Discuss the continuity on 0 ≤ x ≤ 1 & differentiability at x = 0 for the function. f x = x sin 1 x sin 1 x sin 1 x where x ≠ 0 , x ≠ 1 r π & f 0 = f 1 r π = 0 , r = 1 , 2 , 3 , … … .

Given f x = x sin 1 x sin 1 x sin 1 x f 0 + = lim h → 0 f 0 + h = lim h → 0 h sin 1 h sin 1 h sin 1 h = 0 × ( - 1 to 1 ) = 0 ⇒ f 0 = f 0 + = 0 So f is continuous at x = 0 Continuity at x = 1 , f 1 = sin 1 sin 1 sin 1 f 1 - = lim h → 0 1 - h sin 1 1 - h sin 1 1 - h sin 1 1 - h = sin 1 sin 1 sin 1 ⇒ f 1 - = f 1 = Real So, f is continuous at x = 1 Let x = a ∈ 0 , 1 f a = sin 1 a sin 1 1 a sin 1 a f a + = lim h → 0 a + h sin 1 a + h sin 1 a + h sin 1 a + h = sin 1 a sin 1 1 a sin 1 a f a - = lim h → 0 a - h sin 1 a - h sin 1 a - h sin 1 a - h = sin 1 a sin 1 1 a sin 1 a ⇒ f a = f a + = f a - = Real So, f is continuous in 0 , 1 Therefore, f is continuous in 0 ≤ x ≤ 1 . Differentiability at x = 0 f ' 0 + = lim h → 0 f 0 + h - f 0 h = lim h → h sin 1 h sin 1 h sin 1 h - 0 h = lim h → sin 1 h sin 1 h sin 1 h = oscillate between - 1 to 1 ⇒ f is not differentiable at x = 0 Thus, f is continuous in 0 , 1 x = 0 but not differentiable at.

Let f be a function such that f x y = f x · f y ∀ x > 0 , y > 0 . If f 1 + x = 1 + x 1 + g x where lim x → 0 g x = 0 . Find ∫ f x f ' x d x

Given equation is f x y = f x · f y ∀ x > 0 , y > 0 . ⇒ f x = x k satisfies the given equation ∵ f 1 + x = 1 + x 1 + g x Put x = 0 we get f 1 = 1 as lim x → 0 g x = 0 f passes through the point 1 , 1 then ⇒ f x = x and f ' x = 1 Now ∫ f ( x ) f ' ( x ) d x = ∫ x 1 d x = ∫ x d x = x 2 2 + c

If | f p + q - f q | ≤ p q for all p and q ∈ Q & q ≠ 0 , show that ∑ i = 1 k f 2 k - f 2 i ≤ k ( k - 1 ) 2 .

Given, | f p + q - f q | ≤ p q , p and q ∈ Q , q ≠ 0 Now, f 2 k - f 2 i = f 2 k - f 2 k - 1 + f 2 k - 1 - f 2 k - 2 + . . . . . . . . . + f 2 i + 1 - f 2 i ⇒ f 2 k - f 2 i ≤ f 2 k - f 2 k - 1 + f 2 k - 1 - f 2 k - 2 + . . . . . . . f 2 i + 1 - f 2 i . . . i ∵ a + b ≤ a + b Let us consider f 2 k - 1 + 2 k - 1 - f 2 k - 1 ≤ 2 k - 1 2 k - 1 { ∵ Given, | f p + q - f q | ≤ p q } ⇒ f 2 k - f 2 k - 1 ≤ 1 . ∵ x m · x n = x m + n Similarly, f 2 k - 1 - f 2 k - 2 ≤ 1 . . . . . . . . . . . . . . . . . . . . . . . f 2 i + 1 - f 2 i ≤ 1 So, f 2 k - f 2 k - 1 + f 2 k - 1 - f 2 k - 2 + . . . . . . . f 2 i + 1 - f 2 i ≤ 1 + 1 + . . . . . k - i term ⇒ f 2 k - f 2 k - 1 + f 2 k - 1 - f 2 k - 2 + . . . . . . . f 2 i + 1 - f 2 i ≤ k - i From equation i , f 2 k - f 2 i ≤ k - i ⇒ ∑ i = 1 k f 2 k - f 2 i ≤ ∑ i = 1 k k - i ⇒ ∑ i = 1 k f 2 k - f 2 i ≤ k - 1 + k - 2 + . . . . . + 1 ⇒ ∑ i = 1 k f 2 k - f 2 i ≤ k k - 1 2 { ∵ Sum of first n natural number is n n + 1 2 } Hence, proved.