If x is a real number, then log e 4 x - x 2 3 is a real number if and only if

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1 &le; x &le; 3 — Easy A square root is real only when its argument is non-negative, so the logarithm must be $\ge 0$. Since $\ln t\ge 0\iff t\ge 1$, this forces $	frac{4x-x^{2}}{3}\ge 1$, a quadratic inequality to solve. 1 Square-root condition. For $\sqrt{\log_{e}\!\left(	frac{4x-x^{2}}{3}ight)}$ to be real, the log must be $\ge 0$, and $\log_e t\ge 0$ exactly when $t\ge 1$. $$\begin{aligned} &\log_{e}\!\left(\frac{4x-x^{2}}{3}ight)\ge 0\ &\Rightarrow\ \frac{4x-x^{2}}{3}\ge 1 \quad	ext{(}\ln t\ge0\iff t

CAT 2019 Slot 2 — QA Question 19

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