If the function ƒ defined on -13,13 by
fx=1kloge1+3x1-2xwhenx≠0kwhenx=0
is continuous, then k is equal to:
Solve for value of k
Given that the function ƒ defined on -13,13 by fx=1kloge1+3x1-2xwhenx≠0kwhenx=0is continuous
We know that for any continuous function limx→0f(x)=fx-=fx+
So, for the given function, limx→01xloge1+3x1-2x=k
⇒limx→01xloge(1+3x)-loge(1-2x)=k∵logmn=logm-logn⇒limx→0loge(1+3x)x-loge(1-2x)x=k⇒limx→03loge(1+3x)3x+2loge(1-2x)-2x=k⇒3+2=k∵limx→0log1+xx=1⇒5=k
Therefore, f(x) is continuous when k=5.