Given; f(x) is continuous at x = 1 ∴ f(1) = RHL ⇒ f(1)=
lim
x→1
+f(x)→f(x)=
lim
h→0
f(1+h) ⇒ a−b=
lim
h→0
3(1+h)→a−b=3 .....(i) Again, given f{x) is discontinuous at x = 2. ∴ LHL ≠ f(x) ⇒
lim
x→2−
f(x)≠f(2)→
lim
h→0
f(2−h)≠f(2) ⇒
lim
h→0
3(2−h)≠4b−a→6≠4b−a .....(ii) Assume, 6 = 4b - a then from (i) and (ii), we get b = 3. ∴ locus y=3 Which is impossible (∵6≠4b−a) Hence, locus of (a, b) is x - y = 3 excluding the point when it cuts the line y = 3.