Question

Suppose $f\left( x \right)=\left\{ \begin{align}
  & a+\mu x,x<2 \\
 & 6,x=2 \\
 & \mu -ax,x>2 \\
\end{align} \right\}$ and $\displaystyle \lim_{x \to 2}f\left( x \right)=f\left( 2 \right)$. What are the possible values of $a$ and $\mu $?

Answer 1

Hint: Apply the continuity condition of a function f (x) at point x = k given as $\displaystyle \lim_{x \to k^-}f\left( x \right)=\displaystyle \lim_{x \to k^+}f\left( x \right)=f\left( k \right)$ where $\displaystyle \lim_{x \to k^-}f\left( x \right)$ is called the Left Hand Limit (L.H.L), $\displaystyle \lim_{x \to k^+}f\left( x \right)$ is called the Right Hand Limit (R.H.L) and $f\left( k \right)$ is called the value of the function at x = k. Substitute $f\left( 2 \right)=6$ and establish two linear relations between $a$ and $\mu $ to find their values.

Complete step by step solution:
Here we have been provide with the function $f\left( x \right)=\left\{ \begin{align}
  & a+\mu x,x<2 \\
 & 6,x=2 \\
 & \mu -ax,x>2 \\
\end{align} \right\}$ with the condition $\displaystyle \lim_{x \to 2}f\left( x \right)=f\left( 2 \right)$ and we are asked to find the values of $a$ and $\mu $.
Now, we can clearly see that the given function is continuous at x = 2 according to the given conditions in the question. Alternatively, we know that a function is continuous only when we have the condition $\displaystyle \lim_{x \to k^-}f\left( x \right)=\displaystyle \lim_{x \to k^+}f\left( x \right)=f\left( k \right)$ where $\displaystyle \lim_{x \to k^-}f\left( x \right)$ is called the Left Hand Limit (L.H.L), $\displaystyle \lim_{x \to k^+}f\left( x \right)$ is called the Right Hand Limit (R.H.L) and $f\left( k \right)$ is called the value of the function at x = k. Let us find the values of limits one by one.
(i) For the Left Hand Limit we have,
$\Rightarrow $ L.H.L = $\displaystyle \lim_{x \to 2^-}f\left( x \right)$
$\Rightarrow $ L.H.L = $\displaystyle \lim_{x \to 2^-}\left( a+\mu x \right)$
$\Rightarrow $ L.H.L = $\left( a+2\mu \right)$ ……… (1)
(ii) For the Right Hand Limit we have,
$\Rightarrow $ R.H.L = $\displaystyle \lim_{x \to 2^+}f\left( x \right)$
$\Rightarrow $ R.H.L = $\displaystyle \lim_{x \to 2^-}\left( \mu -ax \right)$
$\Rightarrow $ R.H.L = $\left( \mu -2a \right)$ ……… (2)
(iii) For the value of the function we have,
$\Rightarrow f\left( 2 \right)=6$ …….. (3)
So equating the three relations we get,
$\Rightarrow \left( a+2\mu \right)=\left( \mu -2a \right)=6$
Solving the two linear equations for the values of $a$ and $\mu $ we get,
$\therefore a=\dfrac{-6}{5}$ and $\mu =\dfrac{18}{5}$
Hence, the above values of $a$ and $\mu $ are our answer.

Note: You may check the answer by substituting the obtained values of $a$ and $\mu $ in the given function $f\left( x \right)$ and then evaluating the limit at x = 2. You must remember the condition for the existence of a limit around a given point. If any of the limits (R.H.L, L.H.L or value of the function) does not exist or is not a finite value then the limit does not exist at a particular point.