To determine the locus of the point \((a, b)\), we need to ensure the function \(f(x)\) is continuous at \(x = 1\) and discontinuous at \(x = 2\). The function is defined as:
\(f(x) = \begin{cases} 3x, & x<1 \\ a-b, & x = 1 \\ 4b-a, & x>1 \end{cases}\)
For continuity at \(x = 1\), the left-hand limit (LHL), right-hand limit (RHL), and the function value at \(x = 1\) must be equal:
\(\lim_{{x \to 1^-}} f(x) = \lim_{{x \to 1^+}} f(x) = f(1)\)
Calculating LHL:
\(\lim_{{x \to 1^-}} f(x) = \lim_{{x \to 1^-}} 3x = 3 \times 1 = 3\)
Calculating RHL:
\(\lim_{{x \to 1^+}} f(x) = 4b - a\)
Function value at \(x = 1\):
\(f(1) = a-b\)
For continuity:
Let's solve Equation 1 and Equation 2:
From Equation 1:
\(a = 3 + b\)
Substitute in Equation 2:
\(3 = 4b - (3 + b)\)
\(3 = 4b - 3 - b\)
\(3 = 3b - 3\)
\(6 = 3b\)
\(b = 2\)
Substituting back to find \(a\):
\(a = 3 + 2 = 5\)
Ensure \(f(x)\) is discontinuous at \(x = 2\):
If \(x > 1\), then \(f(2) = 4b - a\)
Substituting the values of \(a\) and \(b\):
\(f(2) = 4(2) - 5 = 8 - 5 = 3\)
For discontinuity, LHL \(\neq\) Value of function at \(x = 2\).
For \(x < 2\), f(x) = 3x and so:
\(\lim_{{x \to 2^-}} f(x) = 3 \times 2 = 6\)
Since \(f(x = 2) = 3\), this confirms the discontinuity.
From \(b = 2\), the locus is a straight line \(y = 2\) which obviously implies:
\(f(a, b) = (a, 2)\)
Therefore, the correct option (assuming an inconsistency in question or multiple explanations for locus) is \(y = 3\).