Arithmetic Progression Question (involving modulus inequalities and equations)

Question

OPTIONS:

A) 10

B) 15

C) 25

D) None of the above

Answer 1

ABSOLUTE VALUE FUNCTION:

$ f(x) = |x| = \begin{cases} x &\text{if } x\geq0\\ -x &\text{if } x<0 \\ \end{cases} $

SOLUTION:

---

---

PART I:

First let us find the rational numbers $x$ that make $a,b,c$ an arithmetic progression (AP). $a,b,c$ must be in the form $t_1,t_1+d,t_1+2d$ where $t_1$ is the first term and $d$ is the common difference.

Due to this we can conclude:

$c+a = 2b \\\because t_1 + 2d + t_1 = 2(t_1+d)$

Let $A$ be the set of all such $x$.

$$\therefore A=\left\{x \in \mathbb{Q}: |2x^2-|x|-5|+|x^2-|4x+3|| =8 \right\}$$

We can remove the outer modulus function by splitting A into 4 cases, with their corresponding ranges of $x$, i.e.

$ A = \begin{cases} (2x^2-|x|-5)+(x^2-|4x+3|) =8 &: (2x^2-|x|-5\geq0) \cap (x^2-|4x+3|\geq0)\\ -(2x^2-|x|-5)+(x^2-|4x+3|) =8 &: (2x^2-|x|-5<0) \cap (x^2-|4x+3|\geq0)\\ (2x^2-|x|-5)-(x^2-|4x+3|) =8 &: (2x^2-|x|-5\geq0) \cap (x^2-|4x+3|<0)\\ -(2x^2-|x|-5)-(x^2-|4x+3|) =8 &: (2x^2-|x|-5<0) \cap (x^2-|4x+3|<0)\\ \end{cases} $

Let us find these corresponding ranges of $x$.

LET: $F_1 = x:(2x^2-|x|-5\geq0) \cap (x^2-|4x+3|\geq0)\\ F_2 = x:(2x^2-|x|-5<0) \cap (x^2-|4x+3|\geq0)\\ F_3 = x:(2x^2-|x|-5\geq0) \cap (x^2-|4x+3|<0)\\ F_4 = x:(2x^2-|x|-5<0) \cap (x^2-|4x+3|<0) $

---

$F_1:$

$ (i)\text{ }2x^2-|x|-5 \geq0 = \begin{cases} 2x^2-x-5\geq0 &: x\geq0\\ 2x^2+x-5\geq0 &: x<0 \\ \end{cases} $

CASE $_{F_{1.1.1}}$: $2x^2-x-5\geq0<\\(x-\frac{1+\sqrt{41}}{4})(x-\frac{1-\sqrt{41}}{4})\geq0\\\implies x\leq\frac{1-\sqrt{41}}{4}>, x\geq\frac{1+\sqrt{41}}{4}\\\text{but } x\geq0 \implies x\geq\frac{1+\sqrt{41}}{4}$

$\text{Claim 1 }: 1.75<\frac{1+\sqrt{41}}{4} < 2$

CASE $_{F_{1.1.2}}$: $2x^2+x-5\geq0\\(x+\frac{1+\sqrt{41}}{4})(x+\frac{1-\sqrt{41}}{4})\geq0\\\implies x\leq-\frac{1+\sqrt{41}}{4}>, x\geq\frac{\sqrt{41}-1}{4}\\\text{but } x<0 \implies x<-\frac{1+\sqrt{41}}{4}$

$\implies x\in (-\infty,-\frac{1+\sqrt{41}}{4}]\cup[\frac{1+\sqrt{41}}{4},\infty)$

$ (ii)\text{ }x^2-|4x+3| \geq0 = \begin{cases} x^2-4x-3\geq0 &: x\geq-\frac{3}{4}\\ x^2+4x+3\geq0 &: x<-\frac{3}{4} \\ \end{cases} $

