Question:medium

The largest real value of a for which the equation\( |x+a|+|x−1|=2\) has an infinite number of solutions for x is

Updated On: Jan 15, 2026

  • -1

  • 0

  • 1

  • 2

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The Correct Option is C

Solution and Explanation

To identify the greatest real number \(a\) for which the equation \(|x+a|+|x-1|=2\) yields an infinite set of solutions for \(x\), we must examine the scenarios where the absolute value expressions change signs.
Scenario 1: Both expressions are non-negative:
\(x+a+x-1=2\)
\(2x+a-1=2\)
\(2x+a=3\)

Scenario 2: The first expression is non-negative, and the second is negative:
\(x+a-(x-1)=2\)
\(x+a-x+1=2\)
\(a+1=2\)
\(a=1\)

Scenario 3: Both expressions are negative:
\(-(x+a)-(x-1)=2\)
\(-x-a-x+1=2\)
\(-2x-a+1=2\)
\(-2x-a=1\)

Scenario 4: The first expression is negative, and the second is non-negative:
\(-(x+a)+(x-1)=2\)
\(-x-a+x-1=2\)
\(-a-1=2\)
\(a=-3\)
Now, we analyze the critical points:
1. For \(a>1\), the solution is \(a=3\). However, this contradicts the condition that both expressions are non-negative.
2. For \(a=1\), the solution is \(a=1\). This aligns with the condition for both expressions to be non-negative.
3. For \(a<1\), the solution is \(a=-3\). This contradicts the condition that both expressions are non-negative.
Therefore, the largest real value of \(a\) for which the equation has an infinite number of solutions is \(a = 1\).

Consequently, the correct option is (C): \(1\)

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