For all `'x', x^(2)+2ax+10-3a gt 0`,
Statement 1 : the interval in which 'a' lies is `-5 lt a lt 2`
because
Statement 2 : the sign of coefficient of `x^(2)` and the quadratic expression are same for all R iff the discriminant is negative.

Options

A. Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 2
B. Statement - 1 is True, Statement - 2 is True, Statement - 2 is NOT a correct explanation for statement - 2
C. Statement - 1 is True, Statement - 2 is False
D. Statement - 1 is False, Statement - 2 is True

Answer

To solve the inequality \( x^2 + 2ax + (10 - 3a) > 0 \) for all \( x \in \mathbb{R} \), we need to analyze the quadratic expression and determine the conditions under which it is always positive.

### Step 1: Identify the coefficients
The given quadratic expression can be compared with the standard form \( ax^2 + bx + c \):
- Here, \( a = 1 \)
- \( b = 2a \)
- \( c = 10 - 3a \)