Question:
Function \(f(x)\) is defined as \(f(x) = \begin{cases} 3x, & x<1 \\ a-b, & x = 1 \\ 4b-a, & x>1 \end{cases}\) If \(f(x)\) is continuous at \(x = 1\), but discontinuous at \(x = 2\) then the locus of the point \((a, b)\) is a straight line excluding the point where it cuts the line
Function \(f(x)\) is defined as \(f(x) = \begin{cases} 3x, & x<1 \\ a-b, & x = 1 \\ 4b-a, & x>1 \end{cases}\) If \(f(x)\) is continuous at \(x = 1\), but discontinuous at \(x = 2\) then the locus of the point \((a, b)\) is a straight line excluding the point where it cuts the line
Show Hint
Use continuity conditions and exclusion conditions to find the locus.
Updated On: Apr 23, 2026
- \(y = 3\)
- \(y = 2\)
- \(y = 1\)
- \(y = 0\)
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The Correct Option is A
Solution and Explanation
Step 1: Formula / Definition}
\[ \text{Continuous at } x=1: \lim_{x \to 1^-} f(x) = f(1) = \lim_{x \to 1^+} f(x) \]
Step 2: Calculation / Simplification}
\(\lim_{x \to 1^-} f(x) = 3(1) = 3\)
\(f(1) = a - b\)
\(\lim_{x \to 1^+} f(x) = 4b - a\)
Continuity at \(x=1\): \(3 = a-b = 4b-a \Rightarrow a-b=3\)
Discontinuous at \(x=2\): \(f(2) \neq \lim_{x \to 2^-} f(x)\)
\(4b-a \neq 3(2) = 6\)
From \(a-b=3\), \(a=b+3 \Rightarrow 4b-(b+3) = 3b-3 \neq 6 \Rightarrow b \neq 3\)
Locus: \(a-b=3\) excluding \(b=3\) (intersects \(y=3\))
Step 3: Final Answer
\[ y = 3 \]
\[ \text{Continuous at } x=1: \lim_{x \to 1^-} f(x) = f(1) = \lim_{x \to 1^+} f(x) \]
Step 2: Calculation / Simplification}
\(\lim_{x \to 1^-} f(x) = 3(1) = 3\)
\(f(1) = a - b\)
\(\lim_{x \to 1^+} f(x) = 4b - a\)
Continuity at \(x=1\): \(3 = a-b = 4b-a \Rightarrow a-b=3\)
Discontinuous at \(x=2\): \(f(2) \neq \lim_{x \to 2^-} f(x)\)
\(4b-a \neq 3(2) = 6\)
From \(a-b=3\), \(a=b+3 \Rightarrow 4b-(b+3) = 3b-3 \neq 6 \Rightarrow b \neq 3\)
Locus: \(a-b=3\) excluding \(b=3\) (intersects \(y=3\))
Step 3: Final Answer
\[ y = 3 \]
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