Question

The value of b for which the function 

\[f\left( x \right) = \begin{cases}5x - 4 , & 0 < x \leq 1 \\ 4 x^2 + 3bx , & 1 < x < 2\end{cases}\] is continuous at every point of its domain, is 

पर्याय

• −1  

• 0

• \[\frac{13}{3}\] 

• 1

Answer 1

−1 

Given: 

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

\[\Rightarrow \lim_{x \to 1^+} f\left( x \right) = f\left( 1 \right)\]
\[ \Rightarrow \lim_{h \to 0} f\left( 1 + h \right) = f\left( 1 \right)\]
\[ \Rightarrow \lim_{h \to 0} \left( 4 \left( 1 + h \right)^2 + 3b\left( 1 + h \right) \right) = 5\left( 1 \right) - 4\]
\[ \Rightarrow 4 + 3b = 1\]
\[ \Rightarrow - 3 = 3b\]
\[ \Rightarrow b = - 1\]