CAT 2024 Slot 3 · Quant · Question 20

The relation links f\left( x \right) to f\left( \frac{1}{x}\right), so replacing x by \frac{1}{x} produces a second relation in the same two quantities.

The given relation is f\left( x \right)+2f\left( \frac{1}{x}\right)=3x.

Substituting \frac{1}{x} for x gives f\left( \frac{1}{x}\right)+2f\left( x \right)=\frac{3}{x}.

Multiply this second relation by 2: 2f\left( \frac{1}{x}\right)+4f\left( x \right)=\frac{6}{x}.

Subtracting the first relation from it, the f\left( \frac{1}{x}\right) terms cancel and 3f\left( x \right)=\frac{6}{x}-3x.

Dividing by 3, f\left( x \right)=\frac{2}{x}-x.

Setting f\left( x \right)=3 gives \frac{2}{x}-x=3.

Multiplying through by x, 2-{x}^{2}=3x, which rearranges to {x}^{2}+3x-2=0.

The discriminant is 9+8=17>0, so there are two distinct real roots.

By Vieta's relation, their sum is -\frac{3}{1}=-3 — the second option.

Shortcut. The question asks only for the sum of the roots, so the roots themselves are never needed. Once the equation reaches {x}^{2}+3x-2=0, Vieta gives -3 directly, and the positive discriminant confirms both roots are real.