Question

 Find the values of a and b such that the function defined by
\(f(x)=\left\{\begin{matrix} 5, &if\,x\leq2 \\   ax+b,&if\,2<x<10 \\   21,&if\,x\geq10  \end{matrix}\right.\)

 is a continuous function. 

Answer 1

\(f(x)=\left\{\begin{matrix} 5, &if\,x\leq2 \\   ax+b,&if\,2<x<10 \\   21,&if\,x\geq10  \end{matrix}\right.\)

It is evident that the given function f is defined at all points of the real line. 
If f is a continuous function, then f is continuous at all real numbers. 
In particular,f is continuous at x=2 and x=10 Since f is continuous at x=2, we obtain
\(\lim_{x\rightarrow2^-}\) f(x)=\(\lim_{x\rightarrow2^+}\)f(x)=f(2)
\(\Rightarrow\)\(\lim_{x\rightarrow2^-}\)(5)=\(\lim_{x\rightarrow2^+}\)(ax+b)=5
\(\Rightarrow\)5=2a+b=5
\(\Rightarrow\)2a+b=5   ...(1)
Since f is continuous at x=10, we obtain
\(\lim_{x\rightarrow10^-}\) f(x)=\(\lim_{x\rightarrow10^+}\)f(x)=f(10)
\(\Rightarrow\)\(\lim_{x\rightarrow10^-}\)(ax+b)=\(\lim_{x\rightarrow10^+}\)(21)=21
\(\Rightarrow\)10a+b=21
\(\Rightarrow\)10a+b=21     ....(2)
On subtracting equation (1) from equation (2),
we obtain 8a=16 
\(\Rightarrow\)a=2
By putting a=2 in equation (1), 
we obtain 2×2+b=5 
\(\Rightarrow\)4+b=5
\(\Rightarrow\) b=1

Therefore, the values of a and b for which f is a continuous function are 2 and 1 respectively.