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CAT 2024 Slot 3 — QA Question 9

ModulusEasy

The number of distinct real values of x, satisfying the equation 

max{x, 2} - min{x, 2} = |x + 2| - |x - 2|, is

Answer & solution

Answer: 2

Solution

Medium

Simplify both sides. The left side max⁡{x,2}−min⁡{x,2}\max\{x,2\}-\min\{x,2\} always equals ∣x−2∣|x-2|. The right side ∣x+2∣−∣x−2∣|x+2|-|x-2| is piecewise linear. Equate and count the distinct real solutions.

1

Rewrite the left side. For any two numbers, max⁡−min⁡\max-\min is the absolute difference.

max⁡{x,2}−min⁡{x,2}=∣x−2∣\begin{aligned} &\max\{x,2\}-\min\{x,2\}=|x-2| \end{aligned}
2

Analyse by region. The breakpoints are x=−2x=-2 and x=2x=2. The equation is ∣x−2∣=∣x+2∣−∣x−2∣|x-2|=|x+2|-|x-2|, i.e. 2∣x−2∣=∣x+2∣2|x-2|=|x+2|.

For x≥2: 2(x−2)=x+2⇒x=6(valid, ≥2)⇒ For −2≤x<2: 2(2−x)=x+2⇒x=23(valid)⇒ For x<−2: 2(2−x)=−(x+2)⇒x=6(rejected, not <−2)\begin{aligned} &\text{For }x\ge 2:\ 2(x-2)=x+2\Rightarrow x=6\quad\text{(valid, }\ge2\text{)}\\ &\Rightarrow\ \text{For }-2\le x<2:\ 2(2-x)=x+2\Rightarrow x=\tfrac23\quad\text{(valid)}\\ &\Rightarrow\ \text{For }x<-2:\ 2(2-x)=-(x+2)\Rightarrow x=6\quad\text{(rejected, not }<-2\text{)} \end{aligned}
3

Count distinct solutions. The valid roots are x=23x=\tfrac23 and x=6x=6.

number of distinct real x=2\begin{aligned} &\text{number of distinct real }x=2 \end{aligned}
22
CAT 2024 Slot 3 QA Q9: The number of distinct real values of x, satisfying the equation max{x, 2} - min{x, 2} = |x + 2| - |x - 2|, is — Solution | TheCATExam