A) \[y=3\]
B) \[y=2\]
C) \[y=0\]
D) \[y=1\]
Correct Answer: A
Solution :
Given;\[f(x)\]is continuous at\[x=1\] \[\therefore \]\[f(1)=RHL\] \[\Rightarrow \]\[f(1)=\underset{x\to 1}{\mathop{\lim }}\,+f(x)\Rightarrow f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(1+h)\] \[\Rightarrow \]\[a-b=\underset{h\to 0}{\mathop{\lim }}\,3(1+h)\Rightarrow a-b=3\] ? (i) Again, given\[f(x)\]is discontinuous at\[x=2\]. \[\therefore \] \[LHL\ne f(x)\] \[\Rightarrow \]\[\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f(x)\ne f(2)\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,f(2-h)\ne f(2)\] \[\Rightarrow \]\[\underset{h\to 0}{\mathop{\lim }}\,3(2-h)\ne 4b-a\Rightarrow 6\ne 4b-a\] ... (ii) Assume,\[6=4b-a\] then from (i) and (ii), we get\[b=3\]. \[\therefore \]locus \[y=3\] Which is impossible \[(\because \,\,6\ne 4b-a)\] Hence, locus of\[(a,\,\,b)\]is\[x-y=3\]excluding the point when it cuts the line\[y=3\].You need to login to perform this action.
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