CASE $_{F_{1.2.1}}$: $x^2-4x-3\geq0\\ (x-(2+\sqrt{7})(x-(2-\sqrt{7})\geq0\\ \implies x\leq2-\sqrt{7}, x\geq2+\sqrt{7}\\ \text{but } x\geq-\frac{3}{4}\implies -\frac{3}{4}\leq x\leq 2-\sqrt{7}, x\geq 2+\sqrt{7}$

$\text{Claim 2 }: -\frac{3}{4} < 2-\sqrt{7}$

CASE $_{F_{1.2.2}}$: $x^2+4x+3\geq0<\\(x+3)(x+1)\geq0\\\implies x\leq-3>, x\geq-1\\\text{but } x<-\frac{3}{4} \implies x<-3, -1\leq x<-\frac{3}{4}$

$\implies x\in (-\infty,-3]\cup[-1,2-\sqrt{7}]\cup[2+\sqrt{7}, \infty)$

$F_1: (i)\cap (ii)$ (put pic) $$\implies F_1: (-\infty,-3]\cup[2+\sqrt{7}, \infty)$$

---

Note: if we find the solution set for $f(x)\geq0$, then the solution set for $f(x)<0$ will just be $\mathbb{R} - (x:f(x)\geq0)$. So finding $F_2, F_3$ and $F_4$ will be easy.

---

$F_2:$

$(i) 2x^2-|x|-5 <0\equiv\mathbb{R} - (x:2x^2-|x|-5\geq 0)\implies x\in (-\frac{1+\sqrt{41}}{4},\frac{1+\sqrt{41}}{4}) \\(ii)x^2-|4x+3| \geq0 \implies x\in (-\infty,-3]\cup[-1,2-\sqrt{7}]\cup[2+\sqrt{7}, \infty)$ $$\implies F_2: [-1,2-\sqrt{7}]$$

---

$F_3:$

$(i) 2x^2-|x|-5 \geq0 \implies x\in (-\infty,-\frac{1+\sqrt{41}}{4}]\cup[\frac{1+\sqrt{41}}{4},\infty) \\(ii)x^2-|4x+3| <0 \equiv\mathbb{R} - (x:x^2-|4x+3|\geq 0) \implies x\in (-3,-1)\cup(2-\sqrt{7},2+\sqrt{7})$ $$\implies F_3: \left(-3,-\frac{1+\sqrt{41}}{4}\right]\cup\left[\frac{1+\sqrt{41}}{4}, 2+\sqrt{7}\right)$$

---

$F_4:$

$(i) 2x^2-|x|-5 <0 \implies x\in (-\frac{1+\sqrt{41}}{4},\frac{1+\sqrt{41}}{4}) \\(ii)x^2-|4x+3| <0 \implies x\in (-3,-1)\cup(2-\sqrt{7},2+\sqrt{7})$ $$\implies F_4: \left(-\frac{1+\sqrt{41}}{4},-1\right)\cup\left(2-\sqrt{7},\frac{1+\sqrt{41}}{4}\right)$$

---

$$ \therefore A = \begin{cases} 3x^2-|x|-|4x+3| =13 &\text{if } x\in (-\infty,-3]\cup[2+\sqrt{7}, \infty)\\ x^2-|x|+|4x+3| =3 &\text{if } x\in [-1,2-\sqrt{7}]\\ x^2-|x|+|4x+3| =13 &\text{if } x\in \left(-3,-\frac{1+\sqrt{41}}{4}\right]\cup\left[\frac{1+\sqrt{41}}{4}, 2+\sqrt{7}\right)\\ 3x^2-|x|-|4x+3| =-3 &\text{if } x\in \left(-\frac{1+\sqrt{41}}{4},-1\right)\cup\left(2-\sqrt{7},\frac{1+\sqrt{41}}{4}\right)\\ \end{cases} $$

---

Case$_1$: $x \in F_1 \implies x\in (-\infty,-3]\cup[2+\sqrt{7}, \infty)$

$ 3x^2-|x|-|4x+3| =13 \equiv \begin{cases} 3x^2-(x)-(4x+3) =13 &: x\geq0\\ 3x^2+(x)-(4x+3) =13 &: -\frac{3}{4}\leq x<0\\ 3x^2+(x)+(4x+3) =13 &: x<-\frac{3}{4}\\ \end{cases} $

