Dicuss the continuity of the function, f x = x . sin 1 x , x ≠ 0 0 , x = 0

The given function is f x = x sin 1 x , x ≠ 0 0 , x = 0 at x = 0 Here, f x is said to be continuous at x = 0 , if lim x → 0 - f x = lim x → 0 + f x = f 0 Now, L . H . L = lim x → 0 - f x = lim x → 0 x sin 1 x = 0 × any definite number between - 1 to 1 . { ∵ x → 0 ⇒ 1 x → ∞ ∴ Value of sin 1 x Start oscillating between - 1 to 1 and does approach to a definite number. } L . H . L = 0 … … … … 1 R . H . L = lim x → 0 + f x = lim x → 0 x sin 1 x = 0 × any definite number between - 1 to 1 . R . H . L = 0 … … … … 2 also f 0 = 0 … … … 3 So from equations 1 , 2 and 3 , we get lim x → 0 - f x = lim x → 0 + f x = f 0 Thus, f x is continuous at x = 0 .

Prove that the function f ( x ) = x 2 - 1 ; x ≥ 1 1 - x ; x < 1 is not differentiable at x = 1

To check the differentiability of a function at x = 1 , we need to find L H D & R H D Notice that f ( 1 ) = 1 2 - 1 = 0 R H D = lim h → 0 f ( 1 + h ) - f ( 1 ) h ⇒ R H D = lim h → 0 ( 1 + h ) 2 - 1 - 0 h ⇒ R H D = lim h → 0 h 2 + 2 h h ⇒ R H D = lim h → 0 h + 2 = 2 Now, L H D = lim h → 0 f ( 1 ) - f ( 1 - h ) h ⇒ L H D = lim h → 0 0 - ( 1 - ( 1 - h ) ) h ⇒ L H D = lim h → 0 - h h = - 1 So even though, L H D , R H D are finite, but they are not equal. So the function is not differentiable at x = 1

Prove that the function f ( x ) = x 2 - 1 ; x ≥ 1 1 - x ; x < 1 is not differentiable at x = 1

To check the differentiability of a function at x = 1 , we need to find L H D & R H D Notice that f ( 1 ) = 1 2 - 1 = 0 R H D = lim h → 0 f ( 1 + h ) - f ( 1 ) h ⇒ R H D = lim h → 0 ( 1 + h ) 2 - 1 - 0 h ⇒ R H D = lim h → 0 h 2 + 2 h h ⇒ R H D = lim h → 0 h + 2 = 2 Now, L H D = lim h → 0 f ( 1 ) - f ( 1 - h ) h ⇒ L H D = lim h → 0 0 - ( 1 - ( 1 - h ) ) h ⇒ L H D = lim h → 0 - h h = - 1 So even though, L H D , R H D are finite, but they are not equal. So the function is not differentiable at x = 1

Dicuss the continuity of the function, f x = x . sin 1 x , x ≠ 0 0 , x = 0

The given function is f x = x sin 1 x , x ≠ 0 0 , x = 0 at x = 0 Here, f x is said to be continuous at x = 0 , if lim x → 0 - f x = lim x → 0 + f x = f 0 Now, L . H . L = lim x → 0 - f x = lim x → 0 x sin 1 x = 0 × any definite number between - 1 to 1 . { ∵ x → 0 ⇒ 1 x → ∞ ∴ Value of sin 1 x Start oscillating between - 1 to 1 and does approach to a definite number. } L . H . L = 0 … … … … 1 R . H . L = lim x → 0 + f x = lim x → 0 x sin 1 x = 0 × any definite number between - 1 to 1 . R . H . L = 0 … … … … 2 also f 0 = 0 … … … 3 So from equations 1 , 2 and 3 , we get lim x → 0 - f x = lim x → 0 + f x = f 0 Thus, f x is continuous at x = 0 .

Prove that the function is discontinuous: g ( x ) = 3 x , x < 3 3 , x = 3 x 2 , x > 3

For a function to be continuous at a point, it must be defined at that point, its limit must exist at the point, and the value of the function at that point must equal the value of the limit at that point. ⇒ lim x → 3 g x = g 3 and L . H . L = R . H . L Now, L . H . L = R . H . L Here, lim x → 3 - 3 x = lim x → 3 + x 2 ⇒ 3 × 3 = 3 2 Now, lim x → 3 g x = 9 but g 3 = 3 ⇒ lim x → 3 g x ≠ g 3 Thus, the given function is not continuous.

A particle moves according to the law S = 5 e 4 cos t , find its (a) velocity and (b) acceleration when t = π / 2 .

Given that, the particle moves according to the law S = 5 e 4 cos t We know that, velocity v = d S d t and acceleration a = d 2 S d t 2 . ⇒ v = d d t 5 e 4 cos t = - 5 e 4 sin t ⇒ v t = π 2 = d d t 5 e 4 cos t = - 5 e 4 sin t t = π 2 = - 5 e 4 Now, a = d d t - 5 e 4 sin t = - 5 e 4 cos t a t = π 2 = d d t - 5 e 4 sin t = - 5 e 4 cos t t = π 2 = 0 . Thus the value of velocity and acceleration at t = π 2 is - 5 e 4 and 0 .