Question-8

Consider a function \(f\) defined as: \[ f(x) = \begin{cases} \displaystyle 3ax + b & x < 1, \\ 11 & x = 1,\\ 5ax - 2b & x > 1 \end{cases} \] If \(f\) is continuous at \(x=1\), then find the value of \(a+b\).

Answer

\(5\)

Solution

We have:

\[ f( x) =\begin{cases} {\displaystyle 3ax+b} & x< 1,\\ 11 & x=1,\\ 5ax-2b & x >1 \end{cases} \]

Since \(\displaystyle f\) is continuous at \(\displaystyle x=1\), we have LHL and RHL to be equal to the value of the function at \(\displaystyle x=1\). This gives us the following system of linear equations:

\[ \begin{aligned} 3a+b & =11\\ 5a-2b & =11 \end{aligned} \]

Solving this system, we get \(\displaystyle a=3,b=2\). Therefore, \(\displaystyle a+b=11\).