Subcase$_1$: $x\geq0 \\3x^2-(x)-(4x+3) =13\implies 3x^2 -5x - 16=0 \\\implies x= \frac{5 \pm \sqrt{217}}{6} \\\text{ but } x \in \mathbb{Q}\implies x= \phi$

Subcase$_2$: $-\frac{3}{4}\leq x<0 \\\text{but }F_1\text{ does not contain this interval i.e.} (-\frac{3}{4},0]\cap F_1 = \phi\\\implies x=\phi$

Subcase$_3$: $x<-\frac{3}{4} \\3x^2+(x)+(4x+3) =13\implies 3x^2 +5x - 10=0 \\\implies x= \frac{-5 \pm \sqrt{145}}{6} \\\text{ but } x \in \mathbb{Q}\implies x=\phi$

$$x=\phi$$

---

Case$_2$: $x \in F_2 \implies x\in x\in [-1,2-\sqrt{7}]$

$ x^2-|x|+|4x+3| =3 \equiv \begin{cases} x^2-(x)+(4x+3) =3 &: x\geq0\\ x^2+(x)+(4x+3) =3 &: -\frac{3}{4}\leq x<0\\ x^2+(x)-(4x+3) =3 &: x<-\frac{3}{4}\\ \end{cases} $

Subcase$_1$: $x\geq0 \\\text{but }F_2\text{ does not contain this interval i.e.} [0,\infty)\cap F_2 = \phi\\\implies x=\phi$

Subcase$_2$: $-\frac{3}{4}\leq x<0 \\x^2+(x)+(4x+3) =3\implies x^2+5x=0 \implies x(x+5) =0 \\\implies x= 0,-5 \\\text{ but } -\frac{3}{4}\leq x<0\implies x=\phi$

Subcase$_3$: $x<-\frac{3}{4} \\x^2+(x)-(4x+3) =3\implies x^2 -3x - 6=0 \\\implies x= \frac{3 \pm \sqrt{29}}{2} \\\text{ but } x \in \mathbb{Q}\implies x=\phi$

$$x=\phi$$

---

Case$_3$: $x \in F_3 \implies x\in \left(-3,-\frac{1+\sqrt{41}}{4}\right]\cup\left[\frac{1+\sqrt{41}}{4}, 2+\sqrt{7}\right)$

$ x^2-|x|+|4x+3| =13 \equiv \begin{cases} x^2-(x)+(4x+3) =13 &: x\geq0\\ x^2+(x)+(4x+3) =13 &: -\frac{3}{4}\leq x<0\\ x^2+(x)-(4x+3) =13 &: x<-\frac{3}{4}\\ \end{cases} $

Subcase$_1$: $x\geq 0 \\x^2-(x)+(4x+3) =13\implies x^2 +3x-10=0 \implies (x+5)(x-2)=0\\ x=-5,2 \\\text{ but } x \geq 0 \implies x=2 \\\text{but we must crosscheck if } {2} \in F_3 \text{ and indeed this is true} \\\implies x=2$

Subcase$_2$: $-\frac{3}{4}\leq x<0 \\\text{but }F_3\text{ does not contain this interval } \\\implies x=\phi$

Subcase$_3$: $x<-\frac{3}{4} \\x^2+(x)-(4x+3) =13\implies x^2 -3x - 16=0 \\\implies x= \frac{3 \pm \sqrt{73}}{2} \\\text{ since } x \in \mathbb{Q}\implies x=\phi$

$$x=2$$

---

Case$_4$: $x \in F_4 \implies x\in \left(-\frac{1+\sqrt{41}}{4},-1\right)\cup\left(2-\sqrt{7},\frac{1+\sqrt{41}}{4}\right)$

$ 3x^2-|x|-|4x+3| =-3 \equiv \begin{cases} 3x^2-(x)-(4x+3) =-3 &: x\geq0\\ 3x^2+(x)-(4x+3) =-3 &: -\frac{3}{4}\leq x<0\\ 3x^2+(x)+(4x+3) =-3 &: x<-\frac{3}{4}\\ \end{cases} $

