Question

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

 

Answer 1

Given,  

\[f\left( x \right) = \begin{cases}5x - 4 & 0 < x \leq 1 \\ 4 x^2 + 3bx & 1 < x < 2\end{cases}\]

If  \[f\left( x \right)\]  is continuous at \[x = 1\] , then

\[\lim_{x \to 1^-} f\left( x \right) = \lim_{x \to 1^+} f\left( x \right) = f\left( 1 \right)\]            ...(1)

Now, 

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

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

 

Also,

\[f\left( 1 \right) = 5\left( 1 \right) - 4 = 1\]

\[\lim_{x \to 1^-} f\left( x \right) = \lim_{x \to 1^+} f\left( x \right) = f\left( 1 \right) \left[ \text{ From eq } . (1) \right]\]

\[ \Rightarrow 1 = 4 + 3b = 1\]

\[\Rightarrow 1 = 4 + 3b\]

\[ \Rightarrow - 3 = 3b\]

\[ \Rightarrow b = - 1\]

Thus, for 

\[b = - 1\] , the function 

\[f\left( x \right)\] is continuous at  \[x = 1\] .