Relations & Functions Question - None

Question

If $p$ and $q$ are +ve integers, $f$ is a function defined for +ve numbers and attains only positive values such that $f(xf(y)) = x^p y^q$, then

Accepted Answer

Step 1: Substitute x=1, y=1 into f(xf(y)) = x^py^q

f(f(1)) = 1

Step 2: Let f(1) = c where c > 0. Then f(c) = 1.

Step 3: Substitute x=1 into the original equation:

f(f(y)) = y^q

Step 4: Substitute y=1 into the original equation:

f(xf(1)) = x^p

f(xc) = x^p

Step 5: Let t = xc, so x = t/c. Then:

f(t) = (t/c)^p = t^p/c^p

Step 6: From Step 3, f(f(y)) = y^q. Substituting f(y) = y^p/c^p:

f(y^p/c^p) = y^q

Applying f(t) = t^p/c^p:

(y^p/c^p)^p/c^p = y^q

y^p²/c^p(p+1) = y^q

Step 7: For this to hold for all y > 0: p² = q and c^p(p+1) = 1, so c = 1.

Step 8: Verify: f(xf(y)) = f(xy) = (xy)^p = x^py^p. This equals x^py^q only if p = q.

Combined with p² = q, we get p² = p, so p = 1 and q = 1.

∴ Answer: A (p=1, q=1, and f(x)=x)

Question: