Given f(x) = 4x^3 - 6x^2 cos 2a + 3x sin 2a s | vedclass

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(D) The function $f(x)$ is defined only if the argument of the square root is non-negative,i.e.,$\ln(2a - a^2) \ge 0$.This implies $2a - a^2 \ge 1$,wh

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$f'(1/2) > 0$

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(D) The function $f(x)$ is defined only if the argument of the square root is non-negative,i.e.,$\ln(2a - a^2) \ge 0$.This implies $2a - a^2 \ge 1$,which simplifies to $a^2 - 2a + 1 \le 0$,or $(a - 1)^2 \le 0$.Since the square of a real number cannot be negative,we must have $(a - 1)^2 = 0$,which means $a = 1$.Substituting $a = 1$ into the expression for $f(x)$,we get $f(x) = 4x^3 - 6x^2 \cos 2 + 3x \sin 2 \sin 6$.Now,find the derivative $f'(x) = 12x^2 - 12x \cos 2 + 3 \sin 2 \sin 6$.Evaluating at $x = 1/2$,we get $f'(1/2) = 12(1/4) - 12(1/2) \cos 2 + 3 \sin 2 \sin 6 = 3 - 6 \cos 2 + 3 \sin 2 \sin 6$.Using the identity $\sin 2 \sin 6 = \frac{1}{2}(\cos(2-6) - \cos(2+6)) = \frac{1}{2}(\cos 4 - \cos 8)$,we have $f'(1/2) = 3 - 6 \cos 2 + \frac{3}{2}(\cos 4 - \cos 8)$.Since $\cos 2 \approx -0.416$,$-6 \cos 2 \approx 2.496$. Thus $f'(1/2) = 3 + 2.496 + \dots > 0$.

Given $f(x) = 4x^3 - 6x^2 \cos 2a + 3x \sin 2a \sin 6a + \sqrt{\ln(2a - a^2)}$,then:

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