Let f : R → R be defined as f x = a − b cos 2 x x 2 ; x < 0 x 2 + c x + 2 ; 0 ≤ x ≤ 1 2 x + 1 ; x > 1 If f is continuous everywhere in R and m is the number of points where f is NOT differential then m + a + b + c equals:

Step-by-step solution

At x = 1 , f x is continuous, ⇒ f 1 − = f 1 = f 1 + ⇒ f 1 = 3 + c . . . i ⇒ f 1 + = lim h → 0 2 1 + h + 1 ⇒ f 1 + = lim h → 0 3 + 2 h = 3 . . . i i Using i and i i , ⇒ c = 0 At x = 0 , f x is continuous, ⇒ f 0 − = f 0 = f 0 + . . . i i i ⇒ f 0 = f 0 + = 2 . . . i v So, f 0 − has to be equal to 2 . ⇒ lim h → 0 a − b cos 2 h h 2 ⇒ lim h → 0 a − b 1 − 4 h 2 2 ! + 16 h 4 4 ! + . .. h 2 ⇒ lim h → 0 a − b + b 2 h 2 − 2 3 h 4 . .. h 2 For limit to exist a − b = 0 and limit is 2 b . . . v Using, i i i , i v and v ⇒ a = b =