MCQ
If function $f(x)=\left\{\begin{array}{l}1+x, x \leq 2 \\ k-\frac{x}{2}, x>2\end{array}\right.$, is continuous at $x=2$, then value of $k$ will be :
  • A
    $1$
  • B
    $2$
  • C
    $3$
  • $4$

Answer

Correct option: D.
$4$
value of function at $x=2$
$ f(2) =1+2=3$
$f(2+h) =k-\left(\frac{2+h}{2}\right)$
$f(2-h) =1+(2-h)=(3-h) $
Value of $\text{R.H.L.}$
$f(2+0)=\lim _{h \rightarrow 0} f(2+h) $
$=\lim _{h \rightarrow 0}\left[k-\left(\frac{2+h}{2}\right)\right]$
$ =k-1$
Value of $\text{L.H.L.}$
$f(2-0)=\lim _{h \rightarrow 0} f(2-h)$
$=\lim _{h \rightarrow 0}[3-h]=3$
Function is continuous at $x=2$,
so.$f(2) =\text { R.H.L. }=\text { L.H.L. }$
$\Rightarrow 3 =k-1=3$
$\therefore k=4$
Hence correct option is $(D).$

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