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CAT 2019 Slot 2 — QA Question 19

LogarithmsEasy

If x is a real number, then loge(4x-x23) is a real number if and only if

Answer & solution

  • A

    -3 ≤ x ≤ 3

  • B

    1 ≤ x ≤ 2

  • 1 ≤ x ≤ 3

  • D

    -1 ≤ x ≤ 3

Solution

Easy

A square root is real only when its argument is non-negative, so the logarithm must be ≥0\ge 0. Since ln⁡t≥0  ⟺  t≥1\ln t\ge 0\iff t\ge 1, this forces 4x−x23≥1\tfrac{4x-x^{2}}{3}\ge 1, a quadratic inequality to solve.

1

Square-root condition. For log⁡e ⁣(4x−x23)\sqrt{\log_{e}\!\left(\tfrac{4x-x^{2}}{3}\right)} to be real, the log must be ≥0\ge 0, and log⁡et≥0\log_e t\ge 0 exactly when t≥1t\ge 1.

log⁡e ⁣(4x−x23)≥0⇒ 4x−x23≥1(ln⁡t≥0  ⟺  t≥1)\begin{aligned} &\log_{e}\!\left(\frac{4x-x^{2}}{3}\right)\ge 0\\ &\Rightarrow\ \frac{4x-x^{2}}{3}\ge 1 \quad\text{(}\ln t\ge0\iff t\ge1\text{)} \end{aligned}
2

Solve the quadratic inequality. Clear the denominator and rearrange to a standard form.

4x−x2≥3⇒ x2−4x+3≤0(move all to one side)⇒ (x−1)(x−3)≤0⇒ 1≤x≤3\begin{aligned} &4x-x^{2}\ge 3\\ &\Rightarrow\ x^{2}-4x+3\le 0 \quad\text{(move all to one side)}\\ &\Rightarrow\ (x-1)(x-3)\le 0\\ &\Rightarrow\ 1\le x\le 3 \end{aligned}
1≤x≤31\le x\le 3
CAT 2019 Slot 2 QA Q19: If x is a real number, then log e 4 x - x 2 3 is a real number if and only if — Solution | TheCATExam