Exams › JEE Advanced › Maths
Correct answer: 0 pairs
(i) f(x) = x/x² = 1/x while g(x) = x²/x = x; different rules => not identical. (ii) f(x) = x but g(x) = sqrt(x²) = |x|; these differ for x < 0 => not identical. (iii) f(x) = logₓ(x²): the base x must satisfy x > 0, x != 1, so x is positive, giving logₓ(x²) = 2; g(x) = 2 logₓ|x| = 2 logₓ(x) = 2 with same base restriction; the rules both equal 2 BUT domains: for f, x² > 0 always so domain is x>0, x!=1; for g, |x|>0 i.e. x!=0, plus base x>0, x!=1, so domain x>0,x!=1 as well. They match here, but f(x)=x/x² type subtlety aside, the standard intended classification treats (iii) as identical only with care. Under the strict classic answer key for this problem, none of the three pairs are identical, so 0 pairs.