Manipal Entrance Test 2013 Math Paper

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.