\[ -\sqrt{\frac{1}{2}}<x<\sqrt{\frac{1}{2}}. \]

Exercise 7

Solve the following equation in the real variable \(x\):

\[ \sin^{2}x=2\cos^{2}x-\frac{1}{2} \]

Solution

The equation

\[ \sin^{2}x=2\cos^{2}x-\frac{1}{2} \]

is equivalent to

\[ \begin{array}{l}2\sin^{2}x=4(1-\sin^{2}x)-1\\ \\ 6\sin^{2}x=3\\ \\ \sin x=\pm\sqrt{\frac{1}{2}}.\end{array} \]

The solutions are given by

\[ x=\frac{\pi}{4}+\frac{k\pi}{2},\ \ \ k\in{\bf Z}. \]

Exercise 8

Solve the following inequality in the real variable \(x\):

\[ e^{\sin^{2}x-\sin x}\leq1 \]

Solution

For every \(x\in\R\), \(e^{\sin^{2}x-\sin x}\leq1\) is equivalent to

\[ \sin^{2}x-\sin x\leq0 \]

hence to

\[ 0\leq\sin x\leq1. \]

The solutions are given by

\[ 2k\pi\leq x\leq \pi+2k\pi,\ \ \ k\in{\bf Z} \]

i.e., by

\[ x\in\bigcup_{k\in{\bf Z}}[2k\pi, \pi+2k\pi]. \]

Exercise 9

Solve the following inequality in the real variable \(x\):

\[ 2\left|x^{2}-x\right|>|x| \]

Solution

The inequality \(2\left|x^{2}-x\right|>|x|\) is not satisfied for \(x=0\). For \(x\neq0\), dividing by the positive term \(|x|\), it is equivalent to

\[ 2\left|x-1\right|>1 \]

hence to

\[ |x-1|>1/2. \]

The solutions are given by

\[ (x-1<-1/2, \ x\neq0)\vee (x-1>1/2), \]

\[ (x<1/2, \ x\neq0)\vee x>3/2 \]

hence by

\[ x\in(-\infty,0)\cup(0,1/2)\cup(3/2,+\infty). \]

Exercise 10

Solve the following inequality in the real variable \(x\):

\[ (x+1)^{x^{2}-1}>1 \]

Solution

The function \(f(x)=(x+1)^{x^{2}-1}\) is defined for \(x>-1\). For such \(x\), \(f(x)>1\) is equivalent to

\[ e^{(x^{2}-1)\log(x+1)}>1 \]

hence to

\[ (x^{2}-1)\log(x+1)>0,\ \ \ x>-1. \]

Dividing by the positive term \(x+1\) we obtain

\[ (x-1)\log(x+1)>0,\ \ \ x>-1. \]

The factor \(x-1\) is positive for \(x>1\), zero for \(x=1\), and negative for \(-1<x<1\). The factor \(\log(x+1)\) is positive for \(x>0\), zero for \(x=0\), and negative for \(-1<x<0\). By the rule of signs, the solutions are given by

\[ x\in(-1,0)\cup(1,+\infty). \]

Exercise 11

Solve the following equation in the real variable \(x\):

\[ 3^{|x^{2}-4|}=0 \]

Solution

\(3^{|x^{2}-4|}=0\) has no solutions, since \(3^{y}>0\) for every \(y\in\R\).

Exercise 12

Solve the following inequality in the real variable \(x\):

\[ \frac{\log_{a}(4x-3)}{\log_{a}(2x-1)} > 1 \quad 0<a<1 \]

Solution

Because of the existence conditions of the logarithm and of the denominator, we must impose the system

\[ \begin{cases*} 4x-3 > 0 \\ 2x-1 > 0 \\ \log_{a}(2x-1) \neq 0 \end{cases*} \]

which gives \(\frac{3}{4}<x<1, x>1\). Now moving everything to the left-hand side, finding the least common denominator and using the properties of logarithms, we obtain

\[ \frac{\log_{a}(\frac{4x-3}{2x-1})}{\log_{a}(2x-1)}>0 \]

Studying the sign of the numerator and of the denominator, and keeping in mind that \(a<1\), for the numerator we obtain

\[ \log_{a}\left(\frac{4x-3}{2x-1}\right) > 0 \Longleftrightarrow \frac{4x-3}{2x-1}<1 \Longleftrightarrow \frac{1}{2}<x<1 \]

while for the denominator we obtain

\[ \log_{a}(2x-1)>0 \Longleftrightarrow x<1 \]

Combining the signs, the inequality is satisfied for \(\frac{1}{2}<x<1, x>1\). Intersecting this set of solutions with the existence conditions studied above, we obtain that the inequality is satisfied for every \(x\) belonging to the set \((\frac{3}{4};1) \cup (1,+\infty)\).

Exercise 13

Solve the following inequality in the real variable \(x\):

\[ \sqrt{x^2-3x+5} \leq x+3 \]

Solution

This irrational inequality has the form \(\sqrt{A(x)} \leq B(x)\), with \(A(x)\) and \(B(x)\) polynomials in \(x\). First of all, we must impose the existence condition of the root, i.e., \(x^2-3x+5 \geq 0\). Moreover, since a square root is always positive or zero in \(\mathbb{R}\), we impose \(x+3 \geq 0\). We therefore solve the system:

\[ \begin{cases*} x^2-3x+5 \geq 0 \\ x+3 \geq 0 \\ x^2-3x+5 \leq (x+3)^2 \end{cases*} \]

where, in the last inequality, we were allowed to square both sides, since they are positive. The system reduces to:

\[ \begin{cases*} \forall x \in \mathbb{R} \\ x \geq -3 \\ x \geq -\frac{4}{9} \end{cases*} \]

whose set of solutions is the interval \([-\frac{4}{9}, +\infty)\).

Exercise 14

Solve the following rational inequality in the real variable \(x\):

\[ \frac{x^3-3x^2+2x-6}{x^2-2x-3} > \frac{2}{x+1} \]

Solution

First of all, we must write the inequality in canonical form (0 on the right-hand side). The existence conditions (denominators different from zero) will be automatically included in the sign study. We obtain

\[ \frac{x^3-3x^2+2x-6}{x^2-2x-3} - \frac{2}{x+1} > 0 \]

We factor the numerator and the denominator of the first fraction, obtaining:

\[ \frac{(x^2+2)(x-3)}{(x-3)(x+1)} - \frac{2}{x+1} > 0 \]

Since we can cancel \(x-3\) and obtain an equivalent fraction, provided that \(x \neq 3\) by the existence condition, we have

\[ \frac{x^2+2}{x+1} - \frac{2}{x+1} > 0 \]

that is,

\[ \frac{x^2}{x+1} > 0 \]

The numerator is positive \(\forall x \in \mathbb{R}, x \neq 0\), and the denominator is positive for \(x>-1\). Combining the signs, and taking into account the condition \(x \neq 3\), we obtain the solution set \((-1,0) \cup (0,3) \cup (3,+\infty)\).