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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
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Verified Answer
The correct answer is:
–1
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.
$$
f(x)=\left\{\begin{array}{l}
4 x-5, x \leq 2 \\
x-k, x>2
\end{array}\right.
$$
$\lim _{x \rightarrow 2} f(x)$ 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.
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