Question

The value of a for which the function \[f\left( x \right) = \begin{cases}5x - 4 , & \text{ if } 0 < x \leq 1 \\ 4 x^2 + 3ax, & \text{ if } 1 < x < 2\end{cases}\] is continuous at every point of its domain, is 

Options

• \[\frac{13}{3}\] 

• 1

• 0

• −1

Answer 1

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

If   \[f\left( x \right)\]  is continuous in its domain, then it will be continuous at  \[x = 1\] .

Now,

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

Since f(x) is continuous at x = 1,

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

\[\Rightarrow 4 + 3a = 1\]
\[ \Rightarrow 3a = - 3\]
\[ \Rightarrow a = - 1\]