+1) ∴ Domain x∈R−{0} . . . (ii) ∴ Domain of sech−1(x)+cosech−1(x) is Eqs. (i) and (ii) ⇒‌‌x∈(0,1] ∴‌‌f(0)=sech−1(0)+cosech−1(0)=∞ f(l)=sech−1(l)+cosech−1(l)=0+ln(1+√2) ∴ Range =[(ln(1+√2),∞) Hence, a=ln(1+√2),b=∞