If g(x) = 2f(2x^3 - 3x^2) + f(6x^2 - 4x^3 - 3 | vedclass

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(B) Given $f''(x) > 0$,$f'(x)$ is a strictly increasing function.$g'(x) = 2f'(2x^3 - 3x^2) \cdot (6x^2 - 6x) + f'(6x^2 - 4x^3 - 3) \cdot (12x - 12x^2)

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$\left( - \frac{1}{2},0 \right) \cup \left( 1,\infty \right)$

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(B) Given $f''(x) > 0$,$f'(x)$ is a strictly increasing function.$g'(x) = 2f'(2x^3 - 3x^2) \cdot (6x^2 - 6x) + f'(6x^2 - 4x^3 - 3) \cdot (12x - 12x^2)$$g'(x) = 12(x^2 - x) f'(2x^3 - 3x^2) - 12(x^2 - x) f'(6x^2 - 4x^3 - 3)$$g'(x) = 12x(x - 1) [f'(2x^3 - 3x^2) - f'(6x^2 - 4x^3 - 3)]$For $g'(x) > 0$,we analyze the sign of $12x(x - 1)$ and the difference $[f'(2x^3 - 3x^2) - f'(6x^2 - 4x^3 - 3)]$.Since $f'(x)$ is increasing,$f'(A) > f'(B) \iff A > B$.Let $A = 2x^3 - 3x^2$ and $B = 6x^2 - 4x^3 - 3$.$A - B = (2x^3 - 3x^2) - (6x^2 - 4x^3 - 3) = 6x^3 - 9x^2 + 3 = 3(2x^3 - 3x^2 + 1) = 3(x - 1)^2(2x + 1)$.Thus,$f'(A) - f'(B) > 0$ when $3(x - 1)^2(2x + 1) > 0$,which implies $x > -\frac{1}{2}$ (excluding $x = 1$).Now,$g'(x) > 0$ when $12x(x - 1)$ and $(f'(A) - f'(B))$ have the same sign.Case $I$: $x(x - 1) > 0$ and $x > -\frac{1}{2}$. This occurs when $x \in (1, \infty)$.Case $II$: $x(x - 1) Combining these,$g'(x) > 0$ for $x \in (-\frac{1}{2}, 0) \cup (1, \infty)$.

If $g(x) = 2f(2x^3 - 3x^2) + f(6x^2 - 4x^3 - 3)$,$\forall x \in R$ and $f''(x) > 0$,$\forall x \in R$,then $g'(x) > 0$ for $x$ belonging to

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