Q42Numerical3 Marks20 Nov 2022Consider a function f(x)={mx2−n,x<12,x=1x+n,x>1.f(x)=\begin{cases}mx^2-n,&x<1\\2,&x=1\\x+n,&x>1.\end{cases}f(x)=⎩⎨⎧mx2−n,2,x+n,x<1x=1x>1. If fff is continuous at x=1x=1x=1, then find the value of m+nm+nm+n.Your answerPress Enter to check.
