Question

For what value of λ is the function defined by
\(f(x)=\left\{\begin{matrix} \lambda (x^2-2x) &if\,x\leq0 \\   4x+1&if\,x>0  \end{matrix}\right.\)
continuous at x=0? What about continuity at x=1?

Answer 1

The given function is \(f(x)=\left\{\begin{matrix} \lambda (x^2-2x) &if\,x\leq0 \\   4x+1&if\,x>0  \end{matrix}\right.\)
If f is continuous at x=0,then
\(\lim_{x\rightarrow 0^-}\) f(x)=\(\lim_{x\rightarrow 0^+}\)f(x)=f(0)
⇒\(\lim_{x\rightarrow 0^-}\)\(\lambda(x^2-2x)\)=\(\lim_{x\rightarrow 0^+}\)(4x+1)=\(\lambda(0^2-2\times 0 )\)
⇒\(\lambda(0^2-2\times 0 )\)=4x0+1=0
⇒0=1=0, which is not possible.
Therefore, there is no value of λ for which f is continuous at x=0

At x=1,f(1)=4x+1=4×1+1=5
\(\lim_{x\rightarrow 1}\)(4x+1)=4x1+1=5
∴\(\lim_{x\rightarrow 1}\)f(x)=f(1)
Therefore, for any values of λ,f is continuous at x=1