Question

Write the value of b for which $f\left(x\right)=\left\{\begin{array}{ll}5x-4& 0<x\le 1\\ 4x^2+3bx& 1<x<2\end{array}\right.$ is continuous at x = 1.

Answer 1

Given: $f\left(x\right)=\left\{\begin{array}{l}5x-4,0<x\le 1\\ 4x^2+3bx,1<x<2\end{array}\right.$

If $f\left(x\right)$ is continuous at $x=1$, then
$\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right)=f\left(1\right)$ ...(1)

Now,
$\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(1-h\right)=\underset{h\to 0}{\lim }5\left(1-h\right)-4=5-4=1$

$\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(1+h\right)=\underset{h\to 0}{\lim }4{\left(1+h\right)}^2+3b\left(1+h\right)=4+3b$

Also, $f\left(1\right)=5\left(1\right)-4=1$


$$\begin{gathered} \underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right)=f\left(1\right)\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow 1=4+3b=1 \end{gathered}$$

$$\begin{gathered} \Rightarrow 1=4+3b \\ \Rightarrow -3=3b \\ \Rightarrow b=-1 \end{gathered}$$

Thus, for $b=-1$, the function $f\left(x\right)$ is continuous at $x=1$.