Subcase$_1$: $x\geq0 \\3x^2-(x)-(4x+3) =-3\\ 3x^2 -5x =0 \implies x(3x-5)=0\\ x=0,\frac{5}{3}\\\text{but } x \geq0 \implies x= \frac{5}{3}\\\text{but we must crosscheck if } \frac{5}{3} \in F_4 \text{ i.e. if }\frac{5}{3} \in \left(2-\sqrt{7},\frac{1+\sqrt{41}}{4}\right)\text{ and this is true}\\\implies x=\frac{5}{3}$Claim 3: $\frac{5}{3} < \frac{1+\sqrt{41}}{4}$

Subcase$_2$: $-\frac{3}{4}\leq x<0 \\3x^2+(x)-(4x+3) =-3 \implies 3x^2-3x =0\implies 3x(x-1)=0\\x=0,1 \\\text{but } -\frac{3}{4}\leq x<0 \implies x=0\\\text{but we must crosscheck if } {0} \in F_4 \text{ and indeed this is true} \\\implies x=0$

Subcase$_3$: $x<-\frac{3}{4} \\3x^2+(x)+(4x+3) =-3\implies 3x^2 +5x +6=0 \\\implies x= \frac{-5 \pm \sqrt{-47}}{6} \\\text{ but } x \in \mathbb{Q}\implies x=\phi$

$$x=\frac{5}{3},2$$

---

We have found the rational numbers $x$ satisfying $A$, i.e. $x: a,b,c$ are in AP given $x \in \mathbb{Q}$, $$\therefore A = \left\{0,\frac{5}{3},2\right\}$$

---

---

PART II:

Let us now find the cardinality of the set $S$.

$S = \left\{ x\in \mathbb{Z}: \dfrac{c}{a} \leq 2\right\}$ $$S = \left\{ x\in \mathbb{Z}: \dfrac{|2x^2- |x|-5|}{|x^2-|4x+3||} \leq 2\right\}$$

Just as we did it A, we can split this into four cases:

$ S = x\in \mathbb{Z}: \begin{cases} \dfrac{+(2x^2- |x|-5)}{+(x^2-|4x+3|)} \leq 2 &: x\in F_1\\ \dfrac{-(2x^2- |x|-5)}{+(x^2-|4x+3|)} \leq 2 &: x\in F_2\\ \dfrac{+(2x^2- |x|-5)}{-(x^2-|4x+3|)} \leq 2 &: x\in F_3\\ \dfrac{-(2x^2- |x|-5)}{-(x^2-|4x+3|)} \leq 2 &: x\in F_4\\ \end{cases} $

If we look carefully we can combine cases 1 and 4 as they are essentially the same. Similarly cases 2 and 3 are the same. Hence, we can merge these cases.

but note: $x^2-|4x+3| \neq 0 \implies x \neq -3, -1, 2 \pm \sqrt{7}$ as the denominator can not be $0$.

LET: $G_1 = x:x\in F_1\cap F_4 - \left\{2+\sqrt{7},-3\right\}\\ G_2 = x:x\in F_2\cap F_3 - \left\{2-\sqrt{7},-1\right\} $

$ \implies S = x\in \mathbb{Z}: \begin{cases} \dfrac{2x^2- |x|-5}{x^2-|4x+3|} \leq 2 &: x\in G_1\\ -\dfrac{2x^2- |x|-5}{x^2-|4x+3|} \leq 2 &: x\in G_2\\ \end{cases} $

Note: while solving inequalities we can not cross multiply.

