71. If	x	+	3	= 1 then the value of x3 is
    	3		x

1. 1. 1

   
   
   

   2. 27

   
   
   

   3. 0

   
   
   

   4. –27

   
   
   

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1. View Hint View Answer Discuss in Forum

   x	+	3	= 1
   3		x

   
   

   ⇒	x2 + 9	= 1
   	3x

   
   ⇒  x² + 9 = 3x
   ⇒  x² – 3x + 9 = 0
   ∴  x³ + 3³ = (x + 3) (x² – 3x + 9) = 0
   ⇒  x³ = – 27

   Correct Option: D

   x	+	3	= 1
   3		x

   
   

   ⇒	x2 + 9	= 1
   	3x

   
   ⇒  x² + 9 = 3x
   ⇒  x² – 3x + 9 = 0
   ∴  x³ + 3³ = (x + 3) (x² – 3x + 9) = 0
   ⇒  x³ = – 27

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72. 9x² + 25 – 30x can be expressed as the square of

1. 1. –3x – 5

   
   
   

   2. 3x + 5

   
   
   

   3. 3x – 5

   
   
   

   4. 3x² – 25

   
   
   

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1. View Hint View Answer Discuss in Forum

   9x² + 25 – 30x
   = (3x)² + (5)² – 2 × 3x × 5
   = (3x – 5)²

   Correct Option: C

   9x² + 25 – 30x
   = (3x)² + (5)² – 2 × 3x × 5
   = (3x – 5)²

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73. For real a, b, c if a2 + b2 + c2 = ab + bc + ca, the value of	a + c	is
    	b

1. 1. 3

   
   
   

   2. 1

   
   
   

   3. 2

   
   
   

   4. 0

   
   
   

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1. View Hint View Answer Discuss in Forum

   a² + b² + c² = ab + bc + ca
   ⇒  2a² + 2b² + 2c² – 2ab – 2bc – 2ca = 0
   ⇒  a² + b² – 2ab + b² + c² – 2bc + c² + a² – 2ca = 0
   ⇒  (a – b)² + (b – c)² + (c – a)² = 0
   ∴  a – b = 0 ⇒ a = b
   b – c = 0 ⇒ b = c
   c – a = 0 ⇒ c = a
   ∴  a = b = c
   

   ∴	a + c	=	2a	= 2
   	b		a

   Correct Option: C

   a² + b² + c² = ab + bc + ca
   ⇒  2a² + 2b² + 2c² – 2ab – 2bc – 2ca = 0
   ⇒  a² + b² – 2ab + b² + c² – 2bc + c² + a² – 2ca = 0
   ⇒  (a – b)² + (b – c)² + (c – a)² = 0
   ∴  a – b = 0 ⇒ a = b
   b – c = 0 ⇒ b = c
   c – a = 0 ⇒ c = a
   ∴  a = b = c
   

   ∴	a + c	=	2a	= 2
   	b		a

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74. If   a +	1	= b +	1	= c +	1	, where a ≠ b ≠ c ≠ 0, then the value of a2b2c2
    	b		c		a

1. 1. –1

   
   
   

   2. abc

   
   
   

   3. 0

   
   
   

   4. 1

   
   
   

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1. View Hint View Answer Discuss in Forum

   a +	1	= b +	1	= c +	1
   	b		c		a

   
   

   ⇒	abc + c	=	abc + a	=	abc + b
   	bc		ac		ab

   
   

   ⇒	c	=	a	=	b
   	bc		ac		ab

   
   

   ⇒	1	=	1	=	1
   	b		c		a

   
   ⇒  a = b = c = 1
   ∴  a²b²c² = 1

   Correct Option: D

   a +	1	= b +	1	= c +	1
   	b		c		a

   
   

   ⇒	abc + c	=	abc + a	=	abc + b
   	bc		ac		ab

   
   

   ⇒	c	=	a	=	b
   	bc		ac		ab

   
   

   ⇒	1	=	1	=	1
   	b		c		a

   
   ⇒  a = b = c = 1
   ∴  a²b²c² = 1

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75. Let   x =	√13 + √11	and y =	1	then the value of 3x2 – 5xy + 3y2 is
    	√13 − √11		x

1. 1. 1717

   
   
   

   2. 1177

   
   
   

   3. 1771

   
   
   

   4. 1171

   
   
   

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1. View Hint View Answer Discuss in Forum

   x =	√13 + √11
   	√13 − √11

   
   On rationalising the denominator,
   

   =	√13 + √11	×	√13 + √11
   	√13 − √11		√13 + √11

   
   

   =	(√13 + √11)2
   	(√13)2 − (√11)2

   
   

   =	13 + 11 + 2√143
   	13 − 11

   
   

   =	24 + 2√143	= 12 + √143
   	2

   
   

   ∴  y =	1	=	1
   	x		12 + √143

   
   

   =	1	×	12 − √143
   	12 + √143		12 − √143

   
   

   =	12 − √143	= 12 − √143
   	144 − 143

   
   ∴  x – y = 12 + √143 – 12 + √143 = 2√143
   and
   xy = (12 + √143)(12 – √143)
   = 144 – 143 = 1
   ∴  3x² – 5xy + 3y² = 3x² – 6xy + 3y² + xy
   = 3 (x – y)² + xy
   = 3 (2√143)² + 1
   = 3 × 4 × 143 + 1 = 1716 + 1
   = 1717

   Correct Option: A

   x =	√13 + √11
   	√13 − √11

   
   On rationalising the denominator,
   

   =	√13 + √11	×	√13 + √11
   	√13 − √11		√13 + √11

   
   

   =	(√13 + √11)2
   	(√13)2 − (√11)2

   
   

   =	13 + 11 + 2√143
   	13 − 11

   
   

   =	24 + 2√143	= 12 + √143
   	2

   
   

   ∴  y =	1	=	1
   	x		12 + √143

   
   

   =	1	×	12 − √143
   	12 + √143		12 − √143

   
   

   =	12 − √143	= 12 − √143
   	144 − 143

   
   ∴  x – y = 12 + √143 – 12 + √143 = 2√143
   and
   xy = (12 + √143)(12 – √143)
   = 144 – 143 = 1
   ∴  3x² – 5xy + 3y² = 3x² – 6xy + 3y² + xy
   = 3 (x – y)² + xy
   = 3 (2√143)² + 1
   = 3 × 4 × 143 + 1 = 1716 + 1
   = 1717

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