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If x is a real number, then $\sqrt{\log _{e}\frac{4x-x^{2}}{3}}$ is a real number if and only if

  1. $1\leq x\leq 2$
  2. $-3\leq x\leq 3$
  3. $1\leq x\leq 3$
  4. $-1\leq x\leq 3$
in Quantitative Aptitude 11.5k points 213 1349 2323
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Ans is option (C)

For the value inside the square root to be real, it should be greater than or equal to zero.

$\therefore$   log$_{e}\frac{4x-x^{2}}{3}\geqslant0$

$\Rightarrow$  $\frac{4x-x^{2}}{3}\geqslant1$

$\Rightarrow$  $\frac{4x-x^{2}}{3}-1\geqslant0$

$\Rightarrow$  $\frac{4x-x^{2}-3}{3}\geqslant0$

Multiplying by 3 on both sides inequality, we get   $4x-x^{2}-3\geqslant0$

Multiplying by -1 on both sides inequality, we get   $x^{2}-4x+3\leqslant0$

$\Rightarrow$   $(x-1)(x-3)\leqslant0$     $\Rightarrow$   $x\in[1,3]$

238 points 2 2 4

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