$ \implies S = x\in \mathbb{Z}: \begin{cases} \dfrac{2|4x+3|-|x|-5}{x^2-|4x+3|} \leq 0 &: x\in G_1\\ \dfrac{4x^2-2|4x+3|-|x|-5}{x^2-|4x+3|} \geq 0 &: x\in G_2\\ \end{cases} $

---

Case$_1$: $x \in G_1$

$ \dfrac{2|4x+3|-|x|-5}{x^2-|4x+3|} \leq 0 \equiv \begin{cases} \dfrac{2(4x+3)-x-5}{x^2-4x-3} \leq 0 &: x\geq0\\ \dfrac{2(4x+3)+x-5}{x^2-4x-3} \leq 0 &: -\frac{3}{4}\leq x<0\\ \dfrac{-2(4x+3)+x-5}{x^2+4x+3} \leq 0 &: x<-\frac{3}{4}\\ \end{cases} $

Subcase$_1$: $x\geq 0 \\\dfrac{2(4x+3)-x-5}{x^2-4x-3} \leq 0\implies \dfrac{7x+1}{(x-(2+\sqrt{7}))(x-(2-\sqrt{7}))} \leq 0\\\implies x\in (-\infty, 2-\sqrt{7})\cup \left[-\dfrac{1}{7},2+\sqrt{7}\right) \\\text{but } x\geq 0\\\implies x \in [0,2+\sqrt{7})$

Subcase$_2$: $-\frac{3}{4}\leq x<0 \\\dfrac{2(4x+3)+x-5}{x^2-4x-3} \leq 0\implies \dfrac{9x+1}{(x-(2+\sqrt{7}))(x-(2-\sqrt{7}))} \leq 0\\\implies x\in (-\infty, 2-\sqrt{7})\cup \left[-\dfrac{1}{9},2+\sqrt{7}\right) \\\text{but } -\frac{3}{4}\leq x<0\\\implies x \in \left[-\dfrac{3}{4},2-\sqrt{7}\right) \cup \left[-\dfrac{1}{9},0\right)$

Subcase$_3$: $x<-\frac{3}{4}\\\dfrac{-2(4x+3)+x-5}{x^2+4x+3} \leq 0\implies -\dfrac{7x+11}{(x+1)(x+3)} \leq 0\\\implies -3< x\leq -\frac{11}{7}, x>-1 \\\text{but } x<-\frac{3}{4}\\\implies -3< x\leq -\frac{11}{7}, -1< x< -\frac{3}{4}$

Combining subcases 1,2 and 3:

$x \in \left(-3,-\dfrac{11}{7}\right] \cup \left(-1,2-\sqrt{7}\right) \cup \left[-\dfrac{1}{9},2+\sqrt{7}\right) \\\text{but we must cross check: x must come from the region } G_1$

$$\implies x\in \left(-\dfrac{1+\sqrt{41}}{4}, -\dfrac{11}{7}\right]\cup\left[-\dfrac{1}{9},\dfrac{1+\sqrt{41}}{4}\right)$$

---

Case$_2$: $x \in G_2$

$ \dfrac{4x^2-2|4x+3|-|x|-5}{x^2-|4x+3|} \geq 0 \equiv \begin{cases} \dfrac{4x^2 - 2(4x+3)-x-5}{x^2-4x-3} \geq 0 &: x\geq0\\ \dfrac{4x^2-2(4x+3)+x-5}{x^2-4x-3} \geq 0 &: -\frac{3}{4}\leq x<0\\ \dfrac{4x^2+2(4x+3)+x-5}{x^2+4x+3} \geq 0 &: x<-\frac{3}{4}\\ \end{cases} $

Subcase$_1$: $x\geq 0 \\\dfrac{4x^2 - 9x-11}{x^2-4x-3} \geq 0\implies \dfrac{\left(x-\dfrac{9+\sqrt{257}}{8}\right)\left(x-\dfrac{9-\sqrt{257}}{8}\right)}{(x-(2+\sqrt{7}))(x-(2-\sqrt{7}))} \geq 0\\\implies x\in \left(-\infty, \dfrac{9-\sqrt{257}}{8}\right] \cup \left(2-\sqrt{7}, \dfrac{9+\sqrt{257}}{8}\right) \cup \left(2+\sqrt{7},\infty \right) \\\text{but } x\geq 0\\\implies x \in \left[0,\dfrac{9+\sqrt{257}}{8}\right] \cup \left(2+\sqrt{7},\infty\right)$

