491. If x² + y² + 2x + 1 = 0, then the value of x³¹ + y³⁵ is

1. 1. –1

   
   
   

   2. 0

   
   
   

   3. 1

   
   
   

   4. 2

   
   
   

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1. View Hint View Answer Discuss in Forum

   x² + y² + 2x + 1 = 0
   ⇒  x² + 2x + 1 + y² = 0
   ⇒  (x + 1)² + y² = 0
   ⇒  x + 1 = 0 ⇒ x = –1 and y = 0
   ∴  x³¹ + y³⁵ = –1

   Correct Option: A

   x² + y² + 2x + 1 = 0
   ⇒  x² + 2x + 1 + y² = 0
   ⇒  (x + 1)² + y² = 0
   ⇒  x + 1 = 0 ⇒ x = –1 and y = 0
   ∴  x³¹ + y³⁵ = –1

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492. If  1.5 a = 0.04 b then	b − a	is equal to
     	b + a

1. 1. 73

      77

   
   
   

   2. 77

      33

   
   
   

   3. 2

      75

   
   
   

   4. 75

      2

   
   
   

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1. View Hint View Answer Discuss in Forum

   Tricky Approach
   1.5a = 0.04b
   

   b	=	1.5
   a		0.04

   
   By componendo and dividendo,
   

   b − a	=	1.5 − 0.04
   b + a		1.5 + 0.04

   
   

   =	1.46	=	73
   	1.54		77

   Correct Option: A

   Tricky Approach
   1.5a = 0.04b
   

   b	=	1.5
   a		0.04

   
   By componendo and dividendo,
   

   b − a	=	1.5 − 0.04
   b + a		1.5 + 0.04

   
   

   =	1.46	=	73
   	1.54		77

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493. If   x = (√2 + 1)−1/3, the value of		x3 −	1		is
     			x3

1. 1. 0

   
   
   

   2. −√2

   
   
   

   3. –2

   
   
   

   4. 3√2

   
   
   

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1. View Hint View Answer Discuss in Forum

   x = (√2 + 1)^−1/3
   ⇒  x⁻³ = √2 + 1
   

   ⇒	1	= √2 + 1
   	x3

   
   

   and x3 =	1	=	1(√2 − 1)
   	√2 + 1		(√2 + 1) (√2 − 1)

   
   = √2 − 1
   

   ∴  x3 −	1
   	x3

   
   = √2 − 1 − √2 − 1 = – 2

   Correct Option: C

   x = (√2 + 1)^−1/3
   ⇒  x⁻³ = √2 + 1
   

   ⇒	1	= √2 + 1
   	x3

   
   

   and x3 =	1	=	1(√2 − 1)
   	√2 + 1		(√2 + 1) (√2 − 1)

   
   = √2 − 1
   

   ∴  x3 −	1
   	x3

   
   = √2 − 1 − √2 − 1 = – 2

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494. If	x2 − x + 1	=	2	, then the value of		x −	1		is
     	x2 + x + 1		3				x

1. 1. 4

   
   
   

   2. 5

   
   
   

   3. 6

   
   
   

   4. 8

   
   
   

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1. View Hint View Answer Discuss in Forum

   x2 − x + 1	=	2
   x2 + x + 1		3

   
   

   ⇒	x2 + 1 − x	=	2
   	x2 + x + 1		3

   
   Dividing numerator and denominator by x,
   
   

   ⇒  3		x +	1		− 3 = 2		x +	1		+ 2
   			x					x

   
   

   ⇒  x +	1	= 2 + 3 = 5
   	x

   Correct Option: B

   x2 − x + 1	=	2
   x2 + x + 1		3

   
   

   ⇒	x2 + 1 − x	=	2
   	x2 + x + 1		3

   
   Dividing numerator and denominator by x,
   
   

   ⇒  3		x +	1		− 3 = 2		x +	1		+ 2
   			x					x

   
   

   ⇒  x +	1	= 2 + 3 = 5
   	x

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495. If x, y and z are real numbers such that (x– 3)² + (y – 4)² + (z – 5)² = 0 then (x + y + z) is equal to

1. 1. –12

   
   
   

   2. 0

   
   
   

   3. 8

   
   
   

   4. 12

   
   
   

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1. View Hint View Answer Discuss in Forum

   (x – 3)² + (y – 4)² + (z – 5)² = 0
   ⇒  x – 3 = 0, y – 4 = 0 and z – 5 = 0
   ⇒  x = 3, y = 4 and z = 5
   ∴  x + y + z = 3 + 4 + 5 = 12

   Correct Option: D

   (x – 3)² + (y – 4)² + (z – 5)² = 0
   ⇒  x – 3 = 0, y – 4 = 0 and z – 5 = 0
   ⇒  x = 3, y = 4 and z = 5
   ∴  x + y + z = 3 + 4 + 5 = 12

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