Find all points of discontinuity of f , where f x = x 2 sin 1 x , if x ≠ 0 0 , if x = 0

Given that, the function is f x = x 2 sin 1 x , if x ≠ 0 0 , if x = 0 We know that, f ( 0 ) = 0 We have to determine whether if f is a continuous function. f ( x ) = x 2 sin 1 x at x ≠ 0 Consider R.H.L = lim x → ( 0 + h ) x 2 sin 1 x = lim h → 0 ( 0 + h ) 2 sin 1 0 + h = lim h → 0 h 2 sin 1 h = 0 2 × sin ∞ L.H.L = lim x → ( 0 - h ) x 2 sin 1 x = lim x → 0 ( - h ) 2 sin 1 0 - h L.H.L = lim h → 0 = - h 2 sin 1 h = - 0 2 sin ∞ = 0 R.H.S = lim x → 0 0 = 0 L.H.L = R.H.L = f ( 0 ) So, the given function is continuous. There are no points of discontinuity.

Find d y d x , if y x = x y .

Given that, y x = x y y log x = x log y Differentiating both the sides y · 1 x + logx · dy dx = x · 1 y dy dx + logy ⇒ y x - log y = x y · d y d x - log x d y d x ⇒ y - x log y x = d y d x x - y log x y ⇒ dy dx = y - xlogy x x - ylogx y ⇒ dy dx = y x y - xlogy x - ylogx

Find d y d x , if cos y = x cos ( a + y ) .

Given that, cos y = x cos ( a + y ) Differentiate both sides w.r.t. x . - sin y d y d x = cos ( a + y ) - x sin ( a + y ) d y d x ⇒ dy dx = cos ( a + y ) xsin ( a + y ) - siny = cos ( a + y ) cosy cos ( a + y ) sin ( a + y ) - siny = cos 2 ( a + y ) cos y sin ( a + y ) - sin y cos ( a + y ) Therefore, d y d x = cos 2 ( a + y ) sin a

If x = a ( θ + sin θ ) and y = a ( 1 - cos θ ) , find d 2 y d x 2 at θ = 0 .

Given that x = a ( θ + sin θ ) and y = a ( 1 + cos θ ) d 2 y d x 2 = d d x d y d x = d d x d y d θ · 1 d x d θ d y d θ = - a sin θ d x d θ = a ( 1 + cos θ ) ⇒ d y d x = d y d θ · d θ d x = - a sin θ · 1 a ( 1 + cos θ ) = - sin θ 1 + cos θ = - 2 · sin θ 2 · cos θ 2 2 cos 2 θ 2 = - tan θ 2 ⇒ d y d x = - tan θ 2 = d d x d y d x = - sec 2 θ 2 · 1 2 · d θ d x = - 1 2 · sec 2 θ 2 · 1 a ( 1 + cos θ ) = d 2 y d x 2 = - 1 2 a · sec 2 θ 2 1 + cos θ ⇒ d 2 y d x 2 θ = 0 = - 1 2 a - 1 2 = - 1 4 a

If y = a cos ⁡ ( log ⁡ x ) + b sin ⁡ ( log ⁡ x ) , show that x 2 d 2 y d x 2 + x d y d x + y = 0 .

Given: y = 3 cos ⁡ ( log ⁡ x ) + 4 sin ⁡ ( log ⁡ x ) Differentiate w.r.t x , we get: d y d x = − 3 sin ⁡ ( log ⁡ x ) x + 4 cos ⁡ ( log ⁡ x ) x ⇒ x d y d x = − 3 sin ⁡ ( log ⁡ x ) + 4 cos ⁡ ( log ⁡ x ) Again differentiate w.r.t x , we get: ⇒ x d 2 y d x 2 + d y d x = − 3 cos ⁡ ( log ⁡ x ) x − 4 sin ⁡ ( log ⁡ x ) x ⇒ x 2 d 2 y d x 2 + x d y d x = − ( 3 cos ⁡ ( log ⁡ x ) + 4 sin ⁡ ( log ⁡ x ) ⇒ x 2 d 2 y d x 2 + x d y d x = − y ∵ y = 3 cos ⁡ ( log ⁡ x ) + 4 sin ⁡ ( log ⁡ x ) ⇒ x 2 d 2 y d x 2 + x d y d x + y = 0 Hence proved.

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 5 - Continuity and Differentiability>Assam Board-2018

Embibe Experts Solutions for Chapter: Continuity and Differentiability, Exercise 1: Assam Board-2018

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Continuity and Differentiability, Exercise 1: Assam Board-2018

Attempt the free practice questions on Chapter 5: Continuity and Differentiability, Exercise 1: Assam Board-2018 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.