56. If   x =	√5 + 1	and y =	√5 − 1	, then the value of	x2 + xy + y2	is
    	√5 − 1		√5 + 1		x2 − xy + y2

1. 1. 3

      4

   
   
   

   2. 4

      3

   
   
   

   3. 3

      5

   
   
   

   4. 5

      3

   
   
   

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1. View Hint View Answer Discuss in Forum

   x =	√5 + 1
   	√5 − 1

   
   

   =	(√5 + 1)2
   	(√5 − 1)(√5 + 1)

   
   (Rationalising the denominator)
   

   =	5 + 1 + 2√5
   	5 − 1

   
   

   =	6 + 2√5
   	4

   
   

   =	3 + √5
   	2

   
   

   ∴  y =	√5 − 1	=	3 − √5
   	√5 + 1		2

   
   

   ∴  x + y =	3 + √5	+	3 − √5
   	2		2

   
   

   =	3 + √5 + 3 − √5	= 3
   	2

   
   

   xy =	3 + √5 +	×	3 − √5
   	2		2

   
   

   =	9 − 5	= 1
   	4

   
   

   ∴	x2 + xy + y2	=	(x + y)2 − xy
   	x2 − xy + y2		(x + y)2 − 3xy

   
   

   =	(3)2 − 1	=	9 − 1	=	8	=	4
   	(3)2 − 3		9 − 3		6		3

   Correct Option: B

   x =	√5 + 1
   	√5 − 1

   
   

   =	(√5 + 1)2
   	(√5 − 1)(√5 + 1)

   
   (Rationalising the denominator)
   

   =	5 + 1 + 2√5
   	5 − 1

   
   

   =	6 + 2√5
   	4

   
   

   =	3 + √5
   	2

   
   

   ∴  y =	√5 − 1	=	3 − √5
   	√5 + 1		2

   
   

   ∴  x + y =	3 + √5	+	3 − √5
   	2		2

   
   

   =	3 + √5 + 3 − √5	= 3
   	2

   
   

   xy =	3 + √5 +	×	3 − √5
   	2		2

   
   

   =	9 − 5	= 1
   	4

   
   

   ∴	x2 + xy + y2	=	(x + y)2 − xy
   	x2 − xy + y2		(x + y)2 − 3xy

   
   

   =	(3)2 − 1	=	9 − 1	=	8	=	4
   	(3)2 − 3		9 − 3		6		3

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57. Let 0 < x < 1. Then the correct inequality is
    

1. 1. x < √x < x²

   
   
   

   2. √x < x < x²

   
   
   

   3. x² < x < √x

   
   
   

   4. √x < x² < x

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given,
   0 < x < 1
   ⇒  x.0 < x.x < 1.x
   ⇒  0 < x² < x
   Again, x < 1
   ⇒  √x < 1
   ∴  x² < x < √x

   Correct Option: C

   Given,
   0 < x < 1
   ⇒  x.0 < x.x < 1.x
   ⇒  0 < x² < x
   Again, x < 1
   ⇒  √x < 1
   ∴  x² < x < √x

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58. If a + b = 10 and ab = 21, then the value of (a – b)² is

1. 1. 15

   
   
   

   2. 16

   
   
   

   3. 17

   
   
   

   4. 18

   
   
   

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1. View Hint View Answer Discuss in Forum

   a + b = 10 ;
   ab = 21
   ∴  (a – b)² = (a + b)² – 4ab
   = (10)² – 4 × 21
   = 100 – 84 = 16

   Correct Option: B

   a + b = 10 ;
   ab = 21
   ∴  (a – b)² = (a + b)² – 4ab
   = (10)² – 4 × 21
   = 100 – 84 = 16

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59. If   2x +	1	= 5, the value of	5x	is
    	3x		6x2 + 20x + 1

1. 1. 1

      4

   
   
   

   2. 1

      6

   
   
   

   3. 1

      5

   
   
   

   4. 1

      7

   
   
   

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1. View Hint View Answer Discuss in Forum

   2x +	1	= 5
   	3x

   
   

   ⇒	6x2 + 1	= 5
   	3x

   
   ⇒  6x² + 1 = 15x
   

   ∴	5x
   	6x2 + 20x + 1

   
   

   =	5x
   	6x2 + 1 + 20x

   
   

   =	5x
   	15x + 20x

   
   

   =	5x	=	1
   	35x		7

   Correct Option: D

   2x +	1	= 5
   	3x

   
   

   ⇒	6x2 + 1	= 5
   	3x

   
   ⇒  6x² + 1 = 15x
   

   ∴	5x
   	6x2 + 20x + 1

   
   

   =	5x
   	6x2 + 1 + 20x

   
   

   =	5x
   	15x + 20x

   
   

   =	5x	=	1
   	35x		7

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60. If   x = √a +	1	, y = √a −	1	(a > 0), then the value of (x4 + y4 – 2x2y2) is
    	√a		√a

1. 1. 16

   
   
   

   2. 20

   
   
   

   3. 10

   
   
   

   4. 5

   
   
   

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1. View Hint View Answer Discuss in Forum

   x = √a +	1	,   y = √a −	1
   	√a		√a

   
   

   ∴  x + y = √a +	1	+ √a −	1
   	√a		√a

   
   = 2√a
   

   ∴  x − y = √a +	1	− √a +	1
   	√a		√a

   
   

   =	2
   	√a

   
   Now,
   x⁴ + y⁴ – 2x²y²
   = (x² – y²)²
   = {(x + y) (x – y)}²
   

   	2√a ×	1		2	= 42 = 16
   		√a

   Correct Option: A

   x = √a +	1	,   y = √a −	1
   	√a		√a

   
   

   ∴  x + y = √a +	1	+ √a −	1
   	√a		√a

   
   = 2√a
   

   ∴  x − y = √a +	1	− √a +	1
   	√a		√a

   
   

   =	2
   	√a

   
   Now,
   x⁴ + y⁴ – 2x²y²
   = (x² – y²)²
   = {(x + y) (x – y)}²
   

   	2√a ×	1		2	= 42 = 16
   		√a

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