See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

We have   2 2 4 5 5 0 A(say) sin x sin x f x 2 cost.f t dt cos x cos x        2 2 5 5 sin x sin x f x 2A cos x cos x      2 5 sin x f x 2A 1 cos x    ……..(1) Now,     2 2 4 4 5 4 0 0 sin t sin t A cos t. 2A 1 . dt 2A 1 dt cos t cos t           4 2 2 0 2A 1 tan .sec t dt    Put 2 tant y sec t dt dy     We get     1 2 0 1 A 2A 1 y dy 2A 1 3      3A 2A 1    A 1   Hence from equation (1), we get   2 5 3sin x f x cos x  (A) Clearly,  2 2 5 5 x x 3 3 3 3 2 3sin x Limf x Lim 72 cos x 1 2                      (B) As,  2 5 3sin x f x cos x  So,  f x is periodic with period 2. (C)  2 5 3sin x f x cos x   f ' x  For More Material Join: @JEEAdvanced_2025 AITS-FT-I (Paper-2)-PCM(Sol.)-JEE(Advanced)/2025 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 10     5 2 4 10 cos x. 2sinx.cos x sin x. 5cos x.sin x cos x            f ' 0     M x , y 0    So, equation of normal to the graph of  f x at point M whose abscissa is , is given by x 0  (D) As,  2 5 3sin x f x 0 0 sin x 0 cos x      x n , n I    So, the equation  f x 0  has no root in (0, 3).