2025 CSAT — Q79

For $0 < x < 1$, higher powers shrink the number: $1 > x > x^2 > x^3 > \cdots$, while roots grow it: $\sqrt{x} > x$.

Statement I: $x^2 > x^3$? Since $x^3 = x^2 \cdot x$ and $0 < x < 1$, multiplying $x^2$ by $x$ makes it smaller, so $x^2 > x^3$. TRUE. (Test $x = \tfrac12$: $\tfrac14 > \tfrac18$ ✓.)

Statement II: $x > \sqrt{x}$? For $0 < x < 1$, $\sqrt{x} = x^{1/2} > x^1$ (smaller exponent gives a larger value on $(0,1)$), so in fact $x < \sqrt{x}$. The claim $x > \sqrt{x}$ is FALSE. (Test $x = \tfrac14$: $\sqrt{x} = \tfrac12 > \tfrac14 = x$ ✓ the reverse.)

Only I holds.

Answer: (a) I only.

The single governing fact is the monotonicity of $x^a$ on $(0,1)$ in the exponent: as the exponent grows, the value falls ($x^3 < x^2 < x < \sqrt{x} < 1$). Statement I rides this directly (true); statement II reverses it (false, since $\sqrt{x} > x$). The trap is importing the "$x>1$" intuition where bigger powers mean bigger numbers. A single test value ($x=\tfrac14$, with $\sqrt{x}=\tfrac12$) both confirms I and kills II in seconds — always sanity-check unit-interval power claims with a clean fraction.