Subcase$_2$: $-\frac{3}{4}\leq x<0 \\\dfrac{4x^2 - 7x-11}{x^2-4x-3} \geq 0\implies -\dfrac{\left(x+1\right)\left(4x-11\right)}{(x-(2+\sqrt{7}))(x-(2-\sqrt{7}))} \geq 0\\\implies x\in \left(-\infty, -1\right] \cup \left(2-\sqrt{7}, \dfrac{11}{4}\right) \cup \left(2+\sqrt{7},\infty \right) \\\text{but } -\frac{3}{4}\leq x<0\\\implies 2-\sqrt{7}< x<0$

Subcase$_3$: $x<-\frac{3}{4}\\\dfrac{4x^2 +9x+1}{x^2+4x+3} \geq 0\implies \dfrac{\left(x+\dfrac{9+\sqrt{65}}{8}\right)\left(x+\dfrac{9-\sqrt{65}}{8}\right)}{(x+1)(x+3)} \geq 0\\\implies x\in \left(-\infty, -3\right) \cup \left[-\dfrac{9+\sqrt{65}}{8}, -1\right) \cup \left[-\dfrac{9-\sqrt{65}}{8},\infty \right) \\\text{but } x<-\frac{3}{4}\\\implies x<-3,-\dfrac{9+\sqrt{65}}{8}\leq x<-1$

Combining subcases 1,2 and 3:

$x \in \left(-\infty, -3,\right) \cup \left[-\dfrac{9+\sqrt{65}}{8},-1\right) \cup \left(2-\sqrt{7},\dfrac{9+\sqrt{257}}{8}\right]\cup \left(2+\sqrt{7},\infty\right) \\\text{but we must cross check: x must come from the region } G_2$

$$\implies x\in \left[-\dfrac{9+\sqrt{65}}{8},-\dfrac{1+\sqrt{41}}{4}\right]\cup\left[-\dfrac{1+\sqrt{41}}{4},\dfrac{9+\sqrt{257}}{8}\right]$$

---

Now we must combine the cases $G_1$ and $G_2$ to get the set $S$: $$S=\left\{x \in \mathbb{Z}: x\in \left[-\dfrac{9+\sqrt{65}}{8},-\dfrac{11}{7}\right]\cup \left[-\dfrac{1}{9},\dfrac{9+\sqrt{257}}{8}\right]\right\}$$

We must extract the integers now. Let $pp$ be an arbitrary decimal. We know fractional part of any number lies between 0 to 1. HENCE:

$\implies S=\left\{x \in \mathbb{Z}: x\in \left[-\dfrac{9+8+\{\sqrt{65}\}}{8},-1.pp\right]\cup \left[-0.pp,\dfrac{9+16+\{\sqrt{257}\}}{8}\right]\right\}$ $\implies S=\left\{x \in \mathbb{Z}: x\in \left[-2.pp,-1.pp\right]\cup \left[-0.pp,3.pp\right]\right\}$

$$\implies S=\left\{-2,0,1,2,3\right\}\\\implies |S| = 5$$

---

---

PART III:

Now that we found rational x such that $a,b,c$ are in AP and we know the cardinality of the set $S$, we are in a position to find the first $3|S|$ terms of such an AP i.e. the set $T$.

Let $d$ be the common difference. Also, let $S_{15}$ be the sum of the first $15$ terms.

For $x=0$, we have $a=3,b=4,c=5$ an increasing arithmetic progression.

• This implies $t_1=3,d=1$, so $S_{15}=\frac{15(2t_1+(15-1)d)}{2}=150$.

For $x=\frac 53$, we have $a=\frac{62}{9},b=4,c=\frac{10}{9}$ an decreasing arithmetic progression.

• This implies $t_1=\frac{62}{9},d=-\frac{26}{9}$, so $S_{15}=-200$.

For $x=2$, $a=7,b=4,c=1$ an decreasing arithmetic progression.

• This implies have $t_1=7,d=-3$, so $S_{15}=-210$.

$$\implies T=\left\{-210,-200,150\right\}$$

The only number that divides every element in the set $T$ in the options is 10.

