The true solution set of the inequality sqrt{ | vedclass

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(D) First,evaluate the integrals:$\int_{0}^{x} dz = [z]_{0}^{x} = x$$\int_{0}^{\pi} \sin^2 x dx = \int_{0}^{\pi} \frac{1 - \cos 2x}{2} dx = \left[ \fr

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$(2, 3)$

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(D) First,evaluate the integrals:$\int_{0}^{x} dz = [z]_{0}^{x} = x$$\int_{0}^{\pi} \sin^2 x dx = \int_{0}^{\pi} \frac{1 - \cos 2x}{2} dx = \left[ \frac{x}{2} - \frac{\sin 2x}{4} \right]_{0}^{\pi} = \frac{\pi}{2} - 0 = \frac{\pi}{2}$Substitute these into the inequality:$\sqrt{5x - 6 - x^2} + \frac{\pi}{2} x > x \left( \frac{\pi}{2} \right)$$\sqrt{5x - 6 - x^2} + \frac{\pi x}{2} > \frac{\pi x}{2}$Subtract $\frac{\pi x}{2}$ from both sides:$\sqrt{5x - 6 - x^2} > 0$For the square root to be defined and greater than zero,the expression inside must be strictly positive:$5x - 6 - x^2 > 0$$-(x^2 - 5x + 6) > 0$$x^2 - 5x + 6 $(x - 2)(x - 3) Thus,the solution set is $x \in (2, 3)$.

The true solution set of the inequality $\sqrt{5x-6-x^2} + \left( \frac{\pi}{2} \int_{0}^{x} dz \right) > x \int_{0}^{\pi} \sin^2 x dx$ is:

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