61. The root of the function f(x) = x³ + x – 1 obtained after first iteration on application of Newton Raphson scheme using an initial guess of x₀ = 1 is

1. 1. 0.682

   
   
   

   2. 0.686

   
   
   

   3. 0.750

   
   
   

   4. 1.000

   
   
   

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1. View Hint View Answer Discuss in Forum

   f (x) = x³ + x–1
   f (1) = 1+1 – 1 = 1
   f’(x) = 3x² + 1
   f’ (1) = 3+1=4
   

   x1 = x0 -		f(x0)	= 1 -	1
   		f'(x0)		4

   
   = 1 – 0.25
   x₁ = 0.75

   Correct Option: C

   f (x) = x³ + x–1
   f (1) = 1+1 – 1 = 1
   f’(x) = 3x² + 1
   f’ (1) = 3+1=4
   

   x1 = x0 -		f(x0)	= 1 -	1
   		f'(x0)		4

   
   = 1 – 0.25
   x₁ = 0.75

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62. Solve the equation x = 10 cos(x) using the Newton-Raphson method. The initial guess is x = π / 4. The value of the predicted root after the first iteration, up to second decimal, is _____.

1. 1. 0.75

   
   
   

   2. 0.76

   
   
   

   3. 0.74

   
   
   

   4. 0.78

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   Correct Option: A

   
   

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63. Newton-Raphson method is used to find the roots of the equation, x³ + 2x² + 3x – 1 = 0. If the initial guess is x₀ = 1, then the value of x after 2nd iteration is ________.

1. 1. 0.3043

   
   
   

   2. 0.2043

   
   
   

   3. 1.2043

   
   
   

   4. 1.3043

   
   
   

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1. View Hint View Answer Discuss in Forum

   By Newton-Raphson Method,
   

   1st iteration, x1 = x0 –	f(0)
   	f'(0)

   
   

   = 1 -		f(1)	= 1 -	5	=	1
   		f'(1)		10		2

   
   Where f(x) = x³ + 2x² + 3x – 1 ⇒ f(1) = 5
   f'(x) = 3x² + 4x + 3 ⇒ f'(1) = 10
   

   2nd iteration, x2 = x1		f(x1)	= 0.5 -	f(0.5)	= 0.3043
   		f'(x1)		f'(0.5)

   
   

   Correct Option: A

   By Newton-Raphson Method,
   

   1st iteration, x1 = x0 –	f(0)
   	f'(0)

   
   

   = 1 -		f(1)	= 1 -	5	=	1
   		f'(1)		10		2

   
   Where f(x) = x³ + 2x² + 3x – 1 ⇒ f(1) = 5
   f'(x) = 3x² + 4x + 3 ⇒ f'(1) = 10
   

   2nd iteration, x2 = x1		f(x1)	= 0.5 -	f(0.5)	= 0.3043
   		f'(x1)		f'(0.5)

   
   

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64. The real root of the equation 5x – 2 cosx – 1 = 0 (up to two decimal accuracy) is _______.

1. 1. 0.5424

   
   
   

   2. 0.5423

   
   
   

   3. 1.5424

   
   
   

   4. 1.5423

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let f(x) =5x – 2 cos x – 1
   ⇒ f'(x) = 5 + 2 sin x
   f(0) = – 3; f(1) = 2.9
   By intermediate value theorem roots lie between 0 and 1.
   Let x₀ = 1 rad = 57.32º
   By Newton Raphson method,
   

   Xn+1 = Xn -	f(xn)
   	f'(xn)

   
   

   ⇒ xn+1	2xnsin xn + 2 cos xn + 1
   	5 + 2Sin xn

   
   ⇒ x₁ = 0.5632
   ⇒ x₂ = 0.5425
   ⇒ x₃ = 0.5424

   Correct Option: A

   Let f(x) =5x – 2 cos x – 1
   ⇒ f'(x) = 5 + 2 sin x
   f(0) = – 3; f(1) = 2.9
   By intermediate value theorem roots lie between 0 and 1.
   Let x₀ = 1 rad = 57.32º
   By Newton Raphson method,
   

   Xn+1 = Xn -	f(xn)
   	f'(xn)

   
   

   ⇒ xn+1	2xnsin xn + 2 cos xn + 1
   	5 + 2Sin xn

   
   ⇒ x₁ = 0.5632
   ⇒ x₂ = 0.5425
   ⇒ x₃ = 0.5424

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65. Let X and Y be two independent random variables. Which one of the relations between expectation (n), variance (Var) and covariance (Cov) given below is FALSE?

1. 1. E (XY) = E(X) E(Y)

   
   
   

   2. Coy (X, Y) = 0

   
   
   

   3. Var (X+ Y) = Var (X) + Var (Y)

   
   
   

   4. E(X² Y²) = (E (X))² (E (Y))²

   
   
   

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1. View Hint View Answer Discuss in Forum

   X and Y are independent
   ∴ (a), (b), (c) are true
   Only (d) is odd one
   

   Correct Option: D

   X and Y are independent
   ∴ (a), (b), (c) are true
   Only (d) is odd one
   

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