Question

The value of b for which the function
$f\left(x\right)=\left\{\begin{array}{ll}5x-4,& 0<x\le 1\\ 4x^2+3bx,& 1<x<2\end{array}\right.$is continuous at every point of its domain, is
(a) −1

(b) 0

(c) $\frac{13}{3}$

(d) 1

Answer 1

(a) −1

Given: $f\left(x\right)$ is continuous at every point of its domain. So, it is continuous at $x=1$.

$$\begin{gathered} \Rightarrow \underset{x\to 1^{+}}{\lim }f\left(x\right)=f\left(1\right) \\ \Rightarrow \underset{h\to 0}{\lim }f\left(1+h\right)=f\left(1\right) \\ \Rightarrow \underset{h\to 0}{\lim }\left(4{\left(1+h\right)}^2+3b\left(1+h\right)\right)=5\left(1\right)-4 \\ \Rightarrow 4+3b=1 \\ \Rightarrow -3=3b \\ \Rightarrow b=-1 \end{gathered}$$