Q: If $\large \sqrt{5x – x^2 -6} + \frac{\pi}{2}\int_{0}^{x}dt > x \int_{0}^{\pi} sin^2 t dt $ , then
(A) x ∈ (2 , 3)
(B) x ∈ (- ∞, 2) ∪ (3 , ∞)
(C) x ∈ (5/2 , 3)
(D) none of these
Sol: $\large \sqrt{5x – x^2 -6} + \frac{\pi x}{2} > x \int_{0}^{\pi} \frac{1-cos2x}{2} dt $
$\large \sqrt{5x – x^2 -6} + \frac{\pi x}{2} > \frac{\pi x}{2} + 0 $
$\large \sqrt{5x – x^2 -6} > 0 $
$\large x^2 – 5 x + 6 < 0 $
(x – 2)(x – 3) < 0
⇒ x ∈ (2, 3).
Hence (A) is the correct answer.