Answer 2

The only correct option is A) $10$.

The proofs for the following two claims are written at the end of the answer.

Claim 1 : The only rational numbers $x$ satisfying $t_2-t_1=t_3-t_2$ are $x=0,\frac 53,2$.

Claim 2 : $|S|=5$.

Let $d$ be the common difference. Also, let $S_{15}$ be the sum of the first $15$ terms.

For $x=0$, we have $t_1=3,d=1$, so $S_{15}=\frac{15(2t_1+(15-1)d)}{2}=150$.

For $x=\frac 53$, we have $t_1=\frac{62}{9},d=-\frac{26}{9}$, so $S_{15}=-200$.

For $x=2$, we have $t_1=7,d=-3$, so $S_{15}=-210$.

Now, we want to find options such that all the following three claims are true :

• $150$ is divisible by $(\quad)$.

• $-200$ is divisible by $(\quad)$.

• $-210$ is divisible by $(\quad)$.

The option A) $10$ is correct since all the three claims are true.

The option B) $15$ is wrong since the claim that $-200$ is divisible by $15$ is false.

The option C) $25$ is wrong since the claim that $-210$ is divisible by $25$ is false.

The option D) None of these is wrong since the option A) is correct.

Therefore, the only correct option is A) $10$.

---

Claim 1 : The only rational numbers $x$ satisfying $t_2-t_1=t_3-t_2$ are $x=0,\frac 53,2$.

Proof : $$\begin{align}t_2-t_1=t_3-t_2&\iff \left|x^2 - |4x+3|\right|+\left|2x^2 - \left||x|+5\right|\right|=8 \\\\&\iff \left|x^2 - |4x+3|\right|+\left|2x^2 - |x|-5\right|=8\end{align}$$

• If $x\le -3$, then $x^2 +4x+3+2x^2 +x-5=8$ has no solutions.

• If $-3\lt x\le \frac{-1-\sqrt{41}}{4}$, then solving $-(x^2 +4x+3)+2x^2 +x-5=8$ gives $x=\frac{3-\sqrt{73}}{2}$.

• If $\frac{-1-\sqrt{41}}{4}\lt x\le -\frac 43$, then $-(x^2 +4x+3)-(2x^2 +x-5)=8$ has no solutions.

• If $-\frac 43\lt x\le 2-\sqrt 7$, then $x^2 - 4x-3-(2x^2 +x-5)=8$ has no solutions.

• If $2-\sqrt 7\lt x\le 0$, then solving $-(x^2 - 4x-3)-(2x^2 +x-5)=8$ gives $x=0$.

• If $0\lt x\le \frac{1+\sqrt{41}}{4}$, then solving $-(x^2 - 4x-3)-(2x^2-x-5)=8$ gives $x=\frac 53$.

• If $\frac{1+\sqrt{41}}{4}\lt x\le 2+\sqrt 7$, then solving $-(x^2 -4x-3)+2x^2 - x-5=8$ gives $x=2$.

• If $2+\sqrt 7\le x$, then $x^2 - 4x-3+2x^2 - x-5=8$ has no solutions.

Hence, the only such rational numbers $x$ are $x=0,\frac 53,2$. $\quad\blacksquare$

---

Claim 2 : $|S|=5$

Proof :

$$\begin{align}\frac{t_3}{t_1}\leq2&\iff t_1\not=0\quad\text{and}\quad t_3\le 2t_1 \\\\&\iff x\not=-1,-3,2\pm\sqrt 7\quad\text{and}\quad \left|2x^2 - |x|-5\right|\le 2\left|x^2 - |4x+3|\right|\end{align}$$

• If $x\lt -3$, then $2x^2 +x-5\le 2(x^2 +4x+3)$ has no solutions.

• If $2+\sqrt 7\lt x$, then $2x^2 - x-5\le 2(x^2 - 4x-3)$ has no solutions.

So, the only possible integers $x$ are $x=-2,0,1,2,3,4$.

Checking each $x$ gives that $S=\{-2,0,1,2,3\}$ and $|S|=5$. $\quad\blacksquare$