Let f x = sin ⁡ π x x 2 ,   x > 0 . Let x 1 < x 2 < x 3 < … < x n < … be all the points of local maximum of f and y 1 < y 2 < y 3 < … < y n < … be all the points of local minimum of f . Then which of the following options is/are correct?
D. x n + 1 - x n > 2 for every n
f ' x = 2 x cos π x π x 2 - tan π x x 4 …(i) ⇒ for maxima/minima f ' x = 0 ⇒ cos π x = 0 o r π x 2 = tan π x ∵ cos π x ≠ 0 ∵ tan π x will not be defined ∴ maxima/minima will occur Where tan π x = π x 2 Case I : i For example In x ∈ 2 n , 2 n + 1 2 at point P 2 , x ∈ 2 , 5 2 cos π x > 0 and at P 2 + , tan π x > π x 2 at P 2 - , tan π x π x 2 hence from equation (i) in x ∈ 2 n , 2 n + 1 2 , f ' x goes from positive to negative hence P 2 is maxima Similarly P 2 , P 4 , P 6 … are point of maxima and all lies in x ∈ 2 n , 2 n + 1 2 Case I I In x ∈ 2 n + 1 , 2 n + 3 2 i i for example at P 1 , x ∈ 1 , 3 2 cos π x 0 at P 1 + , tan π x > π x 2 and at P 1 - , tan π x π x 2 hence from equation (i) in x ∈ 2 n + 1 , 2 n + 3 2 , f ' x goes from negative to positive ⇒ so, for the minima y 1 : 1 y 1 3 2 y 2 : 3 y 2 7 2 y 3 : 5 y 3 11 2 and for the maxima x 1 : 2 x 1 5 2 x 2 : 4 x 2 9 2 x 3 : 6 x 3 13 2 Hence x n > y n … a ⇒ now, x 1 > y 1 tan π x 1 > tan π y 1 tan π x 1 > tan π + π y 1 tan π x 1 > tan π 1 + y 1 π x 1 > π 1 + y 1 x 1 > 1 + y 1 Similarly, x n > 1 + y n x n - y n > 1 …(b) x n - y n > 1 ⇒ y 2 > x 1 tan π y 2 > tan π x 1 tan π y 2 > tan π + π x 1 ⇒ y 2 > x 1 + 1 ⇒ y n + 1 > x n + 1 Therefore x n y n + 1 x n + 1 ⇒ y n + 1 - x n > 1 …(c) Now consider x n + 1 - x n = x n + 1 - y n + 1 + y n + 1 - x n From (b) and (c) x n + 1 - x n > 2
Related: Mathematics — Application of Derivatives · All PYQ Banks