Given, f(x)=loge(6−|x2+x−6|) The function f(x) is defined, if (6−|x2+x−6|)>0 ⇒‌‌|x2+x−6|<6 ⇒−6<x2+x−6<6 if x2+x−6<6 ⇒‌‌x2+x−12<0 ⇒‌‌(x+4)(x−3)<0
∴x∈(−4,3) ....(i) Now, if $-6⇒x2+x>0⇒x(x+1)>0
∴‌‌x∈(0,∞)‌‌... (ii) From Eqs. (i) and (ii), we get x∈(0,3) ∵f(x) has only two integral values. ∴‌‌x=1,2