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(B) Given the function $f(x) = x^3 + 2ax^2 + 3(a+1)x + 5$.For $f(x)$ to be strictly increasing in its entire domain,its derivative $f'(x)$ must be gre

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$(-\frac{3}{4}, 3)$

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(B) Given the function $f(x) = x^3 + 2ax^2 + 3(a+1)x + 5$.For $f(x)$ to be strictly increasing in its entire domain,its derivative $f'(x)$ must be greater than or equal to $0$ for all $x \in \mathbb{R}$,and $f'(x)$ should not be zero for all $x$ in any interval.First,find the derivative: $f'(x) = 3x^2 + 4ax + 3(a+1)$.Since $f'(x)$ is a quadratic expression with a positive leading coefficient $(3 > 0)$,for $f'(x) \geq 0$ to hold for all $x$,the discriminant $D$ must be less than or equal to $0$.$D = (4a)^2 - 4(3)(3(a+1)) \leq 0$.$16a^2 - 36(a+1) \leq 0$.Divide by $4$: $4a^2 - 9(a+1) \leq 0$.$4a^2 - 9a - 9 \leq 0$.Factor the quadratic: $(4a+3)(a-3) \leq 0$.This inequality holds when $a$ lies between the roots: $a \in [-\frac{3}{4}, 3]$.Since the question asks for strictly increasing,we consider the interval where $f'(x) > 0$ for all $x$ except possibly at isolated points,which corresponds to $a \in (-\frac{3}{4}, 3)$.

The set of all real values of $a$ such that the real valued function $f(x) = x^3 + 2ax^2 + 3(a+1)x + 5$ is strictly increasing in its entire domain is

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