Question

If the function $\{\begin{array}{c}x,\text{if}x\le 1\\ cx+k,\text{if}1<x<4\\ -2x,\text{if}x\ge 4\end{array}$
is contionus everywhere, then the value of $c$ and $k$ are respectively :

• A. $-3,-5$
• B. $-3,5$
• C. $-3,-4$
• D. $-3,4$
• E. $-3,3$

Answer 1

Answer: (D)

$-3,4$

$\{\begin{array}{c}x,\text{if}x\le 1\\ cx+k,\text{if}1<x<4\\ -2x,\text{if}x\ge 4\end{array}$

Since, $f\left(x\right)$ is continous everywhere.

$\therefore \underset{x\to 1}{lim}f\left(x\right)=f\left(1\right)$ and $\underset{x\to 4}{lim}f\left(x\right)=f\left(4\right)$

$\therefore \underset{x\to 1}{lim}cx+k=1$ and $\underset{x\to 4}{lim}cx+k=-2\times 4$

$\Rightarrow c+k=1$ ......(1)

$4c+k=-8$ .....(2)

Subtracting equations (1) and (2), we get

$3c=-9$

$\Rightarrow c=-3$

Put $c=-3$ in equation $1$ We get,

$\therefore k=4$