The given function is
f(x)={kx+1,if x≤5
3x-5,if x>5 at x=5
The given function f is continuous at x=5,if f is defined at x=5 and if the value of the f at x=5 equals the limit of f at x=5.
It is evident that f is defined at x=5 and f(5)=kx+1=5k+1
limx→5- f(x)=limx→5+f(x)=f(5)
⇒limx→5-(kx+1)=limx→5+(3x-5)=5k+1
⇒5k+1=15-5=5k+1
⇒5k+1=10
⇒5k=9
k=9/5
Therefore, the required value of k is 9/5