Question

Plzz answer...... the following question​

Answer 1

Answer:

ANSWER A HAI

BECAUSE IT IS ALWAYS WRITTEN IN [ ] THIS TYPE OF BRACKETS

Answer 2

Answer:

None of the options is correct.

The correct answer is [0,1].

Step-by-step explanation:

Let, y = √(x-1)(3-x)

on squaring we get,

y^2 = (x-1)(3-x)

${y}^{2} = (x - 1)(3 - x) \\ = > {y}^{2} = { - x}^{2} + 4x - 3 \\ = > {x}^{2} - 4x + {y}^{2} + 3 = 0 \\ = > x = \frac{4 ± \sqrt{16 - 4(1)( {y}^{2} + 3) } }{2} \\ = > for \: x \: to \: be \: real \: 16 - 4( {y}^{2} + 3) \geqslant 0 \\ = > 4( {y}^{2} + 3) \leqslant 16 \\ = > {y}^{2} + 3 \leqslant 4 \\ = > {y}^{2} \leqslant 1 \\ = > y \leqslant 1 \: and \: y \geqslant - 1 \\ = > - 1 \leqslant y \leqslant 1$

But, since square root can never give negative value, so, 0 ≤ y ≤ 1.

Therefore, range of the given expression is [0,1].

Please mark it as brainliest.