Discuss the continuity & differentiability of the function f ( x ) = sin x + sin | x | , x ∈ R . Draw a rough sketch of the graph of f x .

Given, f ( x ) = sin x + sin | x | , x = x x ≥ 0 - x - x < 0 Therefore, f ( x ) = sin x + sin x = 2 sin x x ≥ 0 sin x + sin - x = sin x - sin x = 0 x < 0 Continuity at x = 0 , f ( 0 ) = 0 LHL = lim x → 0 - f ( x ) = 0 RHL = lim x → 0 + 2 sin x = 0 Therefore, f ( x ) is continuous everywhere. LHD = lim x → 0 - f ' ( x ) = 0 RHD = lim x → 0 + f ' ( x ) = lim x → 0 + 2 cos x = 2 Since, LHD ≠ RHD Therefore, not differentiable at x = 0 .

Discuss the continuity & differentiability of the function f ( x ) = sin x + sin | x | , x ∈ R . Draw a rough sketch of the graph of f x .

Given, f ( x ) = sin x + sin | x | , x = x x ≥ 0 - x - x < 0 Therefore, f ( x ) = sin x + sin x = 2 sin x x ≥ 0 sin x + sin - x = sin x - sin x = 0 x < 0 Continuity at x = 0 , f ( 0 ) = 0 LHL = lim x → 0 - f ( x ) = 0 RHL = lim x → 0 + 2 sin x = 0 Therefore, f ( x ) is continuous everywhere. LHD = lim x → 0 - f ' ( x ) = 0 RHD = lim x → 0 + f ' ( x ) = lim x → 0 + 2 cos x = 2 Since, LHD ≠ RHD Therefore, not differentiable at x = 0 .

Find the set of values of m for which f ( x ) = x m sin 1 x x > 0 0 x = 0 (a) is discontinuous at x = 0 (b) is continuous but not derivabie at x = 0

Given, f ( x ) = x m sin 1 x x > 0 0 x = 0 (a) Hence, lim x → 0 + x m sin 1 x = lim x → 0 + h h m sin 1 h For discontinuity, f ( 0 + ) ≠ f ( 0 ) ⇒ h m sin 1 h ≠ 0 Case 1: for m > 0 ⇒ lim x → 0 + f ( 0 + ) = 0 Case 2: for m < 0 ⇒ lim x → 0 + f ( 0 - ) ≠ 0 Hence, f ( x ) is discontinuous for the set of values at x = 0 is m ∈ ( − ∞ , 0 ) (b) LHL = lim x → 0 - f ( x ) = lim h → 0 f ( 0 - h ) = lim h → 0 - h m sin - 1 h = lim h → 0 - - h m sin 1 h ⇒ 0 × k ; - 1 ≤ k ≤ 1 ⇒ 0 RHL = lim x → 0 + f ( x ) = lim h → 0 f ( 0 + h ) = lim h → 0 0 + h m sin 1 0 + h = lim h → 0 h m sin 1 h ⇒ 0 × k ' ; - 1 ≤ k ' ≤ 1 ⇒ 0 Therefore, LHL = f ( x ) = RHL, hence f ( x ) is continuous at x = 0 . Now, for Differentiability at x = 0 LHD at x = 0 = lim x → 0 - f ( x ) - f ( 0 ) x - 0 = lim h → 0 - f ( 0 - h ) - f ( 0 ) ( 0 - h ) - 0 = lim h → 0 - - h m sin - 1 h h = lim h → 0 - - - h m - 1 sin 1 h = Not defined RHD at x = 0 = lim x → 0 + f ( x ) - f ( 0 ) x - 0 = lim h → 0 + f ( 0 + h ) - f ( 0 ) ( 0 + h ) - 0 = lim h → 0 + h m sin 1 h h = lim h → 0 + h m - 1 sin 1 h = 0 Since, ( LHD at x = 0 ) ≠ ( RHD at x = 0 ) Hence, f ( x ) is continuous at x = 0 but not differentiable.

Find the set of values of m for which f x = x m sin 1 x x > 0 0 x = 0 , is continuous but not derivabie at x = 0

LHL = lim x → 0 - f ( x ) = lim h → 0 f ( 0 - h ) = lim h → 0 - h m sin - 1 h = lim h → 0 - - h m sin 1 h ⇒ 0 × k ; - 1 ≤ k ≤ 1 ⇒ 0 RHL = lim x → 0 + f ( x ) = lim h → 0 f ( 0 + h ) = lim h → 0 0 + h m sin 1 0 + h = lim h → 0 h m sin 1 h ⇒ 0 × k ' ; - 1 ≤ k ' ≤ 1 ⇒ 0 Therefore, LHL = f ( x ) = RHL, hence f ( x ) is continuous at x = 0 . Now, for Differentiability at x = 0 LHD at x = 0 = lim x → 0 - f ( x ) - f ( 0 ) x - 0 = lim h → 0 - f ( 0 - h ) - f ( 0 ) ( 0 - h ) - 0 = lim h → 0 - - h m sin - 1 h h = lim h → 0 - - - h m - 1 sin 1 h = Not defined RHD at x = 0 = lim x → 0 + f ( x ) - f ( 0 ) x - 0 = lim h → 0 + f ( 0 + h ) - f ( 0 ) ( 0 + h ) - 0 = lim h → 0 + h m sin 1 h h = lim h → 0 + h m - 1 sin 1 h = 0 Since, ( LHD at x = 0 ) ≠ ( RHD at x = 0 ) Hence, f ( x ) is continuous at x = 0 but not differentiable.

