The largest real value of a for which the equation\( |x+a|+|x−1|=2\) has an infinite number of solutions for x is
-1
0
1
2
The Correct Option is C
Approach Solution - 1
To find the largest real value of \(a\) for which the equation \(|x+a|+|x-1|=2\) has an infinite number of solutions for \(x\), we need to analyze the possible cases when the absolute value expressions change their signs.
Case 1: When both expressions are positive:
\(x+a+x-1=2\)
\(2x+a-1=2\)
\(2x+a=3\)
Case 2: When the first expression is positive and the second is negative:
\(x+a-(x-1)=2\)
\(x+a-x+1=2\)
\(a+1=2\)
\(a=1\)
Case 3: When both expressions are negative:
\(-(x+a)-(x-1)=2\)
\(-x-a-x+1=2\)
\(-2x-a+1=2\)
\(-2x-a=1\)
Case4: When the first expression is negative and the second is positive:
\(-(x+a)+(x-1)=2\)
\(-x-a+x-1=2\)
\(-a-1=2\)
\(a=-3\)
Now, let's analyze the critical points:
1. When \(a>1\), we have the solution \(a=3\), but it does not satisfy the condition that both expressions should be positive.
2. When \(a=1\), we have the solution \(a=1\), which satisfies the condition for both expressions to be positive.
3. When \(a<1\), we have the solution \(a=-3\), but it does not satisfy the condition that both expressions should be positive.
Hence, the largest real value of "a" for which the equation has an infinite number of solutions is \(a = 1\).
So, the correct option is (C): \(1\)
Approach Solution -2
It is stated in the question that there are an endless number of solutions to the equation \(|x+a|+|x−1|=2\) for any value of x. When x in \(|x+a|\) and x in \(|x-1|\) cancel out, this is feasible.
Case I:
\(x + a < 0, x - 1 ≥ 0\)
\(- a - x + x - 1 = 2\)
\(a = -3\)
Case II:
\(x + a ≥ 0\) and \(x - 1 < 0\)
\(x + a - x + 1 = 2\)
\(a = 1\)
A has a maximum value of 1.
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