Integrate : If $ \sqrt{5x – x^2 -6} + \frac{\pi}{2}\int_{0}^{x}dt > x \int_{0}^{\pi} sin^2 t dt $ , then

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.