Q. 22 - Calculus

Question

If possible, find constants a and b so that the function f that follows is continuous and differentiable everywhere. If it is not possible, explain why not.

$\mathrm{f}\left(\mathrm{x}\right) = \left\{\begin{array}{l}\mathrm{ax}-\mathrm{b}, \mathrm{if} \mathrm{x}<2\\ {\mathrm{bx}}^2 + 1, \mathrm{if} \mathrm{x}\ge 2\end{array}\right.$

Step-by-Step Solution

Answer

$\mathrm{Ans}: \mathrm{b}=\frac{1}{5} \& \mathrm{a}=\frac{4}{5}$

Step 1. Given information is:

$\mathrm{f}\left(\mathrm{x}\right) = \left\{\begin{array}{l}\mathrm{ax}-\mathrm{b}, \mathrm{if} \mathrm{x}<2\\ {\mathrm{bx}}^2 + 1, \mathrm{if} \mathrm{x}\ge 2\end{array}\right.$

Step 2. Calculating a and b

$$\begin{gathered} \mathrm{Differentiating} \mathrm{the} \mathrm{function}, \\ \mathrm{f}\text{'}\left(\mathrm{x}\right) = \left\{\begin{array}{l}\mathrm{a}, \mathrm{if} \mathrm{x}<2\\ 2\mathrm{bx}, \mathrm{if} \mathrm{x}\ge 2\end{array}\right. \\ \mathrm{Since} \mathrm{it} \mathrm{is} \mathrm{continuous} \mathrm{and} \mathrm{differentiable} \mathrm{therefore} \mathrm{left} \mathrm{hand} \mathrm{derivative} \mathrm{at} \mathrm{x}=2, \\ \mathrm{is} \mathrm{a} \mathrm{and} \mathrm{right} \mathrm{hand} \mathrm{derivative} \mathrm{at} \mathrm{x}=2 \mathrm{is} \mathrm{f}\text{'}\left(2\right)=4\mathrm{b}. \\ \mathrm{Therefore}, \\ \mathrm{a}=4\mathrm{b} .....\left(1\right) \\ \mathrm{Now} \mathrm{the} \mathrm{function} \mathrm{is} \mathrm{continuous} \mathrm{at} \mathrm{x}=2 \\ \mathrm{Therefore}, \\ \underset{\mathrm{x}\to 2^{-}}{\lim }\mathrm{ax}+\mathrm{b} = \underset{\mathrm{x}\to 2^{+}}{\lim }{\mathrm{bx}}^2+1 \\ 2\mathrm{a}+\mathrm{b} = 4\mathrm{b}+1 \\ 2\left(4\mathrm{b}\right)+\mathrm{b} = 4\mathrm{b}+1 [\mathrm{From} (1\left)\right] \\ 5\mathrm{b} = 1 \\ \mathrm{b} = \frac{1}{5} \\ \mathrm{Put} \mathrm{value} \mathrm{of} \mathrm{b} \mathrm{in} \left(1\right) \\ \mathrm{a} = \frac{4}{5} \end{gathered}$$