Question:
If x is a real number, then $\sqrt{\log _{e} \frac{4 x-x^{2}}{3}}$ is a real number if and only if
- $1 \leq x \leq 2$
- $-3 \leq x \leq 3$
- $1 \leq x \leq 3$
- $-1 \leq x \leq 3$
Correct Answer: Option: 3
The expression will be real only if ${{\log }_{e}}\frac{4x-{{x}^{2}}}{3}\ge 0$
Or $\frac{4x-{{x}^{2}}}{3}\ge {{e}^{0}}$
$\Rightarrow \frac{4x-{{x}^{2}}}{3}\ge 1$
$\Rightarrow 4x-{{x}^{2}}\ge 3$
$\Rightarrow {{x}^{2}}-4x+3\le 0$
$\Rightarrow \left( x-1 \right)\left( x-3 \right)\le 0$
$1 \leq x \leq 3$
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