Domain of y = sqrt{log _{10} frac{3x - x^2}{2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For the function $y = \sqrt{\log _{10} \frac{3x - x^2}{2}}$ to be defined, the expression inside the square root must be non-negative: $\log _{10}

Vedclass · Accepted Answer

$1 \leq x \leq 2$

Vedclass · Answer

(C) For the function $y = \sqrt{\log _{10} \frac{3x - x^2}{2}}$ to be defined, the expression inside the square root must be non-negative: $\log _{10} \left( \frac{3x - x^2}{2} \right) \geq 0$ Since $\log _{10} 1 = 0$, we have: $\frac{3x - x^2}{2} \geq 1$ $3x - x^2 \geq 2$ $x^2 - 3x + 2 \leq 0$ $(x - 1)(x - 2) \leq 0$ This inequality holds for $x \in [1, 2]$. Additionally, the argument of the logarithm must be positive: $\frac{3x - x^2}{2} > 0$ $x(3 - x) > 0$ $x(x - 3) This holds for $x \in (0, 3)$. The intersection of $x \in [1, 2]$ and $x \in (0, 3)$ is $x \in [1, 2]$.

Domain of $y = \sqrt{\log _{10} \frac{3x - x^2}{2}}$ is

Similar Questions

Is it true that $x = e^{\log x}$ for all real $x$?

Function $f:(2, \infty) \rightarrow R$ defined by $f(x) = x^2 - 4x + 5$. The range of $f$ is $=$ . . . . . . .

The range of the function $f(x) = \frac{x-3}{5-x}, x \neq 5$ is

The domain of the real valued function $f(x) = \sqrt{9 - \sqrt{x^2 - 144}}$ is

The domain of the function $\sqrt{\log({x^2} - 6x + 6)}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module