If $f(x)=\left\{\begin{array}{cc}4 x-5, & x \leq 2 \\ x-k, & x>2\end{array}\right.$ then the value of ' $k$…

If $f(x)=\left\{\begin{array}{cc}4 x-5, & x \leq 2 \\ x-k, & x>2\end{array}\right.$ then the value of ' $k$ ' if $\lim _{x \rightarrow 2} f(x)$ may exist is equal to
  1. –1
  2. –2
  3. 1
  4. 2

Solution

Given, $ f(x)=\left\{\begin{array}{l} 4 x-5, x \leq 2 \\ x-k, x>2 \end{array}\right. $ $\lim _{x \rightarrow 2} f(x)$ is exists when LHL at $x=2=$ RHL at $x=2$ $ \begin{aligned} \therefore \rightarrow 2 \quad \lim _{x \rightarrow 2^{-}} f(x) & =\lim _{x \rightarrow 2^{+}} f(x) \\ \lim _{x \rightarrow 2^{-}} 4 x-5 & =\lim _{x \rightarrow 2^{+}} x-k \\ 4(2)-5 & =2-k \\ 3 & =2-k \Rightarrow k=-1 \end{aligned} $ Hence, option (1) is correct

Asked in: AP EAMCET 2020 (22 Sep Shift 2)

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