Let the function f : R → R be defined by f x = x 3 − x 2 + x − 1 sin x and let g : R → R be an arbitrary function. Let f g : R → R be the product function defined by f g x = f x g x . Then which of the following statements is/are TRUE?
C. If g is differentiable at x = 1 , then f g is differentiable at x = 1
Differentiability of f g at x = 1 Left-hand derivative : f g ' 1 - = lim h → 0 f g 1 - h − f g 1 - h = lim h → 0 1 - h 3 - 1 - h 2 - h sin 1 - h g 1 - h − 0 - h = lim h → 0 1 - h 2 + sin 1 - h g 1 - h   . . . . . . . i Right-hand derivative : f g ' 1 + = lim h → 0 f g 1 + h − f g 1 h = lim h → 0 1 + h 3 - 1 + h 2 + h sin 1 + h g 1 + h − 0 h = lim h → 0 1 + h 2 + sin 1 + h g 1 + h   . . . . . . . i i If g is continuous at x = 1 , then lim h → 0 g 1 + h = lim h → 0 g 1 - h = g 1   . . . . . . . . . i i i From equations i ,   i i   &   i i i , we get lim h → 0 f g ' 1 + = lim h → 0 f g ' 1 - = 1 + sin 1 g 1 ∴ f g is differentiable at x = 1 . So, option A is correct. Now, from equations i   &   i i , we can say that for f g to be differentiable, we need only g 1 + h = g 1 - h . But for g to be continuous, we need g 1 + h = g 1 - h = g 1 . So, option B is incorrect. Now, if g is differentiable at x = 1 and f is already differentiable at x = 1 as f ' 1 - = f ' 1 + = 1 + sin 1 , so product of two differentiable functions is also differentiable. ∴ f g is differentiable at x = 1 . So, option C is correct. Now, from option B , if f g is differentiable at x = 1 , we cannot guarantee g to be continuous at x = 1 . So, we also cannot guarantee g to be differentiable at x = 1 . So, option D is incorrect.
Related: Mathematics — Continuity and Differentiability · All PYQ Banks