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