question_answer

Find the value of k for which \[f(x)=\left\{ \begin{matrix}    kx+5, & \text{when}\,\,x\le 2  \\    x-1, & \text{when}\,\,x>2  \\ \end{matrix} \right.\]             Is continuous at x = 2.

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 \[k=-\,2.\]