CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

The values of $$b$$ for which the function
$$f(x)=\begin{cases} 5x-4,\quad 0<x\le 1 \\ 4{ x }^{ 2 }+3bx,\quad 1<x<2 \end{cases}$$ is continuous at every point of its domain is


A
1
loader
B
0
loader
C
1
loader
D
133
loader

Solution

The correct option is B $$-1$$
Given, $$f(x)=\begin{cases} 5x-4,\quad 0<x\le 1 \\ 4{ x }^{ 2 }+3bx,\quad 1<x<2 \end{cases}$$ is continuous at every point of its domain.

$$f(x)$$ is continuous.

$$\Rightarrow L.H.L=R.H.L=f(1)\\$$
$$\Rightarrow \lim _{ x\rightarrow 1 }{ \quad 5x-4 } =\lim _{ x\rightarrow 1 }{ \quad 4{ x }^{ 2 }+3bx } = 1\\$$
$$\Rightarrow 1=4+3b=1\\$$
$$\Rightarrow 3b=-3\\$$
$$\therefore b=-1\\$$
Hence, option A.

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image