Answer:
At x = 2, \[f(2)=k(2)+5=2k+5\] \[\underset{x\to \,\,{{2}^{+}}}{\mathop{\lim }}\,\,\,f(x)=\underset{h\to \,\,0}{\mathop{\lim }}\,\,\,f(2+h)\] \[=\underset{h\to \,\,0}{\mathop{\lim }}\,\,\,[(2+h)-1]\] \[=\underset{h\to \,\,0}{\mathop{\lim }}\,\,\,(1+h)=1\] \[\underset{x\to \,\,{{2}^{-}}}{\mathop{\lim }}\,\,\,f(x)=\underset{h\to \,\,0}{\mathop{\lim }}\,\,\,f(2-h)\] \[=\underset{h\to \,\,0}{\mathop{\lim }}\,\,\,\{k(2-h)+5\}.\] \[=\underset{h\to \,\,0}{\mathop{\lim }}\,\,\,\{2k+5)-kh\}=2k+5\] Now, \[\underset{x\to \,\,2}{\mathop{\lim }}\,\,\,f(x)\] exists only when \[2k+5=\] 1 i.e. when \[k=-\,2.\] When \[k=-\,2,\] then \[\underset{x\to \,\,2}{\mathop{\lim }}\,\,\,f(x)=f(2)=1\] Hence, f(x) is continous at x = 2, when \[k=-\,2.\]
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