The given function is f(x)={kx+1, if x≤53x−5, if x>5
The
given function f is continuous at x=5, if f is defined at x=5 and
if the value of 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=95
Therefore, the required value of k is 95