Let f ( 0 ) = 0 and f ' ( 0 ) = 1 . For a positive integer k , show that lim x → 0 1 x f x + f x 2 + … . . f x k = 1 + 1 2 + 1 3 + … … … + 1 k

Given: f 0 = 0 And, f ' 0 = lim h → 0 f 0 + h - f 0 h = 1 Now, lim x → 0 1 x f x + f x 2 + … . . f x k = lim x → 0 f x - f 0 x + 1 2 f x 2 - f 0 x 2 + 1 3 f x 3 - f 0 x 3 + . . . . + 1 k f x k - f 0 x k = f ' 0 + 1 2 f ' 0 + 1 3 f ' 0 + … … + 1 k f ' 0 = f ' 0 1 + 1 2 + 1 3 + … … . + 1 k = 1 + 1 2 + 1 3 + … … . + 1 k Hence proved

The function f is defined by y = f ( x ) , where x = 2 t - | t | , y = t 2 + t | t | , t ∈ R . Draw the graph of f for the interval - 1 ≤ x ≤ 1 . Also discuss its continuity & differentiability at x = 0 .

Given, x = 2 t - | t | , y = t 2 + t 2 = 2 t 2 Now, let t ≥ 0 , then x = 2 t − t = t , y = t 2 + t 2 = 2 t 2 ∴ y = 2 x 2 , x ≥ 0 [ ∵ t ≥ 0 ⇒ x ≥ 0 ] for, t ≤ 0 , x = 2 t + t = 3 t , y = t 2 − t 2 = 0 ∴ t = x 3 and t ≤ 0 ⇒ x ≤ 0 ∴ t = x 3 and t ≤ 0 ⇒ x ≤ 0 Therefore, the function is defined by, f ( x ) = 2 x 2 , x ≥ 0 and f ( x ) = y = 0 , x ≤ 0 The function is continuous at x = 0 , f ( 0 + 0 ) = lim h → 0 2 ( 0 + h ) 2 = 0 f ( 0 − 0 ) = 0 and f ( 0 ) = 0 It is also differentiable at x = 0 , RHD = f ' ( 0 ) = lim h → 0 2 ( 0 + h ) 2 - 0 h = lim h → 0 2 h = 0 LHD = f ' ( 0 ) = lim h → 0 0 - 0 - h = 0 Thus the function is continuous and differentiable for all x ∈ R Above is the graph.

Let f x = x e - 1 x + 1 x ; x ≠ 0 0 ; x = 0 , test the continuity & differentiability at x = 0 .

We have, f x = x e - 1 x + 1 x ; x ≠ 0 0 ; x = 0 ⇒ f x = x e - 2 x ; x > 0 0 ; x = 0 x ; x < 0 Now, f ' 0 + = lim h → 0 0 + h e - 2 0 + h - 0 h = lim h → 0 e - 2 h = lim h → 0 1 e 2 / h = 0 f ' 0 - = lim h → 0 0 - h - 0 - h = 1 Since, LHD ≠ RHD , hence f x is non derivable at x = 0 but continuous as RHD and LHD exist and are finite.

Examine for continuity & differentiability the points x = 1 and x = 2 , the function f defined by f x = x x , 0 ≤ x < 2 x - 1 x , 2 ≤ x ≤ 3 where x is greatest integer less than or equal to x .

Given: f x = x x , 0 ≤ x < 2 x - 1 x , 2 ≤ x ≤ 3 ⇒ f x = 0 , 0 ≤ x < 1 x , 1 ≤ x < 2 2 x - 1 , 2 ≤ x < 3 3 x - 1 , x = 3 Now, f ' 1 + = lim h → 0 ( 1 + h ) - 1 h = 1 f ' 1 - = lim h → 0 0 - 1 - h → ∞ Hence, f ( x ) is non-derivable at x = 1 and discontinuous. Again f ' 2 + = lim h → 0 2 ( 2 + h - 1 ) - 2 h = lim h → 0 2 + 2 h - 2 h = 2 f ' 2 - = lim h → 0 2 - h - 2 - h = lim h → 0 - h - h = 1 Since, LHD ≠ RHD , therefore f x is non-derivable at x = 2 but continuous at x = 2 as LHD & RHD are finite.

A function f : R → R , where R is a set of real numbers, satisfying the equation f x + y 3 = f x + f y + f 0 3 for all x , y in R . If the function is differentiable at x = 0 , then show that it is differentiable for all x in R .

Given, f x + y 3 = f x + f y + f 0 3 … 1 It is given that, f x is differentiable at x = 0 . Differentiating the equation (1) w.r.t. x , keeping y , as constant, we get, f ' x + y 3 · 1 3 = 1 3 · f ' x + 0 + 0 ⇒ f ' x + y 3 = f ' x … 2 Replace, x → 0 & y → 3 x in the equation (2), we get, f ' 0 + 3 x 3 = f ' 0 ⇒ f ' x = f ' 0 = m (say) ⇒ f x = m x + c , where c is the constant of integration. (Integrating both sides). Hence, f x is differentiable for all x ∈ R , because it is a straight line.