66. Starting from x₀ = 1, one step of NewtonRaphson method in solving the equation x³ + 3x – 7 = 0 gives the next value (x₁) as

1. 1. x₁ = 0.5

   
   
   

   2. x₁ = 1.406

   
   
   

   3. x₁ = 1.5

   
   
   

   4. x₁ = 2

   
   
   

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1. View Hint View Answer Discuss in Forum

   By N-R method, xn+1 = xn -	f(x)	, x0 = 1
   	f'(x)

   
   f(x) = x³ + 3x - 7
   ⇒ f(1) = -3,
   

   ⇒ x1 = x0 -	f(x0)
   	f'(x0)

   
   ⇒ f(x) = 3x² + 3,
   ⇒ f(1) = 6,
   

   x4 = 1 -	-3	= 1 - (-0.5) =1.5
   	6

   
   

   Correct Option: C

   By N-R method, xn+1 = xn -	f(x)	, x0 = 1
   	f'(x)

   
   f(x) = x³ + 3x - 7
   ⇒ f(1) = -3,
   

   ⇒ x1 = x0 -	f(x0)
   	f'(x0)

   
   ⇒ f(x) = 3x² + 3,
   ⇒ f(1) = 6,
   

   x4 = 1 -	-3	= 1 - (-0.5) =1.5
   	6

   
   

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67. The values of a function f(x) are tabulated below:
    
    Using Newton's forward difference formula, the cubic polynomial that can be fitted to the above data, is

1. 1. 2x³ + 7x² – 6x + 2

   
   
   

   2. 2x³ - 7x² + 6x – 2

   
   
   

   3. x³ – 7x² – 6x² + 1

   
   
   

   4. 2x³ - 7x² + 6x + 1

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   Difference table as
   
   

   We take x0 = 0, p =	x - 0	= x [∵ k = 1]
   	k

   
   using Newton’s forward interpolation formulae, we get
   

   f(x) = f(0) +		x	Δ f(0) +	x(x-1)
   		1		1.2

   
   

   ∆²f(0) +	x(x-1)(x-2)	∆3f(0)
   	1.2.3

   
   

   = 1 + x(1) +		x(x-1)	(-2) +	x(x-1)(x-2)	(12)
   		2		6

   
   = 1 + x + (x – x²) + 2x(x² – 3x + 2)
   = 1 + x + x – x² + 2x³ – 6x + 4x
   = 2x³ – 7x² + 6x + 1

   Correct Option: D

   
   Difference table as
   
   

   We take x0 = 0, p =	x - 0	= x [∵ k = 1]
   	k

   
   using Newton’s forward interpolation formulae, we get
   

   f(x) = f(0) +		x	Δ f(0) +	x(x-1)
   		1		1.2

   
   

   ∆²f(0) +	x(x-1)(x-2)	∆3f(0)
   	1.2.3

   
   

   = 1 + x(1) +		x(x-1)	(-2) +	x(x-1)(x-2)	(12)
   		2		6

   
   = 1 + x + (x – x²) + 2x(x² – 3x + 2)
   = 1 + x + x – x² + 2x³ – 6x + 4x
   = 2x³ – 7x² + 6x + 1

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68. A box contains 4 red balls and 6 black balls. Three balls are selected randomly from the box one after another, without replacement. The probability that the selected set contains one, red ball and two black balls is

1. 1. 1

      20

   
   
   

   2. 1

      12

   
   
   

   3. 3

      10

   
   
   

   4. 1

      2

      
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given :		Red		Black
   		4		6

   
   Selection will be RBB or BBR of BRB
   

   Probability of selecting RBB =		4	×	6	×	5
   		10		9		8

   
   

   Probability of selecting BBR =		6	×	5	×	4
   		10		9		8

   
   

   Probability of selecting BRB =		6	×	4	×	5
   		10		9		8

   
   P (Red = 1) = sum of above three probabilities = 0.5

   Correct Option: D

   Given :		Red		Black
   		4		6

   
   Selection will be RBB or BBR of BRB
   

   Probability of selecting RBB =		4	×	6	×	5
   		10		9		8

   
   

   Probability of selecting BBR =		6	×	5	×	4
   		10		9		8

   
   

   Probability of selecting BRB =		6	×	4	×	5
   		10		9		8

   
   P (Red = 1) = sum of above three probabilities = 0.5

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69. A box contains 2 washers, 3 nuts and 4 bolts. Items are drawn from the box at random one at a time without replacement. The probability of drawing 2 washers first followed by 3 nuts and subsequently the 4 bolts is

1. 1. 2

      315

   
   
   

   2. 1

      630

   
   
   

   3. 1

      1260

   
   
   

   4. 1

      2520

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given: Washers – 2, Nuts – 3, Bolts – 4,
   Total– 9 objects
   Hence probability of drawing two washers from 9 objects
   

   =	2C2	=	1
   	9C2		36

   
   Probability of drawing, 3 nuts out of remaining 7 objects is
   

   =	4C4	= 1
   	4C4

   
   Probability of drawing 4 bolts out of remaining 4 bolts
   

   ∴ Required Probability =		1	×	1	× 1 =	1
   		36		35		1260

   
   

   Correct Option: C

   Given: Washers – 2, Nuts – 3, Bolts – 4,
   Total– 9 objects
   Hence probability of drawing two washers from 9 objects
   

   =	2C2	=	1
   	9C2		36

   
   Probability of drawing, 3 nuts out of remaining 7 objects is
   

   =	4C4	= 1
   	4C4

   
   Probability of drawing 4 bolts out of remaining 4 bolts
   

   ∴ Required Probability =		1	×	1	× 1 =	1
   		36		35		1260

   
   

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70. If three coins are tossed simultaneously, the probability of getting at least one head is

1. 1. 1

      8

   
   
   

   2. 3

      8

   
   
   

   3. 1

      2

   
   
   

   4. 7

      8

      
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let probability of getting atleast one head = P(H) then P(at least one head) = 1 – P (no head)
   ⇒ P(H) = 1– P(all tails)
   But in all cases, 2³ = 8
   

   ∴ P(H) = 1 -		1	=	7
   		8		8

   
   Alternately: Probability of getting at least one head
   
   

   Correct Option: D

   Let probability of getting atleast one head = P(H) then P(at least one head) = 1 – P (no head)
   ⇒ P(H) = 1– P(all tails)
   But in all cases, 2³ = 8
   

   ∴ P(H) = 1 -		1	=	7
   		8		8

   
   Alternately: Probability of getting at least one head
   
   

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