Question

Domain of $y = \sqrt{\log_{10} \left( \frac{3x - x^2}{2} \right)}$ is

• A. x<1
• B. 2<x
• C. 1≤x≤2
• D. 2<x<3

Answer 1

The Correct Option is C

We are given the function \( y = \sqrt{\log_{10} \left( \frac{3x - x^2}{2} \right)} \). To find the domain, we need to ensure that the expression inside the square root is non-negative (since the square root of a negative number is undefined in the real number system) and that the argument of the logarithm is positive (since the logarithm of a non-positive number is undefined). Step 1: Conditions for the logarithmic function For the logarithmic function \( \log_{10} \left( \frac{3x - x^2}{2} \right) \) to be defined, the argument \( \frac{3x - x^2}{2} \) must be positive. Therefore, we need: \[ \frac{3x - x^2}{2}>0 \] Multiplying both sides by 2 (which does not change the inequality since 2 is positive): \[ 3x - x^2>0 \] This is a quadratic inequality. Factoring the quadratic expression: \[ x(3 - x)>0 \] The roots of the equation \( x(3 - x) = 0 \) are \( x = 0 \) and \( x = 3 \). Now, solve the inequality \( x(3 - x)>0 \) using a sign chart. The expression is positive between the roots, so: \[ 0<x<3 \] Step 2: Conditions for the square root For the square root to be defined, the logarithmic function must be non-negative. That is, we require: \[ \log_{10} \left( \frac{3x - x^2}{2} \right) \geq 0 \] This implies: \[ \frac{3x - x^2}{2} \geq 1 \] Multiplying both sides by 2: \[ 3x - x^2 \geq 2 \] Rearranging the terms: \[ x^2 - 3x + 2 \leq 0 \] Factoring the quadratic inequality: \[ (x - 1)(x - 2) \leq 0 \] The solution to this inequality is: \[ 1 \leq x \leq 2 \] Step 3: Combine the two conditions We need to combine the conditions \( 0<x<3 \) (from the logarithmic function) and \( 1 \leq x \leq 2 \) (from the square root condition). The intersection of these two intervals is: \[ 1 \leq x \leq 2 \] Conclusion Thus, the domain of the function is: \[ \boxed{1 \leq x \leq 2} \]