Let f 1 :   R → R ,   f 2 - π 2 , π 2   → R ,   f 3 : - 1 ,   e π 2 - 2 → R and f 4 :   ℝ → ℝ   be functions defined by (i) f 1 x = sin ⁡ 1 - e - x 2 (ii) f 2 ( x ) = sin x tan - 1 x ,   x ≠ 0             1 ,   x = 0 , where the inverse trigonometric function tan − 1 x assume

Step-by-step solution

(i) f 1 x = sin 1 - e - x 2 . Clearly f 1 ( x ) is continuous at x = 0 and f 0 = 0 f 1 ′ 0 = lim x → 0 sin 1 - e - x 2 1 - e - x 2 · 1 - e - x 2 x 2 · x x = 1 · 1 · lim x → 0 x x , which does not exist. So it is not differentiable at x = 0 . So, P → 2 (ii) f 2 x = sin x tan - 1 x , x ≠ 0 0 x = 0 lim x → 0 + sin x x x tan - 1 x = 1 and lim x → 0 − − sin x x × x tan − 1 x = − 1 ⇒ f 2 x is not continuous at x = 0 So Q &#85