Question

If the function f\left( x \right)=\left\\{ \begin{matrix} a\left| \pi -x \right|+1,x\le 5 \\\ b\left| \pi -x \right|+3,x>5 \\\ \end{matrix} \right. is continuous at $x=5$, then find the value of $a-b$?
(a) $\dfrac{2}{5-\pi }$,
(b) $\dfrac{2}{\pi -5}$,
(c) $\dfrac{2}{\pi +5}$,
(d) $\dfrac{-2}{\pi +5}$.

Answer 1

We start solving the problem by recalling the condition of a continuous function at a given value of x. We then find the left-hand limit for the function $f\left( x \right)$ at $x=5$ by using the properties of the modulus function. We then find the right-hand limit for the function $f\left( x \right)$ at $x=5$ by using the properties of the modulus function. We equate both these limits and make the necessary calculations to find the desired result.

Complete step-by-step solution
According to the problem, we have given that the function f\left( x \right)=\left\\{ \begin{matrix} a\left| \pi -x \right|+1,x\le 5 \\\ b\left| \pi -x \right|+3,x>5 \\\ \end{matrix} \right. is continuous at $x=5$. We need to find the value of $a-b$.
We know that if the function $g\left( x \right)$ is continuous at $x=a$, then the condition to be satisfied is $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,g\left( x \right)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,g\left( x \right)=g\left( a \right)$ ---(1).
Let us find the left-hand limit at $x=5$ for the given function $f\left( x \right)$.
So, we have $\underset{x\to {{5}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{5}^{-}}}{\mathop{\lim }}\,\left( a\left| \pi -x \right|+1 \right)$ ---(2).
Let us recall the definition of $\left| a-x \right|$. The modulus function $\left| a-x \right|$ is defined as \left| a-x \right|=\left\\{ \begin{matrix} a-x, x < a \\\ 0, x = a \\\ -\left( a-x \right), x > a \\\ \end{matrix} \right..
We know that the value of $\pi$ is $3.14$, which is less than 5.
So, we get the function $\left| \pi -x \right|$ as $-\left( \pi -x \right)$. We use this as equation (2).
$\Rightarrow \underset{x\to {{5}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{5}^{-}}}{\mathop{\lim }}\,\left( -a\left( \pi -x \right)+1 \right)$.
$\Rightarrow \underset{x\to {{5}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{5}^{-}}}{\mathop{\lim }}\,\left( ax-a\pi +1 \right)$.
We know that $\underset{x\to a}{\mathop{\lim }}\,\left( cx+d \right)=c\left( a \right)+d$.
$\Rightarrow \underset{x\to {{5}^{-}}}{\mathop{\lim }}\,f\left( x \right)=5a-a\pi +1$ ---(3).
Let us find the right-hand limit at $x=5$ for the given function $f\left( x \right)$.
So, we have $\underset{x\to {{5}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{5}^{+}}}{\mathop{\lim }}\,\left( b\left| \pi -x \right|+3 \right)$ ---(4).
Let us recall the definition of $\left| a-x \right|$. The modulus function $\left| a-x \right|$ is defined as \left| a-x \right|=\left\\{ \begin{matrix} a-x, x < a \\\ 0, x = a \\\ -\left( a-x \right), x > a \\\ \end{matrix} \right..
We know that the value of $\pi$ is $3.14$, which is less than 5.
So, we get the function $\left| \pi -x \right|$ as $-\left( \pi -x \right)$. We use this equation (4).
$\Rightarrow \underset{x\to {{5}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{5}^{+}}}{\mathop{\lim }}\,\left( -b\left( \pi -x \right)+3 \right)$.
$\Rightarrow \underset{x\to {{5}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{5}^{+}}}{\mathop{\lim }}\,\left( bx-b\pi +3 \right)$.
We know that $\underset{x\to a}{\mathop{\lim }}\,\left( cx+d \right)=c\left( a \right)+d$.
$\Rightarrow \underset{x\to {{5}^{+}}}{\mathop{\lim }}\,f\left( x \right)=5b-b\pi +3$ ---(5).
From equation (1), we have $\underset{x\to {{5}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{5}^{+}}}{\mathop{\lim }}\,f\left( x \right)$.
From equations (3) and (5), we get $5a-a\pi +1=5b-b\pi +3$.
$\Rightarrow a\left( 5-\pi \right)=b\left( 5-\pi \right)+3-1$.
$\Rightarrow a\left( 5-\pi \right)-b\left( 5-\pi \right)=2$.
$\Rightarrow \left( a-b \right)\left( 5-\pi \right)=2$.
$\Rightarrow a-b=\dfrac{2}{5-\pi }$.
We have found the value of $a-b$ as $\dfrac{2}{5-\pi }$.
$\therefore$ The correct option for the given problem is (a).

Note: We can also equate any of the limits to the value of $f\left( 5 \right)$, which will be the same as the left-hand limit. We need not always check left-hand and right-hand limits of the functions that are not varied between the intervals. We should not make calculation mistakes while solving this problem. We should always have left-hand and right-hand limits of the given function as modulus, greatest integer, and fractional part functions.