Limit Continuity And Differentiability Ques 92

Answer:

Correct Answer: 92.(b)

Solution:

92. Given function.

$$ f(x)=\begin{array}{ll} a|\pi - x| + 1, & x \leq 5 \\ b|x-\pi|+3, & x>5 \end{array} $$

and it is also given that $f(x)$ is continuous on the interval

Clearly,

$$ f(5)=a(5-\pi)+1 $$

$$ \begin{alignedat} \lim _{x \rightarrow 5^{-}} f(x) & =\lim _{h \rightarrow 0}[a|\pi-(5-h)|+1] \\ & =a(5-\pi)+1 \end{aligned} $$

and $\lim _{x \rightarrow 5^{+}} f(x)=\lim _{h \rightarrow 0}[b|(5+h)-\pi|+3]$

$$ =b(5-\pi)+3 $$

$\because$ Function $f(x)$ is continuous at $x=5$.

$$ \begin{aligned} & \therefore f(5)=\lim _{x \rightarrow 5^{+}} f(x)=\lim _{x \rightarrow 5^{-}} f(x) \\ & \Rightarrow \quad a(5-\pi)+1=b(5-\pi)+3 \\ & \Rightarrow \quad(a-b)(5-\pi)=2 \\ & \Rightarrow \quad a-b=\frac{2}{5-\pi} \end{aligned} $$