Question

Find the value of k such that the function ‘f’ defined by $f(x) = \left\{ \begin{gathered}
  \dfrac{{k\cos x}}{{\pi - 2x}},x \ne \dfrac{\pi }{2} \\
  3,x = \dfrac{\pi }{2} \\
\end{gathered} \right.$ is continuous at $x = \dfrac{\pi }{2}$

Answer 1

Hint: The function is said to be continuous at any given point means the function is defined at that point. We will find the right-hand limit and the left-hand limit of the function and they must be equal to the value of the function at the given point.

Complete step-by-step answer:
We have given a piecewise function:
$f(x) = \left\{ \begin{gathered}
  \dfrac{{k\cos x}}{{\pi - 2x}},x \ne \dfrac{\pi }{2} \\
  3,x = \dfrac{\pi }{2} \\
\end{gathered} \right.$
The goal of the problem is to value $k$, that makes the given function continuous.
As defined in the function if $x \ne \dfrac{\pi }{2}$, then the function is taken as:
$f\left( x \right) = \dfrac{{k\cos x}}{{\pi - 2x}}$
First, we find the left-hand limit of the given function at$x = \dfrac{\pi }{2}$. That is,
$\mathop {LHL = \lim }\limits_{h \to 0} f\left( {\dfrac{\pi }{2} - h} \right)$
Now choose the function at this argument:
$LHL = \mathop {\lim }\limits_{h \to 0} \dfrac{{k\cos \left( {\dfrac{\pi }{2} - h} \right)}}{{\pi - 2\left( {\dfrac{\pi }{2} - h} \right)}}$
Now, we can apply the following trigonometric identity:
$\cos \left( {\dfrac{\pi }{2} - h} \right) = \sinh $
After putting the value we get:
\[LHL = \mathop {\lim }\limits_{h \to 0} \dfrac{{k\sinh }}{{2h}}\]
Now, apply the following standard result of limits:
$\mathop {\lim }\limits_{h \to 0} \dfrac{{\sinh }}{h} = 1$
After applying the result, we get the left hand limit as:
$LHL = \dfrac{k}{2}$
Now, we need to show that the left hand limit of the function is equal to the value of the function at the given point. SO, find the value of the function at $x = \dfrac{\pi }{2}$. As given in the problem that when $x = \dfrac{\pi }{2}$, then the function is given as $f\left( x \right) = 3$. Now, set the value of the function equal to the left hand limit.
$LHL = {\text{Value of the function at }}\dfrac{\pi }{2}$
Substitute the values:
$\dfrac{k}{2} = 3$
Now, we solve the equation and get the value of $k$.
$k = 3 \times 2 = 6$
So, the given function is continuous when $k = 6$.

Note: While finding the right hand or the left-hand limit we must choose the correct function as the given function is a piecewise function. Any function has a limit at any point if the left hand limit and the right hand limit of the function is equal at that point.