Question
The value of b for which the function $\text{f(x)}=\begin{cases}5\text{x}-4,&0<\text{x}\leq1\\4\text{x}^2+3\text{bx},&1<\text{x}<2\end{cases}$ is continuous at every point of its domain, is:
- -1
- 0
- $\frac{13}{3}$
- 1
Solution:
Given, f(x) is continuous at every point of its domain. So, it is continuous at x = 1.
$\Rightarrow\lim\limits_{\text{x}\rightarrow1^{+}}\text{f}\text{(x)}=\text{f}(1)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\text{f}(1+\text{h})=\text{f}(1)$
$\Rightarrow\lim\limits_{\text{h}\rightarrow0}\Big(4(1+\text{h})^2+3\text{b}(1+\text{h})\Big)=5(1)-4$
$\Rightarrow4+3\text{b}=1$
$\Rightarrow-3=3\text{b}$
$\Rightarrow \text{b} = -1$
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