The true solution set of the inequality,sqrt{ | vedclass

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(D) Given inequality: $\sqrt{5x - 6 - x^2} + \frac{\pi}{2} \int_{0}^{x} dz > x \int_{0}^{\pi} \sin^2 x \, dx$Step $1$: Evaluate the integrals.$\int_{0

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$(2, 3)$

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(D) Given inequality: $\sqrt{5x - 6 - x^2} + \frac{\pi}{2} \int_{0}^{x} dz > x \int_{0}^{\pi} \sin^2 x \, dx$Step $1$: Evaluate the integrals.$\int_{0}^{x} dz = [z]_{0}^{x} = x$$\int_{0}^{\pi} \sin^2 x \, dx = \int_{0}^{\pi} \frac{1 - \cos(2x)}{2} \, dx = [\frac{x}{2} - \frac{\sin(2x)}{4}]_{0}^{\pi} = \frac{\pi}{2}$Step $2$: Substitute the values into the inequality.$\sqrt{5x - 6 - x^2} + \frac{\pi}{2} (x) > x (\frac{\pi}{2})$$\sqrt{5x - 6 - x^2} + \frac{\pi x}{2} > \frac{\pi x}{2}$$\sqrt{5x - 6 - x^2} > 0$Step $3$: Solve for the domain of the square root.For the expression to be defined and greater than $0$,we need $5x - 6 - x^2 > 0$.$-(x^2 - 5x + 6) > 0$$x^2 - 5x + 6 $(x - 2)(x - 3) Thus,$x \in (2, 3)$.

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

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