1. If the sum of all the elements of a \( 3 \times 3 \) scalar matrix is 9 , then the product of all its elements i \( \begin{array}{llll}\text { (a) } 0 & \text { (b) } 9 & \text { (c) } 27\end{array} \) 2. Let \( f: R, \rightarrow[-5, a c) \) be defined \( a s f(x)=9 x^{2}+6 x-5 \), where \( R \), is the set of all non-negative re.

Let's address each problem one by one. --- ### **Problem 1** **Statement:** If the sum of all the elements of a \( 3 \times 3 \) scalar matrix is 9, then the product of all its elements is: \[ \begin{array}{llll} \text{(a) } 0 & \text{(b) } 9 & \text{(c) } 27 \end{array} \] **Solution:** A **scalar matrix** is a diagonal matrix where all the diagonal elements are equal, and the off-diagonal elements are zero. A general form of a \( 3 \times 3 \) scalar matrix is: \[ \begin{bmatrix} k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k \\ \end{bmatrix} \] **Sum of All Elements:** \[ \text{Sum} = k + 0 + 0 + 0 + k + 0 + 0 + 0 + k = 3k \] Given that the sum is 9: \[ 3k = 9 \implies k = 3 \] **Product of All Elements:** \[ \text{Product} = k \times 0 \times 0 \times 0 \times k \times 0 \times 0 \times 0 \times k = 0 \] Since the off-diagonal elements are zero, the entire product becomes zero regardless of the value of \( k \). **Answer:** **(a) 0** --- ### **Problem 2** **Statement:** Let \( f: \mathbb{R} \rightarrow [-5, a) \) be defined as \( f(x) = 9x^{2} + 6x - 5 \), where \( \mathbb{R} \) is the set of all non-negative real numbers. **Solution:** First, let's determine the range of the function \( f(x) = 9x^{2} + 6x - 5 \) for \( x \geq 0 \). 1. **Finding the Minimum Value:** - The function is a quadratic equation that opens upwards (since the coefficient of \( x^2 \) is positive). - The vertex of the parabola gives the minimum value. The vertex occurs at: \[ x = -\frac{b}{2a} = -\frac{6}{2 \times 9} = -\frac{1}{3} \] - However, since \( x \geq 0 \), the minimum value within the domain occurs at \( x = 0 \): \[ f(0) = 9(0)^2 + 6(0) - 5 = -5 \] 2. **Behavior as \( x \) Increases:** - As \( x \) increases beyond 0, \( f(x) \) increases without bound because it's a quadratic function opening upwards. - Therefore, \( f(x) \) approaches infinity as \( x \) approaches infinity. **Conclusion:** The range of \( f(x) \) for \( x \geq 0 \) is: \[ [-5, \infty) \] This implies that \( a = \infty \). **Answer:** The range of \( f(x) \) is \([-5, \infty)\), so \( a = \infty \). --- ### **Summary of Answers** 1. **Problem 1 Answer:** (a) 0 2. **Problem 2 Answer:** The range of \( f(x) \) is \([-5, \infty)\), so \( a = \infty \).